Characterizations of zero-dimensional complete intersections

Abstract

Given a 0-dimensional subscheme ${\mathbb X}$ of a projective space ${\mathbb P}^n_K$ over a field K, we characterize in different ways whether ${\mathbb X}$ is the complete intersection of n hypersurfaces. Besides a generalization of the notion of a Cayley–Bacharach scheme, these characterizations involve the Kähler and the Dedekind different of the homogeneous coordinate ring of ${\mathbb X}$ or its Artinian reduction. We also characterize arithmetically Gorenstein schemes in novel ways and bring in further tools such as the module of regular differential forms, the fundamental class, and the Jacobian module of ${\mathbb X}$. Throughout we strive to work over an arbitrary base field K and keep the scheme ${\mathbb X}$ as general as possible, thereby improving several known characterizations.

1 Introduction

Given a 0-dimensional scheme ${\mathbb X}$ in a projective space ${\mathbb P}^n_K$ over a field K, it is an interesting question to characterize algebraically and geometrically when ${\mathbb X}$ is the complete intersection of n hypersurfaces. In general, characterizations of complete intersection rings using differentials, differents, and complementary modules have been studied by R. Berger, E. Kunz, H.-J. Nastold, G. Scheja, U. Storch, and others in the 1960s and 1970s. Many of these results we collected and unified in the book (Kunz 1986) which we use as our reference. In this paper we are specifically interested in zero-dimensional subschemes of projective spaces, for which the homogeneous coordinate ring is a 1-dimensional standard graded Cohen Macaulay ring.

In the case $n=2$, i.e., for subschemes of the projective plane, Davis and Maroscia (1984) characterized complete intersections via the Cayley–Bacharach property and the symmetry of the Hilbert function of ${\mathbb X}$. Later it turned out that these conditions characterize arithmetically Gorenstein schemes for arbitrary $n\ge 2$ (see Davis et al. 1985; Kreuzer 1992). The attempt to refine this characterization by showing that ${\mathbb X}$ is the zero-set of a section of a vector bundle and then forcing the vector bundle to split into a direct sum of line bundles led to rather complicated and unwieldy conditions (see Kreuzer 1992 for the case $n=3$ and Kreuzer et al. 2000 for the general case).

On the algebraic side, Wiebe (1969) proved for 0-dimensional local rings that they are complete intersections if and only if the 0-th Fitting ideal of the maximal ideal is non-zero. If we assume that the 0-dimensional scheme ${\mathbb X}$ is contained in the affine space $D_+(X_0)$, this characterization can be applied to the Artinian reduction $R/\langle x_0\rangle$ of the homogeneous coordinate ring $R=K[X_0,\ldots ,X_n]/I_{\mathbb X}$ of ${\mathbb X}$. Similarly, assuming $\text {char}(K)=0$, Scheja and Storch (1975) characterized 0-dimensional local complete intersections by the non-vanishing of the Kähler different, i.e., the 0-th Fitting ideal of the Kähler differential module. Unfortunately, these characterizations do not allow finer distinctions such as the characterization of arithmetically Gorenstein schemes ${\mathbb X}$.

The main idea in this paper is to combine these two approaches and to use the Kähler differents of the algebras $R/K[x_0]$ and R / K together with some geometric properties to characterize 0-dimensional complete intersection schemes. For instance, we prove the following theorem which provides one possible answer to the question posed by Griffiths and Harris (1978): if ${\mathbb X}$ is a Cayley–Bacharach scheme, when is ${\mathbb X}$ a complete intersection?

Theorem 1.1

Let ${\mathbb X}$ be a smooth 0-dimensional subscheme of ${\mathbb P}^n_K$. Then ${\mathbb X}$ is a complete intersection if and only if ${\mathbb X}$ is a Cayley–Bacharach scheme and the Hilbert function of the Kähler different ${\vartheta }_{\mathbb X}$ of ${\mathbb X}$ satisfies $\mathop {\text {HF}}\nolimits _{{\vartheta }_{\mathbb X}}(r_{\mathbb X}) \ne 0$, where $r_{\mathbb X}$ is the regularity index of the Hilbert function of ${\mathbb X}$.

As a consequence, we will see that this condition is also equivalent to the Kähler different being a principal ideal generated by a homogeneous non-zerodivisor of degree $r_{\mathbb X}$. The latter characterization is the graded version of a result given by Lenstra (1993) in the local case which in turn is a slight sharpening of a result used by Wiles (1995). Along the way, we prove a number of further characterizations of smooth 0-dimensional complete intersections.

Let us describe the contents of this paper in more detail. In Sect. 2 we recall the definition of the Kähler different. In fact, we introduce three Kähler differents which will be used later on. Assuming that ${\mathbb X}$ is contained in $D_+(X_0)$, as we always do, the residue class $x_0$ of $X_0$ in the homogeneous coordinate ring $R=K[X_0,\ldots ,X_n]/I_{\mathbb X}$ of ${\mathbb X}$ is a non-zerodivisor. Then we define the Kähler different ${\vartheta }_{{\mathbb X}} = F_0(\Omega ^1_{R/K[x_0]})$, the reduced Kähler different $\bar{\vartheta }_{\mathbb X}= F_0(\Omega ^1_{\overline{R}/K})$, where $\overline{R}=R/\langle x_0\rangle$, and the higher Kähler different ${\vartheta }^{(1)}_{\mathbb X}= F_1(\Omega ^1_{R/K})$. After discussing the basic properties of these differents, we have a closer look at the Hilbert function of ${\vartheta }_{\mathbb X}$. Using some examples, we see that this is a tricky invariant and that even its eventual value, the Hilbert polynomial of ${\vartheta }_{\mathbb X}$, is in general difficult to determine.

Section 3 deals with the Dedekind different of a 0-dimensional scheme ${\mathbb X}$ . To define it, we need to assume that ${\mathbb X}$ is locally Gorenstein. We make the construction in Herzog et al. (1971) explicit and embed the canonical module of ${\mathbb X}$ in its homogeneous coordinate ring to get the Dedekind complementary module whose inverse ideal is the Dedekind different. Notice that this construction depends on the choice of a system of traces for the local rings of ${\mathbb X}$. However, if ${\mathbb X}$ is smooth, we can use the canonical traces and get a well-defined result. The section ends with some relations between the Kähler and the Dedekind different of ${\mathbb X}$.

Many characterizations of 0-dimensional complete intersections use the Cayley–Bacharach property. In Sect. 4 we generalize the notion of a Caylay–Bacharach scheme (CB-scheme) to the case of a not necessarily reduced scheme ${\mathbb X}$ defined over an arbitrary base field K. Many concepts such as the degree of a point in ${\mathbb X}$ require careful adjustments in this generality. We also characterize CB-schemes via the existence of particular elements in the first homogeneous component of the Dedekind complementary module and use this result to generalize the characterization of arithmetically Gorenstein schemes via the Cayley–Bacharach property and the symmetry of their Hilbert function (see Kreuzer 1992).

The heart of the paper is contained in Sect. 5 where we characterize 0-dimensional complete intersections in several ways. The first characterization generalizes the above-mentioned result by Scheja and Storch (1975) and uses the non-vanishing of the reduced Kähler different. A second criterion uses a single value of the Hilbert function of ${\vartheta }_{\mathbb X}$ to distinguish complete intersections from arithmetically Gorenstein schemes. The third characterization answers the question of P. Griffiths and J. Harris for smooth 0-dimensional schemes by requiring a CB-scheme to have $\mathop {\text {HF}}\nolimits _{{\vartheta }_{\mathbb X}}(r_{\mathbb X})\ne 0$. If we replace the Kähler different by the Dedekind different, we get a characterization of 0-dimensional arithmetically Gorenstein schemes, as Proposition 5.8 shows.

In the last section we present some characterizations which use the first Kähler different ${\vartheta }^{(1)}_{\mathbb X}$ of R / K. After collecting some properties of this different and its Hilbert function, we show that it can be used to characterize smooth 0-dimensional complete intersections in the same way as with ${\vartheta }_{\mathbb X}$ by the Cayley–Bacharach property and one non-zero value of its Hilbert function. Finally, we introduce the module of regular differential forms $\Omega _{\mathbb X}$ and the fundamental class $c_{\mathbb X}:\; \Omega ^1_{R/K} \longrightarrow \Omega _{\mathbb X}$ of ${\mathbb X}$. Then smooth 0-dimensional complete intersections are also characterized by $c_{\mathbb X}(\Omega ^1_{R/K}) = {\vartheta }^{(1)}_{\mathbb X}\, \Omega _{\mathbb X}$. Lastly, smooth arithmetically Gorenstein schemes are characterized by the Cayley–Bacharach property and the dimension of their Jacobian module $J_{\mathbb X}= \Omega _{\mathbb X}/ c_{\mathbb X}(\Omega ^1_{R/K})$.

Unless mentioned explicitly otherwise, we adhere to the definitions and notation introduced in the books (Kreuzer and Robbiano 2000, 2005). The examples in this paper were calculated using a package implemented by the second author in the computer algebra system The ApCoCoA Team (2007).

2 Kähler differents of zero-dimensional schemes

In this paper we always work in the following setting. Let K be an arbitrary field, and let $P=K[X_0,\ldots ,X_n]$ be the polynomial ring in $n+1$ indeterminates over K, equipped with the standard grading. Then P is the homogeneous coordinate ring of projective n-space ${\mathbb P}^n_K$.

Our main object of study is a (non-empty) 0-dimensional subscheme ${\mathbb X}$ of ${\mathbb P}^n_K$. Its homogeneous vanishing ideal in P is denoted by $I_{\mathbb X}$. Then $R=P/I_{\mathbb X}$ is the homogeneous coordinate ring of ${\mathbb X}$. It is a 1-dimensional standard graded Cohen–Macaulay K-algebra. Its homogeneous maximal ideal will be denoted by ${\mathfrak {m}}$.

Assumption In the following we assume that no point of the support of ${\mathbb X}$ is contained in the hypersurface at infinity $\mathcal {Z}(X_0)$. Consequently, the residue class $x_0$ of $X_0$ in R is a non-zerodivisor.

The ring $\overline{R}=R/\langle x_0\rangle$ is called the Artinian reduction of R. It is a 0-dimensional, local K-algebra, and hence a finite dimensional K-vector space of dimension $d=\text {deg}({\mathbb X})$. The maximal ideal of $\overline{R}$ is denoted by $\overline{\mathfrak {m}}$. It follows that R is a finite free $K[x_0]$-module of rank d. The modules of Kähler differentials of the three algebras R / K, $\overline{R}/K$, and $R/K[x_0]$ are related as follows (cf. Kunz 1986).

Proposition 2.1

The element $x_0$ is a non-zerodivisor for the module $\Omega ^1_{R/K[x_0]}$, and we have $\Omega ^1_{R/K[x_0]}/x_0\, \Omega ^1_{R/K[x_0]} \cong \Omega ^1_{\overline{R}/K}$. Furthermore, there exists a canonical homogeneous exact sequence

$$\begin{aligned} 0 \;\longrightarrow \; R dx_0 \;\longrightarrow \; \Omega ^1_{R/K} \;\longrightarrow \; \Omega ^1_{R/K[x_0]} \;\longrightarrow \; 0 \end{aligned}$$

Now the Fitting ideals of these three Kähler differential modules are given the following names. (For some basic properties of Fitting ideals we refer the reader to Kunz 1986, Appendix D.)

Definition 2.2

• (a) The ideal ${\vartheta }_{R/K[x_0]}=F_0(\Omega ^1_{R/K[x_0]})$ of R is called the Kähler different of ${\mathbb X}$ (or of $R/K[x_0]$). For short, we also write ${\vartheta }_{\mathbb X}$ instead of ${\vartheta }_{R/K[x_0]}$.

• (b) The ideal ${\vartheta }_{\overline{R}/K} = F_0(\Omega ^1_{\overline{R}/K})$ of $\overline{R}$ is called the reduced Kähler different of ${\mathbb X}$. We also write $\bar{\vartheta }_{\mathbb X}$ instead of ${\vartheta }_{\overline{R}/K}$ for short.

• (c) The ideal ${\vartheta }^{(1)}_{R/K} = F_1(\Omega ^1_{R/K})$ is called the higher Kähler different of ${\mathbb X}$ (or of R / K). This ideal is also denoted by ${\vartheta }^{(1)}_{\mathbb X}$ for short.

Recall that the scheme ${\mathbb X}$ is called a complete intersection if its homogeneous vanishing ideal can be generated by n homogeneous polynomials, i.e., if there exist homogeneous polynomials $F_1,\ldots ,F_n\in P$ such that $I_{\mathbb X}=\langle F_1,\ldots ,F_n\rangle$. It is well-known that the Kähler different is given by ${\vartheta }_{\mathbb X}= \langle \frac{\partial (F_1,\ldots ,F_n)}{\partial (x_1,\ldots ,x_n)} \rangle _R$ in this case, i.e., it is the principal ideal generated by the Jacobian determinant of $(F_1,\ldots ,F_n)$. A similar description can be given for $\bar{\vartheta }_{\mathbb X}$.

Given a finitely generated graded module M over a standard graded K-algebra, we let $\mathop {\text {HF}}\nolimits _{\!M}:\; \mathbb {Z} \longrightarrow \mathbb {N}$ be the map defined by $\mathop {\text {HF}}\nolimits _{\!M}(i)=\dim _K(M_i)$. Recall that this map is called the Hilbert function of M. This is an integer function of polynomial type, i.e., it agrees with the value of the Hilbert polynomial $\text {HP}_M(i)$ in large degrees $i\gg 0$. The number $\text {ri}(M)=\min \{i\ge 0\mid \mathop {\text {HF}}\nolimits _M(j)=\text {HP}_M(j)$ for all $j\ge i\}$ is called the regularity index of M. Instead of $\mathop {\text {HF}}\nolimits _{\!R}$, we also write $\mathop {\text {HF}}\nolimits _{\mathbb X}$ and call it the Hilbert function of ${\mathbb X}$.

Remark 2.3

The Hilbert function of ${\mathbb X}$ satisfies $\mathop {\text {HF}}\nolimits _{\mathbb X}(i)=0$ for $i<0$ and

$$\begin{aligned} 1=\mathop {\text {HF}}\nolimits _{\mathbb X}(0) \;<\; \mathop {\text {HF}}\nolimits _{\mathbb X}(1) \;<\; \cdots \;<\; \mathop {\text {HF}}\nolimits _{\mathbb X}(r_{\mathbb X}) = d =\mathop {\text {HF}}\nolimits _{\mathbb X}(r_{\mathbb X}+1) = \cdots \end{aligned}$$

for some number $r_{\mathbb X}\ge 0$ which is called the regularity index of ${\mathbb X}$.

In the rest of this section we collect some general results and examples illustrating properties of the Hilbert function of the Kähler different of ${\mathbb X}$. Later we will see that the reduced Kähler different is much less interesting (it vanishes most of the time) and that one can prove similar things for the Hilbert function of the higher Kähler different (see Sect. 6).

Remark 2.4

Since $x_0$ is a non-zerodivisor of R and ${\vartheta }_{\mathbb X}$ is an ideal in R, the Hilbert function $\mathop {\text {HF}}\nolimits _{{\vartheta }_{\mathbb X}}$ is non-decreasing. Therefore it has an initial degree $\min \{ i\in {\mathbb Z}\mid \mathop {\text {HF}}\nolimits _{{\vartheta }_{\mathbb X}}(i)\ne 0\}$ and an eventual value $\mathop {\text {HF}}\nolimits _{{\vartheta }_{\mathbb X}}(j)$ for $j\gg 0$ which is also known as the Hilbert polynomial of ${\vartheta }_{\mathbb X}$ and denoted by $\text {HP}_{{\vartheta }_{\mathbb X}}$.

Recall that the minimal prime divisors of $I_{\mathbb X}$ are homogeneous prime ideals which correspond to points in ${\mathbb P}^n_K= \text {Proj}(P)$. The set of these points is called the support of ${\mathbb X}$ and will be denoted by $\text {Supp}({\mathbb X})=\{p_1,\ldots ,p_s\}$. To each point $p_i$ we have the associated local ring ${\mathcal {O}}_{{\mathbb X},p_i}$, its homogeneous vanishing ideal $\mathfrak {P}_i$ in P, and the associated homogeneous ideal $\mathfrak {p}_i$ in R.

Definition 2.5

• (a) A point $p_i\in \text {Supp}({\mathbb X})$ is called a reduced point of ${\mathbb X}$, if ${\mathcal {O}}_{{\mathbb X},p_i}$ is a reduced ring.

• (b) A point $p_i\in \text {Supp}({\mathbb X})$ is called a smooth point of ${\mathbb X}$, or ${\mathbb X}$ is called smooth at $p_i$, if ${\mathcal {O}}_{{\mathbb X},p_i}/K$ is a finite separable field extension.

• (c) We say that ${\mathbb X}$ is reduced (resp. smooth) if it is reduced (resp. smooth) at all points of its support.

Clearly, if $p_i$ is a smooth point of ${\mathbb X}$ then it is a reduced point of ${\mathbb X}$. The converse is true if K is a perfect field (cf. Kunz 1986, Propositions 5.18 and 7.12).

If ${\mathbb X}$ is a smooth complete intersection, then we know the Hilbert function of ${\vartheta }_{\mathbb X}$. More generally, if ${\mathbb X}$ is reduced then the Hilbert function of ${\vartheta }_{\mathbb X}$ can be described in terms of the subset ${\mathbb Y}$ of smooth points of ${\mathbb X}$, as the following proposition shows. Notice that for ${\mathbb Y}=\emptyset$ we have $I_{{\mathbb Y}/{\mathbb X}}={\mathfrak {m}}$.

Proposition 2.6

Let ${\mathbb X}$ be a 0-dimensional reduced complete intersection in ${\mathbb P}^n_K$. We write $I_{{\mathbb X}} = \langle F_1, \ldots , F_n \rangle$, where $F_j \in P$ is a homogeneous polynomial of degree $d_j$ for $j=1,\ldots ,n$. (Recall that we have $r_{\mathbb X}= \sum _{i=1}^n d_i - n$ in this case.) Let ${\mathbb Y}$ be the subscheme of ${\mathbb X}$ consisting of all smooth points in $\text {Supp}({\mathbb X})$. Then we have

$$\begin{aligned} \mathop {\text {HF}}\nolimits _{{\vartheta }_{\mathbb X}}(i) = {\left\{ \begin{array}{ll} \mathop {\text {HF}}\nolimits _{{\mathbb Y}}(i-r_{\mathbb X}) &{} \text{ if } {\mathbb Y}\ne \emptyset \\ 0 &{}\text{ if } {\mathbb Y}=\emptyset \end{array}\right. } \end{aligned}$$

for all $i\in {\mathbb Z}$.

Proof

Let $I_{{\mathbb Y}/{\mathbb X}}$ be the ideal of ${\mathbb Y}$ in R and put $\Delta := \frac{\partial (F_1,\ldots ,F_n)}{\partial (x_1,\ldots ,x_n)}$. Then we have $\vartheta _{\mathbb X}=\langle \Delta \rangle _R$. In the case ${\mathbb Y}=\emptyset$, we have $\Delta \in \mathfrak {p}_1\cap \cdots \cap \mathfrak {p}_s = \langle 0\rangle$ by Kunz (1986, Theorem 10.12). This implies ${\vartheta }_{\mathbb X}= \langle 0\rangle$.

In the case ${\mathbb Y}\ne \emptyset$, we use Kunz (1986, Theorem 10.12) again to conclude that the element $\Delta + I_{{\mathbb Y}/{\mathbb X}}$ is a non-zerodivisor of $R_{{\mathbb Y}}=R/I_{{\mathbb Y}/{\mathbb X}}$ and that we have $\Delta \in \mathfrak {p}_j$ for all $j\in \{1,\ldots ,s\}$ such that $p_j\notin \text {Supp}({\mathbb Y})$. Now we fix the degree $i \ge 0$ and suppose $\mathop {\text {HF}}\nolimits _{{\mathbb Y}}(i) = t$. Let $\{g_1+I_{{\mathbb Y}/{\mathbb X}},\ldots ,g_t+I_{{\mathbb Y}/{\mathbb X}}\}$ be a K-basis of the vector space $(R_{{\mathbb Y}})_i$. Then the set $\{\Delta \cdot g_1+I_{{\mathbb Y}/{\mathbb X}},\ldots ,\Delta \cdot g_t+I_{{\mathbb Y}/{\mathbb X}}\} \subseteq (R_{{\mathbb Y}})_{i+r_{{\mathbb X}}}$ is K-linearly independent. It follows that the vector space $(\Delta \cdot R)_{i+r_{{\mathbb X}}}$ has K-dimension greater than or equal to t, in other words, we have $\mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}}(i+r_{{\mathbb X}}) \ge \mathop {\text {HF}}\nolimits _{{\mathbb Y}}(i)$.

On the other hand, we observe that $\Delta \cdot h=0$ in R for every homogeneous element $h\in I_{{\mathbb Y}/{\mathbb X}}{\setminus }\{0\}$, since ${\mathbb X}$ is reduced. For every $f\in R_i$, we write $f=a_1g_1+\cdots +a_tg_t+h$ for some $a_1,\ldots ,a_t\in K$ and $h \in (I_{{\mathbb Y}/{\mathbb X}})_i$. Then $\Delta \cdot f = \Delta \cdot (a_1g_1+\cdots +a_tg_t+h) = a_1\Delta \cdot g_1+\cdots +a_t\Delta \cdot g_t \in \langle \Delta \cdot g_1,\ldots ,\Delta \cdot g_t \rangle _K$ (as $\Delta \cdot h=0$ in R). Thus $(\vartheta _{\mathbb X})_{i+r_{{\mathbb X}}} = (\Delta \cdot R)_{i+r_{{\mathbb X}}} \subseteq \langle \Delta \cdot g_1, \ldots , \Delta \cdot g_t \rangle _K$, and hence $\mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}}(i+r_{{\mathbb X}}) \le t = \mathop {\text {HF}}\nolimits _{{\mathbb Y}}(i)$. Therefore the conclusion follows. $\square$

In particular, if ${\mathbb X}$ is smooth, this proposition simplifies as follows.

Corollary 2.7

Let ${\mathbb X}$ be a smooth 0-dimensional complete intersection in ${\mathbb P}^n_K$.

• (a) The Kähler different of ${\mathbb X}$ is given by ${\vartheta }_{\mathbb X}= \big \langle \frac{\partial (F_1,\ldots ,F_n)}{\partial (x_1,\ldots ,x_n)} \big \rangle _R$, where $\frac{\partial (F_1,\ldots ,F_n)}{\partial (x_1,\ldots ,x_n)}$ is a homogeneous non-zerodivisor of R of degree $r_{{\mathbb X}}=\sum _{i=1}^nd_i-n$.

• (b) The Hilbert function of ${\vartheta }_{\mathbb X}$ satisfies $\mathop {\text {HF}}\nolimits _{{\vartheta }_{\mathbb X}}(i)=\mathop {\text {HF}}\nolimits _{{\mathbb X}}(i-r_{{\mathbb X}})$ for all $i \in {\mathbb Z}$.

Remark 2.8

In the setting of the proposition, the regularity index of the Kähler different is $r_{\mathbb X}+ r_{{\mathbb Y}}$. If we remove the condition that ${\mathbb X}$ is reduced, then we have $\mathop {\text {HF}}\nolimits _{{\vartheta }_{\mathbb X}}(i) \ge \mathop {\text {HF}}\nolimits _{{\mathbb Y}}(i - r_{\mathbb X})$ for all $i \in {\mathbb Z}$. This follows from the first part of the proof of the proposition.

The following example shows that, even if ${\mathbb X}$ is a complete intersection, the eventual value of the Hilbert function of ${\vartheta }_{\mathbb X}$ may not be equal to $\text {deg}({\mathbb X})$.

Example 2.9

Let K be a field with $\text {char}(K) \notin \{2, 3\}$, and let ${\mathbb X}\subseteq {\mathbb P}^2_K$ be the 0-dimensional complete intersection defined by the ideal $I_{{\mathbb X}} = \langle F, G \rangle$, where we have $F=X_1(X_1-2X_0)(X_1+2X_0)$ and $G=(X_2-X_0)(X_1^2+X_2^2-4X_0^2)$. Then $\text {deg}({\mathbb X}) = 9$ and $\text {Supp}({\mathbb X}) = \{p_1, \ldots , p_7\}$, where $p_1=(1:0:1)$, $p_2=(1:0:2)$, $p_3=(1:0:-2)$, $p_4=(1:2:1)$, $p_5=(1:2:0)$, $p_6=(1:-2:1)$, and $p_7=(1:-2:0)$. By $\mathfrak {P}_j$ we denote the associated homogeneous prime ideal in P of $p_j$ for $j=1,\ldots ,7$. The homogeneous primary decomposition of the ideal $I_{\mathbb X}$ is $I_{{\mathbb X}} = I_1 \cap \cdots \cap I_7$, where $I_j=\mathfrak {P}_j$ for $j \ne 5, 7$, $I_5 = \langle X_1-2X_0, X_2^2 \rangle$, and $I_7 = \langle X_1+2X_0, X_2^2 \rangle$. This means that ${\mathbb X}$ is not reduced at $p_5$ and $p_7$, and so ${\mathbb X}$ is not smooth at those points. In this case we have $\vartheta _{\mathbb X}= \big \langle \, \tfrac{\partial (F,G)}{\partial (x_1,x_2)} \, \big \rangle =\langle \, 4x_{0}x_{1}^{2}x_{2} - 16x_{0}^{2}x_{2}^{2} - 3x_{1}^{2}x_{2}^{2} - 2x_{0}x_{2}^{3} + 6x_{2}^{4} \, \rangle$. In particular, the Jacobian determinant $\frac{\partial (F,G)}{\partial (x_1,x_2)}$ is a zerodivisor of R. This shows that the smoothness of ${\mathbb X}$ in the above corollary is a necessary hypothesis. Furthermore, we have

$$\begin{aligned} \begin{array}{lll} \mathop {\text {HF}}\nolimits _{{\mathbb X}} &{}:&{} 1\ 3\ 6\ 8\ 9\ 9\cdots \\ \mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}} &{} :&{} 0\ 0\ 0\ 0\ 1\ 3\ 6\ 7\ 7\cdots \end{array} \end{aligned}$$

Hence the Hilbert function of ${\vartheta }_{\mathbb X}$ can stabilize at a value $\ne \text {deg}({\mathbb X})$.

The next example shows that, in general, the initial degree of the Hilbert function of ${\vartheta }_{\mathbb X}$ can be less than $r_{\mathbb X}$.

Example 2.10

Let ${\mathbb X}\subseteq {\mathbb P}^2_{{\mathbb Q}}$ be the set of five ${\mathbb Q}$-rational points: $p_1 = (1:1:0)$, $p_2 = (1:1:1)$, $p_3=(1:-1:1)$, $p_4=(1:2:1)$, and $p_5=(1:-2:1)$. We have $\mathop {\text {HF}}\nolimits _{{\mathbb X}}:\ 1\ 3\ 4\ 5\ 5\cdots$ and $r_{{\mathbb X}}=3$. Moreover, we have $\mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}}:\ 0\ 0\ 1\ 1\ 3\ 4\ 5\ 5 \cdots$ and $\vartheta _{\mathbb X}= \langle x_0x_1 - x_1x_2, x_1^2x_2^2 - \tfrac{8}{5}x_2^4, x_1^3x_2 - \frac{5}{2}x_1x_2^3\rangle$. Thus, in this case, the initial degree of the Hilbert function of $\vartheta _{{\mathbb X}}$ is less than $r_{{\mathbb X}}=3$.

The following condition will play an important role in this paper.

Definition 2.11

A 0-dimensional scheme ${\mathbb X}\subseteq {\mathbb P}^n_K$ is called arithmetically Gorenstein if R is a Gorenstein ring.

Note that if ${\mathbb X}$ is a complete intersection then it is arithmetically Gorenstein, but the converse is not true in general, as the next example shows. Moreover, for an arithmetically Gorenstein scheme ${\mathbb X}$, later results will show that the initial degree of the Hilbert function of ${\vartheta }_{\mathbb X}$ is at least $r_{\mathbb X}$. The next example also illustrates that this initial degree can be strictly higher.

Example 2.12

Let ${\mathbb X}\subseteq {\mathbb P}^3_{{\mathbb F}_7}$ be the following set of five distinct ${\mathbb F}_7$-rational points on the twisted cubic curve: $p_1= (1:0:0:0)$, $p_2=(1:1:1:1)$, $p_3=(1:-1:1:-1)$, $p_4=(1:2:4:8)$, and $p_5=(8:4:2:1)$. We have $\mathop {\text {HF}}\nolimits _{{\mathbb X}}:\ 1\ 4\ 5\ 5\cdots$ and $r_{\mathbb X}=2$. An application of Theorem 7 in Geramita and Orecchia (1981) shows that ${\mathbb X}$ is arithmetically Gorenstein. Moreover, a calculation gives us $\vartheta _{\mathbb X}= \langle x_{2}x_{3}^{2} -3x_{3}^{3}, x_{1}x_{3}^{2}, x_{0}x_{3}^{2} -3x_{3}^{3},x_{0}^{3} \rangle$ and $\mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}}:\ 0\ 0\ 0\ 4\ 5\ 5\cdots$. Hence ${\mathbb X}$ is not a complete intersection and the initial degree of $\mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}}$ is $3>r_{{\mathbb X}}$.

As mentioned above, the Kähler different of a complete intersection ${\mathbb X}$ is a principal ideal. The following example shows that the Kähler different can be principal without ${\mathbb X}$ being a complete intersection.

Example 2.13

Let ${\mathbb X}\subseteq {\mathbb P}^2_{{\mathbb Q}}$ be the set consisting of six ${\mathbb Q}$-rational points: $p_1=(1:0:0)$, $p_2 = (1:0:1)$, $p_3 = (1:1:0)$, $p_4 = (1:1:1)$, $p_5 = (1:2:0)$, and $p_6 = (1:2:1)$, and let ${\mathbb Y}$ be the fat point scheme defined by the saturated homogeneous ideal

$$\begin{aligned} I_{{\mathbb Y}} = \mathfrak {P}_1\cap \cdots \cap \mathfrak {P}_4 \cap \mathfrak {P}_5^2 \cap \mathfrak {P}_6^2. \end{aligned}$$

Then ${\mathbb Y}$ is an almost complete intersection, i.e., $I_{{\mathbb Y}}$ is minimally generated by three homogeneous polynomials in P. In particular, ${\mathbb Y}$ is not a complete intersection. But in this case $\vartheta _{\mathbb Y}$ is the principal ideal generated by the homogeneous polynomial

$$\begin{aligned} x_{0}^{5} -\tfrac{15}{4}x_{0}^{2}x_{1}^{3} + \tfrac{55}{16}x_{0}x_{1}^{4} -\tfrac{13}{16}x_{1}^{5} + \tfrac{3}{4}x_{1}^{3}x_{2}^{2} -\tfrac{7}{2}x_{1}^{2}x_{2}^{3} + 5x_{1}x_{2}^{4} -2x_{2}^{5}. \end{aligned}$$

Moreover, the Hilbert function of $\vartheta _{\mathbb Y}$ is $\mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb Y}}:\ 0\ 0\ 0\ 0\ 0\ 1\ 3\ 4\ 4\cdots$ and its regularity index satisfies $\text {ri}(\vartheta _{\mathbb Y}) = 7 < 8= 2r_{\mathbb Y}$.

In the last part of this section we collect some results about the eventual value of $\mathop {\text {HF}}\nolimits _{{\vartheta }_{\mathbb X}}$, i.e., the Hilbert polynomial of ${\vartheta }_{\mathbb X}$.

Remark 2.14

If ${\mathbb X}$ is a fat point scheme in ${\mathbb P}^n_K$, then the Hilbert polynomial of the Kähler different $\vartheta _{\mathbb X}$ is exactly the number of reduced points of the scheme ${\mathbb X}$, and we have $\text {ri}(\vartheta _{\mathbb X}) \le n\, r_{{\mathbb X}}$ (see Kreuzer et al. 2015, Theorem 2.5).

Apart from some other special cases, to exactly determine the Hilbert polynomial of the Kähler different for an arbitrary 0-dimensional subscheme ${\mathbb X}$ of ${\mathbb P}^n_K$ is not an easy task. Hence we try at least to find (possibly sharp) bounds for it.

Proposition 2.15

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional scheme, and let ${\mathbb X}_{\mathrm {sm}}$ be the set of smooth points in its support $\text {Supp}({\mathbb X})=\{p_1,\ldots ,p_s\}$. Then we have

$$\begin{aligned} {\textstyle \sum \limits _{p_j\in {\mathbb X}_\mathrm{{sm}}}} \dim _K({\mathcal {O}}_{{\mathbb X},p_j}) \le \text {HP}_{\vartheta _{\mathbb X}} \le \text {deg}({\mathbb X})-(s-\#{\mathbb X}_\mathrm{{sm}}). \end{aligned}$$

Proof

Let $\mathfrak {P}_j\subseteq P$ be the associated prime ideal of $p_j$ for $j=1,\ldots ,s$, and set

$$\begin{aligned} I := { \textstyle \bigcap \limits _{p_j\in \text {Supp}({\mathbb X}){\setminus }{\mathbb X}_{\mathrm {sm}}} } \mathfrak {P}_j. \end{aligned}$$

It follows from Kunz (1986, Theorem 10.12) that $\vartheta _{\mathbb X}\subseteq \mathfrak {p}_j= \mathfrak {P}_j/I_{{\mathbb X}}$ for every point $p_j \in \text {Supp}({\mathbb X}) {\setminus } {\mathbb X}_{\mathrm {sm}}$. Hence we get $\vartheta _{\mathbb X}\subseteq I/I_{{\mathbb X}}$, and consequently

$$\begin{aligned} \mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}}(i) \le \mathop {\text {HF}}\nolimits _{I/I_{{\mathbb X}}}(i) = \mathop {\text {HF}}\nolimits _{{\mathbb X}}(i)-\mathop {\text {HF}}\nolimits _{P/I}(i) \le \text {deg}({\mathbb X}) - \text {deg}({\mathbb Y}) \end{aligned}$$

for all $i \in {\mathbb N}$, where ${\mathbb Y}$ is the 0-dimensional subscheme of ${\mathbb P}^n_K$ defined by I. Observe that the scheme ${\mathbb Y}$ has degree $\text {deg}({\mathbb Y}) \ge s - \#{\mathbb X}_{\mathrm {sm}}$. Thus we obtain

$$\begin{aligned} \text {HP}_{\vartheta _{\mathbb X}} \le \text {deg}({\mathbb X}) - \text {deg}({\mathbb Y}) \le \text {deg}({\mathbb X}) - (s - \#{\mathbb X}_{\mathrm {sm}}). \end{aligned}$$

Next we prove the first inequality of $\text {HP}_{\vartheta _{\mathbb X}}$. If ${\mathbb X}_{\mathrm {sm}} = \emptyset$, then there is nothing to prove, since we always have $\text {HP}_{\vartheta _{\mathbb X}} \ge 0$. Now let us consider the case $\#{\mathbb X}_{\mathrm {sm}}\ge 1$. W.l.o.g. we may assume that ${\mathbb X}_{\mathrm {sm}} = \{p_{1}, \ldots , p_{\varrho }\}$ where $\varrho := \#{\mathbb X}_{\mathrm {sm}}$. Then Kunz (1986, Theorem 10.12) implies $\vartheta _{\mathbb X}\nsubseteq \mathfrak {p}_j$ for all $j = 1, \ldots , \varrho$. It follows from Homogeneous Prime Avoidance (see for instance Kreuzer and Robbiano 2005, Proposition 5.6.22) that there exists a homogeneous element $h \in (\vartheta _{\mathbb X})_m {\setminus }\{0\}$ for some $m \ge 0$ such that $h \notin \bigcup _{j=1}^{\varrho }(\mathfrak {p}_j)_m$. This implies that $h_{p_j} \ne 0$ in ${\mathcal {O}}_{{\mathbb X},p_j}$ for $j = 1,\ldots ,\varrho$. Let $j \in \{1, \ldots , \varrho \}$, let $\varkappa _j = \dim _K({\mathcal {O}}_{{\mathbb X},p_j})$, and let $\{e_{j1}, \ldots , e_{j\varkappa _j}\}$ be a K-basis of ${\mathcal {O}}_{{\mathbb X},p_j}$. For any non-zero element $a \in {\mathcal {O}}_{{\mathbb X},p_j}$, it is not difficult to verify that $\{a e_{j1}, \ldots , a e_{j\varkappa _j}\}$ is a K-basis of ${\mathcal {O}}_{{\mathbb X},p_j}$, then so is $\{h_{p_j}e_{jk}e_{j1},\ldots ,h_{p_j}e_{jk}e_{j\varkappa _j}\}$, where $1\le k \le \varkappa _j$.

Now we consider the isomorphism of K-vector spaces $\imath : R_{r_{{\mathbb X}}} \rightarrow \prod _{j=1}^s {\mathcal {O}}_{{\mathbb X},p_j}$ given by $\imath (f) = (f_{p_1},\ldots ,f_{p_s})$, where $f_{p_j}\in {\mathcal {O}}_{{\mathbb X},p_j}$ is the germ of f at $p_j$ for $j=1,\ldots ,s$ (cf. Kreuzer 1994, Lemma 1.1). For all $j = 1, \ldots , \varrho$ and for all $k_j = 1, \ldots , \varkappa _j$, we let $f_{jk_j} = \imath ^{-1}((0, \ldots , 0, e_{jk_j}, 0, \ldots , 0)) \in R_{r_{{\mathbb X}}}$. Then we get

$$\begin{aligned} \langle hf_{11}, \ldots , hf_{1\varkappa _1}, \ldots , hf_{\varrho 1}, \ldots , hf_{\varrho \varkappa _{\varrho }} \rangle _K \subseteq (\vartheta _{\mathbb X})_{r_{{\mathbb X}}+m} \subseteq R_{r_{{\mathbb X}}+m}. \end{aligned}$$

We show that $\{\, hf_{11}, \ldots , hf_{1\varkappa _1}, \ldots , hf_{\varrho 1}, \ldots ,hf_{\varrho \varkappa _{\varrho }} \,\}$ is K-linearly independent. Remark that for $j_1, j_2 \in \{1,\ldots ,\varrho \}$ and for $k_{i}\in \{1,\ldots ,\varkappa _{j_i}\}$, where $i = 1, 2$, we have $f_{j_1k_1} \cdot f_{j_2k_2} \ne 0$ if $j_1 = j_2$ and $f_{j_1k_1} \cdot f_{j_2k_2} = 0$ if $j_1 \ne j_2$, and $hf_{j_1k_1}^2 \ne 0$ in $R_{2r_{{\mathbb X}}+m}$. Suppose for contradiction that there are $c_{11}, \ldots , c_{1\varkappa _1}, \ldots , c_{\varrho 1}, \ldots , c_{\varrho \varkappa _{\varrho }} \in K$, not all equal to zero, such that $\sum _{j=1}^{\varrho }\sum _{k_j=1}^{\varkappa _j} c_{jk_j}h f_{jk_j} = 0$. W.l.o.g. we may assume $c_{11} \ne 0$. We then have

$$\begin{aligned} hf_{11}^2=\tfrac{1}{c_{11}}\left( {\textstyle \sum \limits _{k_1=2}^{\varkappa _1}} c_{1k_1}hf_{1k_1}f_{11} + {\textstyle \sum \limits _{ j=2}^{\varrho } \sum \limits _{ k_j=1}^{\varkappa _j} } c_{jk_j}hf_{jk_j}f_{11}\right) =\tfrac{1}{c_{11}} {\textstyle \sum \limits _{k_1=2}^{\varkappa _1}} c_{1k_1}hf_{11}f_{1k_1}. \end{aligned}$$

Thus, in ${\mathcal {O}}_{{\mathbb X},p_1}$, we get the equality $h_{p_1}e_{11}^2 = \tfrac{1}{c_{11}}\sum _{k_1=2}^{\varkappa _1} c_{1k_1}h_{p_1}e_{11}e_{1k_1}$, in contradiction to the fact that $\{h_{p_1}e_{11}^2,h_{p_1}e_{11}e_{12},\ldots ,h_{p_1}e_{11}e_{1\varkappa _1}\}$ is a K-basis of ${\mathcal {O}}_{{\mathbb X},p_1}$. Therefore we obtain

$$\begin{aligned} \text {HP}_{\vartheta _{\mathbb X}}\ge & {} \dim _K \langle hf_{11}, \ldots , hf_{1\varkappa _1}, \ldots , hf_{\varrho 1}, \ldots , hf_{\varrho \varkappa _{\varrho }} \rangle _K \\= & {} {\textstyle \sum \limits _{j=1}^{\varrho }} \varkappa _j = {\textstyle \sum \limits _{p_j \in {\mathbb X}_{\mathrm {sm}}} }\dim _K({\mathcal {O}}_{{\mathbb X},p_j}) \end{aligned}$$

and the proposition is completely proved. $\square$

Clearly, the lower bound for $\text {HP}_{\vartheta _{\mathbb X}}$ is attained for a smooth scheme ${\mathbb X}$. Our next example shows that the upper bound for $\text {HP}_{\vartheta _{\mathbb X}}$ is also sharp.

Example 2.16

Let us go back to Example 2.9. The scheme ${\mathbb X}$ is a complete intersection with $\text {deg}({\mathbb X})=9$, and it is not smooth at two points $p_5$ and $p_7$. In this case we have $\varrho = 5$ and

$$\begin{aligned} \text {HP}_{\vartheta _{\mathbb X}} = \text {deg}({\mathbb X})-(s-\varrho ) = 7 > 5 = \varrho = {\textstyle \sum \limits _{p_j \in {\mathbb X}_{\mathrm {sm}}} } \dim _K({\mathcal {O}}_{{\mathbb X},p_j}) \end{aligned}$$

and $\text {ri}(\vartheta _{\mathbb X}) = 7 < 8=2r_{{\mathbb X}}$.

3 Dedekind differents of zero-dimensional schemes

In this section we continue to use the notation introduced above. Recall that the graded R-module $\omega _R = \underline{\text {Hom}}_{K[x_0]}(R,K[x_0])(-1)$ is called the canonical module of R. It is a finitely generated graded R-module with Hilbert function $\mathop {\text {HF}}\nolimits _{\omega _R}(i)=\text {deg}({\mathbb X})-\mathop {\text {HF}}\nolimits _{\mathbb X}(-i)$ for all $i\in {\mathbb Z}$. (For further details about this module, we refer to Kreuzer 1994.)

It is known that, for a locally Gorenstein scheme ${\mathbb X}$, one can embed the canonical module of R as a fractional ideal into its homogeneous ring of quotients (see Herzog et al. 1971). Subsequently, we need to make this construction explicit. The presentation follows the construction given in Kreuzer (2000), generalizing it to the case at hand.

In a first step, we want to describe the homogeneous ring of quotients $Q^h(R)$ of R. It is defined as the localization of R with respect to the set of all homogeneous non-zerodivisors of R. In view of Kreuzer (1994, Lemma 1.1), there is a homogeneous injection of degree zero

$$\begin{aligned} \widetilde{\imath }: R \rightarrow \widetilde{R} := {\textstyle \prod \limits _{j=1}^s} {\mathcal {O}}_{{\mathbb X},p_j}[T_j] \hookrightarrow {\textstyle \prod \limits _{j=1}^s} {\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]\qquad \qquad \qquad \qquad \quad (*) \end{aligned}$$

given by $\widetilde{\imath }(f)=(f_{p_1}T_1^i,\ldots ,f_{p_s}T_s^i)$, where $f\in R_i$ for $i\ge 0$, and where $T_1,\ldots ,T_s$ are indeterminates with $\text {deg}(T_1)=\cdots =\text {deg}(T_s)=1$. In particular, we have $\widetilde{\imath }(x_0)=(T_1,\ldots ,T_s)$. Now we can describe $Q^h(R)$ as follows.

Proposition 3.1

The map $\widetilde{\imath }$ extends to an isomorphism of graded R-modules

$$\begin{aligned} \Psi : Q^h(R) \mathop {\longrightarrow }\limits ^{\sim } {\textstyle \prod \limits _{j=1}^s} {\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}], \end{aligned}$$

where for every element $f/g\in Q^h(R)$ with $f \in R_k$ and a non-zerodivisor $g\in R_l$ we have

$$\begin{aligned} \Psi (\frac{f}{g}) = \frac{\widetilde{\imath }(f)}{\widetilde{\imath }(g)} = \left( \frac{f_{p_1}}{g_{p_1}} T_1^{k-l}, \ldots , \frac{f_{p_s}}{g_{p_s}} T_s^{k-l} \right) . \end{aligned}$$

In particular, we have $Q^h(R) \cong R_{x_0}$.

Proof

For a non-zerodivisor $g \in R_i$, the element $g_{p_j} \in {\mathcal {O}}_{{\mathbb X},p_j}$ is a unit element for all $j = 1, \ldots , s$ (see Kreuzer 1998, Lemma 1.5). Let $f/g \in Q^h(R)$ with $f \in R_k$ and a non-zerodivisor $g \in R_l$. Then $\widetilde{\imath }(f) = (f_{p_1}T_1^k,\ldots ,f_{p_s}T_s^k)$ and $\widetilde{\imath }(g) = (g_{p_1}T_1^l,\ldots ,g_{p_s}T_s^l)$, so we get

$$\begin{aligned} \Psi \left( \frac{f}{g} \right) = \frac{\widetilde{\imath }(f)}{\widetilde{\imath }(g)} = \left( \frac{f_{p_1}}{g_{p_1}} T_1^{k-l}, \ldots , \frac{f_{p_s}}{g_{p_s}} T_s^{k-l} \right) \in {\textstyle \prod \limits _{j=1}^s} {\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]. \end{aligned}$$

Thus the map $\Psi : Q^h(R) \rightarrow \prod _{j=1}^s{\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]$ is well defined. It is clearly true that $\Psi$ is R-linear, homogeneous of degree zero. If $\Psi (\frac{f}{g}) = 0$, then $\frac{f_{p_j}}{g_{p_j}}=0\in {\mathcal {O}}_{{\mathbb X},p_j}$ for all $j=1,\ldots ,s$. This implies $f_{p_j} = 0$ for all $j=1,\ldots ,s$, and so $f=0$, since the map $\widetilde{\imath }$ is injective. Hence the map $\Psi$ is an injection. Now we show that the map $\Psi$ is surjective. Let $(g_1,\ldots ,g_s)\in \prod _{j=1}^s{\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]$. For $i \gg 0$ we have $\dim _K(R_i) = \text {deg}({\mathbb X}) = \dim _K(\prod _{j=1}^s{\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}])_i$. Thus, for $i \gg 0$, $(T_1^ig_1, \ldots , T_s^ig_s)$ is of the form $\Psi (f)$ for some $f \in R$. Therefore the element $(g_1, \ldots , g_s)$ is the image of $f/x_0^i$, and the claim follows. $\square$

Recall that a trace map of a finite algebra T / S is a T-basis of the module $\text {Hom}_S(T,S)$. The second task we want to tackle is the construction of a trace map for the algebra $Q^h(R) / K[x_0,x_0^{-1}]$. For this we need to restrict our attention to a special class of 0-dimensional schemes.

Definition 3.2

A 0-dimensional scheme ${\mathbb X}\subseteq {\mathbb P}^n_K$ is called locally Gorenstein if at each point $p_j\in \text {Supp}({\mathbb X})$ the local ring ${\mathcal {O}}_{{\mathbb X},p_j}$ is a Gorenstein ring.

The next proposition says that in the locally Gorenstein case the desired trace map exists.

Proposition 3.3

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional locally Gorenstein scheme with $\text {Supp}({\mathbb X}) = \{p_1, \ldots , p_s\}$, and let $L_0 = K[x_0,x_0^{-1}]$. Then the following statements hold true.

• (a) The algebra $Q^h(R) / L_0$ has a homogeneous trace map $\sigma$ of degree zero.

• (b) The map $\Sigma : Q^h(R)\rightarrow \underline{\text {Hom}}_{L_0}(Q^h(R),L_0)$ given by $\Sigma (1)=\sigma$ is an isomorphism of graded $Q^h(R)$-modules.

• (c) A homogeneous element $\sigma ' \in \underline{\text {Hom}}_{L_0}(Q^h(R),L_0)$ is a trace map of the algebra $Q^h(R) / L_0$ if and only if there exists a unit $u\in Q^h(R)$ such that $\sigma ' = u\cdot \sigma$.

Proof

According to Proposition 3.1, we may identify $Q^h(R) = \prod _{j=1}^s {\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]$. Then we have

$$\begin{aligned} \underline{\text {Hom}}_{L_0}(Q^h(R),L_0)= & {} \underline{\text {Hom}}_{L_0}\left( {\textstyle \prod \limits _{j=1}^s} {\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}], L_0 \right) \\\cong & {} {\textstyle \prod \limits _{j=1}^s} \underline{\text {Hom}}_{L_0}({\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}], L_0) \\\cong & {} {\textstyle \prod \limits _{j=1}^s} \underline{\text {Hom}}_{L_0}(L_0\otimes _K{\mathcal {O}}_{{\mathbb X},p_j}, L_0) \\\cong & {} {\textstyle \prod \limits _{j=1}^s} L_0 \otimes _K \text {Hom}_{K}({\mathcal {O}}_{{\mathbb X},p_j}, K). \end{aligned}$$

Since ${\mathbb X}$ is locally Gorenstein, the algebra ${\mathcal {O}}_{{\mathbb X},p_j}/K$ is a finite Gorenstein algebra for every $j\in \{1,\ldots ,s\}$. It then follows from Kunz (1986, E.16) that there is a trace map $\overline{\sigma }_j\in \text {Hom}_{K}({\mathcal {O}}_{{\mathbb X},p_j},K)$ such that $\text {Hom}_{K}({\mathcal {O}}_{{\mathbb X},p_j},K)={\mathcal {O}}_{{\mathbb X},p_j}\cdot \overline{\sigma }_j$ for $j=1,\ldots ,s$. By Kunz (1986, F.16), the map $\sigma _j=\overline{\sigma }_j\otimes \text {id}_{L_0}: {\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]\rightarrow K[T_j,T_j^{-1}]\cong L_0$ is a homogeneous trace map of degree zero of the algebra ${\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]/L_0$. Hence the trace system $\sigma =(\sigma _1,\ldots ,\sigma _s)$ satisfies $\underline{\text {Hom}}_{L_0}(Q^h(R),L_0) = Q^h(R)\cdot \sigma$, and claim (a) follows.

Since (c) follows from (b), it remains to prove claim (b). To this end, we show that $\text {Ann}_{Q^h(R)}(\sigma ) = \langle 0\rangle$. Assume that $f\cdot \sigma = 0$ for some homogeneous element $f\in Q^h(R)$. We have $f\cdot \sigma (g)=\sigma (fg)=g\cdot \sigma (f)=0$ for all $g\in Q^h(R)$. This implies $\alpha (f)=0$ for every $\alpha \in \underline{\text {Hom}}_{L_0}(Q^h(R),L_0)$. Since the algebra $R/K[x_0]$ is free of rank $\text {deg}({\mathbb X})$ and $Q^h(R) \cong R_{x_0} \cong R\otimes _{K[x_0]}L_0$, it follows that the algebra $Q^h(R)/L_0$ is also free of rank $\text {deg}({\mathbb X})$. Let $d = \text {deg}({\mathbb X})$, let $\{b_1, \ldots , b_d\}$ be a $L_0$-basis of $Q^h(R)$, and let $\{b_1^*,\ldots ,b_d^*\}$ be the dual basis of $\{b_1, \ldots , b_d\}$. We write $f = \sum _{j=1}^d g_jb_j \in Q^h(R)$ with $g_1,\ldots ,g_d \in L_0$. Then $g_j = b_j^*(\sum _{j=1}^d g_jb_j) = b_j^*(f) = 0$ for all $j = 1, \ldots , d$. Hence we obtain $f=0$, and so $\text {Ann}_{Q^h(R)}(\sigma ) = \langle 0 \rangle$. $\square$

When the scheme ${\mathbb X}$ is smooth, the algebra $Q^h(R) / K[x_0,x_0^{-1}]$ has a canonical trace map.

Corollary 3.4

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional smooth scheme, let Q(R) denote the full ring of quotients of R, and set $L_0:=K[x_0,x_0^{-1}]$.

• (a) The canonical trace map $\text {Tr}_{Q^h(R)/L_0}$ is a homogeneous trace map of degree zero of $Q^h(R)/L_0$. In particular, $Q^h(R)/L_0$ is étale.

• (b) The canonical trace map $\text {Tr}_{Q(R)/K(x_0)}$ is a trace map of $Q(R)/K(x_0)$. In particular, $Q(R)/K(x_0)$ is étale.

Proof

For $j=1,\ldots ,s$, the algebra ${\mathcal {O}}_{{\mathbb X},p_j}/K$ is a finite separable field extension, and hence the canonical trace map $\text {Tr}_{{\mathcal {O}}_{{\mathbb X},p_j}/K}$ (see Kunz 1986, F.3) is a trace map of ${\mathcal {O}}_{{\mathbb X},p_j}/K$. If we choose $\overline{\sigma }_j = \text {Tr}_{{\mathcal {O}}_{{\mathbb X},p_j}/K}$ for $j = 1, \ldots , s$ in the construction of the trace map $\sigma$ in Proposition 3.3(a), then $\sigma$ is exactly the canonical trace map $\text {Tr}_{Q^h(R)/L_0}$. The additional claim of (a) follows by Kunz (1986, F.8).

For (b), it suffices to show that $\text {Tr}_{Q(R)/K(x_0)}$ is a trace map of $Q(R)/K(x_0)$. Since ${\mathbb X}$ is reduced, it follows from Kunz (1985, III, Proposition 4.23) and Bourbaki (1989, V, §1, Proposition 9) that

$$\begin{aligned} Q(R) \cong {\textstyle \prod \limits _{j=1}^s} Q(R/\mathfrak {p}_j) \cong {\textstyle \prod \limits _{j=1}^s} {\mathcal {O}}_{{\mathbb X},p_j}(T_j). \end{aligned}$$

As above, ${\mathcal {O}}_{{\mathbb X},p_j}/K$ is a finite separable field extension, and so ${\mathcal {O}}_{{\mathbb X},p_j}$ and $K(x_0)$ are linearly disjoint over K (cf. Morandi 1996, V, Section 20). This implies ${\mathcal {O}}_{{\mathbb X},p_j}\otimes _KK(x_0)\cong {\mathcal {O}}_{{\mathbb X},p_j}K(x_0)={\mathcal {O}}_{{\mathbb X},p_j}(x_0)$. By letting $\Gamma = \prod _{j=1}^s {\mathcal {O}}_{{\mathbb X},p_j}$, we have the isomorphism $Q(R) \cong K(x_0)\otimes _K \Gamma$. Notice that $\text {Tr}_{\Gamma /K} = (\text {Tr}_{{\mathcal {O}}_{{\mathbb X},p_1}/K},\ldots ,\text {Tr}_{{\mathcal {O}}_{{\mathbb X},p_s}/K})$ is a trace map of $\Gamma /K$, and $\text {Tr}_{Q(R)/K(x_0)}= \mathrm {id}_{K(x_0)}\otimes _K \text {Tr}_{\Gamma /K}$ (cf. Kunz 1986, F.5). Therefore $\text {Tr}_{Q(R)/K(x_0)}$ is a trace map of $Q(R)/K(x_0)$, as desired. $\square$

Now we are ready to introduce the Dedekind complementary module and the Dedekind different for a 0-dimensional locally Gorenstein scheme as follows.

Definition 3.5

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional locally Gorenstein scheme, let $L_0= K[x_0,x_0^{-1}]$, let $\sigma$ be a fixed trace map of $Q^h(R)/L_0$. Then there is an injective homomorphism of graded R-modules

$$\begin{aligned} \Phi : \omega _R(1)&\hookrightarrow&\underline{\text {Hom}}_{L_0}(Q^h(R),L_0) = Q^h(R)\cdot \sigma \xrightarrow {\Sigma ^{-1}} Q^h(R)\\ \varphi\longmapsto & {} \varphi \otimes \text {id}_{L_0} \end{aligned}$$

The image of $\Phi$ is a homogeneous fractional R-ideal $\mathfrak {C}^{\sigma }_{\mathbb X}$ of L, it is called the Dedekind complementary module of ${\mathbb X}$ (or of $R/K[x_0]$) with respect to $\sigma$. Its inverse,

$$\begin{aligned} \delta ^{\sigma }_{\mathbb X}= (\mathfrak {C}^{\sigma }_{\mathbb X})^{-1} = \{\, f \in Q^h(R) \, \mid \, f \cdot \mathfrak {C}^{\sigma }_{\mathbb X}\subseteq R \, \}, \end{aligned}$$

is called the Dedekind different of ${\mathbb X}$ (or of $R/K[x_0]$) with respect to $\sigma$.

In the next remark we collect some basic properties of the Dedekind complementary module.

Remark 3.6

• (a) It follows from the isomorphism $\mathfrak {C}^{\sigma }_{{\mathbb X}} \cong \omega _R(1)$ and Kreuzer (1994, Proposition 1.3) that the graded R-module $\mathfrak {C}^{\sigma }_{{\mathbb X}}$ is finitely generated and

  $$\begin{aligned} \mathop {\text {HF}}\nolimits _{\mathfrak {C}^{\sigma }_{{\mathbb X}}}(i) = \text {deg}({\mathbb X}) - \mathop {\text {HF}}\nolimits _{{\mathbb X}}(-i-1)\quad \text {for all} \ i\in {\mathbb Z}. \end{aligned}$$

• (b) A system of generators of $\mathfrak {C}^{\sigma }_{{\mathbb X}}$ can be computed as follows. Let $<_{\tau }$ be a degree-compatible term ordering on the set of terms ${\mathbb T}^n$ of $K[X_1,\ldots ,X_n]$, and let $d = \text {deg}({\mathbb X})$. Then ${\mathbb T}^n{\setminus }\text {LT}_{\tau }(I_{{\mathbb X}}^{\mathrm {deh}}) = \{\, T_1, \ldots , T_d \,\}$ with $T_j = X_1^{\alpha _{j1}}\cdots X_n^{\alpha _{jn}}$ and $\alpha _j=(\alpha _{j1}, \ldots , \alpha _{jn}) \in {\mathbb N}^n$ for $j = 1, \ldots , d$. W.l.o.g. we assume that $T_1<_{\tau }\cdots <_{\tau }T_m$. Let $t_j = T_j + I_{{\mathbb X}} \in R$ and set $\text {deg}(t_j) := \text {deg}(T_j) = n_j$ for $j = 1, \ldots , d$. Then we have $n_1 \le \cdots \le n_d \le r_{{\mathbb X}}$ and the set $\{\, t_1, \ldots , t_d \,\}$ is a $K[x_0]$-basis of R (cf. Kreuzer and Robbiano 2005, Theorem 4.3.22). Let $\{t_1^*, \ldots , t_d^*\}$ be the dual basis of $\{t_1, \ldots , t_d\}$, and let $g_j = \Phi (t_j^*)$ for $j = 1, \ldots , d$. We get $\mathfrak {C}^{\sigma }_{{\mathbb X}} = \langle g_1, \ldots , g_d \rangle _{K[x_0]} \subseteq Q^h(R)$.

• (c) When ${\mathbb X}$ is smooth, we also denote the Dedekind complementary module (respectively, the Dedekind different) with respect to the canonical trace map by $\mathfrak {C}_{\mathbb X}$ (respectively, $\delta _{\mathbb X}$).

Further properties of the Dedekind different of ${\mathbb X}$ are given in our next proposition.

Proposition 3.7

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional locally Gorenstein scheme, and let $\sigma$ be a trace map for $Q^h(R)/L_0$.

• (a) The Dedekind different $\delta ^{\sigma }_{\mathbb X}$ is a homogeneous ideal of R and $x_0^{2r_{{\mathbb X}}} \in \delta ^{\sigma }_{\mathbb X}$.

• (b) The Hilbert function of $\delta ^{\sigma }_{\mathbb X}$ satisfies $\mathop {\text {HF}}\nolimits _{\delta ^{\sigma }_{\mathbb X}}(i) = 0$ for $i < 0$, $\mathop {\text {HF}}\nolimits _{\delta ^{\sigma }_{\mathbb X}}(i) = \text {deg}({\mathbb X})$ for $i \ge 2r_{{\mathbb X}}$, and

  $$\begin{aligned} 0 \le \mathop {\text {HF}}\nolimits _{\delta ^{\sigma }_{\mathbb X}}(0) \le \cdots \le \mathop {\text {HF}}\nolimits _{\delta ^{\sigma }_{\mathbb X}}(2r_{{\mathbb X}}) = \text {deg}({\mathbb X}). \end{aligned}$$

• (c) The regularity index of $\delta ^{\sigma }_{\mathbb X}$ satisfies $r_{{\mathbb X}} \le \text {ri}(\delta ^{\sigma }_{\mathbb X}) \le 2r_{{\mathbb X}}$.

Proof

By Remark 3.6(a), we have $\mathop {\text {HF}}\nolimits _{\mathfrak {C}^{\sigma }_{{\mathbb X}}}(0) = \text {deg}({\mathbb X})-\mathop {\text {HF}}\nolimits _{{\mathbb X}}(-1) = \text {deg}({\mathbb X}) = \mathop {\text {HF}}\nolimits _{Q^h(R)}(0)$. This implies $R_0 \subseteq (\mathfrak {C}^{\sigma }_{{\mathbb X}})_0 = Q^h(R)_0$, and in particular, $1 \in \mathfrak {C}^{\sigma }_{{\mathbb X}}$. Hence $\delta ^{\sigma }_{\mathbb X}$ is a homogeneous ideal of R.

Now let us write $\mathfrak {C}^{\sigma }_{{\mathbb X}} = \langle g_1, \ldots , g_d \rangle _{K[x_0]} \subseteq Q^h(R)$ as in Remark 3.6(b). Here $g_j$ is homogeneous of degree $\text {deg}(g_j) = -n_j$ (since $\Phi$ is homogeneous of degree zero). We claim that, for $j \in \{1, \ldots , d\}$, there is a homogeneous element $g'_j \in R_{r_{{\mathbb X}}}$ such that $g_j = x_0^{-r_{{\mathbb X}}-n_j} g'_j \in \mathfrak {C}^{\sigma }_{\mathbb X}$. Indeed, since $g_j \in Q^h(R) \cong R_{x_0}$, there exist $g''_j \in R$ and $d_j \in {\mathbb N}$ such that $g_j = x_0^{-d_j} g''_j$. If $\text {deg}(g''_j) = d_j - n_j \le r_{{\mathbb X}}$, then we set $g'_j = x_0^{r_{{\mathbb X}} - d_j + n_j} g''_j \in R_{r_{{\mathbb X}}}$. If $\text {deg}(g''_j) = d_j - n_j > r_{{\mathbb X}}$, then we write $g''_j = x_0^{d_j - n_j - r_{{\mathbb X}}} g'_j$ for some $g'_j\in R_{r_{{\mathbb X}}}$, since $R_i = x_0^{i-r_{{\mathbb X}}} R_{r_{{\mathbb X}}}$ for all $i \ge r_{{\mathbb X}}$. Thus we get $g_j = x_0^{-r_{{\mathbb X}}-n_j} g'_j$, as claimed. Consequently, we have $\mathfrak {C}^{\sigma }_{\mathbb X}= \langle x_0^{-r_{{\mathbb X}}-n_1}g'_1, \ldots , x_0^{-r_{{\mathbb X}}-n_m}g'_m \rangle _R$. Now it is easy to see that $x_0^{2r_{{\mathbb X}}} \in \delta ^{\sigma }_{\mathbb X}$, since $n_j \le r_{{\mathbb X}}$ and $x_0^{2r_{{\mathbb X}}}\cdot (x_0^{-r_{{\mathbb X}}-n_j}g'_j) = x_0^{r_{{\mathbb X}}-n_j}g'_j \in R_{2r_{{\mathbb X}}-n_j}$ for all $j = 1, \ldots , d$. Hence claim (a) follows.

Next we shall prove claim (b). It is clear that $\mathop {\text {HF}}\nolimits _{\delta ^{\sigma }_{\mathbb X}}(i)=0$ for $i<0$ and $\mathop {\text {HF}}\nolimits _{\delta ^{\sigma }_{\mathbb X}}(i) \le \mathop {\text {HF}}\nolimits _{\delta ^{\sigma }_{\mathbb X}}(i+1)$ for all $i\in {\mathbb Z}$, since $\delta ^{\sigma }_{\mathbb X}$ is a homogeneous ideal of R by (a). Notice that $\mathop {\text {HF}}\nolimits _{{\mathbb X}}(i)=d=\text {deg}({\mathbb X})$ for all $i \ge r_{{\mathbb X}}$ and $\mathop {\text {HF}}\nolimits _{\delta ^{\sigma }_{\mathbb X}}(i) \le \mathop {\text {HF}}\nolimits _{{\mathbb X}}(i)$ for all $i \in {\mathbb Z}$. So, the Hilbert function of $\delta ^{\sigma }_{\mathbb X}$ satisfies $\mathop {\text {HF}}\nolimits _{\delta ^{\sigma }_{\mathbb X}}(i) \le d$ for all $i \in {\mathbb Z}$. We write $\mathfrak {C}^{\sigma }_{\mathbb X}= \langle x_0^{-r_{{\mathbb X}}-n_1}g'_1, \ldots , x_0^{-r_{{\mathbb X}}-n_d}g'_d \rangle _R$ with $g'_1, \ldots , g'_d \in R_{r_{{\mathbb X}}}$ as above, and let $\{f_1, \ldots , f_d\}$ be a K-basis of $R_{r_{{\mathbb X}}}$. Then $f_ig'_j \in R_{2r_{{\mathbb X}}}$. There is $\widetilde{f}_{ij}\in R_{r_{{\mathbb X}}}$ such that $f_ig'_j = x_0^{r_{{\mathbb X}}}\widetilde{f}_{ij}$ for all $i,j\in \{1,\ldots ,d\}$. Thus $(x_0^{r_{{\mathbb X}}}f_i) \cdot (x_0^{-r_{{\mathbb X}}-n_j} g'_j) = x_0^{-n_j} f_ig'_j = x_0^{r_{{\mathbb X}} - n_j} \widetilde{f}_{ij} \in R_{2r_{{\mathbb X}} - n_j}$ for all i, j. It follows that $\{x_0^{r_{{\mathbb X}}}f_1, \ldots , x_0^{r_{{\mathbb X}}}f_d\} \subseteq (\delta ^{\sigma }_{\mathbb X})_{2r_{{\mathbb X}}} \subseteq R_{2r_{{\mathbb X}}}$. On the other hand, we see that

$$\begin{aligned} d = \mathop {\text {HF}}\nolimits _{{\mathbb X}}(2r_{{\mathbb X}}) = \dim _K \langle x_0^{r_{{\mathbb X}}}f_1, \ldots , x_0^{r_{{\mathbb X}}}f_d \rangle _K \le \mathop {\text {HF}}\nolimits _{\delta ^{\sigma }_{\mathbb X}}(2r_{{\mathbb X}}) \le \mathop {\text {HF}}\nolimits _{{\mathbb X}}(2r_{{\mathbb X}}) = d. \end{aligned}$$

Therefore we obtain the equalities $\mathop {\text {HF}}\nolimits _{\delta ^{\sigma }_{\mathbb X}}(i) = d = \text {deg}({\mathbb X})$ for all $i \ge 2r_{{\mathbb X}}$.

Finally, claim (c) is an immediate consequence of the claims (a) and (b). $\square$

In the last part of this section we present some relations between the Kähler different and the Dedekind different of a locally Gorenstein 0-dimensional scheme ${\mathbb X}$.

Proposition 3.8

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional smooth scheme. Then we have

$$\begin{aligned} \delta _{\mathbb X}^n \subseteq \vartheta _{\mathbb X}\subseteq \delta _{\mathbb X}. \end{aligned}$$

Proof

Notice that the algebra $R/K[x_0]$ is free of rank $\text {deg}({\mathbb X})$, so it is flat. By Corollary 3.4, we know that the algebra $Q^h(R)/K[x_0,x_0^{-1}]$ is étale. Then Kunz (1986, G.11) yields $\delta _{\mathbb X}=\delta _N$, where $\delta _N$ is the Noether different of $R/K[x_0]$ (as defined in Kunz 1986, G.1). Thus the desired chain of inclusions follows from Kunz (1986, Proposition 10.18). $\square$

Let us point out that, if ${\mathbb X}$ is a smooth complete intersection, then Proposition 5.2 below and Kunz (1986, Proposition 10.17) show that $\vartheta _{{\mathbb X}}= \delta _{\mathbb X}$. Moreover, the above inclusions can be false if the scheme ${\mathbb X}$ is not a smooth scheme. The following example implies that, in general, the Kähler and Dedekind differents do not agree even when ${\mathbb X}$ is a complete intersection.

Example 3.9

Let ${\mathbb X}$ be the 0-dimensional complete intersection given in Example 2.9. We know already that ${\mathbb X}$ is not smooth at $p_5$ and $p_7$ and that the Kähler different is a principal ideal generated by a non-zero homogeneous element of degree $r_{{\mathbb X}}$. In this case the canonical trace map $\text {Tr}_{{\mathcal {O}}_{{\mathbb X},p_j}/K} = \mathrm {id}_K$ is a trace map of ${\mathcal {O}}_{{\mathbb X},p_j}/K$ for $j \ne 5,7$. Moreover, we observe that ${\mathcal {O}}_{{\mathbb X},p_5} \cong K[X_1,X_2]/\langle X_1-2,X_2^2\rangle = K\oplus x_2K$, and so $\{1,x_2\}$ is a K-basis of ${\mathcal {O}}_{{\mathbb X},p_5}$. In particular, $\overline{\sigma }_5: {\mathcal {O}}_{{\mathbb X},p_5} \rightarrow K$ given by $\overline{\sigma }_5(x_2)=1$ and $\overline{\sigma }_5(1)=0$ is a trace map of ${\mathcal {O}}_{{\mathbb X},p_5}/K$. The trace map $\overline{\sigma }_7: {\mathcal {O}}_{{\mathbb X},p_7} \rightarrow K$ of ${\mathcal {O}}_{{\mathbb X},p_7}/K$ can be found in the same way. Using these trace maps, we get a trace system $\sigma : Q^h(R)\rightarrow K[x_0,x_0^{-1}]$ of the algebra $Q^h(R)/K[x_0,x_0^{-1}]$. Now we compute the Dedekind different and get

$$\begin{aligned} \delta ^\sigma _{\mathbb X}= \langle x_{1}^{4} -4x_{0}x_{1}^{2}x_{2} -32x_{0}^{2}x_{2}^{2} + 2x_{1}^{2}x_{2}^{2} -4x_{0}x_{2}^{3} + 12x_{2}^{4} \rangle . \end{aligned}$$

Note that the homogeneous element $x_{1}^{4} -4x_{0}x_{1}^{2}x_{2} -32x_{0}^{2}x_{2}^{2} + 2x_{1}^{2}x_{2}^{2} -4x_{0}x_{2}^{3} + 12x_{2}^{4}$ is a non-zerodivisor of R. So, we have $\text {HP}_{\vartheta _{\mathbb X}} = 7< 9 =\text {HP}_{\delta ^\sigma _{\mathbb X}} = \text {HP}_{(\delta ^\sigma _{\mathbb X})^n}$. Hence we get $(\delta ^\sigma _{\mathbb X})^n \nsubseteq \vartheta _{\mathbb X}$ and $\vartheta _{\mathbb X}\nsubseteq \delta ^\sigma _{\mathbb X}$.

The last inclusion in the proposition can be a strict inclusion for 0-dimensional arithmetically Gorenstein schemes, as our next example demonstrates.

Example 3.10

Let ${\mathbb X}= \{p_1,\ldots ,p_5\} \subseteq {\mathbb P}^3_{{\mathbb F}_7}$ be the set of five distinct K-rational points given in Example 2.12. We know that ${\mathbb X}$ is arithmetically Gorenstein, but not a complete intersection. We also have $\delta _{\mathbb X}= \langle x_{0}^{2} -2x_{0}x_{3} -3x_{1}x_{3} -2x_{3}^{2}\rangle$ and $\vartheta _{{\mathbb X}} = \langle x_{2}x_{3}^{2} -3x_{3}^{3}, x_{1}x_{3}^{2}, x_{0}x_{3}^{2} -3x_{3}^{3},x_{0}^{3} \rangle$. Thus we get $\vartheta _{\mathbb X}\subsetneq \delta _{\mathbb X}$.

Finally, we relate the Hilbert functions of the Kähler different and the Dedekind different of a smooth scheme ${\mathbb X}$ as follows.

Corollary 3.11

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional smooth scheme. Then we have $\text {HP}_{\vartheta _{\mathbb X}} = \text {HP}_{\delta _{\mathbb X}} = \text {deg}({\mathbb X})$ and $r_{{\mathbb X}} \le \text {ri}(\vartheta _{\mathbb X}) \le (n+1)r_{{\mathbb X}}$.

Proof

The equalities of Hilbert polynomials follow from Propositions 3.7 and 3.8. Also, it is clear that $r_{{\mathbb X}} \le \text {ri}(\vartheta _{\mathbb X})$. Since ${\mathbb X}$ is smooth, we can argue similarly as in the proof of Proposition 2.15 to get a homogeneous element $h \in (\vartheta _{\mathbb X})_m {\setminus } \bigcup _{j=1}^{s}(\mathfrak {p}_j)_m$ for some $m \ge 0$, where $\mathfrak {p}_j$ is the homogeneous prime ideal of R corresponding to $p_j \in \text {Supp}({\mathbb X})$. So, h is a non-zerodivisor of R. According to Geramita and Maroscia (1984, Proposition 1.1), we can find a minimal system $\{F_1, \ldots , F_r\}$ of generators of $I_{{\mathbb X}}$ such that $\text {deg}(F_j) \le r_{{\mathbb X}}+1$ for all $j = 1, \ldots , r$. Hence $\vartheta _{\mathbb X}$ is generated in degree $\le nr_{{\mathbb X}}$. If $m > nr_{{\mathbb X}}$ and $\langle (\vartheta _{\mathbb X})_{nr_{{\mathbb X}}}\rangle _R \subseteq \bigcup _{j=1}^{s}\mathfrak {p}_j$, then $\langle (\vartheta _{\mathbb X})_{nr_{{\mathbb X}}}\rangle _R \subseteq \mathfrak {p}_j$ for some $j \in \{1, \ldots , s\}$, and hence the element h cannot exist. Thus h can be chosen such that $\text {deg}(h) = m \le nr_{{\mathbb X}}$. Moreover, if $\{f_1, \ldots , f_{\text {deg}({\mathbb X})}\}$ is a K-basis of $R_{r_{{\mathbb X}}}$, then $\{hf_1, \ldots , hf_{\text {deg}({\mathbb X})}\}$ is a K-basis of $(\vartheta _{\mathbb X})_{r_{{\mathbb X}}+m}$. Therefore we have $\text {ri}(\vartheta _{\mathbb X}) \le (n+1)r_{{\mathbb X}}$. $\square$

4 Cayley–Bacharach Schemes

In many previous characterizations of 0-dimensional complete intersections, a particular geometric condition has played a leading role: the Cayley–Bacharach property. In this section we define and study the notion of a Cayley–Bacharach scheme in a substantially more general context than has been done hitherto. We work with an arbitrary 0-dimensional subscheme ${\mathbb X}$ of a projective space ${\mathbb P}^n_K$ over an arbitrary base field K. As in the previous sections, let the support of ${\mathbb X}$ be given by $\text {Supp}({\mathbb X}) = \{p_1, \ldots , p_s\}$. We denote the residue field of ${\mathbb X}$ at $p_j$ by $K(p_j) ={\mathcal {O}}_{{\mathbb X},p_j}/{\mathfrak {m}}_{{\mathbb X},p_j}$, and set $\varkappa _j = \dim _K K(p_j)$.

Definition 4.1

A subscheme ${\mathbb Y}\subseteq {\mathbb X}$ is called a $p_j$-subscheme if the following conditions are satisfied:

• (a) ${\mathcal {O}}_{{\mathbb Y},p_k} = {\mathcal {O}}_{{\mathbb X},p_k}$ for $k \ne j$.

• (b) The map ${\mathcal {O}}_{{\mathbb X},p_j}\twoheadrightarrow {\mathcal {O}}_{{\mathbb Y},p_j}$ is an epimorphism.

A $p_j$-subscheme ${\mathbb Y}\subseteq {\mathbb X}$ is called maximal if $\text {deg}({\mathbb Y}) = \text {deg}({\mathbb X})-\varkappa _j$.

Notice that a maximal $p_j$-subscheme of a 0-dimensional scheme ${\mathbb X}$ in ${\mathbb P}_K^n$ which has K-rational support, i.e., for which all closed points of ${\mathbb X}$ are K-rational, is nothing but a subscheme ${\mathbb Y}\subseteq {\mathbb X}$ of degree $\text {deg}({\mathbb Y}) = \text {deg}({\mathbb X})-1$ with ${\mathcal {O}}_{{\mathbb Y},p_j} \ne {\mathcal {O}}_{{\mathbb X},p_j}$.

Proposition 4.2

Let $\Gamma = \prod _{j=1}^s {\mathcal {O}}_{{\mathbb X},p_j}$, and let $\mathfrak {G}({\mathcal {O}}_{{\mathbb X},p_j})=\text {Ann}_{{\mathcal {O}}_{{\mathbb X},p_j}}({\mathfrak {m}}_{{\mathbb X},p_j})$ be the socle of ${\mathcal {O}}_{{\mathbb X},p_j}$. There is a 1-1 correspondence

$$\begin{aligned} \left\{ \begin{array}{l} \text {maximal} \; p_j-subschemes\\ \text {of the scheme} {\mathbb X}\end{array} \right\}&\longleftrightarrow \left\{ \begin{array}{l} \text {ideals}\; \langle (0,\ldots ,0,s_j,0,\ldots ,0)\rangle _{\Gamma } \subseteq \Gamma \\ \text {with} \; s_j\in \mathfrak {G}({\mathcal {O}}_{{\mathbb X},p_j}){\setminus }\{0\} \end{array} \right\} \end{aligned}$$

Proof

Let ${\mathbb Y}\subseteq {\mathbb X}$ be a maximal $p_j$-subscheme, let $I_{{\mathbb Y}/{\mathbb X}}$ denote the saturated ideal of ${\mathbb Y}$ in R, and let $\alpha _{{\mathbb Y}/{\mathbb X}} = \min \{ i\in {\mathbb N}\mid (I_{{\mathbb Y}/{\mathbb X}})_i \ne \langle 0\rangle \}$. Furthermore, we let $\{e_{j1},\ldots ,e_{j\varkappa _j}\}\subseteq {\mathcal {O}}_{{\mathbb X},p_j}$ be such that their residue classes form a K-basis of $K(p_j)$, and let $f_{{\mathbb Y}} \in (I_{{\mathbb Y}/{\mathbb X}})_{\alpha _{{\mathbb Y}/{\mathbb X}}}{\setminus }\{0\}$. Since $R_i = x_0^{i-r_{{\mathbb X}}}R_{r_{{\mathbb X}}}$ for $i \ge r_{{\mathbb X}}$, we get from Kreuzer (1994, Lemma 1.2) that $\alpha _{{\mathbb Y}/{\mathbb X}}\le r_{{\mathbb X}}$. Using the map $\widetilde{\imath }$ given by $(*)$, we write $\widetilde{\imath }(f_{{\mathbb Y}}) = (0,\ldots ,0,s_jT_j^{\alpha _{{\mathbb Y}/{\mathbb X}}},0,\ldots ,0) \in \widetilde{R}$. Clearly, we have $s_j \ne 0$.

We claim that $s_j \in \mathfrak {G}({\mathcal {O}}_{{\mathbb X},p_j})$. Indeed, if otherwise, then there is an element $a \in {\mathfrak {m}}_{{\mathbb X},p_j}$ such that $as_j \ne 0$. Suppose there are $c_1, \ldots , c_{\varkappa _j+1} \in K$ such that

$$\begin{aligned} c_1e_{j1}s_j + \cdots + c_{\varkappa _j}e_{j\varkappa _j}s_j + c_{\varkappa _j+1}as_j = (c_1e_{j1} + \cdots + c_{\varkappa _j}e_{j\varkappa _j} + c_{\varkappa _j+1}a)s_j = 0. \end{aligned}$$

If $c_1\overline{e}_{j1} + \cdots + c_{\varkappa _j}\overline{e}_{j\varkappa _j} \ne 0$ in $K(p_j)$, then $c_1e_{j1} + \cdots + c_{\varkappa _j}e_{j\varkappa _j}$ is a unit element, so is $c_1e_{j1} + \cdots + c_{\varkappa _j}e_{j\varkappa _j}+c_{\varkappa _j+1}a$ (as $a \in {\mathfrak {m}}_{{\mathbb X},p_j}$). It follows from the above equality that $s_j = 0$, it is impossible. So, we must have $c_1\overline{e}_{j1} + \cdots + c_{\varkappa _j}\overline{e}_{j\varkappa _j} = 0$. This implies $c_1 = \cdots = c_{\varkappa _j} = 0$, since $\{\overline{e}_{j1}, \ldots , \overline{e}_{j\varkappa _j}\}$ is a K-basis of $K(p_j)$. From this we deduce $c_{\varkappa _j+1}as_j=0$, hence $c_{\varkappa _j+1}=0$ (as $as_j \ne 0$). Therefore the set $\{e_{j1}s_j, \ldots , e_{j\varkappa _j}s_j, as_j\}$ is K-linearly independent. Let

$$\begin{aligned} f_{a} = \widetilde{\imath }^{-1}((0, \ldots , 0, aT_j^{r_{{\mathbb X}}}, 0, \ldots , 0)) \quad \text {and} \quad f_{as_j} = \widetilde{\imath }^{-1}((0, \ldots , 0, as_jT_j^{r_{{\mathbb X}}}, 0, \ldots , 0)). \end{aligned}$$

Then $f_a, f_{as_j} \in R_{r_{\mathbb X}}$ and $x_0^{\alpha _{{\mathbb Y}/{\mathbb X}}}f_{as_j} = f_af_{{\mathbb Y}}$, and so Kreuzer (1994, Lemma 1.2) yields $f_{as_j} \in \langle f_{{\mathbb Y}} \rangle _R^{\mathrm {sat}} \subseteq I_{{\mathbb Y}/{\mathbb X}}$. Similarly, we have

$$\begin{aligned} f_{jk_j}=\widetilde{\imath }^{-1}((0, \ldots , 0, e_{jk_j}s_jT_j^{r_{{\mathbb X}}}, 0, \ldots , 0)) \in (I_{{\mathbb Y}/{\mathbb X}})_{r_{{\mathbb X}}} \end{aligned}$$

for $k_j = 1, \ldots , \varkappa _j$. Thus we get

$$\begin{aligned} \dim _K(I_{{\mathbb Y}/{\mathbb X}})_{r_{{\mathbb X}}} \ge \dim _K\big \langle f_{j1},\ldots ,f_{j\varkappa _j},f_{as_j}\big \rangle _K = \varkappa _j+1 \end{aligned}$$

and hence $\text {deg}({\mathbb Y})<\text {deg}({\mathbb X})-\varkappa _j$, a contradiction.

Next we consider $f \in (I_{{\mathbb Y}/{\mathbb X}})_{i}{\setminus }\{0\}$ with $i \ge \alpha _{{\mathbb Y}/{\mathbb X}}$. The previous claim also tells us that $f_{p_j} \in \mathfrak {G}({\mathcal {O}}_{{\mathbb X},p_j}){\setminus }\{0\}$. If $f_{p_j} \in \mathfrak {G}({\mathcal {O}}_{{\mathbb X},p_j}){\setminus } \langle s_{j}\rangle _{{\mathcal {O}}_{{\mathbb X},p_j}}$, then it is not difficult to check that $\{\, f_{p_j}, e_{j1}s_{j}, \ldots , e_{j\varkappa _j}s_{j} \,\}$ is K-linearly independent. This implies $\text {deg}({\mathbb Y}) < \text {deg}({\mathbb X}) - \varkappa _j$, and it is impossible. Hence we have $f_{p_j} \in \langle s_{j}\rangle _{{\mathcal {O}}_{{\mathbb X},p_j}}$.

Let $g\in R_i{\setminus }\{0\}$ with $i\ge \alpha _{{\mathbb Y}/{\mathbb X}}$ be such that $\widetilde{\imath }(g)=(0, \ldots , 0, g_{p_j}T_j^{i}, 0, \ldots , 0)$ and $g_{p_j} \in \langle s_{j}\rangle _{{\mathcal {O}}_{{\mathbb X},p_j}}$. We are able to write $g_{p_j} = as_j$ for some $a \in {\mathcal {O}}_{{\mathbb X},p_j}{\setminus }{\mathfrak {m}}_{{\mathbb X},p_j}$. Using a similar argument as the previous part we get $g \in \langle f_{{\mathbb Y}}\rangle _R^{\mathrm {sat}}\subseteq I_{{\mathbb Y}/{\mathbb X}}$. Therefore the image of $I_{{\mathbb Y}/{\mathbb X}}$ in $\Gamma = \prod _{j=1}^s {\mathcal {O}}_{{\mathbb X},p_j}$ is $\langle (0,\ldots ,0,s_j,0,\ldots ,0)\rangle _{\Gamma }$ with $s_j \in \mathfrak {G}({\mathcal {O}}_{{\mathbb X},p_j}){\setminus }\{0\}$, as was to be shown.

Conversely, let $(0,\ldots ,0,s_j,0,\ldots ,0)\in \Gamma$ with $s_j \in \mathfrak {G}({\mathcal {O}}_{{\mathbb X},p_j}){\setminus }\{0\}$, and let $f=\widetilde{\imath }^{-1}((0, \ldots , 0, s_jT_j^{r_{{\mathbb X}}}, 0, \ldots , 0)) \in R_{r_{{\mathbb X}}}$. We set ${\mathbb Y}:= {\mathcal {Z}}(f)\subseteq {\mathbb X}$. Then we have $I_{{\mathbb Y}/{\mathbb X}} = \langle f \rangle _R^{\text {sat}}$. Obviously, the scheme ${\mathbb Y}$ is a $p_j$-subscheme of ${\mathbb X}$. It suffices to prove $\text {deg}({\mathbb Y})=\text {deg}({\mathbb X})-\varkappa _j$. Let $f_{jk_j} = \widetilde{\imath }^{-1}((0,\ldots ,0,e_{jk_j}s_jT_j^{r_{{\mathbb X}}},0,\ldots ,0)) \in R_{r_{{\mathbb X}}}$ and $g_{jk_j} = \widetilde{\imath }^{-1}((0,\ldots ,0,e_{jk_j}T_j^{r_{{\mathbb X}}},0,\ldots ,0)) \in R_{r_{{\mathbb X}}}$ for $k_j=1,\ldots ,\varkappa _j$. We see that $x_0^{r_{{\mathbb X}}}f_{jk_j}=g_{jk_j}f$ for every $k_j\in \{1,\ldots ,\varkappa _j\}$, and so Kreuzer (1994, Lemma 1.2) implies $f_{jk_j}\in (I_{{\mathbb Y}/{\mathbb X}})_{r_{{\mathbb X}}}$. Thus we get the inequality

$$\begin{aligned} \dim _K(I_{{\mathbb Y}/{\mathbb X}})_{r_{{\mathbb X}}} \ge \dim _K\langle f_{j1},\ldots ,f_{j\varkappa _j}\rangle _K=\varkappa _j. \end{aligned}$$

Moreover, for $h \in (I_{{\mathbb Y}/{\mathbb X}})_{r_{{\mathbb X}}}{\setminus }\{0\}$, there is a number $m\in {\mathbb N}$ such that $x_0^mh \in \langle f \rangle _R$. This clearly forces $h_{p_j} = as_j$ for some $a \in {\mathcal {O}}_{{\mathbb X},p_j}{\setminus }{\mathfrak {m}}_{{\mathbb X},p_j}$ and $h_{p_k}=0$ for $k\ne j$. Let us write $as_j = \sum _{k_j=1}^{\varkappa _j}c_{jk_j}e_{jk_j}s_j$ for some $c_{j1}, \ldots , c_{j\varkappa _j}\in K$. Then $\widetilde{\imath }(h)=\widetilde{\imath }(\sum _{k_j=1}^{\varkappa _j}c_{jk_j}f_{jk_j})$. Since the map $\widetilde{\imath }$ is injective, we have $h=\sum _{k_j=1}^{\varkappa _j}c_{jk_j}f_{jk_j}\in \langle f_{j1}, \ldots , f_{j\varkappa _j} \rangle _K$. This implies $\dim _K(I_{{\mathbb Y}/{\mathbb X}})_{r_{{\mathbb X}}} \le \varkappa _j$, and therefore this inequality becomes an equality. Hence we obtain $\dim _K(I_{{\mathbb Y}/{\mathbb X}})_{i+r_{{\mathbb X}}}=\varkappa _j$ for $i\ge 0$ or $\text {deg}({\mathbb Y})=\text {deg}({\mathbb X})-\varkappa _j$, as desired. $\square$

Corollary 4.3

A 0-dimensional scheme ${\mathbb X}\subseteq {\mathbb P}^n_K$ contains a subscheme of degree $\text {deg}({\mathbb X})-1$ if and only if it has a K-rational point.

Proof

Due to Proposition 4.2, it suffices to show that if ${\mathbb Y}\subseteq {\mathbb X}$ is a subscheme of degree $\text {deg}({\mathbb X})-1$ with ${\mathcal {O}}_{{\mathbb Y},p_j} \ne {\mathcal {O}}_{{\mathbb X},p_j}$, then $p_j$ is a K-rational point. Suppose that $p_j$ is not K-rational, i.e., $\dim _K K(p_j)=\varkappa _j \ge 2$. We may write $K(p_j) = K\oplus Kv_1\oplus \cdots \oplus Kv_{\varkappa _j-1}$, where $\{1,v_1,\ldots ,v_{\varkappa _j-1}\}$ is a K-basis of $K(p_j)$. Let $f \in (I_{{\mathbb Y}/{\mathbb X}})_{r_{{\mathbb X}}}{\setminus }\{0\}$. Then $\widetilde{\imath }(f)=(0,\ldots ,0,s_{j}T^{r_{\mathbb X}},0,\ldots ,0)$ for some $s_{j}\in {\mathcal {O}}_{{\mathbb X},p_j}{\setminus }\{0\}$. It is not difficult to check that $s_{j}, v_1s_{j}$ are K-linear independent. By setting $g:=\widetilde{\imath }^{-1}((0,\ldots ,0,v_1s_jT^{r_{\mathbb X}},0,\ldots ,0))\in R_{r_{{\mathbb X}}}$, we have $g\in I_{{\mathbb Y}/{\mathbb X}}$ and $\dim _K \langle f,g \rangle _K = 2$. Hence we get $2=\dim _K\langle f,g \rangle _K \le \dim _K(I_{{\mathbb Y}/{\mathbb X}})_{r_{\mathbb X}}=1$, a contradiction. $\square$

Let ${\mathbb Y}$ be a maximal $p_j$-subscheme of ${\mathbb X}$, and let $s_j\in \mathfrak {G}({\mathcal {O}}_{{\mathbb X},p_j}){\setminus }\{0\}$ be a socle element corresponding to a non-zero element of $(I_{{\mathbb Y}/{\mathbb X}})_{\alpha _{{\mathbb Y}/{\mathbb X}}}$. We also say that $s_j$ is a socle element of ${\mathcal {O}}_{{\mathbb X},p_j}$ corresponding to ${\mathbb Y}$. Let $\{e_{j1}, \ldots , e_{j\varkappa _j}\}\subseteq {\mathcal {O}}_{{\mathbb X},p_j}$ be elements whose residue classes form a K-basis of $K(p_j)$. For $a \in {\mathcal {O}}_{{\mathbb X},p_j}$, we set

$$\begin{aligned} \mu (a):=\min \{\, i\in {\mathbb N}\mid (0,\ldots ,0,aT_j^{i},0,\ldots ,0) \in \widetilde{\imath }(R) \,\} \end{aligned}$$

and

$$\begin{aligned} \mu _{{\mathbb Y}/{\mathbb X}} := \max \{\, \mu (e_{jk_j}s_{j}) \mid k_j = 1, \ldots , \varkappa _j \,\}. \end{aligned}$$

Lemma 4.4

The number $\mu _{{\mathbb Y}/{\mathbb X}}$ depends neither on the choice of the socle element $s_j$ nor on the specific choice of $\{e_{j1},\ldots ,e_{j\varkappa _j}\}$.

Proof

First we show that $\mu (a+b)\le \max \{\, \mu (a), \mu (b) \,\}$ for all $a,b\in {\mathcal {O}}_{{\mathbb X},p_j}$. It suffices to consider the case $\mu (a) \le \mu (b)$. Let

$$\begin{aligned} f = \widetilde{\imath }^{-1}((0,\ldots ,0,aT_j^{\mu (a)},0,\ldots ,0)) \ \text {and} \ g = \widetilde{\imath }^{-1}((0,\ldots ,0,bT_j^{\mu (b)},0,\ldots ,0)). \end{aligned}$$

Then we have $f,g\in R$ and

$$\begin{aligned} \widetilde{\imath }(x_0^{\mu (b)-\mu (a)}f+g) = (0,\ldots ,0,(a+b)T_j^{\mu (b)},0,\ldots ,0). \end{aligned}$$

It follows that $\mu (a+b)\le \mu (b)$.

Now let $s'_j \in \mathfrak {G}({\mathcal {O}}_{{\mathbb X},p_j}){\setminus }\{0\}$ such that $s'_j = as_j$ for some unit $a \in {\mathcal {O}}_{{\mathbb X},p_j}$. Note that $s_j = a^{-1}s'_j$. We set $d_j := \max \{\, \mu (e_{jk_j}s_{j}) \mid k_j = 1, \ldots , \varkappa _j \,\}$ and $d'_j := \max \{\, \mu (e_{jk_j}s'_{j}) \mid k_j = 1, \ldots , \varkappa _j \,\}$. We want to prove the equality $d_j = d'_j$. By symmetry, we need only show that $d'_j \le d_j$. For $l\in \{1,\ldots ,\varkappa _j\}$, we write $e_{jl}a = a^l_{j1}e_{j1} + \cdots + a^l_{j\varkappa _j}e_{j\varkappa _j} \mod {\mathfrak {m}}_{{\mathbb X},p_j}$ for $a^l_{j1}, \ldots , a^l_{j\varkappa _j}\in K$, not all equal to zero. We deduce $e_{jl}s'_j = e_{jl}as_j = a^l_{j1}e_{j1}s_j + \cdots + a^l_{j\varkappa _j}e_{j\varkappa _j}s_j$. Thus we have

$$\begin{aligned} \mu (e_{jl}s'_j)= & {} \mu (a^l_{j1}e_{j1}s_j + \cdots + a^l_{j\varkappa _j}e_{j\varkappa _j}s_j) \\\le & {} \max \{\, \mu (a^l_{jk_j}e_{jk_j}s_j) \mid k_j = 1, \ldots ,\varkappa _j \,\}\\= & {} \max \{\, \mu (e_{jk_j}s_j) \mid k_j = 1, \ldots ,\varkappa _j \,\} =d_j. \end{aligned}$$

Therefore we obtain $d'_j = \max \{\, \mu (e_{jk_j}s'_j) \mid k_j = 1, \ldots , \varkappa _j \,\} \le d_j$.

Finally, let $\{e'_{j1},\ldots , e'_{j\varkappa _j}\} \subseteq {\mathcal {O}}_{{\mathbb X},p_j}$ be another set whose residue classes form a K-basis of $K(p_j)$, and let $d''_j := \max \{\, \mu (e'_{jk_j}s_j) \mid k_j = 1, \ldots , \varkappa _j \,\}$. We can argue similarly as above to get $\mu (e'_{jk_j}s_j) \le d_j$ for $k_j \in \{1,\ldots ,\varkappa _j\}$. This implies $d''_j \le d_j$, and hence $d''_j = d_j$. $\square$

Using this lemma, we can now generalize the definition of the degree of a point in ${\mathbb X}$ as follows.

Definition 4.5

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional scheme. For $p_j\in \text {Supp}({\mathbb X})$, the degree of $p_j$ in ${\mathbb X}$ is defined as

$$\begin{aligned} \text {deg}_{{\mathbb X}}(p_j) := \min \{\, \mu _{{\mathbb Y}/{\mathbb X}} \; \big | \; {\mathbb Y}\ \text {is a maximal}\; p_j{-}\mathrm{subscheme}\; \mathrm{of} \;{\mathbb X}\,\}. \end{aligned}$$

Let us check that this definition agrees with the usual one in the case of a scheme ${\mathbb X}$ with K-rational support, and thus for instance in the case of an algebraically closed base field K.

Remark 4.6

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional scheme with $\text {Supp}({\mathbb X})=\{p_1,\ldots ,p_s\}$.

• (a) If ${\mathbb X}$ has K-rational support then $\varkappa _1 = \cdots = \varkappa _s = 1$. In this case we have

  $$\begin{aligned} \text {deg}_{{\mathbb X}}(p_j) = \min \{\, \alpha _{{\mathbb Y}}/{{\mathbb X}} \,\mid \, {\mathbb Y}\ \text {is a maximal}\;p_j{-}\hbox {subscheme}\; \mathrm{of} \;{\mathbb X}\,\}. \end{aligned}$$

  If, in addition, ${\mathbb X}$ is reduced, then

  $$\begin{aligned} \text {deg}_{{\mathbb X}}(p_j) = \alpha _{{\mathbb X}{\setminus }\{p_j\}/{\mathbb X}} = \min \{i\in {\mathbb N}\mid (I_{{\mathbb X}{\setminus }\{p_j\}/{\mathbb X}})_i \ne \langle 0\rangle \} \end{aligned}$$

  for all $j = 1, \ldots , s$. In other words, the degree $\text {deg}_{{\mathbb X}}(p_j)$ agrees with the degree of a point in the sense of Geramita et al. (1993, Definition 2.1).

• (b) We have $0 \le \text {deg}_{{\mathbb X}}(p_j) \le r_{{\mathbb X}}$ for all $j=1,\ldots ,s$. In particular, if ${\mathbb X}$ is a reduced scheme which has K-rational support, then there always exists a point $p_j \in {\mathbb X}$ with maximal degree $\text {deg}_{{\mathbb X}}(p_j) = r_{{\mathbb X}}$ (cf. Geramita et al. 1993, Proposition 1.14).

In case all points of $\text {Supp}({\mathbb X})$ have the maximum possible degree $r_{{\mathbb X}}$, we have the following notion.

Definition 4.7

A 0-dimensional scheme ${\mathbb X}\subseteq {\mathbb P}^n_K$ is called a Cayley–Bacharach scheme (in short, CB-scheme) if every point $p_j\in \text {Supp}({\mathbb X})$ has the maximum possible degree $\text {deg}_{{\mathbb X}}(p_j)=r_{{\mathbb X}}$.

First of all, we give an example which shows that a 0-dimensional scheme ${\mathbb X}\subseteq {\mathbb P}^n_K$ which does not have K-rational support can be a CB-scheme.

Example 4.8

Let $K=\mathbb {Q}$, and let ${\mathbb X}$ be the 0-dimensional subscheme of ${\mathbb P}^2_K$ of degree 14 with support $\text {Supp}({\mathbb X})=\{p_1, \ldots ,p_{12}\}$, where $p_1=(1:0:0)$, $p_2=(1:1:0)$, $p_3=(1:1:1)$, $p_4=(1:0:1)$, $p_5=(1:-1:1)$, $p_6=(1:1:-1)$, $p_7=(1:0:-1)$, $p_8=(1:2:0)$, $p_9=(1:2:1)$, $p_{10}=(1:2:-1)$, $p_{11}$ corresponds to $\mathfrak {P}_{11} = \langle 2X_0^2+X_1^2, X_2 \rangle$, and $p_{12}$ corresponds to $\mathfrak {P}_{12} = \langle X_1, X_0^2+7X_2^2 \rangle$. Clearly, ${\mathbb X}$ does not have ${\mathbb Q}$-rational support, since two points $p_{11}$ and $p_{12}$ are not K-rational. A calculation gives us

$$\begin{aligned} \begin{array}{ll} \mathop {\text {HF}}\nolimits _{{\mathbb X}}: &{} 1\ 3\ 6\ 10\ 14\ 14\cdots \\ \mathop {\text {HF}}\nolimits _{{\mathbb X}{\setminus }\{p_j\}}: &{} 1\ 3\ 6\ 10\ 13\ 13\cdots \quad (j=1, \ldots ,10) \\ \mathop {\text {HF}}\nolimits _{{\mathbb X}{\setminus }\{p_{11}\}}: &{} 1\ 3\ 6\ 10\ 12\ 12\cdots \\ \mathop {\text {HF}}\nolimits _{{\mathbb X}{\setminus }\{p_{12}\}}: &{} 1\ 3\ 6\ \ 9\ 12\ 12\cdots . \end{array} \end{aligned}$$

We have $\alpha _{{\mathbb X}{\setminus }\{p_j\}/{\mathbb X}}=r_{{\mathbb X}{\setminus }\{p_j\}}=r_{{\mathbb X}}=4$ for $j = 1, \ldots , 11$. This implies $\text {deg}_{{\mathbb X}}(p_j)=4$ for $j = 1, \ldots , 11$. We also see that $\alpha _{{\mathbb X}{\setminus }\{p_{12}\}/{\mathbb X}}=3<r_{{\mathbb X}{\setminus }\{p_{12}\}}=r_{{\mathbb X}}=4$ and $\mathop {\text {HF}}\nolimits _{I_{{\mathbb X}{\setminus }\{p_{12}\}/{\mathbb X}}}(3)=1 <\varkappa _{12}=\dim _{{\mathbb Q}}{\mathcal {O}}_{{\mathbb X},p_{12}}=2$. Let $\{e_{12\,1}, e_{12\,2}\}$ be the K-basis of ${\mathcal {O}}_{{\mathbb X},p_{12}}$ given by $e_{12,1} = 1$ and $e_{12, 2}= x_2$. Here we use the isomorphism ${\mathcal {O}}_{{\mathbb X},p_{12}} \cong K[X_1,X_2]/\langle X_1,1+7X_2^2 \rangle = K \oplus x_2K$. Then we find $\mu (e_{12,1}) = 3$ and $\mu (e_{12, 2})=4$. This implies $\text {deg}_{{\mathbb X}}(p_{12})=4$. Hence ${\mathbb X}$ is a CB-scheme.

Our next remark points out an important difference between the more general definition of a CB-scheme given here and the classical definition based on hypersurfaces passing through all points of ${\mathbb X}$ but one.

Remark 4.9

Given a 0-dimensional scheme ${\mathbb X}\subseteq {\mathbb P}^n_K$, we consider the following two statements:

• (a) The scheme ${\mathbb X}$ is a CB-scheme.

• (b) Every hypersurface of degree $r_{{\mathbb X}}-1$ which contains all but one point of ${\mathbb X}$ automatically contains ${\mathbb X}$.

Clearly, if ${\mathbb X}$ has K-rational support, then the statements (a) and (b) are equivalent. In general, we observe that (b) implies (a), but (a) does not imply (b). For example, the reduced 0-dimensional scheme ${\mathbb X}\subseteq {\mathbb P}^2_{{\mathbb Q}}$ given in Example 4.8 is a CB-scheme. But $\alpha _{{\mathbb X}{\setminus }\{p_{12}\}/{\mathbb X}}=3 < r_{{\mathbb X}{\setminus }\{p_{12}\}}=r_{{\mathbb X}}=4$, so statement (b) is not satisfied.

The following proposition gives a simple criterion for detecting whether a given 0-dimensional scheme ${\mathbb X}\subseteq {\mathbb P}^n_K$ is a CB-scheme.

Proposition 4.10

A 0-dimensional scheme ${\mathbb X}\subseteq {\mathbb P}^n_K$ is a CB-scheme if and only if, for all $p_j\in \text {Supp}({\mathbb X})$, every maximal $p_j$-subscheme ${\mathbb Y}\subseteq {\mathbb X}$ satisfies

$$\begin{aligned} \dim _K(I_{{\mathbb Y}/{\mathbb X}})_{r_{{\mathbb X}}-1} < \varkappa _j. \end{aligned}$$

Proof

We always have $\dim _K(I_{{\mathbb Y}_j/{\mathbb X}})_{i} \le \varkappa _j$ for $i \ge 0$. Also, $\dim _K({\mathcal {I}}_{{\mathbb Y}/{\mathbb X}})_{r_{{\mathbb X}}-1} = \varkappa _j$ if and only if $\mu _{{\mathbb Y}/{\mathbb X}} \le r_{{\mathbb X}}-1$. This is equivalent to $\text {deg}_{{\mathbb X}}(p_j)\le r_{{\mathbb X}}-1$. Hence the conclusion follows. $\square$

Let us see an example for the applicability of this proposition.

Example 4.11

Let ${\mathbb X}\subseteq {\mathbb P}^2_{{\mathbb Q}}$ be the 0-dimensional scheme of degree 8 with support $\text {Supp}({\mathbb X})=\{p_1, \ldots ,p_7\}$, where $p_1 = (1:0:0)$, $p_2 = (1:1:0)$, $p_3=(1:1:1)$, $p_4=(1:0:1)$, $p_5=(1:-1:1)$, $p_6=(1:2:3)$, and $p_{7}$ corresponds to $\mathfrak {P}_{7} = \langle 2X_0^2 + X_1^2, X_2 \rangle$. We have $\varkappa _1 = \cdots = \varkappa _6 = 1$ and $\varkappa _7 = 2$. The Hilbert functions of ${\mathbb X}$ and its subschemes are computed as follows

$$\begin{aligned} \begin{array}{ll} \mathop {\text {HF}}\nolimits _{{\mathbb X}}: &{} 1\ 3\ 6\ 8\ 8\ldots \\ \mathop {\text {HF}}\nolimits _{{\mathbb X}{\setminus }\{p_j\}}: &{} 1\ 3\ 6\ 7\ 7\cdots \quad (j=1, \ldots ,6) \\ \mathop {\text {HF}}\nolimits _{{\mathbb X}{\setminus }\{p_{7}\}}: &{} 1\ 3\ 6\ 6\ 6\ldots . \end{array} \end{aligned}$$

From this we deduce $(I_{{\mathbb X}{\setminus }\{p_j\}/{\mathbb X}})_{r_{{\mathbb X}}-1} = (I_{{\mathbb X}{\setminus }\{p_j\}/{\mathbb X}})_{2} = \langle 0 \rangle$ for all $j = 1, \ldots , 7$. By Proposition 4.10, ${\mathbb X}$ is a CB-scheme.

Next let us consider the subscheme ${\mathbb Y}= {\mathbb X}{\setminus } \{p_6\}$ of ${\mathbb X}$. The support of ${\mathbb Y}$ is $\text {Supp}({\mathbb Y}) = \{p_1, \ldots , p_5, p_7\}$. The Hilbert functions of ${\mathbb Y}$ and its subschemes are

$$\begin{aligned} \begin{array}{ll} \mathop {\text {HF}}\nolimits _{{\mathbb Y}}: &{} 1\ 3\ 6\ 7\ 7\ldots \\ \mathop {\text {HF}}\nolimits _{{\mathbb Y}{\setminus }\{p_j\}}: &{} 1\ 3\ 5\ 6\ 6\ldots \quad (j=1,3,5) \\ \mathop {\text {HF}}\nolimits _{{\mathbb Y}{\setminus }\{p_j\}}: &{} 1\ 3\ 6\ 6\ 6\ldots \quad (j=2,4) \\ \mathop {\text {HF}}\nolimits _{{\mathbb Y}{\setminus }\{p_{7}\}}: &{} 1\ 3\ 5\ 5\ 5\ldots . \end{array} \end{aligned}$$

We see that $\dim _K (I_{{\mathbb Y}{\setminus }\{p_j\}/{\mathbb Y}})_{r_{{\mathbb Y}}-1} = \dim _K (I_{{\mathbb Y}{\setminus }\{p_j\}/{\mathbb Y}})_{2} = 1 = \dim _K{\mathcal {O}}_{{\mathbb Y},p_j}$ for $j = 1, 3, 5$. Thus the subscheme ${\mathbb Y}$ is not a CB-scheme by Proposition 4.10.

A key result for characterizing 0-dimensional schemes which are CB-schemes is Kreuzer (1994, Theorem 2.4) which shows that this property is equivalent to the existence of special elements in the initial homogeneous component of the canonical module. In our more general setting, this result can be generalized as follows.

Proposition 4.12

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional locally Gorenstein scheme with support $\text {Supp}({\mathbb X})=\{p_1, \ldots , p_s\}$, and let $\sigma$ be a trace map of $Q^h(R)/K[x_0,x_0^{-1}]$. Then ${\mathbb X}$ is a CB-scheme if and only if for every $j \in \{1, \ldots , s\}$ there exists a non-zero element $g^*_j \in (\mathfrak {C}^\sigma _{\mathbb X})_{-r_{\mathbb X}}$ such that $g^*_j = x_0^{-2r_{{\mathbb X}}}\widetilde{g}^*_j$ with $\widetilde{g}^*_j \in R_{r_{{\mathbb X}}}$ and $(\widetilde{g}^*_j)_{p_j} \in {\mathcal {O}}_{{\mathbb X},p_j}{\setminus } {\mathfrak {m}}_{{\mathbb X},p_j}$.

Proof

Since ${\mathbb X}$ is locally Gorenstein, there is for each point $p_j$ a uniquely determined maximal $p_j$-subscheme ${\mathbb Y}_j \subseteq {\mathbb X}$ corresponding to a socle element $s_j \in \mathfrak {G}({\mathcal {O}}_{{\mathbb X},p_j})$ of ${\mathcal {O}}_{{\mathbb X},p_j}$. Let $\{e_{j1}, \ldots , e_{j\varkappa _j}\} \subseteq {\mathcal {O}}_{{\mathbb X},p_j}$ be such that whose residue classes form a K-basis of $K(p_j)$. For $k_j\in \{1,\ldots ,\varkappa _j\}$, we set

$$\begin{aligned} f^*_{jk_j} := \widetilde{\imath }^{-1}((0,\ldots ,0, e_{jk_j}s_{j}T_j^{\mu (e_{jk_j}s_{j})},0,\ldots ,0)). \end{aligned}$$

Since ${\mathbb X}$ is a CB-scheme, there is for each $j\in \{ 1,\ldots ,s\}$ an index $k_j \in \{1, \ldots , \varkappa _j\}$ such that $f^*_{jk_j} \in R_{r_{\mathbb X}} {\setminus } x_0R_{r_{{\mathbb X}}-1}$. W.l.o.g. we assume that $f^*_{j1} \in R_{r_{\mathbb X}} {\setminus } x_0R_{r_{{\mathbb X}}-1}$ for $j = 1, \ldots , s$. Let us fix an index $j \in \{1, \ldots , s\}$. Then we can define a K-linear map $\overline{\varphi }_j: R_{r_{{\mathbb X}}} \rightarrow K$ such that $\overline{\varphi }_j(x_0R_{r_{{\mathbb X}}-1}) = 0$ and $\overline{\varphi }_j(f^*_{j1}) \ne 0$. By Kreuzer (1994, Lemma 1.5), we may lift $\overline{\varphi }_j$ to obtain a $K[x_0]$-linear map $\varphi _j: R \rightarrow K[x_0]$ of degree $-r_{{\mathbb X}}$, i.e., $\varphi _j$ is an element of $\underline{\text {Hom}}_{K[x_0]}(R,K[x_0])_{-r_{{\mathbb X}}}$ such that $\varphi _j|_{R_{r_{{\mathbb X}}}} = \overline{\varphi }_j$, especially, $\varphi _j(f^*_{j1}) \ne 0$.

Given a homogeneous element $f \in (I_{{\mathbb Y}_j/{\mathbb X}})_{r_{\mathbb X}}{\setminus }\{0\}$, we proceed to show that $f\cdot \varphi _j \ne 0$. Obviously, we have $I_{{\mathbb Y}_j/{\mathbb X}} = \langle f \rangle ^{\mathrm {sat}} = \langle f^*_{j1} \rangle ^{\mathrm {sat}}$. This implies that $x_0^kf^*_{j1} \in \langle f \rangle$ for some $k \ge 0$, so we may write $x_0^{k}f^*_{j1} = fh$ for some $h \in R_k{\setminus }\{0\}$. Consequently, $(f\cdot \varphi _j)(h) = \varphi _j(hf) = \varphi _j(x_0^{k} f^*_{j1}) = x_0^{k} \varphi _j(f^*_{j1}) \ne 0$. From this we conclude $f\cdot \varphi _j \ne 0$ for all $f \in (I_{{\mathbb Y}_j/{\mathbb X}})_{r_{{\mathbb X}}}{\setminus }\{0\}$.

Since $\underline{\text {Hom}}_{K[x_0]}(R,K[x_0]) \cong \mathfrak {C}^\sigma _{{\mathbb X}}=\Phi (\omega _R(1))$, where $\Phi$ is the monomorphism of graded R-modules in Definition 3.5, we find $g^*_j = \Phi (\varphi _j) \in (\mathfrak {C}^\sigma _{{\mathbb X}})_{-r_{{\mathbb X}}}$ such that $f\cdot g^*_j \ne 0$ for all $f \in (I_{{\mathbb Y}_j/{\mathbb X}})_{r_{{\mathbb X}}}{\setminus }\{0\}$. By Proposition 3.7, we have $x_0^{2r_{{\mathbb X}}} \in \delta ^\sigma _{\mathbb X}$, so we may write $g^*_j = x_0^{-2r_{{\mathbb X}}}\widetilde{g}^*_j \in (\mathfrak {C}^\sigma _{{\mathbb X}})_{-r_{{\mathbb X}}}$ with $\widetilde{g}^*_j \in R_{r_{{\mathbb X}}} {\setminus } \{0\}$. Then, for $k_j = 1, \ldots , \varkappa _j$, we get $f^*_{jk_j}\cdot \widetilde{g}^*_j \ne 0$, and so $e_{jk_j}s_j\cdot (\widetilde{g}^*_j)_{p_j} \ne 0$ in ${\mathcal {O}}_{{\mathbb X},p_j}$. Since $s_j \in \mathfrak {G}({\mathcal {O}}_{{\mathbb X},p_j})$, we must have $(\widetilde{g}^*_j)_{p_j} \notin {\mathfrak {m}}_{{\mathbb X},p_j}$. Therefore, for every $j \in \{1, \ldots , s\}$, we have constructed a non-zero element $g^*_j \in (\mathfrak {C}^\sigma _{{\mathbb X}})_{-r_{{\mathbb X}}}$ such that $g^*_j = x_0^{-2r_{{\mathbb X}}}\widetilde{g}^*_j$ with $\widetilde{g}^*_j \in R_{r_{{\mathbb X}}}$ and $(\widetilde{g}^*_j)_{p_j} \in {\mathcal {O}}_{{\mathbb X},p_j}{\setminus }{\mathfrak {m}}_{{\mathbb X},p_j}$.

Conversely, we assume for contradiction that the scheme ${\mathbb X}$ is not a CB-scheme, i.e., $\text {deg}_{{\mathbb X}}(p_j) < r_{{\mathbb X}}$ for some $j \in \{1, \ldots , s\}$. For such an j, we let $g^*_j \in (\mathfrak {C}^\sigma _{{\mathbb X}})_{-r_{{\mathbb X}}}$ such that $g^*_j = x_0^{-2r_{{\mathbb X}}}\widetilde{g}^*_j$ with $\widetilde{g}^*_j \in R_{r_{{\mathbb X}}}$ and $(\widetilde{g}^*_j)_{p_j} \notin {\mathfrak {m}}_{{\mathbb X},p_j}$, and let $\varphi _j = \Phi ^{-1}(g^*_j)$. Clearly, $f \cdot \varphi _j \ne 0$ for all $f \in (I_{{\mathbb Y}_j/{\mathbb X}})_{r_{{\mathbb X}}}{\setminus }\{0\}$. We set $f_{j1} := x_0^{r_{{\mathbb X}}-\text {deg}(f^*_{j1})}f^*_{j1} \in (I_{{\mathbb Y}_j/{\mathbb X}})_{r_{{\mathbb X}}}{\setminus }\{0\}$. Let $i \ge 0$, and let $h \in R_i$ be a non-zero homogeneous element. If $hf_{j1} = 0$, then $(f_{j1}\cdot \varphi )(h) = 0$. Suppose that $hf_{j1} \ne 0$. In this case we write $hf_{j1} = \sum _{k_j=1}^{\varkappa _j}c_{jk_j} x_0^{r_{{\mathbb X}}+i-\text {deg}(f^*_{jk_j})}f^*_{jk_j}$ for some $c_{j1}, \ldots , c_{j\varkappa _j} \in K$. By assumption, we have $\text {deg}(f^*_{jk_j}) < r_{{\mathbb X}}$, and so $\varphi (f^*_{jk_j}) = 0$ for all $k_j = 1, \ldots , \varkappa _j$. Thus

$$\begin{aligned} (f_{j1}\cdot \varphi )(h)= & {} \varphi (hf_{j1}) = \varphi \left( {\textstyle \sum \limits _{k_j=1}^{\varkappa _j}} c_{jk_j}x_0^{r_{{\mathbb X}}+i-\text {deg}(f^*_{jk_j})}f^*_{jk_j}\right) \\= & {} {\textstyle \sum \limits _{k_j=1}^{\varkappa _j}} c_{jk_j}x_0^{r_{{\mathbb X}}+i-\text {deg}(f^*_{jk_j})}\varphi (f^*_{jk_j}) = 0. \end{aligned}$$

Hence we obtain $f_{j1}\cdot \varphi = 0$, a contradiction. $\square$

This characterization is related to Kreuzer (1994, Theorem 2.4) as follows.

Remark 4.13

In the setting of Proposition 4.12, if there is a homogeneous element $g \in (\mathfrak {C}^\sigma _{\mathbb X})_{-r_{{\mathbb X}}}$ such that $\text {Ann}_R(g) = \langle 0 \rangle$ then ${\mathbb X}$ is a CB-scheme. The converse holds true if the field K is infinite.

Our next example shows that the converse of the preceding remark may fail if the base field is finite.

Example 4.14

Let ${\mathbb X}\subseteq {\mathbb P}^n_{{\mathbb F}_2}$ be the set consisting of three points $p_1=(1:1:0)$, $p_2=(1:0:1)$, and $p_3=(1:1:1)$. We have $\mathop {\text {HF}}\nolimits _{{\mathbb X}}:\ 1\ 3\ 3\ldots$ and $r_{{\mathbb X}}=1$. It is not difficult to check that ${\mathbb X}$ is a CB-scheme. A calculation gives us $(\mathfrak {C}_{\mathbb X})_{-1} = \langle g_1, g_2\rangle _{{\mathbb F}_2}$, where $g_1=x_0^{-2}x_1$ and $g_2=x_0^{-2}x_2$. If $g\in (\mathfrak {C}_{\mathbb X})_{-1}$, then g is one of three forms: $g_1$, $g_2$, and $g_1+g_2$. We see that $x_{0} + x_{1} \in \text {Ann}_R(g_1)$, $x_{0} + x_{2} \in \text {Ann}_R(g_2)$, and $x_{0} + x_{1} + x_{2} \in \text {Ann}_R(g_1+g_2)$. Thus $(\mathfrak {C}_{\mathbb X})_{-1}$ cannot contain an element g such that $\text {Ann}_R(g)=\langle 0 \rangle$.

The most widely known class of 0-dimensional schemes which are CB-schemes are arithmetically Gorenstein schemes. The following characterization of arithmetically Gorenstein schemes is a generalization of Kreuzer (1992, Theorem 1.1) where the base field K was assumed to be algebraically closed.

Theorem 4.15

A 0-dimensional subscheme ${\mathbb X}$ of ${\mathbb P}^n_K$ is arithmetically Gorenstein if and only if it is a locally Gorenstein Cayley–Bacharach scheme and its Hilbert function is symmetric, i.e., we have $\mathop {\text {HF}}\nolimits _{\mathbb X}(i)+\mathop {\text {HF}}\nolimits _{\mathbb X}(r_{\mathbb X}-i-1)=\text {deg}({\mathbb X})$ for all $i\in {\mathbb Z}$.

Proof

If ${\mathbb X}$ is arithmetically Gorenstein, then it is also locally Gorenstein. By Goto and Watanabe (1978, Proposition 2.1.3), R is a Gorenstein ring if and only if $\omega _R \cong R(d)$ for some $d \in {\mathbb Z}$. Since $\mathfrak {C}^\sigma _{\mathbb X}\cong \omega _R(1)$ and $\mathop {\text {HF}}\nolimits _{\mathfrak {C}^\sigma _{\mathbb X}}(i) = \text {deg}({\mathbb X}) - \mathop {\text {HF}}\nolimits _{{\mathbb X}}(-i-1)$ for all $i\in {\mathbb Z}$, this is equivalent to $\mathfrak {C}^\sigma _{\mathbb X}\cong R(r_{{\mathbb X}})$. Consequently, the Hilbert function $\mathop {\text {HF}}\nolimits _{{\mathbb X}}$ is symmetric. Moreover, there is an element $g\in (\mathfrak {C}^\sigma _{\mathbb X})_{-r_{\mathbb X}}$ such that $\mathfrak {C}^\sigma _{\mathbb X}= \langle g\rangle _R$ and $\text {Ann}_R(g)=\langle 0\rangle$. Hence ${\mathbb X}$ is a CB-scheme by Proposition 4.12.

Conversely, suppose that ${\mathbb X}$ is a locally Gorenstein CB-scheme and its Hilbert function is symmetric. We have $\mathop {\text {HF}}\nolimits _{\mathfrak {C}^\sigma _{\mathbb X}}(-r_{\mathbb X}) = \text {deg}({\mathbb X}) - \mathop {\text {HF}}\nolimits _{{\mathbb X}}(r_{{\mathbb X}}-1) = 1$. It follows that the K-vector space $(\mathfrak {C}^\sigma _{\mathbb X})_{-r_{\mathbb X}}$ is generated by one element $g = x_0^{-2r_{{\mathbb X}}} \widetilde{g}$, where $\widetilde{g} \in R_{r_{{\mathbb X}}}{\setminus }\{0\}$. Since ${\mathbb X}$ is a CB-scheme, Proposition 4.12 implies that the element $\widetilde{g}_{p_j}$ is a unit of ${\mathcal {O}}_{{\mathbb X},p_j}$ for every $j \in \{1, \ldots , s\}$. Thus $\widetilde{g}$ is a non-zerodivisor of R (cf. Kreuzer 1994, Lemma 1.1]), and hence $\text {Ann}_R(g) = \langle 0 \rangle$. Because $\mathop {\text {HF}}\nolimits _{\mathbb X}$ is symmetric, we must have $\mathfrak {C}^\sigma _{\mathbb X}= \langle g\rangle _R \cong R(r_{\mathbb X})$. Therefore Goto and Watanabe (1978, Proposition 2.1.3) yields that ${\mathbb X}$ is arithmetically Gorenstein, as wanted. $\square$

5 Characterizations of zero-dimensional complete intersections

In this section we discuss some characterizations of 0-dimensional complete intersection schemes ${\mathbb X}$ in ${\mathbb P}^n_K$ using their Kähler and Dedekind differents. Before we begin, let us examine the relations between several versions of the definition of a complete intersection.

Recall that a local ring $(S,{\mathfrak {m}})$ is called a complete intersection if it is Noetherian and its ${\mathfrak {m}}$-adic completion $\widehat{S}$ is a quotient of a regular local ring A by an ideal generated by an A-regular sequence. It is well known (cf. Bruns and Herzog 1993, Theorem 2.3.3) that if S is a Noetherian local ring and $S=A/I$ with a regular local ring A, then S is a complete intersection if and only if I is generated by an A-regular sequence. For more properties of complete intersection rings, we refer to Bruns and Herzog (1993, Section 2.3).

Definition 5.1

Given a ring S and an algebra T / S, we say that T / S is locally a complete intersection if for all $\mathfrak {P}\in \text {Spec}(T)$ the algebra $T_{\mathfrak {P}}/S_{\mathfrak {p}}$ with $\mathfrak {p}= \mathfrak {P}\cap S$ is flat and the local ring $T_{\mathfrak {P}}/\mathfrak {p}T_{\mathfrak {P}}$ is a complete intersection.

At this point we can describe 0-dimensional complete intersection schemes in the following ways.

Proposition 5.2

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional scheme. Then the following statements are equivalent.

• (a) The scheme ${\mathbb X}$ is a complete intersection.

• (b) The algebra $R/K[x_0]$ is locally a complete intersection.

• (c) The local ring $\overline{R}=R/\langle x_0 \rangle$ is a complete intersection.

Proof

Let $\{F_1, \ldots , F_r\}$ be a minimal homogeneous system of generators of $I_{{\mathbb X}}$, where $r\ge n$. If ${\mathbb X}$ is a complete intersection, then $r=n$ and $\{F_1, \ldots , F_n\}$ is a P-regular sequence. Thus the algebra $R/K[x_0]$ is locally a complete intersection by Kunz (1986, C.7) and we have “(a)$\Rightarrow$(b)”. “(b)$\Rightarrow$(c)” follows from the observation that $\langle x_0 \rangle _{K[x_0]}={\mathfrak {m}}\cap K[x_0]$ and $R_{{\mathfrak {m}}}/\langle x_0 \rangle R_{{\mathfrak {m}}}=(R/\langle x_0 \rangle )_{\overline{{\mathfrak {m}}}} =\overline{R}_{\overline{{\mathfrak {m}}}} = \overline{R}$, where $\overline{{\mathfrak {m}}}={\mathfrak {m}}/ \langle x_0 \rangle$ is the maximal ideal of $\overline{R}$.

It remains to prove “(c)$\Rightarrow$(a)”. Let $\mathfrak {M}$ denote the homogeneous maximal ideal $\langle X_0,\ldots ,X_n\rangle _P$ of P. Observe that if $\{F_1, \ldots , F_r, X_0\}$ is a minimal homogeneous system of generators of $I_{{\mathbb X}} + \langle X_0 \rangle$ then we have

$$\begin{aligned} \overline{R} = \overline{R}_{\overline{{\mathfrak {m}}}} \cong (P/I_{{\mathbb X}}+\langle X_0 \rangle )_{\overline{{\mathfrak {m}}}} \cong P_{\mathfrak {M}} / (\langle F_1, \ldots , F_r, X_0 \rangle )_{\mathfrak {M}}. \end{aligned}$$

Since $\overline{R}$ is a complete intersection, the set $\{F_1, \ldots , F_r, X_0\}$ is a $P_{\mathfrak {M}}$-regular sequence (see Bruns and Herzog 1993, Theorem 2.1.2). Now Kunz (1986, Lemma C.28) implies that $\{F_1, \ldots , F_r, X_0\}$ is a P-regular sequence, and hence $r=n$ or ${\mathbb X}$ is a complete intersection. Therefore it suffices to show that $\{F_1, \ldots , F_r, X_0\}$ is a minimal homogeneous system of generators of $I_{{\mathbb X}} + \langle X_0 \rangle$. Clearly, we have $X_0 \notin I_{{\mathbb X}}$. If there is an index $i\in \{1,\ldots ,r\}$ such that $F_i \in \langle F_1,\ldots , F_{i-1}, F_{i+1},\ldots ,F_r,X_0\rangle$, then we get a representation $F_i = \sum _{j\ne i} G_jF_j+GX_0$ where $G_j\in P$ is a homogeneous polynomial of degree $\text {deg}(F_i)-\text {deg}(F_j)$ for $j\ne i$ and $G\in P$ is a homogeneous polynomial of degree $\text {deg}(F_i)-1$ (cf. Kreuzer and Robbiano 2000, Corollary 1.7.11). This implies $GX_0 = F_i - \sum _{j\ne i} G_jF_j \in I_{{\mathbb X}}$, and so $G \in I_{{\mathbb X}}$ (as $x_0$ is a non-zerodivisor of R). Thus there are homogeneous polynomials $H_1,\ldots , H_r\in P$ such that $G=\sum _{j=1}^r H_jF_j$ and $\text {deg}(H_j)=\text {deg}(G)-\text {deg}(F_j)$. Note that $H_i=0$ (as $\text {deg}(G) < \text {deg}(F_i)$). Hence we have $F_i=\sum _{j\ne i} (G_j+ H_jX_0)F_j$, in contradiction to the minimality of $\{F_1,\ldots , F_r\}$. $\square$

The following characterization of 0-dimensional complete intersections in Proposition 5.4 generalizes a result of G. Scheja and U. Storch (see Scheja and Storch 1975, p. 187) where the characteristic of the base field was assumed to be zero, but higher dimension was allowed. Our proof follows essentially the argument given in Scheja and Storch (1975), but offers some simplifications and clarifications. The following lemma makes some arguments in Zariski and Samuel (1958, Ch. 4, Thm. 34) explicit.

Lemma 5.3

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional scheme, let $\{F_1,\ldots ,F_n\}$ be a P-regular sequence in $I_{{\mathbb X}}$, and let $\mathfrak {M} =\langle X_0,\ldots ,X_n\rangle$. Then the colon ideal $\langle X_0,F_1,\ldots ,F_n \rangle : \mathfrak {M}$ is the smallest ideal in P that properly contains $\langle X_0,F_1,\ldots ,F_n \rangle$.

Proof

By localizing at $\mathfrak {M}$ and applying Kunz (1986, C.27), it suffices to prove that the ideal $\mathfrak {q}:\mathfrak {M}_{\mathfrak {M}}$ is the smallest ideal in $P_{\mathfrak {M}}$ that properly contains $\mathfrak {q}$, where $\mathfrak {q} = \langle X_0,F_1,\ldots ,F_n \rangle _{P_\mathfrak {M}}$. Since $P_{\mathfrak {M}}$ is a regular local ring and $\{X_0,F_1,\ldots ,F_n\}$ is a regular sequence, the residue class ring $P_{\mathfrak {M}}/\mathfrak {q}$ is a complete intersection, and so it is a Gorenstein ring. In particular, we have $\dim _K ((\mathfrak {q}:\mathfrak {M}_{\mathfrak {M}})/\mathfrak {q})=1$.

Now let J be an ideal properly containing $\mathfrak {q}$. Then we have $J \cap (\mathfrak {q}:\mathfrak {M}_{\mathfrak {M}}) \nsubseteq \mathfrak {q}$. Suppose that k is the smallest exponent such that $J\cdot \mathfrak {M}_{\mathfrak {M}}^{k} \subseteq \mathfrak {q}$. Clearly, $k\ge 1$ and $J\cdot \mathfrak {M}_{\mathfrak {M}}^{k-1} \subseteq (\mathfrak {q}:\mathfrak {M}_{\mathfrak {M}}) \cap J$ and $J\cdot \mathfrak {M}_{\mathfrak {M}}^{k-1} \nsubseteq \mathfrak {q}$. Because $\dim _K ((\mathfrak {q}:\mathfrak {M}_{\mathfrak {M}})/\mathfrak {q})=1$, there are no ideals between $\mathfrak {q}$ and $\mathfrak {q}:\mathfrak {M}_{\mathfrak {M}}$. Moreover, the intersection of J with $\mathfrak {q}:\mathfrak {M}_{\mathfrak {M}}$ is different from $\mathfrak {q}$, and therefore J contains $\mathfrak {q}:\mathfrak {M}_{\mathfrak {M}}$. $\square$

Now we are ready to characterize 0-dimensional complete intersections using the non-vanishing of the reduced Kähler different.

Proposition 5.4

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional scheme, let $\{F_1,\ldots ,F_r\}$ be a minimal homogeneous system of generators of the vanishing ideal $I_{{\mathbb X}}$, and suppose $\text {char}(K)> \max \{\text {deg}({\mathbb X}),\text {deg}(F_1),\ldots ,\text {deg}(F_r)\}$. Then ${\mathbb X}$ is a complete intersection if and only if $\overline{\vartheta }_{\mathbb X}$ is non-zero.

Proof

If ${\mathbb X}$ is a complete intersection, then $\vartheta _{\mathbb X}=\delta _{\mathbb X}$ is a principal ideal generated by the Jacobian determinant $h = \det (\tfrac{\partial F_j}{\partial x_i})$. According to Kunz (1986, F.20), there is a trace map $\sigma : R \rightarrow K[x_0]$ associated with the presentation $R = P/\langle F_1,\ldots ,F_n\rangle$. By Kunz (1986, F.23), the canonical trace map $\text {Tr}_{R/K[x_0]}: R\rightarrow K[x_0]$ satisfies $\text {Tr}_{R/K[x_0]} = h \cdot \sigma$. Let $d = \text {deg}({\mathbb X})$, let $\{t_1, \ldots , t_d\}$ be the $K[x_0]$-basis of R as in Remark 3.6(b), and let $\{t'_1, \ldots ,t'_d\}$ be the dual $K[x_0]$-basis of R to $\{t_1, \ldots , t_d\}$ w.r.t. $\sigma$ (see Kunz 1986, F.11). For $k = 1, \ldots , d$, we see that

$$\begin{aligned} \left( \left( { \textstyle \sum \limits _{j=1}^d } \text {Tr}_{R/K[x_0]}(t_j)t'_j\right) \cdot \sigma \right) (t_k)= & {} \sigma \left( { \textstyle \sum \limits _{j=1}^d } \text {Tr}_{R/K[x_0]}(t_j)t'_jt_k\right) ={\textstyle \sum \limits _{j=1}^d} \text {Tr}_{R/K[x_0]}(t_j)\sigma (t'_jt_k)\\= & {} {\textstyle \sum \limits _{j=1}^d} \text {Tr}_{R/K[x_0]}(t_j)\delta _{jk} = \text {Tr}_{R/K[x_0]}(t_k). \end{aligned}$$

This implies $h \cdot \sigma = \text {Tr}_{R/K[x_0]} = (\sum _{j=1}^d \text {Tr}_{R/K[x_0]}(t_j)t'_j)\cdot \sigma$. Hence we get the equality $h = \sum _{j=1}^d \text {Tr}_{R/K[x_0]}(t_j)t'_j$. We may assume $t_1=1$. Then $\text {Tr}_{R/K[x_0]}(t_1) = d \ne \text {char}(K)$, and so $h = d t'_1+ \sum _{j=2}^d \text {Tr}_{R/K[x_0]}(t_j)t'_j$. Consequently, we have $\vartheta _{{\mathbb X}} = \langle h \rangle _R \nsubseteq \langle x_0 \rangle _R$. In particular, $\overline{\vartheta }_{\mathbb X}$ is non-zero.

Conversely, suppose that ${\mathbb X}$ is not a complete intersection, i.e., that $r>n$. For every subset $\{F_{i_1},\ldots ,F_{i_n}\}$ of $\{F_1,\ldots ,F_r\}$ consisting of n elements, we want to show that $\tfrac{\partial (F_{i_1},\ldots ,F_{i_n})}{\partial (X_1,\ldots ,X_n)} \in \langle X_0\rangle + I_{{\mathbb X}} =: J$. It suffices to do this for $\{F_1,\ldots ,F_n\}$.

W.l.o.g., let $F_1,\ldots ,F_t$ be a P-regular sequence of maximal length in $\langle F_1,\ldots ,F_n\rangle$. In the case $t=n$, Euler’s rule yields

$$\begin{aligned} \text {deg}(F_j)F_j = {\textstyle \sum \limits _{i=0}^n} \tfrac{\partial F_j}{\partial X_i} X_i \end{aligned}$$

for $j=1,\ldots ,n$. Then Lemma 5.3 implies

$$\begin{aligned} \tfrac{\partial (X_0,F_1,\ldots ,F_n)}{\partial (X_0,\ldots ,X_n)} = \tfrac{\partial (F_1,\ldots ,F_n)}{\partial (X_1,\ldots ,X_n)} \in \langle X_0,F_1,\ldots ,F_n\rangle : {\mathfrak {M}}\subseteq J. \end{aligned}$$

Now we consider the case $t<n$. It is well-known that all associated prime ideals of $\langle X_0, F_1,\ldots ,F_t\rangle$ have the same height $t+1$. So, we may assume that $X_{t+1}$ is not contained in any associated prime ideal of $\langle X_0,F_1,\ldots ,F_t\rangle$. Let $G_{t+1}$ be a homogeneous polynomial in the intersection of the associated prime ideals of $\langle X_0,F_1,\ldots ,F_t\rangle$ which does not contain $F_{t+1}$. For $k>1$, the elements $X_0,F_1,\ldots ,F_t, F_{t+1} + G_{t+1}^kX_{t+1}^{k+1}$ form a P-regular sequence. Repeating this process, we can construct a P-regular sequence $X_0,F_1,\ldots ,F_t, F_{t+1} + G_{t+1}^kX_{t+1}^{k+1},\ldots ,F_n + G_n^kX_n^{k+1}$ eventually.

We set $J_k = \langle X_0,F_1,\ldots ,F_t, F_{t+1} + G_{t+1}^kX_{t+1}^{k+1}, \ldots , F_n + G_n^kX_n^{k+1}\rangle$. If $G_n^kX_n^k \in J_k$, then

$$\begin{aligned} G_n^kX_n^k = X_0H_0+F_1H_1+\cdots +(F_n + G_n^kX_n^{k+1})H_n \end{aligned}$$

for some $H_0,\ldots ,H_n \in P$. Since $\{X_0,F_1,\ldots ,F_t, F_n\}$ is not a P-regular sequence, there exist $H, H'_0,\ldots ,H'_t \in P$ such that $H\notin \langle X_0,F_1,\ldots ,F_t\rangle$ and

$$\begin{aligned} H F_n = X_0H'_0+F_1H'_1+\cdots F_tH'_t. \end{aligned}$$

Hence we get

$$\begin{aligned} H(F_n + G_n^kX_n^{k+1})&= (X_nH_0H + H'_0)X_0+ (X_nH_1H+H'_1)F_1 +\cdots \\&\quad + (X_nH_tH+H'_t)F_t+ HX_nH_{t+1}(F_{t+1}+G_{t+1}^kX_{t+1}^{k+1})+\cdots \\&\quad + X_nH_nH (F_n + G_n^kX_n^{k+1}). \end{aligned}$$

This contradicts the fact that $X_0,F_1,\ldots ,F_t, F_{t+1} + G_{t+1}^kX_{t+1}^{k+1},\ldots ,F_n + G_n^kX_n^{k+1}$ form a P-regular sequence. Thus we have $G_n^kX_n^k\notin J_k$.

Consequently, we get $J_k : \mathfrak {M} \subseteq J_k + \langle G_n^kX_n^k\rangle$ by Lemma 5.3. Let $d_j = \text {deg}(F_j)$ for $j=1,\ldots ,n$ and let

$$\begin{aligned} \mathcal {M} = {\scriptstyle \begin{pmatrix} 1 &{}\quad \tfrac{\partial F_1}{\partial X_0} &{}\quad \!\!\cdots \!\! &{}\quad \tfrac{\partial F_t}{\partial X_0} &{}\quad \tfrac{\partial F_{t+1}}{d_{t+1}\partial X_0} &{}\quad \!\!\cdots \!\! &{}\quad \tfrac{\partial F_n}{d_n\partial X_0} \\ 0 &{}\quad \tfrac{\partial F_1}{\partial X_1} &{}\quad \!\!\cdots \!\! &{}\quad \tfrac{\partial F_t}{\partial X_1} &{}\quad \tfrac{\partial F_{t+1}}{d_{t+1}\partial X_1} &{}\quad \!\!\cdots \!\! &{}\quad \tfrac{\partial F_n}{d_n\partial X_1} \\ \vdots &{}\quad \vdots &{}\quad \!\!\cdots \!\! &{}\quad \vdots &{}\quad \vdots &{}\quad \!\!\cdots \!\! &{}\quad \vdots \\ 0 &{}\quad \tfrac{\partial F_1}{\partial X_{t+1}} &{}\quad \!\!\cdots \!\! &{}\quad \tfrac{\partial F_t}{\partial X_{t+1}} &{}\quad \tfrac{\partial F_{t+1}}{d_{t+1}\partial X_{t+1}}+G_{t+1}^kX_{t+1}^k&{}\quad \!\!\cdots \!\! &{}\quad \tfrac{\partial F_n}{d_n\partial X_{t+1}} \\ \vdots &{}\quad \vdots &{}\quad \!\!\cdots \!\! &{}\quad \vdots &{}\quad \vdots &{}\quad \!\!\cdots \!\! &{}\quad \vdots \\ 0 &{}\quad \tfrac{\partial F_1}{\partial X_{n}} &{}\quad \!\!\cdots \!\! &{}\quad \tfrac{\partial F_t}{\partial X_{n}} &{}\quad \tfrac{\partial F_{t+1}}{d_{t+1}\partial X_{n}}&{}\quad \!\!\cdots \!\! &{}\quad \tfrac{\partial F_n}{d_n\partial X_{n}}+G_{n}^kX_{n}^k \end{pmatrix}.} \end{aligned}$$

We shall show that $\det (\mathcal {M}) \in J_k:\mathfrak {M}$, and hence $\det (\mathcal {M}) \in J_k + \langle G_n^kX_n^k\rangle$. Let $\mathcal {M}_i$ denote the i-th row of the matrix $\mathcal {M}$ for $i=1,\ldots ,n+1$, and let

$$\begin{aligned} \mathcal {V} := (X_0, d_1F_1,\ldots ,d_tF_t, F_{t+1}\!+\!G_{t+1}^kX_{t+1}^{k+1}, \ldots ,F_{n}+G_{n}^kX_{n}^{k+1}). \end{aligned}$$

We use Euler’s rule and calculate

$$\begin{aligned} {\textstyle \sum \limits _{j=0}^n}X_j\cdot \mathcal {M}_{j+1}= & {} \left( X_0, {\textstyle \sum \limits _{j=0}^n}\tfrac{\partial F_1}{\partial X_j} X_j,\ldots , {\textstyle \sum \limits _{j=0}^n}\tfrac{\partial F_t}{\partial X_j}X_j, \tfrac{1}{d_{t+1}}{\textstyle \sum \limits _{j= 0}^n} \tfrac{\partial F_{t+1}}{\partial X_j} X_j + G_{t+1}^kX_{t+1}^{k+1},\right. \\&\quad \left. \ldots ,\tfrac{1}{d_{n}}{\textstyle \sum \limits _{j=0}^n} \tfrac{\partial F_{n}}{\partial X_j} X_j+G_{n}^kX_{n}^{k+1}\right) \\= & {} (X_0, d_1F_1,\ldots ,d_tF_t, F_{t+1}\!+\!G_{t+1}^kX_{t+1}^{k+1},\ldots , F_{n}\!+\!G_{n}^kX_{n}^{k+1}). \end{aligned}$$

Thus we obtain $\mathcal {V} = {\textstyle \sum \nolimits _{j=0}^n}X_j\cdot \mathcal {M}_{j+1}$. Therefore we have

$$\begin{aligned} X_{i-1} \cdot \det (\mathcal {M}) = \det {\scriptstyle \begin{pmatrix} \mathcal {M}_1\\ \vdots \\ \mathcal {M}_{i-1}\\ X_{i-1}\cdot \mathcal {M}_{i} \\ \mathcal {M}_{i+1}\\ \vdots \\ \mathcal {M}_n \end{pmatrix}} = \det {\scriptstyle \begin{pmatrix} \mathcal {M}_1\\ \vdots \\ \mathcal {M}_{i-1}\\ \mathcal {V} \\ \mathcal {M}_{i+1}\\ \vdots \\ \mathcal {M}_n \end{pmatrix}} \in J_k \end{aligned}$$

for $i=1,\ldots ,n+1$, and hence we get $\det (\mathcal {M}) \in J_k:\mathfrak {M} \subseteq J_k + \langle G_n^kX_n^k\rangle$. Furthermore, we have $\det (\mathcal {M}) - \tfrac{1}{d_{t+1}\cdots d_n}\cdot \tfrac{\partial (F_1,\ldots ,F_n)}{\partial (X_1,\ldots ,X_n)} \in \langle G_{t+1}^kX_{t+1}^k,\ldots ,G_{n}^kX_{n}^k\rangle$, and so $\tfrac{\partial (F_1,\ldots ,F_n)}{\partial (X_1,\ldots ,X_n)} \in J_k + \langle G_{t+1}^kX_{t+1}^k,\ldots ,G_{n}^kX_{n}^k\rangle \subseteq J + \langle G_{t+1}^kX_{t+1}^k,\ldots ,G_{n}^kX_{n}^k\rangle$ for all $k>1$. Therefore, by the Krull Intersection Theorem (cf. Eisenbud 1995, Corollary 5.4), we obtain $\tfrac{\partial (F_1,\ldots ,F_n)}{\partial (X_1,\ldots ,X_n)} \in J$, as desired. $\square$

The characterization given in the preceding proposition is nice in the sense that it only uses the reduced Kähler different. But we are also looking for a refined version which lets us distinguish between complete intersections and arithmetically Gorenstein schemes. If we know already that ${\mathbb X}$ is arithmetically Gorenstein, then we can use the following characterization.

Proposition 5.5

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional smooth scheme which is arithmetically Gorenstein. Then the following conditions are equivalent.

• (a) The scheme ${\mathbb X}$ is a complete intersection.

• (b) The Hilbert function of $\vartheta _{\mathbb X}$ satisfies $\mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}}(r_{\mathbb X}) \ne 0$.

• (c) We have $\vartheta _{\mathbb X}=\delta _{{\mathbb X}}$.

Proof

(a)$\Rightarrow$(b): This follows from Corollary 2.7.

(b)$\Rightarrow$(c): Since ${\mathbb X}$ is arithmetically Gorenstein, as in the proof of Theorem 4.15 we find an element $g \in (\mathfrak {C}_{\mathbb X})_{-r_{\mathbb X}}$ such that $\mathfrak {C}_{\mathbb X}= \langle g \rangle _R$ and $\text {Ann}_R(g) = \langle 0 \rangle$. Note that g is a unit of $Q^h(R)$ and $h = g^{-1} \in (\delta _{\mathbb X})_{r_{\mathbb X}}$. For $f \in (\delta _{\mathbb X})_i$ with $i < r_{{\mathbb X}}$, we see that $f\cdot g \in R_{i-r_{{\mathbb X}}} = \langle 0 \rangle$, and hence $f=0$. If $f \in (\delta _{\mathbb X})_i$ with $i \ge r_{{\mathbb X}}$, then $f_1=f\cdot g \in R_{i-r_{{\mathbb X}}}$. This implies $(f-f_1h)\cdot g = 0$, and consequently $f = f_1h$. So, we get $\delta _{\mathbb X}= \langle h \rangle _R$.

Furthermore, we have

$$\begin{aligned} \langle 0 \rangle \ne (\vartheta _{\mathbb X})_{r_{\mathbb X}}\subseteq (\delta _{\mathbb X})_{r_{\mathbb X}} = \langle h \rangle _K, \end{aligned}$$

and so $h\in \vartheta _{\mathbb X}$. Therefore we obtain $\vartheta _{\mathbb X}=\delta _{{\mathbb X}}$.

(c)$\Rightarrow$(a): Suppose that $\vartheta _{\mathbb X}=\delta _{{\mathbb X}}$. Since ${\mathbb X}$ is arithmetically Gorenstein, we argue as above to get $\vartheta _{\mathbb X}= \langle h \rangle _R$ for some non-zerodivisor $h \in R_{r_{{\mathbb X}}}$. In particular, $\vartheta _{\mathbb X}$ is an invertible ideal. Moreover, if we have $\Omega ^1_{Q(R)/K[x_0]} = \langle 0 \rangle$, where Q(R) is the full ring of quotients of R, then it follows from Kunz (1986, Theorem 10.14) that the algebra $R/K[x_0]$ is locally a complete intersection, and hence ${\mathbb X}$ is a complete intersection by Proposition 5.2. So, it suffices to prove that $\Omega ^1_{Q(R)/K[x_0]} = \langle 0 \rangle$. According to Corollary 3.4, the algebra $Q(R)/K(x_0)$ is étale, free of rank $\text {deg}({\mathbb X})$. Thus  Kunz (1986, Proposition 6.8) yields $\Omega ^1_{Q(R)/K(x_0)} = \langle 0 \rangle$. Additionally, it is not difficult to see that $\text {Ker}(K(x_0)\otimes _{K[x_0]} K(x_0) \mathop {\rightarrow }\limits ^{\mu } K(x_0)) = \langle f\otimes 1 -1\otimes f \mid f\in K(x_0) \rangle _{K(x_0)\otimes _{K[x_0]} K(x_0)} = \langle 0 \rangle$, so $\Omega ^1_{K(x_0)/K[x_0]}= \langle 0 \rangle$. On the other hand, we have $Q(R)\cong K(x_0)\otimes _K \prod _{j=1}^s{\mathcal {O}}_{{\mathbb X},p_j}$ (as ${\mathbb X}$ is smooth). This implies

$$\begin{aligned} Q(R)\cong K(x_0)\otimes _{K[x_0]}\left( K[x_0]\otimes _K {\textstyle \prod \limits _{j=1}^s} {\mathcal {O}}_{{\mathbb X},p_j}\right) \cong K(x_0)\otimes _{K[x_0]}\widetilde{R} \end{aligned}$$

where $\widetilde{R}=\prod _{j=1}^s{\mathcal {O}}_{{\mathbb X},p_j}[T_j]$ and $T_1, \ldots , T_s$ are indeterminates. By Kunz (1986, Formulas 4.4), we obtain $\Omega ^1_{Q(R)/K(x_0)} \cong K(x_0) \otimes _{K[x_0]}\Omega ^1_{\widetilde{R}/K[x_0]}$ and

$$\begin{aligned} \Omega ^1_{Q(R)/K[x_0]}\cong & {} K(x_0)\otimes _{K[x_0]}\Omega ^1_{\widetilde{R}/K[x_0]} \oplus \widetilde{R}\otimes _{K[x_0]}\Omega ^1_{K(x_0)/K[x_0]}\\\cong & {} \Omega ^1_{Q(R)/K(x_0)}\oplus \widetilde{R}\otimes _{K[x_0]}\Omega ^1_{K(x_0)/K[x_0]} = \langle 0 \rangle . \end{aligned}$$

This completes the proof. $\square$

Now we present our main result of this section. It answers a question posed in Griffiths and Harris (1978) and Davis and Maroscia (1984): CB-scheme $+$ (?) $=$ Complete intersection? in the case of smooth 0-dimensional subscheme ${\mathbb X}$ of ${\mathbb P}^n_K$. In other words, we replace the assumption that ${\mathbb X}$ is arithmetically Gorenstein by the weaker assumption that ${\mathbb X}$ is a CB-scheme and show again that the non-vanishing of a single homogeneous component of the Kähler different characterizes 0-dimensional complete intersections.

Theorem 5.6

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a smooth 0-dimensional scheme. Then ${\mathbb X}$ is a complete intersection if and only if it is a CB-scheme and $\mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}}(r_{{\mathbb X}}) \ne 0$.

Proof

Suppose that the scheme ${\mathbb X}$ is a complete intersection. Then ${\mathbb X}$ is arithmetically Gorenstein. It follows from Goto and Watanabe (1978, Proposition 2.1.3) that $\mathfrak {C}_{\mathbb X}\cong R(r_{{\mathbb X}})$, and so there exists a homogeneous element $g \in (\mathfrak {C}_{\mathbb X})_{-r_{{\mathbb X}}}$ such that $\mathfrak {C}_{\mathbb X}= \langle g \rangle _R$ and $\text {Ann}_R(g) = \langle 0 \rangle$. By Proposition 4.12, the scheme ${\mathbb X}$ is a CB-scheme. Moreover, it follows from Corollary 2.7 that $\mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}}(r_{{\mathbb X}}) \ne 0$.

Now we prove the converse. Since $\vartheta _{\mathbb X}\subseteq \delta _{\mathbb X}$ and $\mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}}(r_{\mathbb X}) \ne 0$, we have $\mathop {\text {HF}}\nolimits _{\delta _{\mathbb X}}(r_{\mathbb X}) \ne 0$. We let h be a non-zero element in $(\delta _{\mathbb X})_{r_{\mathbb X}}$. Then there is a point $p_j \in \text {Supp}({\mathbb X})$ such that $h_{p_j} \ne 0$ in ${\mathcal {O}}_{{\mathbb X},p_j}$. Because ${\mathbb X}$ is a CB-scheme, Proposition 4.12 implies that there is an element $g^*_j \in (\mathfrak {C}_{\mathbb X})_{-r_{\mathbb X}}{\setminus }\{0\}$ such that $g^*_j = x_0^{-2r_{{\mathbb X}}} \widetilde{g}^*_j$ with $\widetilde{g}^*_j \in R_{r_{{\mathbb X}}}$ and $(\widetilde{g}^*_j)_{p_j} \ne 0$. In ${\mathcal {O}}_{{\mathbb X},p_j}$, we have $h_{p_j} \cdot (\widetilde{g}^*_j)_{p_j} \ne 0$. This implies $h \cdot \widetilde{g}^*_j \ne 0$, and then $h \cdot g^*_j \ne 0$ (as $x_0$ is a non-zerodivisor of R). Thus there is an element $c\in K {\setminus } \{0\}$ such that $c = h \cdot g^*_j \in R_0 =K$. In particular, h is a non-zerodivisor of R and $\text {Ann}_R(g^*_j)=\langle 0 \rangle$. W.l.o.g. we may assume that $c = 1$.

Next we prove that $\mathfrak {C}_{\mathbb X}= \langle g^*_j \rangle _R$. Let $i \ge 0$ and let $g \in (\mathfrak {C}_{\mathbb X})_{i-r_{{\mathbb X}}}$. Then we have $g\cdot h \in R_{i}$. Set $f=g\cdot h \in R_i$. We have $g\cdot h = f = fh \cdot g^*_j$, and so $(x_0^{2r_{{\mathbb X}}}\cdot g - x_0^{2r_{{\mathbb X}}}f \cdot g^*_j) h =0$. Since $x_0^{2r_{{\mathbb X}}} \in \delta _{\mathbb X}$ and $\text {Ann}_R(h)=\langle 0\rangle$, we have $x_0^{2r_{{\mathbb X}}}\cdot g - x_0^{2r_{{\mathbb X}}}f \cdot g^*_j = x_0^{2r_{{\mathbb X}}}\cdot (g - f g^*_j) = 0$. The fact that $x_0$ is a non-zerodivisor on $\mathfrak {C}_{\mathbb X}$ implies $g = fg^*_j$. Thus we get $\mathfrak {C}_{\mathbb X}= \langle g^*_j \rangle _R$, as claimed.

Consequently, we have $\mathfrak {C}_{\mathbb X}\cong R(r_{\mathbb X})$, since $\text {Ann}_R(g^*_j)=\langle 0 \rangle$. Hence the scheme ${\mathbb X}$ is arithmetically Gorenstein by Goto and Watanabe (1978, Proposition 2.1.3). Therefore an application of Proposition 5.5 yields that ${\mathbb X}$ is a complete intersection, as we wanted to show. $\square$

The following corollary is an immediate consequence of Theorem 5.6. This corollary can be deduced from Kunz (1986, Theorem 9.5), and its corresponding local version is given in Lenstra (1993, Proposition 1).

Corollary 5.7

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional smooth scheme. Then ${\mathbb X}$ is a complete intersection if and only if $\vartheta _{\mathbb X}$ is a principal ideal of R generated by a non-zero homogeneous element of degree $r_{{\mathbb X}}$.

Proof

Suppose that $\vartheta _{\mathbb X}= \langle h\rangle _R$ for some element $h\in R_{r_{{\mathbb X}}}{\setminus }\{0\}$. Since $\text {HP}_{\vartheta _{\mathbb X}} = \text {deg}({\mathbb X})$ by Corollary 3.11, the element h must be a non-zerodivisor of R. Clearly, $\mathop {\text {HF}}\nolimits _{\mathfrak {C}_{\mathbb X}}(-r_{{\mathbb X}}) = \mathop {\text {HF}}\nolimits _{{\mathbb X}}(r_{{\mathbb X}}) - \mathop {\text {HF}}\nolimits _{{\mathbb X}}(r_{{\mathbb X}}-1) \ge 1$, and so there exists a non-zero element $g \in (\mathfrak {C}_{\mathbb X})_{-r_{{\mathbb X}}}$. We can write $g = x_0^{-2r_{{\mathbb X}}}\widetilde{g}$ for some $\widetilde{g} \in R_{r_{{\mathbb X}}}$. Since $x_0$ and h are non-zerodivisors of R, we have

$$\begin{aligned} 0\ne h\cdot g = x_0^{-2r_{{\mathbb X}}}h\widetilde{g} \in (\vartheta _{\mathbb X})_{r_{\mathbb X}}(\mathfrak {C}_{\mathbb X})_{-r_{\mathbb X}} \subseteq R_0. \end{aligned}$$

It follows that $h\cdot g\in K{\setminus }\{0\}$, and hence $\text {Ann}_{R}(g) = \langle 0 \rangle$. Thus Proposition 4.12 yields that ${\mathbb X}$ is a CB-scheme, and therefore ${\mathbb X}$ is a complete intersection by Theorem 5.6. The other implication follows from Corollary 2.7. $\square$

We conclude this section with the following criterion for 0-dimensional arithmetically Gorenstein schemes.

Proposition 5.8

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional locally Gorenstein scheme. Then the following conditions are equivalent.

• (a) The scheme ${\mathbb X}$ is arithmetically Gorenstein.

• (b) There exists an element $h \in R_{r_{{\mathbb X}}} {\setminus } \{0\}$ such that $\delta _{{\mathbb X}} = \langle h \rangle _R$.

• (c) There exists an element $h \in (\delta _{{\mathbb X}})_{r_{{\mathbb X}}}$ with $\text {Ann}_R(h) = \langle 0 \rangle$.

• (d) The scheme ${\mathbb X}$ is a CB-scheme and $\mathop {\text {HF}}\nolimits _{\delta _{{\mathbb X}}}(r_{{\mathbb X}}) \ne 0$.

If these conditions are satisfied, then the Hilbert function of the Dedekind different satisfies $\mathop {\text {HF}}\nolimits _{\delta _{{\mathbb X}}}(i)=\mathop {\text {HF}}\nolimits _{{\mathbb X}}(i-r_{{\mathbb X}})$ for all $i \in {\mathbb Z}$, and $\text {ri}(\delta _{{\mathbb X}}) = 2r_{{\mathbb X}}$.

Proof

(a)$\Rightarrow$(b): This follows from the proof of “(b)$\Rightarrow$(c)” of Proposition 5.5.

(b)$\Rightarrow$(c): Assume that $\delta _{\mathbb X}= \langle h \rangle _R$ with $h \in R_{r_{{\mathbb X}}} {\setminus } \{0\}$. According to Proposition 3.7, there is an element $f \in R_{r_{{\mathbb X}}}$ such that $fh = x_0^{2r_{{\mathbb X}}} \in \delta _{\mathbb X}$, and therefore $\text {Ann}_R(h) = \langle 0 \rangle$.

(c)$\Rightarrow$(d): Suppose that there is $h \in (\delta _{\mathbb X})_{r_{\mathbb X}}$ with $\text {Ann}_R(h) = \langle 0 \rangle$. Obviously, we have $\mathop {\text {HF}}\nolimits _{\delta _{\mathbb X}}(r_{\mathbb X}) \ne 0$. Since $\mathop {\text {HF}}\nolimits _{\mathfrak {C}_{\mathbb X}}(-r_{\mathbb X}) \ge 1$, there is a non-zero homogeneous element $g \in (\mathfrak {C}_{\mathbb X})_{-r_{{\mathbb X}}}$. Note that $\delta _{\mathbb X}\mathfrak {C}_{\mathbb X}\subseteq R$. So, we can argue similarly as in the proof of Corollary 5.7 to get $\text {Ann}_{R}(g) = \langle 0 \rangle$. Hence ${\mathbb X}$ is a CB-scheme by Proposition 4.12.

(d)$\Rightarrow$(a): This follows easily from the proof of Theorem 5.6.

The additional claim follows from the fact that $\delta _{\mathbb X}\cong R(-r_{{\mathbb X}})$. $\square$

6 Characterizations using higher Kähler differents

Previously, we mainly considered the Kähler different ${\vartheta }_{{\mathbb X}} = F_0 (\Omega ^1_{R/K[x_0]})$ of the algebra $R/K[x_0]$ to study complete intersections, arithmetically Gorenstein schemes, etc. But what about the algebra R / K? Since R is 1-dimensional, it is natural to consider the higher Kähler different ${\vartheta }^{(1)}_{\mathbb X}= F^1(\Omega ^1_{R/K})$ in this case. After exhibiting relations between this higher Kähler different and the differents studied earlier and collecting some results about its Hilbert function, we shall show that it, too, can be used to characterize 0-dimensional complete intersections in a nice way.

As in the previous sections, we let ${\mathbb X}$ be a 0-dimensional subscheme of ${\mathbb P}^n_K$ such that $\text {Supp}({\mathbb X})\cap {\mathcal {Z}}(X_0) = \emptyset$. Recall that the first Kähler different $\vartheta ^{(1)}_{\mathbb X}$ of the algebra R / K is the first Fitting ideal of $\Omega ^1_{R/K}$ and that we called it the higher Kähler different of ${\mathbb X}$. Using Kunz (1986, Proposition 4.19), we can compute $\vartheta ^{(1)}_{\mathbb X}$ as follows.

Proposition 6.1

Let $\{F_1,\ldots ,F_r\}$ be a homogeneous set of generators of $I_{\mathbb X}$. For every subset $S=\{i_1,\ldots ,i_n\}$ of $\{1,\ldots ,r\}$ and every $j\in \{0,\ldots ,n\}$, we define $\Delta _{S,j} := \tfrac{ \partial (F_{i_1},\ldots ,F_{i_n}) }{\partial (x_0,\ldots ,\hat{x}_j,\ldots ,x_n) }$. Then we have

$$\begin{aligned} \vartheta ^{(1)}_{{\mathbb X}} \;=\; \langle \; \Delta _{S,j} \mid S\subseteq \{1,\ldots ,r\},\, \#S=n,\, j\in \{0,\ldots ,n\} \;\rangle . \end{aligned}$$

The following lemma provides useful relations between the Kähler differents of $R/K[x_0]$ and of R / K.

Lemma 6.2

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional scheme. Then we have

$$\begin{aligned} x_0\vartheta ^{(1)}_{\mathbb X}= \vartheta _{\mathbb X}{\mathfrak {m}}\subseteq \vartheta _{\mathbb X}\subseteq \vartheta ^{(1)}_{\mathbb X}. \end{aligned}$$

Proof

Analogous to Kreuzer et al. (2015, Lemma 2.2). $\square$

By applying the lemma, we can give bounds for the Hilbert polynomial and the regularity index of the Kähler different $\vartheta ^{(1)}_{\mathbb X}$, as the following proposition shows.

Proposition 6.3

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional scheme, and let ${\mathbb X}_{\mathrm {sm}}$ be the set of smooth points of ${\mathbb X}$ in $\text {Supp}({\mathbb X})=\{p_1, \ldots ,p_s\}$. Then we have

$$\begin{aligned} {\textstyle \sum \limits _{p_j\in {\mathbb X}_{\mathrm {sm}}}} \dim _K({\mathcal {O}}_{{\mathbb X},p_j}) \le \text {HP}_{\vartheta ^{(1)}_{\mathbb X}} = \text {HP}_{\vartheta _{\mathbb X}} \le \text {deg}({\mathbb X}) - (s - \# {\mathbb X}_{\mathrm {sm}}) \end{aligned}$$

and $\text {ri}(\vartheta _{\mathbb X})-1 \le \text {ri}(\vartheta ^{(1)}_{\mathbb X}) \le \text {ri}(\vartheta _{\mathbb X}).$

Proof

By Lemma 6.2, we have $x_0\vartheta ^{(1)}_{\mathbb X}\subseteq \vartheta _{\mathbb X}\subseteq \vartheta ^{(1)}_{\mathbb X}$. This implies the equalities $\text {HP}_{x_0\vartheta ^{(1)}_{\mathbb X}} = \text {HP}_{\vartheta ^{(1)}_{\mathbb X}} = \text {HP}_{\vartheta _{\mathbb X}}$, since $x_0$ is a non-zerodivisor of R. Hence Proposition 2.15 yields the bounds for $\text {HP}_{\vartheta ^{(1)}_{\mathbb X}}$. Now we prove the claimed inequalities between regularity indices. Obviously, we have $\text {ri}(\vartheta ^{(1)}_{\mathbb X}) \le \text {ri}(\vartheta _{\mathbb X})$. It follows from the inclusion $x_0\vartheta ^{(1)}_{\mathbb X}\subseteq \vartheta _{\mathbb X}$ that $\mathop {\text {HF}}\nolimits _{\vartheta ^{(1)}_{\mathbb X}}(i) \le \mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}}(i+1)$ for every $i \in {\mathbb Z}$. Consequently, we get $\text {ri}(\vartheta _{\mathbb X})-1 \le \text {ri}(\vartheta ^{(1)}_{\mathbb X}) \le \text {ri}(\vartheta _{\mathbb X})$. $\square$

If ${\mathbb X}\subseteq {\mathbb P}^n_K$ is a 0-dimensional smooth scheme, then the Kähler different $\vartheta ^{(1)}_{\mathbb X}$ satisfies $\text {HP}_{\vartheta ^{(1)}_{\mathbb X}}=\text {deg}({\mathbb X})$, and we have $\text {ri}(\vartheta ^{(1)}_{\mathbb X})\le \text {ri}(\vartheta _{\mathbb X}) \le r_{{\mathbb X}}(n+1)$ by Corollary 3.11 and the preceding proposition. Furthermore, we have the following particular values of the Hilbert function of ${\vartheta }^{(1)}_{\mathbb X}$.

Corollary 6.4

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a reduced 0-dimensional complete intersection.

• (a) Assume that ${\mathbb X}$ contains no smooth point in its support. Then we have $\mathop {\text {HF}}\nolimits _{\vartheta ^{(1)}_{\mathbb X}}(i) = 0$ for all $i\in {\mathbb Z}$.

• (b) Assume that ${\mathbb X}$ contains at least one smooth point in its support. Let ${\mathbb Y}\subseteq {\mathbb X}$ be the subscheme defined by $I_{{\mathbb Y}} = \bigcap \nolimits _{p_j\in \text {Supp}({\mathbb X}):\,p_j \mathrm {\ smooth}} \mathfrak {P}_j$. Then, for all $i\in {\mathbb Z}$, we have

  $$\begin{aligned} \mathop {\text {HF}}\nolimits _{\vartheta ^{(1)}_{\mathbb X}}(i) = {\left\{ \begin{array}{ll} 0 &{} \quad \text {if}\,\, i < r_{\mathbb X}, \\ \mathop {\text {HF}}\nolimits _{{\mathbb Y}}(i+1-r_{\mathbb X}) &{} \quad \text {if}\,\, i \ge r_{\mathbb X}. \end{array}\right. } \end{aligned}$$

  In particular, we have $\text {ri}(\vartheta ^{(1)}_{\mathbb X}) = \text {ri}(\vartheta _{\mathbb X})-1 = r_{\mathbb X}+r_{{\mathbb Y}}-1$ in this case.

Proof

Since (a) follows immediately from the proposition, we prove (b). Let $I_{{\mathbb X}} = \langle F_1, \ldots , F_n \rangle$ and set $\Delta _0 = \frac{\partial (F_1,\ldots ,F_n)}{\partial (x_1,\ldots ,x_n)}$. Then $\vartheta _{\mathbb X}= \langle \Delta _0 \rangle _R$. It follows from Lemma 6.2 that $x_0\vartheta ^{(1)}_{\mathbb X}= \Delta _0{\mathfrak {m}}$. Since $x_0$ is a non-zerodivisor of R, the Hilbert function of $\vartheta ^{(1)}_{\mathbb X}$ satisfies $\mathop {\text {HF}}\nolimits _{\vartheta ^{(1)}_{\mathbb X}}(i) = \mathop {\text {HF}}\nolimits _{\Delta _0{\mathfrak {m}}}(i+1)$ for all $i \in {\mathbb Z}$. If $i < r_{\mathbb X}$, then $0 \le \mathop {\text {HF}}\nolimits _{\vartheta ^{(1)}_{\mathbb X}}(i) \le \mathop {\text {HF}}\nolimits _{{\mathfrak {m}}}(i+1-r_{\mathbb X})=0$, and so $\mathop {\text {HF}}\nolimits _{\vartheta ^{(1)}_{\mathbb X}}(i)=0$. For $i \ge r_{\mathbb X}$, we see that $\mathop {\text {HF}}\nolimits _{\vartheta ^{(1)}_{\mathbb X}}(i) = \mathop {\text {HF}}\nolimits _{\Delta _0{\mathfrak {m}}}(i+1) = \mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}}(i+1)$. Furthermore, Proposition 2.6 yields that $\mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}}(i)= \mathop {\text {HF}}\nolimits _{{\mathbb Y}}(i-r_{\mathbb X})$ for all $i \in {\mathbb Z}$. This implies $\mathop {\text {HF}}\nolimits _{\vartheta ^{(1)}_{\mathbb X}}(i) = \mathop {\text {HF}}\nolimits _{{\mathbb Y}}(i+1-r_{\mathbb X})$ for all $i \ge r_{\mathbb X}$. Hence the claim follows. $\square$

Corollary 6.5

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional smooth scheme which is a complete intersection. Then we have

$$\begin{aligned} \mathop {\text {HF}}\nolimits _{\vartheta ^{(1)}_{\mathbb X}}(i) = {\left\{ \begin{array}{ll} 0 &{} \quad \text {if}\,\, i < r_{{\mathbb X}}, \\ \mathop {\text {HF}}\nolimits _{{\mathbb X}}(i+1-r_{{\mathbb X}}) &{} \quad \text {if}\,\,i \ge r_{{\mathbb X}}. \end{array}\right. } \end{aligned}$$

In particular, we have $\text {ri}(\vartheta ^{(1)}_{\mathbb X}) = \text {ri}(\vartheta _{\mathbb X})-1 = 2r_{{\mathbb X}}-1$.

Now we present a criterion for a smooth 0-dimensional scheme ${\mathbb X}$ to be a complete intersection which uses one value of the Hilbert function of the Kähler different $\vartheta ^{(1)}_{\mathbb X}$.

Theorem 6.6

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional smooth scheme. Then the following conditions are equivalent.

• (a) The scheme ${\mathbb X}$ is a complete intersection.

• (b) The scheme ${\mathbb X}$ is a CB-scheme and $\mathop {\text {HF}}\nolimits _{\vartheta ^{(1)}_{\mathbb X}}(r_{{\mathbb X}}) \ne 0$.

• (c) We have $x_0\vartheta ^{(1)}_{\mathbb X}\cdot \mathfrak {C}_{\mathbb X}= {\mathfrak {m}}$.

Proof

(a)$\Leftrightarrow$(b): According to Lemma 6.2, we know that $\mathop {\text {HF}}\nolimits _{\vartheta ^{(1)}_{\mathbb X}}(r_{{\mathbb X}}) \ne 0$ if and only if $\mathop {\text {HF}}\nolimits _{\vartheta _{\mathbb X}}(r_{{\mathbb X}}) \ne 0$. Hence the claim follows from Theorem 5.6.

(a)$\Rightarrow$(c): If ${\mathbb X}$ is a complete intersection and $\{F_1,\ldots ,F_n\}$ is a homogeneous regular sequence generating $I_{\mathbb X}$, then we let $\Delta _0 := \tfrac{\partial (F_1, \ldots ,F_n)}{\partial (x_1, \ldots ,x_n)}$. By Corollary 2.7, we get $\vartheta _{\mathbb X}= \langle \Delta _0 \rangle _R$, $\mathfrak {C}_{\mathbb X}= \langle \Delta _0^{-1} \rangle _R$, and $\Delta _0$ is a non-zerodivisor of R. We also have $x_0\vartheta ^{(1)}_{\mathbb X}= \Delta _0{\mathfrak {m}}$ by Lemma 6.2. Then, multiplying by $\mathfrak {C}_{\mathbb X}$, we obtain $x_0\vartheta ^{(1)}_{\mathbb X}\cdot \mathfrak {C}_{\mathbb X}= {\mathfrak {m}}$.

(c)$\Rightarrow$(a): Suppose that $x_0\vartheta ^{(1)}_{\mathbb X}\cdot \mathfrak {C}_{\mathbb X}={\mathfrak {m}}$. It follows from the equality $x_0\vartheta ^{(1)}_{\mathbb X}= \vartheta _{\mathbb X}{\mathfrak {m}}$ that $\vartheta _{\mathbb X}\mathfrak {C}_{\mathbb X}{\mathfrak {m}}= {\mathfrak {m}}$. Furthermore, since $\vartheta _{\mathbb X}$ is a subideal of $\delta _{\mathbb X}$, this implies

$$\begin{aligned} \vartheta _{\mathbb X}\mathfrak {C}_{\mathbb X}\subseteq \delta _{\mathbb X}\mathfrak {C}_{\mathbb X}\subseteq R. \end{aligned}$$

If $\vartheta _{\mathbb X}\mathfrak {C}_{\mathbb X}\subsetneq R$, then $\vartheta _{\mathbb X}\mathfrak {C}_{\mathbb X}$ is a homogeneous ideal of R contained in ${\mathfrak {m}}$ and

$$\begin{aligned} \langle 0 \rangle \subseteq {\mathfrak {m}}= \vartheta _{\mathbb X}\mathfrak {C}_{\mathbb X}{\mathfrak {m}}\subseteq \langle 0 \rangle + {\mathfrak {m}}^2. \end{aligned}$$

By Nakayama’s lemma (cf. Kreuzer and Robbiano 2000, Proposition 1.7.15), we have ${\mathfrak {m}}= \langle 0 \rangle$, which is impossible. Thus we must have $\vartheta _{\mathbb X}\mathfrak {C}_{\mathbb X}= R$. Consequently, $\mathfrak {C}_{\mathbb X}$ is invertible and $\vartheta _{\mathbb X}= \delta _{\mathbb X}$. So, the scheme ${\mathbb X}$ is arithmetically Gorenstein and $\vartheta _{\mathbb X}= \delta _{\mathbb X}$. Therefore Proposition 5.5 yields that ${\mathbb X}$ is a complete intersection. $\square$

In the last part of this section we characterize 0-dimensional complete intersections using the image of the canonical map $\Psi :\; \Omega ^1_{R/K} \longrightarrow \Omega ^1_{Q^h(R)/K}$. Given a 0-dimensional scheme ${\mathbb X}\subseteq {\mathbb P}^n_K$ with $\text {Supp}({\mathbb X})=\{p_1, \ldots ,p_s\}$, we have the isomorphism $Q^h(R) \cong \prod _{j=1}^s{\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]$ (see Proposition 3.1). Our next lemma gives us a smoothness criterion for ${\mathbb X}$ in terms of the module of Kähler differentials $\Omega ^1_{Q^h(R)/K}$ of the K-algebra $Q^h(R)$.

Lemma 6.7

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional scheme with $\text {Supp}({\mathbb X})=\{p_1, \ldots ,p_s\}$. Then ${\mathbb X}$ is smooth if and only if $\Omega ^1_{Q^h(R)/K} \!\cong \prod _{j=1}^s{\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]dT_j$.

Proof

On account of Kunz (1986, Corollary 4.8), we have

$$\begin{aligned} \Omega ^1_{Q^h(R)/K} \cong {\textstyle \prod \limits _{j=1}^s}\Omega ^1_{{\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]/K}. \end{aligned}$$

Also, Kunz (1986, Formula 4.4(b)) implies

$$\begin{aligned} \Omega ^1_{{\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]/K}= & {} \Omega ^1_{{\mathcal {O}}_{{\mathbb X},p_j}\otimes _KK[T_j,T_j^{-1}]/K}\\= & {} {\mathcal {O}}_{{\mathbb X},p_j}\otimes _K\Omega ^1_{K[T_j,T_j^{-1}]/K} \oplus K[T_j,T_j^{-1}]\otimes _K\Omega ^1_{{\mathcal {O}}_{{\mathbb X},p_j}/K}\\= & {} {\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]dT_j\oplus K[T_j,T_j^{-1}]\otimes _K\Omega ^1_{{\mathcal {O}}_{{\mathbb X},p_j}/K}. \end{aligned}$$

It follows that $\Omega ^1_{{\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]/K}={\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]dT_j$ if and only if $\Omega ^1_{{\mathcal {O}}_{{\mathbb X},p_j}/K} = 0$. This is equivalent to the condition that $p_j$ is a smooth point of ${\mathbb X}$ (see Kunz 1986, Theorem 7.14). Therefore $\Omega ^1_{Q^h(R)/K} \!\cong \prod _{j=1}^s{\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]dT_j$ if and only if the scheme ${\mathbb X}$ is smooth. $\square$

For the remainder of this section, we assume that the scheme ${\mathbb X}$ is smooth. Using the lemma and the isomorphism $Q^h(R) \cong \prod _{j=1}^s{\mathcal {O}}_{{\mathbb X},p_j}[T_j,T_j^{-1}]$, we see that $\Omega ^1_{Q^h(R)/K} = Q^h(R)\, dx_0$ is a free $Q^h(R)$-module of rank one and that $dx_i = \frac{x_i}{x_0} dx_0$ for all $i=0,\ldots ,n$. Furthermore, letting $L_0=K[x_0,x_0^{-1}]$, the canonical trace $\text {Tr}=\text {Tr}_{Q^h(R)/L_0}:\; Q^h(R) \longrightarrow L_0$ is a $Q^h(R)$-basis of $\underline{\text {Hom}}_{L_0}(Q^h(R),L_0)$ and induces a homogeneous $L_0$-linear map of degree zero

$$\begin{aligned} \text {Tr}^\Omega :\; \Omega ^1_{Q^h(R)/K} \;\longrightarrow \; \Omega ^1_{L_0/K} \end{aligned}$$

such that $\text {Tr}^\Omega (f\,dx_0) = \text {Tr}(f)\, dx_0$ for $f\in Q^h(R)$.

Definition 6.8

In the setting defined above, the set

$$\begin{aligned} \Omega _{\mathbb X}\;= \{ \omega \in \Omega ^1_{Q^h(R)/K} \mid \text {Tr}^\Omega (R\,\omega ) \subseteq \Omega ^1_{K[x_0]/K} \} \end{aligned}$$

is clearly a graded R-module. It is called the module of regular differential forms of R / K (or of ${\mathbb X}$).

This module was introduced by Kunz (1975, (1978) and later generalized and extended by Kunz and Waldi (1988). In our setting, it has the following properties.

Proposition 6.9

Let ${\mathbb X}$ be a smooth 0-dimensional subscheme of ${\mathbb P}^n_K$.

• (a) The module of regular differential forms of ${\mathbb X}$ satisfies $\Omega _{\mathbb X}= \mathfrak {C}_{\mathbb X}\, dx_0$.

• (b) The image of the canonical map $\Psi :\; \Omega ^1_{R/K} \longrightarrow \Omega ^1_{Q^h(R)/K}$ is contained in $\Omega _{\mathbb X}$. Thus it induces an R-linear map $c_{\mathbb X}:\; \Omega ^1_{R/K} \longrightarrow \Omega _{\mathbb X}$ which is called the fundamental class of ${\mathbb X}$.

Proof

Let $f \in Q^h(R)$ be a non-zero homogeneous element. We see that $w = f dx_0 \in \Omega _{\mathbb X}$ if and only if $\text {Tr}^\Omega (gf dx_0) = \text {Tr}(gf) dx_0 \in K[x_0] dx_0$ for all $g \in R$. This is equivalent to $(f\cdot \text {Tr})(g) \in K[x_0]$ for all $g \in R$, i.e., to $f\cdot \text {Tr}\in \omega _R(1)$. Under the injection $\Phi$ given in Definition 3.5, this is also equivalent to $f \in \mathfrak {C}_{\mathbb X}$, and consequently claim (a) follows.

Next we prove claim (b). In $\Omega ^1_{Q^h(R)/K}$, we have $dx_i = \frac{x_i}{x_0}dx_0$ for $i=0,\ldots ,n$. Since $\Omega ^1_{R/K} = Rdx_0+\cdots +Rdx_n$, we have $\Psi (\Omega ^1_{R/K}) = \tfrac{{\mathfrak {m}}}{x_0} dx_0$. Moreover, we see that $\mathop {\text {HF}}\nolimits _{\mathfrak {C}_{\mathbb X}}(0)=\text {deg}({\mathbb X})-\mathop {\text {HF}}\nolimits _{{\mathbb X}}(-1)=\text {deg}({\mathbb X})$, and so we have $x_0(\mathfrak {C}_{\mathbb X})_0 = x_0(Q^h(R))_0 = (Q^h(R))_1$. Thus we get ${\mathfrak {m}}\subseteq x_0\mathfrak {C}_{\mathbb X}$. Therefore, by part (a), we obtain $\Psi (\Omega ^1_{R/K}) = \tfrac{{\mathfrak {m}}}{x_0} dx_0 \subseteq \mathfrak {C}_{\mathbb X}dx_0 = \Omega _{{\mathbb X}}$, as we wanted to show. $\square$

In the preceding proof we showed that $c_{\mathbb X}(\Omega _{R/K}^1)=\tfrac{{\mathfrak {m}}}{x_0}dx_0$. The kernel of the fundamental class can be described as follows.

Proposition 6.10

We have $\text {Ker}(c_{\mathbb X})=\big \{ \, \sum _{i=0}^n g_idx_i \in \Omega ^1_{R/K} \, \mid \, \sum _{i=0}^n g_ix_i = 0 \, \big \}$.

Proof

For an element $w = \sum _{i=0}^n g_idx_i \in \Omega ^1_{R/K}$, $c_{\mathbb X}(w) =(\frac{1}{x_0}\sum _{i=0}^n g_ix_i) dx_0$. So, $c_{\mathbb X}(w) =0$ if and only if $\frac{1}{x_0}\sum _{i=0}^n g_ix_i =0$. Since $x_0$ is a non-zerodivisor for $Q^h(R)$, this is equivalent to $\sum _{i=0}^n g_ix_i = 0$. Hence the conclusion follows. $\square$

Next we use the image of the fundamental class to characterize 0-dimensional complete intersections.

Theorem 6.11

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional smooth scheme. Then ${\mathbb X}$ is a complete intersection if and only if $c_{\mathbb X}(\Omega _{R/K}^1) = \vartheta ^{(1)}_{\mathbb X}\Omega _{{\mathbb X}}$.

Proof

Assume that ${\mathbb X}$ is a complete intersection. Let $\{F_1,\ldots ,F_n\}$ be a homogeneous regular sequence generating $I_{\mathbb X}$. We set $\Delta _j:=\frac{\partial (F_1, \ldots ,F_n)}{\partial (x_0, \ldots ,\widehat{x_j}, \ldots ,x_n)}$ for $j = 0, \ldots , n$. In $\Omega ^1_{R/K}$, there are relations

$$\begin{aligned} \frac{\partial F_{i}}{\partial x_0}dx_0+\frac{\partial F_{i}}{\partial x_1}dx_1 +\cdots +\frac{\partial F_{i}}{\partial x_n}dx_n=0 \quad \text {for}\; i=1, \ldots , n. \end{aligned}$$

By Cramer’s Rule, we have $\Delta _0dx_j = (-1)^{n+1-j} \Delta _j dx_0$ for $j=1,\ldots ,n$. Thus we deduce from $\Omega ^1_{R/K}=Rdx_0+\cdots +Rdx_n$ and $\mathfrak {C}_{\mathbb X}=\langle \Delta _0^{-1} \rangle _R$ that

$$\begin{aligned} c_{\mathbb X}(\Omega ^1_{R/K})= & {} \left\langle \frac{\Delta _0}{\Delta _0}, \frac{\Delta _1}{\Delta _0}, \ldots , \frac{\Delta _n}{\Delta _0} \right\rangle _R dx_0\\= & {} \vartheta ^{(1)}_{\mathbb X}\langle \Delta _0^{-1} \rangle _R dx_0 = \vartheta ^{(1)}_{\mathbb X}\Omega _{{\mathbb X}}. \end{aligned}$$

Conversely, suppose $c_{\mathbb X}(\Omega _{R/K}^1) = \vartheta ^{(1)}_{\mathbb X}\Omega _{{\mathbb X}}$. We know that $c_{\mathbb X}(\Omega _{R/K}^1) = \frac{{\mathfrak {m}}}{x_0} dx_0$. Thus we get the equality ${\mathfrak {m}}dx_0 = x_0\vartheta ^{(1)}_{\mathbb X}\mathfrak {C}_{{\mathbb X}} dx_0$. Since $\text {Ann}_{R}(dx_0) = \langle 0 \rangle$, we obtain ${\mathfrak {m}}= x_0\vartheta ^{(1)}_{\mathbb X}\mathfrak {C}_{\mathbb X}$. Therefore the conclusion follows from Theorem 6.6. $\square$

From the proof of this theorem we also get a characterization of arithmetically Gorenstein schemes which is based on the following observations.

Remark 6.12

Let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional smooth scheme.

• (a) There is an exact sequence of graded R-modules

  $$\begin{aligned} 0\longrightarrow \text {Ker}(c_{\mathbb X})\longrightarrow \Omega _{R/K}^1 \longrightarrow \Omega _{{\mathbb X}} \longrightarrow \Omega _{{\mathbb X}}/c_{\mathbb X}(\Omega _{R/K}^1) \longrightarrow 0. \end{aligned}$$

  The R-module $J_{{\mathbb X}}=\Omega _{{\mathbb X}}/\Psi (\Omega _{R/K}^1)$ is also known as the Jacobian module of ${\mathbb X}$.

• (b) Notice that $c_{\mathbb X}(\Omega _{R/K}^1) \cong {\mathfrak {m}}$ and $\Omega _{{\mathbb X}} \cong \mathfrak {C}_{\mathbb X}(1)$. Thus the Hilbert function of $J_{\mathbb X}$ is given by

  $$\begin{aligned} \mathop {\text {HF}}\nolimits _{J_{{\mathbb X}}}(i)=\text {deg}({\mathbb X})-\mathop {\text {HF}}\nolimits _{{\mathbb X}}(-i) -\mathop {\text {HF}}\nolimits _{{\mathfrak {m}}}(i) \end{aligned}$$

  for all $i\in {\mathbb Z}$. Hence $J_{\mathbb X}$ is a finite dimensional K-vector space with

  $$\begin{aligned}&\dim _K(J_{{\mathbb X}}) = (r_{{\mathbb X}}-1)\text {deg}({\mathbb X}) + 1 + {\textstyle \sum \limits _{i=0}^{r_{{\mathbb X}}-1}} (\text {deg}({\mathbb X})-\mathop {\text {HF}}\nolimits _{{\mathbb X}}(i)\\&\quad -\mathop {\text {HF}}\nolimits _{{\mathbb X}}(r_{{\mathbb X}}-i-1)). \end{aligned}$$

Finally, we characterize arithmetically Gorenstein schemes using the dimension of their Jacobian module as follows.

Corollary 6.13

Let K be an infinite field, and let ${\mathbb X}\subseteq {\mathbb P}^n_K$ be a 0-dimensional smooth scheme. Then the following conditions are equivalent.

• (a) The scheme ${\mathbb X}$ is arithmetically Gorenstein.

• (b) The scheme ${\mathbb X}$ is a CB-scheme and $\dim _K(J_{{\mathbb X}})=(r_{{\mathbb X}}-1)\text {deg}({\mathbb X})+1$.

Proof

If ${\mathbb X}$ is an arithmetically Gorenstein scheme, Theorem 4.15 implies $\mathop {\text {HF}}\nolimits _{{\mathbb X}}(i)+\mathop {\text {HF}}\nolimits _{{\mathbb X}}(r_{{\mathbb X}}-i-1)=\text {deg}({\mathbb X})$ for $i=0,\ldots ,r_{{\mathbb X}}-1$. Hence Remark 6.12(b) yields the equality $\dim _K(J_{{\mathbb X}})= (r_{{\mathbb X}}-1)\text {deg}({\mathbb X})+1$. Moreover, Proposition 5.8 shows that ${\mathbb X}$ is a CB-scheme.

Conversely, assume that ${\mathbb X}$ is a CB-scheme and $\dim _K(J_{{\mathbb X}})=(r_{{\mathbb X}}-1)\text {deg}({\mathbb X})+1$. Since K is infinite, Remark 4.13 tells us that there is an element $g\in (\mathfrak {C}_{\mathbb X})_{-r_{{\mathbb X}}}$ such that $\text {Ann}_R(g) = \langle 0 \rangle$. Thus $R(-r_{{\mathbb X}})\cong R \cdot g \subseteq \mathfrak {C}_{\mathbb X}$. This implies

$$\begin{aligned} \mathop {\text {HF}}\nolimits _{{\mathbb X}}(i)\le \mathop {\text {HF}}\nolimits _{\mathfrak {C}_{\mathbb X}}(i-r_{{\mathbb X}})=\text {deg}({\mathbb X})-\mathop {\text {HF}}\nolimits _{{\mathbb X}}(r_{{\mathbb X}}-i-1) \end{aligned}$$

for all $i \in {\mathbb Z}$. Since $\dim _K(J_{{\mathbb X}})=(r_{{\mathbb X}}-1)\text {deg}({\mathbb X})+1$, we deduce $\sum _{i=0}^{r_{{\mathbb X}}-1} (\text {deg}({\mathbb X})-\mathop {\text {HF}}\nolimits _{{\mathbb X}}(i)-\mathop {\text {HF}}\nolimits _{{\mathbb X}}(r_{{\mathbb X}}-i-1))=0$. Hence we must have $\mathop {\text {HF}}\nolimits _{{\mathbb X}}(i)=\text {deg}({\mathbb X})-\mathop {\text {HF}}\nolimits _{{\mathbb X}}(r_{{\mathbb X}}-i-1)$ for all $i\in {\mathbb Z}$. Observe that $\text {deg}({\mathbb X})-\mathop {\text {HF}}\nolimits _{{\mathbb X}}(r_{{\mathbb X}}-1) =\mathop {\text {HF}}\nolimits _{{\mathbb X}}(0)=1$ and $\mathop {\text {HF}}\nolimits _{\mathfrak {C}_{\mathbb X}}(i-r_{{\mathbb X}}) =\text {deg}({\mathbb X})-\mathop {\text {HF}}\nolimits _{{\mathbb X}}(r_{{\mathbb X}}-i-1)=\mathop {\text {HF}}\nolimits _{{\mathbb X}}(i) =\mathop {\text {HF}}\nolimits _{\langle g \rangle _R}(i-r_{{\mathbb X}})$. Thus we get $\mathfrak {C}_{\mathbb X}=\langle g \rangle _R$, and therefore ${\mathbb X}$ is arithmetically Gorenstein. $\square$