Graded Annihilators and Uniformly F-Compatible Ideals

Abstract

Let R be a commutative (Noetherian) local ring of prime characteristic p that is F-pure. This paper is concerned with comparison of three finite sets of radical ideals of R, one of which is only defined in the case when R is F-finite (that is, is finitely generated when viewed as a module over itself via the Frobenius homomorphism). Two of the afore-mentioned three sets have links to tight closure, via test ideals. Among the aims of the paper are a proof that two of the sets are equal, and a proposal for a generalization of I. M. Aberbach’s and F. Enescu’s splitting prime.

1 Introduction

Throughout the paper, let $(R,\mathfrak {m})$ be a commutative (Noetherian) local ring of prime characteristic p having maximal ideal $\mathfrak {m}$. In recent years, the study of R-modules with a Frobenius action has assisted in the development of the theory of tight closure over R. An R-module with a Frobenius action can be viewed as a left module over the Frobenius skew polynomial ring over R, and such left modules will play a central role in this paper.

The Frobenius skew polynomial ring over R is described as follows. Throughout, f:R→R denotes the Frobenius ring homomorphism, for which f(r)=r ^p for all r∈R. The Frobenius skew polynomial ring over R is the skew polynomial ring R[x,f] associated to R and f in the indeterminate x; as a left R-module, R[x,f] is freely generated by (x ⁱ) _i≥0, and so consists of all polynomials ${\sum }_{i = 0}^{n} r_{i} x^{i}$, where n≥0 and r ₀,…,r _n ∈R; however, its multiplication is subject to the rule x r=f(r)x=r ^p x for all r∈R.

We can think of R[x,f] as a positively-graded ring $R[x,f] = \bigoplus _{n=0}^{\infty } R[x,f]_{n}$, where R[x,f] _n =R x ⁿ for n≥0. The graded annihilator of a left R[x,f]-module H is the largest graded two-sided ideal of R[x,f] that annihilates H; it is denoted by gr-ann _R[x,f] H.

Let G be a left R[x,f]-module that is x-torsion-free in the sense that x g=0 for g∈G, only when g= 0. Then $\text {gr-ann}_{R[x,f]}G = \mathfrak {b} R[x,f]$, where $\mathfrak {b} = (0:_{R}G)$ is a radical ideal. See [11, Lemma 1.9]. We shall use $\mathcal {I}(G)$ (or $\mathcal {I}_{R}(G)$) to denote the set of R-annihilators of the R[x,f]-submodules of G; we shall refer to the members of $\mathcal {I}(G)$ as the G-special R-ideals. For a graded two-sided ideal $\mathfrak {B}$ of R[x,f], we denote by $\text {ann}_{G}(\mathfrak {B})$ or $\text {ann}_{G}\mathfrak {B}$ the R[x,f]-submodule of G consisting of all elements of G that are annihilated by $\mathfrak {B}$. Also, we shall use $\mathcal {A}(G)$ to denote the set of special annihilator submodules of G, that is, the set of R[x,f]-submodules of G of the form $\text {ann}_{G}(\mathfrak {A})$, where $\mathfrak {A}$ is a graded two-sided ideal of R[x,f]. In [11, §1], the present author showed that there is a sort of ‘Galois’ correspondence between $\mathcal {I}(G)$ and $\mathcal {A}(G)$. In more detail, there is an order-reversing bijection, ${\Delta } : \mathcal {A}(G) \longrightarrow \mathcal {I}(G)$ given by

$${\Delta} : N \longmapsto \left(\text{gr-ann}_{R[x,f]}N\right)\cap R = (0:_{R}N). $$

The inverse bijection, ${\Delta }^{-1} : \mathcal {I}(G) \longrightarrow \mathcal {A}(G),$ also order-reversing, is given by

$${\Delta}^{-1} : \mathfrak{b} \longmapsto \text{ann}_{G}\left(\mathfrak{b} R[x,f])\right). $$

We shall be mainly concerned in this paper with the situation where R is F-pure. We remind the reader what this means. For $j \in \mathbb {N}$ (the set of positive integers) and an R-module M, let M ^(j) denote M considered as a left R-module in the natural way and as a right R-module via f ^j, the jth iterate of the Frobenius ring homomorphism. Then R is F-pure if, for every R-module M, the natural map M→R ⁽¹⁾⊗ _R M (which maps m∈M to 1⊗m) is injective.

Note that R ^(j)≅R x ^j as (R,R)-bimodules. Let $i\in \mathbb {N}_{0}$, the set of non-negative integers. When we endow R x ⁱ and R x ^j with their natural structures as (R,R)-bimodules (inherited from their being graded components of R[x,f]), there is an isomorphism of (left) R-modules ϕ:R x ^i+j⊗ _R M→≅R x ⁱ⊗ _R (R x ^j⊗ _R M) for which ϕ(r x ^i+j⊗m)=r x ⁱ⊗(x ^j⊗m) for all r∈R and m∈M. It follows that R is F-pure if and only if the left R[x,f]-module R[x,f]⊗ _R M is x-torsion-free for every R-module M. This means that, when R is F-pure, there is a good supply of natural x-torsion-free left R[x,f]-modules.

In fact, we shall use Φ (or Φ _R when it is desirable to specify which ring is being considered) to denote the functor R[x,f]⊗ _R ∙ from the category of R-modules (and all R-homomorphisms) to the category of all $\mathbb {N}_{0}$-graded left R[x,f]-modules (and all homogeneous R[x,f]-homomorphisms). For an R-module M, we shall identify Φ(M) with $\bigoplus _{n\in \mathbb {N}_{0}}Rx^{n}\otimes _{R}M$, and (usually) identify its 0th component R⊗ _R M with M, in the obvious ways.

Let E be the injective envelope of the simple R-module $R/\mathfrak {m}$. We shall be concerned with Φ(E), the $\mathbb {N}_{0}$-graded left R[x,f]-module $\bigoplus _{n\in \mathbb {N}_{0}}Rx^{n}\otimes _{R}E$. Assume now that R is F-pure. In [12, Corollary 4.11], the present author proved that the set $\mathcal {I}({\Phi }(E))$ is a finite set of radical ideals of R; in [11, Theorem 3.6 and Corollary 3.7], he proved that $\mathcal {I}({\Phi }(E))$ is closed under taking primary (prime in this case) components; and in [14, Corollary 2.8], he proved that the big test ideal $\widetilde {\tau }(R)$ of R (for tight closure) is equal to the smallest member of $\mathcal {I}({\Phi }(E))$ that meets R ^∘, the complement in R of the union of the minimal prime ideals of R.

Let $\mathfrak {a} \in \mathcal {I}({\Phi }(E))$ (with $\mathfrak {a} \neq R$), still in the F-pure case. The special annihilator submodule $\text {ann}_{\Phi (E)}(\mathfrak {a} R[x,f])$ of Φ(E) corresponding to $\mathfrak {A}$ inherits a natural structure as a graded left module over the Frobenius skew polynomial ring $(R/\mathfrak {a})[x,f]$, and its 0th component is contained in $(0:_{E}\mathfrak {a})$. As $R/\mathfrak {a}$-module, the latter is isomorphic to the injective envelope of the simple $R/\mathfrak {a}$-module. Motivated by results in [14, §3] in the case where R is complete, and by work of K. Schwede in [10, §5] in the F-finite case, we say that $\mathfrak {A}$ is fully Φ(E)-special if (it is Φ(E)-special and) its 0th component is exactly $(0:_{E}\mathfrak {a})$. The main result of this paper is that a Φ(E)-special ideal of R is always fully Φ(E)-special provided that R is an (F-pure) homomorphic image of an excellent regular local ring of characteristic p. When R satisfies this condition, corollaries can be drawn from that main result: we shall establish an analogue of [14, Theorem 3.1] and, in particular, show that $R/\mathfrak {a}$ is F-pure whenever $\mathfrak {A}$ is a proper Φ(E)-special ideal of R.

Along the way, we shall show that, in the case where R is F-finite as well as F-pure, the set $\mathcal {I}({\Phi }(E))$ of Φ(E)-special ideals of R is equal to the set of uniformly F-compatible ideals of R, introduced by K. Schwede in [10, §3]. An ideal $\mathfrak {B}$ of R is said to be uniformly F-compatible if, for every j>0 and every ϕ∈Hom _R (R ^(j),R), we have $\phi (\mathfrak {b}^{(j)}) \subseteq \mathfrak {b}$. In [10, Corollary 5.3 and Corollary 3.3], Schwede proved that there are only finitely many uniformly F-compatible ideals of R and that they are all radical; in [10, Proposition 4.7 and Corollary 4.8], he proved that the set of uniformly F-compatible ideals is closed under taking primary (prime in this case) components; in [10, Theorem 6.3], Schwede proved that the big test ideal $\widetilde {\tau }(R)$ of R is equal to the smallest uniformly F-compatible ideal of R that meets R ^∘; and in [10, Remark 4.4 and Proposition 4.7], he proved that there is a unique largest proper uniformly F-compatible ideal of R, and that is prime and equal to the splitting prime of R discovered and defined by I. M. Aberbach and F. Enescu [1, §3].

Thus, in the F-finite F-pure case, the set of uniformly F-compatible ideals of R has properties similar to some properties of $\mathcal {I}({\Phi }(E))$. Are the two sets the same? We shall, during the course of the paper, show that the answer is ‘yes’. It should be emphasized, however, that Schwede only defined uniformly F-compatible ideals in the F-finite case, whereas the majority of this paper is devoted to the study of fully Φ(E)-special ideals in the (F-pure but) not necessarily F-finite case.

We shall use the notation of this Introduction throughout the remainder of the paper. In particular, R will denote a local ring of prime characteristic p having maximal ideal $\mathfrak {m}$. We shall sometimes use the notation $(R,\mathfrak {m})$ just to remind the reader that R is local. The completion of R will be denoted by $\widehat {R}$. We shall only assume that R is reduced, or F-pure, or F-finite, when there is an explicit statement to that effect; also E will continue to denote $E_{R}(R/\mathfrak {m})$. We continue to use $\mathbb {N}$, respectively $\mathbb {N}_{0}$, to denote the set of all positive, respectively non-negative, integers.

For $j \in \mathbb {N}_{0}$, the jth component of an $\mathbb {N}_{0}$-graded left R[x,f]-module G will be denoted by G _j .

2 Fully Φ(E)-Special Ideals

We remind the reader that we usually identify the 0th component of ${\Phi }(E) = \bigoplus _{n\in \mathbb {N}_{0}}Rx^{n}\otimes _{R}E$ with E in the obvious natural way. For an ideal $\mathfrak {A}$ of R, we have, with this convention, that the 0th component of $\text {ann}_{\Phi (E)}(\mathfrak {a} R[x,f])$ is contained in $(0:_{E}\mathfrak {a})$.

Lemma 2.1

Assume that $(R,\mathfrak {m})$ is F-pure; let $\mathfrak {A}$ be an ideal of R. Then the 0th component $(\text {ann}_{\Phi (E)}(\mathfrak {a} R[x,f]))_{0}$ of $\text {ann}_{\Phi (E)}(\mathfrak {a} R[x,f])$ contains $(0:_{E}\mathfrak {a})$ if and only if $\mathfrak {A}$ is Φ(E)-special and $(\text {ann}_{\Phi (E)}(\mathfrak {a} R[x,f]))_{0} = (0:_{E}\mathfrak {a})$.

Proof

Only the implication ‘$\Rightarrow$’ needs proof.

Assume that $(0:_{E}\mathfrak {a}) \subseteq (\text {ann}_{\Phi (E)}(\mathfrak {a} R[x,f]))_{0}$. Since $\text {ann}_{\Phi (E)}(\mathfrak {a} R[x,f])$ is an R[x,f]-submodule of Φ(E), it follows that $\text {ann}_{\Phi (E)}(\mathfrak {a} R[x,f])$ contains the image J of the map

$${\Phi}((0:_{E}\mathfrak{a})) = R[x,f]\otimes_{R}(0:_{E}\mathfrak{a}) \longrightarrow R[x,f]\otimes_{R}E = {\Phi}(E) $$

induced by inclusion. Let $\mathfrak {B}$ be the radical ideal of R for which $\text {gr-ann}_{R[x,f]}J = \mathfrak {b} R[x,f]$, so that $\mathfrak {b} = (0:_{R}J)$. As $J \subseteq \text {ann}_{\Phi (E)}(\mathfrak {a} R[x,f])$, we must have $\mathfrak {a} \subseteq \mathfrak {b}$. Furthermore, $\mathfrak {B}$ annihilates $(0:_{E}\mathfrak {a}) \cong \text {Hom}_{R}(R/\mathfrak {a},E)$, and since an R-module and its Matlis dual have the same annihilator, we also have $\mathfrak {b} \subseteq \mathfrak {a}$. Thus $\mathfrak {a} = \mathfrak {b}$ is the R-annihilator of an R[x,f]-submodule of Φ(E), and so $\mathfrak {a} \in \mathcal {I}({\Phi }(E))$.

Finally, note that an $e \in (\text {ann}_{\Phi (E)}(\mathfrak {a} R[x,f]))_{0}$ must be annihilated by $\mathfrak {A}$, and so lies in $(0:_{E}\mathfrak {a})$. □

Definition 2.2

Assume that $(R,\mathfrak {m})$ is F-pure; let $\mathfrak {A}$ be an ideal of R. We say that $\mathfrak {A}$ is fully Φ(E)-special if the equivalent conditions of Lemma 2.1 are satisfied.

Thus $\mathfrak {A}$ is fully Φ(E)-special if and only if $(0:_{E}\mathfrak {a}) \subseteq (\text {ann}_{\Phi (E)}(\mathfrak {a} R[x,f]))_{0}$, and then $\mathfrak {A}$ is Φ(E)-special and we have the equality $(0:_{E}\mathfrak {a}) = (\text {ann}_{\Phi (E)}(\mathfrak {a} R[x,f]))_{0}.$

To facilitate the presentation of some examples of Φ(E)-special ideals that are fully Φ(E)-special, we review next the theory of S-tight closure, where S is a multiplicatively closed subset of R. This theory was developed in [14]. The special case of the theory in which S=R ^∘ is the ‘classical’ tight closure theory of M. Hochster and C. Huneke [2].

Reminders 2.3

Let H be a left R[x,f]-module and let S be a multiplicatively closed subset of R.

• (i) We define the internal S-tight closure of zero in H, denoted by Δ^S(H), to be the R[x,f]-submodule of H given by

  $${\Delta}^{S}(H)= \left\{ h \in H : \text{~there exists~} s \in S \text{~with~} sx^{n}h = 0 \text{~for all~} n \gg 0 \right\}. $$

  When M is an R-module and we take the graded left R[x,f]-module Φ(M)=R[x,f]⊗ _R M for H, the R[x,f]-submodule Δ^S(Φ(M)) of Φ(M) is graded, and we refer to its 0th component as the S-tight closure of 0 in M, or the tight closure with respect to S of 0 in M, and denote it by $0^{*,S}_{M}$. See [14, §1].

• (ii) By [14, Example 1.3(ii)], we have, for an R-module M,

  $${\Delta}^{S}(R[x,f]\otimes_{R}M) = 0^{*,S}_{M} \oplus 0^{*,S}_{Rx \otimes_{R}M} \oplus {\cdots} \oplus 0^{*,S}_{Rx^{n} \otimes_{R}M}\oplus \cdots. $$

• (iii) Recall that an S-test element for R is an element s∈S such that, for every R-module M and every $j \in \mathbb {N}_{0}$, the element s x ^j annihilates 1⊗m∈(Φ(M))₀ for every $m \in 0^{*,S}_{M}$. The ideal of R generated by all the S-test elements for R is called the S-test ideal of R, and denoted by τ ^S(R).

Reminders 2.4

Suppose that $(R,\mathfrak {m})$ is F-pure. Let S be a multiplicatively closed subset of R. Recall that the set $\mathcal {I}({\Phi }(E))$ of Φ(E)-special R-ideals is finite; let $\mathfrak {b}^{S,{\Phi }(E)}$ denote the intersection of all the minimal members of the set

$$\left\{ \mathfrak{p} \in \text{Spec}(R) \cap \mathcal{I}({\Phi}(E)) : \mathfrak{p} \cap S \neq \emptyset\right\}. $$

Thus $\mathfrak {b}^{S,{\Phi }(E)}$ is the smallest member of $\mathcal {I}({\Phi }(E))$ that meets S.

• (i) By [14, Theorem 2.6], the set $S \cap \mathfrak {b}^{S,{\Phi }(E)}$ is (non-empty and) equal to the set of S-test elements for R.

• (ii) Thus there exists an S-test element for R.

• (iii) Furthermore, ${\Delta }^{S}({\Phi }(E)) = \text {ann}_{\Phi (E)}(\mathfrak {b}^{S,{\Phi }(E)}R[x,f])$ and $(0:_{R}{\Delta }^{S}({\Phi }(E))) = \mathfrak {b}^{S,{\Phi }(E)}$, by [14, Proposition 1.5].

• (iv) By [14, Proposition 2.10(v)], we have $\mathfrak {b}^{S,{\Phi }(E)} = (0:_{R}0^{*,S}_{E})$.

Lemma 2.5

(Sharp [14, Corollary 2.8])

Suppose that $(R,\mathfrak {m})$ is F-pure. Let S be the complement in R of the union of finitely many prime ideals.

Then the S-test ideal τ ^S (R) is equal to $\mathfrak {b}^{S,{\Phi }(E)}$ , the smallest member of the finite set $\mathcal {I}({\Phi }(E))$ that meets S.

We shall also use the following result from [14].

Theorem 2.6

(Sharp [14, Theorem 2.12]) Suppose that $(R,\mathfrak {m})$ is F-pure. Let $\mathfrak {a} \in \mathcal {I}({\Phi }(E))$ . Then there exists a multiplicatively closed subset S of R such that $\mathfrak {A}$ is the S-test ideal of R. Moreover, S can be taken to be the complement in R of the union of finitely many prime ideals.

We are now able to give examples of fully Φ(E)-special ideals because the next result shows that, when $(R,\mathfrak {m})$ is complete and F-pure, a Φ(E)-special ideal of R is automatically fully Φ(E)-special.

Proposition 2.7

Suppose that $(R,\mathfrak {m})$ is complete and F-pure. Then every Φ(E)-special ideal of R is fully Φ(E)-special.

Proof

Let $\mathfrak {A}$ be a Φ(E)-special ideal of R. If $\mathfrak {a} = R$, then

$$(0:_{E}\mathfrak{a}) = 0 \subseteq \text{ann}_{\Phi(E)}(\mathfrak{a} R[x,f]) $$

and $\mathfrak {A}$ is fully Φ(E)-special. We therefore assume that $\mathfrak {A}$ is proper.

By Theorem 2.6 and [14, Corollary 2.8], there exist finitely many prime ideals $\mathfrak {p}_{1}, \ldots , \mathfrak {p}_{n}$ of R such that, if we set $S := R\setminus \bigcup _{i=1}^{n}\mathfrak {p}_{i}$, then $\mathfrak {A}$ is the S-test ideal of R, that is $\mathfrak {a} = \tau ^{S}(R) = \mathfrak {b}^{S,{\Phi }(E)},$ where the notation is as in 2.3(iii) and 2.4. Therefore, by 2.3(ii) and 2.4(iii),

$$\begin{array}{@{}rcl@{}} 0^{*,S}_{E} \oplus 0^{*,S}_{Rx \otimes_{R}E} \oplus {\cdots} \oplus 0^{*,S}_{Rx^{n} \otimes_{R}E}\oplus {\cdots} & = & {\Delta}^{S}({\Phi}(E))\\ & =& \text{ann}_{\Phi(E)}(\mathfrak{b}^{S,{\Phi}(E)}R[x,f]). \end{array} $$

Now, we know that $\mathfrak {b}^{S,{\Phi }(E)} = (0:_{R}0^{*,S}_{E})$, by 2.4(iv). Since R is complete, it follows from Matlis duality (see, for example, [15, p. 154]) that $0^{*,S}_{E} = (0:_{E}\mathfrak {b}^{S,{\Phi }(E)})$. We have thus shown that $(0:_{E}\mathfrak {a}) = R \otimes _{R}(0:_{E}\mathfrak {a}) \subseteq (\text {ann}_{\Phi (E)}(\mathfrak {a} R[x,f]))_{0}$. Thus $\mathfrak {A}$ is fully Φ(E)-special. □

Next, we develop some theory for fully Φ(E)-special ideals.

Lemma 2.8

Suppose that $(R,\mathfrak {m})$ is F-pure, and let $\mathfrak {A}$ be a fully Φ(E)-special ideal of R. Then $\mathfrak {A}$ is radical and every associated prime of $\mathfrak {A}$ is also fully Φ(E)-special.

Proof

We can assume that $\mathfrak {A}$ is proper. Since $\mathfrak {A}$ is Φ(E)-special, it must be radical. Let $\mathfrak {a} = \mathfrak {p}_{1} \cap {\cdots } \cap \mathfrak {p}_{t}$ be the minimal primary (prime in this case) decomposition of $\mathfrak {A}$, and let i∈{1,…,t}.

Since $\mathfrak {A}$ is fully Φ(E)-special, $(0:_{E}\mathfrak {a}) \subseteq (\text {ann}_{\Phi (E)}(\mathfrak {a} R[x,f]))_{0}$. Let $e \in (0:_{E}\mathfrak {p}_{i})$ and let $r \in \mathfrak {p}_{i}$. We show that r x ⁿ annihilates the element 1⊗e of the 0th component of Φ(E). There exists

$$a \in \bigcap^{t}_{\stackrel{\scriptstyle j=1}{j\neq i}} \mathfrak{p}_{j} \setminus \mathfrak{p}_{i}. $$

Now $(0:_{E}\mathfrak {p}_{i}) = a(0:_{E}\mathfrak {p}_{i})$, because multiplication by a provides a monomorphism of $R/\mathfrak {p}_{i}$ into itself and E is injective. Therefore $e = ae^{\prime }$ for some $e^{\prime } \in (0:_{E}\mathfrak {p}_{i})$. Therefore $rx^{n}\otimes e = rx^{n}\otimes ae^{\prime } = ra^{p^{n}}x^{n} \otimes e^{\prime } = 0$ since $ra^{p^{n}} \in \mathfrak {a}$ and

$$(0:_{E}\mathfrak{p}_{i})\subseteq (0:_{E}\mathfrak{a}) \subseteq \text{ann}_{\Phi(E)}(\mathfrak{a} R[x,f]).$$

Therefore $(0:_{E}\mathfrak {p}_{i})\subseteq (\text {ann}_{\Phi (E)}(\mathfrak {p}_{i} R[x,f]))_{0}$ and $\mathfrak {p}_{i}$ is fully Φ(E)-special. □

Proposition 2.9

Suppose that $(R,\mathfrak {m})$ is F-pure. Let $(\mathfrak {a}_{\lambda })_{\lambda \in {\Lambda }}$ be a non-empty family of fully Φ(E)-special ideals of R. Then ${\sum }_{\lambda \in {\Lambda }}\mathfrak {a}_{\lambda }$ is again fully Φ(E)-special.

Proof

Set $\mathfrak {a} := {\sum }_{\lambda \in {\Lambda }}\mathfrak {a}_{\lambda }$, and observe that $\mathfrak {a} R[x,f] = {\sum }_{\lambda \in {\Lambda }}(\mathfrak {a}_{\lambda } R[x,f])$. By assumption, we have $(0:_{E}\mathfrak {a}_{\lambda }) \subseteq \text {ann}_{\Phi (E)}(\mathfrak {a}_{\lambda } R[x,f])$ for all λ∈Λ. It follows that

$$\begin{array}{@{}rcl@{}} (0:_{E}\mathfrak{a}) & = &\left(0:_{E}{\sum}_{\lambda \in {\Lambda}}\mathfrak{a}_{\lambda}\right) = \bigcap_{\lambda \in {\Lambda}}\left(0:_{E}\mathfrak{a}_{\lambda}\right)\\ & \subseteq& \bigcap_{\lambda \in {\Lambda}}(\text{ann}_{\Phi(E)}(\mathfrak{a}_{\lambda} R[x,f]))_{0}\\ & =& \left(\text{ann}_{\Phi(E)}\left({\sum}_{\lambda \in {\Lambda}}\left(\mathfrak{a}_{\lambda} R[x,f]\right)\right)\right)_{0} = (\text{ann}_{\Phi(E)}(\mathfrak{a} R[x,f]))_{0}. \end{array} $$

Therefore $\mathfrak {a} := {\sum }_{\lambda \in {\Lambda }}\mathfrak {a}_{\lambda }$ is fully Φ(E)-special. □

Corollary 2.10

Suppose that $(R,\mathfrak {m})$ is F-pure. Then R has a unique largest fully Φ(E)-special proper ideal, and this is prime.

Proof

The zero ideal is fully Φ(E)-special, and so it follows from Proposition 2.9 that the sum $\mathfrak {B}$ of all the fully Φ(E)-special proper ideals of R is fully Φ(E)-special (and contained in $\mathfrak {m}$), and so is the unique largest fully Φ(E)-special proper ideal of R. Also $\mathfrak {B}$ must be prime, since all the associated primes of $\mathfrak {B}$ are fully Φ(E)-special, by Lemma 2.8. □

In what follows, we shall have cause to pass between R and its completion. Note that if R is F-pure, then so too is $\widehat {R}$, by Hochster and Roberts [3, Corollary 6.13]. The following technical lemma will be helpful.

Lemma 2.11

(See [13, Lemma 4.3]) There is a unique way of extending the R-module structure on $E := E_{R}(R/\mathfrak {m})$ to an ${\widehat R}$ -module structure. Recall that, as an $\widehat {R}$ -module, $E \cong E_{\widehat {R}}(\widehat {R}/\widehat {\mathfrak {m}})$.

Since each element of Φ _R (E)=R[x,f]⊗ _R E is annihilated by some power of $\mathfrak {m}$ , the left R[x,f]-module structure on Φ _R (E) can be extended in a unique way to a left ${\widehat R}[x,f]$ -module structure.

The map $\beta : {\Phi }_{R}(E) = R[x,f]\otimes _{R}E \longrightarrow \widehat {R}[x,f]\otimes _{\widehat {R}}E = {\Phi }_{\widehat {R}}(E)$ for which

$$\beta(rx^{i} \otimes h) = rx^{i} \otimes h \quad \text{for all~} r \in R,\ i \in \mathbb{N}_{0} \text{~and~} h \in E $$

is a homogeneous $\widehat {R}[x,f]$ -isomorphism.

Since each element of Φ _R (E) is annihilated by some power of $\mathfrak {m}$ , it follows that a subset of Φ _R (E) is an R[x,f]-submodule if and only if it is an $\widehat {R}[x,f]$ -submodule. Consequently,

$$\mathcal{I}_{R}({\Phi}_{R}(E)) = \left\{ \mathfrak{B} \cap R : \mathfrak{B} \in \mathcal{I}_{\widehat{R}}({\Phi}_{\widehat{R}}(E)) \right\}. $$

Lemma 2.12

Suppose that $(R,\mathfrak {m})$ is F-pure, and let $\mathfrak {A}$ be an ideal of R. Then $\mathfrak {a} \widehat {R}$ is a fully ${\Phi }_{\widehat {R}}(E)$ -special ideal of $\widehat {R}$ if and only if $\mathfrak {A}$ is a fully Φ _R (E)-special ideal of R.

Proof

By Lemma 2.11, when we extend the left R[x,f]-module structure on Φ _R (E), in the unique way possible, to a left ${\widehat R}[x,f]$-module structure, $E \cong E_{\widehat {R}}(\widehat {R}/\widehat {\mathfrak {m}})$ as $\widehat R$-modules and ${\Phi }_{R}(E) \cong {\Phi }_{\widehat {R}}(E)$ as left ${\widehat R}[x,f]$-modules. The claim therefore follows from the facts that

$$\text{ann}_{{\Phi}_{R}(E)}(\mathfrak{a} R[x,f]) = \text{ann}_{{\Phi}_{R}(E)}((\mathfrak{a} \widehat R)\widehat R[x,f])$$

and $(0:_{E} \mathfrak {a}) =(0:_{E} \mathfrak {a}\widehat R)$. □

3 The Case Where R Is an F-Pure Homomorphic Image of an Excellent Regular Local Ring of Characteristic p

The main aim of this section is to prove that, when R is an F-pure homomorphic image of an excellent regular local ring of characteristic p, every Φ(E)-special ideal of R is a fully Φ(E)-special ideal. This will enable us to extend some results obtained in [14, §3] about an F-pure complete local ring to an F-pure homomorphic image of an excellent regular local ring of characteristic p. We begin the section with a lemma that is derived from a result of G. Lyubeznik [5, Lemma 4.1].

Lemma 3.1

Let $(S,\mathfrak {M})$ be a complete regular local ring of characteristic p, and let $\mathfrak {B}$ be a proper, non-zero ideal of S. Denote $E_{S}(S/\mathfrak {M})$ by E _S , and let S[x,f] denote the Frobenius skew polynomial ring over S. Let $n \in \mathbb {N}$.

Since S is regular, S ⁽ⁿ⁾ is faithfully flat over S, and we identify $Sx^{n}\otimes _{S}(0:_{E_{S}}\mathfrak {B})$ as an S-submodule of Sx ⁿ ⊗ _S E _S in the natural way. Let a ₁ ,…,a _d be a regular system of parameters for S. Consider the S-isomorphism δ _n :Sx ⁿ ⊗ _S E _S →≅E _S of [11, 4.2(iii)], for which (with the notation used in the statement of that result)

$$\delta_{n} \left( bx^{n} \otimes \left[\frac{s}{(a_{1}{\ldots} a_{d})^{j}}\right] \right) = \left[\frac{bs^{p^{n}}}{(a_{1}{\ldots} a_{d})^{jp^{n}}}\right] \quad \text{~for all~} b,s \in S \text{~and~} j \in \mathbb{N}_{0}. $$

The isomorphism δ _n maps

• (i) $Sx^{n}\otimes _{S}(0:_{E_{S}}\mathfrak {B})$ onto $(0:_{E_{S}}\mathfrak {B}^{[p^{n}]})$ , and

• (ii) $\mathfrak {B}(Sx^{n}\otimes _{S}(0:_{E_{S}}\mathfrak {B}))$ onto $(0:_{E_{S}}(\mathfrak {B}^{[p^{n}]}:\mathfrak {B}))$ .

Proof

• (i) Use of the analogue of Lyubeznik [5, Lemma 4.1] for the functor S x ⁿ⊗ _S ∙ shows that the Matlis dual of $Sx^{n}\otimes _{S}(0:_{E_{S}}\mathfrak {B})$ is S-isomorphic to $Sx^{n} \otimes _{S} (S/\mathfrak {B}) \cong S/\mathfrak {B}^{[p^{n}]}$. Since each S-module has the same annihilator as its Matlis dual, we thus see that $Sx^{n}\otimes _{S}(0:_{E_{S}}\mathfrak {B})$ has annihilator $\mathfrak {B}^{[p^{n}]}$. As S is complete, $T = (0:_{E_{S}}(0:_{S}T))$ for each submodule T of E _S , by Matlis duality (see, for example, [15, p. 154]). It therefore follows that

  $$\delta_{n}(Sx^{n}\otimes_{S}(0:_{E_{S}}\mathfrak{B})) =(0:_{E_{S}}\mathfrak{B}^{[p^{n}]}). $$

• (ii) Set $N := Sx^{n}\otimes _{S}(0:_{E_{S}}\mathfrak {B})$. Similar reasoning shows that

  $$\delta_{n}(\mathfrak{B} N) = (0:_{E_{S}}(0:_{S}\mathfrak{B} N)) = (0:_{E_{S}}((0:_{S}N) : \mathfrak{B})) = (0:_{E_{S}}(\mathfrak{B}^{[p^{n}]}:\mathfrak{B})). $$

□

Proposition 3.2

Suppose that $R = S/\mathfrak {A}$ , where $(S,\mathfrak {M})$ is a regular local ring of characteristic p, and $\mathfrak {A}$ is a proper ideal of S. Assume also that R is F-pure. Let $\mathfrak {B}$ be a proper ideal of R; let $\mathfrak {B}$ be the unique ideal of S that contains $\mathfrak {A}$ and is such that $\mathfrak {B}/\mathfrak {A} = \mathfrak {b}$.

Then $\mathfrak {B}$ is fully Φ(E)-special if and only if $(\mathfrak {A}^{[p^{n}]}:\mathfrak {A}) \subseteq (\mathfrak {B}^{[p^{n}]}:\mathfrak {B})$ for all $n \in \mathbb {N}$.

Note

In the F-finite case, this result is already known and due to K. Schwede [10, Proposition 3.11 and Lemma 5.1].

Proof

If $\mathfrak {A} = 0$, then R is regular, so that its big test ideal is R itself (by [6, Theorem 8.8], for example) and the only proper Φ(E)-special ideal of R is 0; also, $(0^{[p^{n}]}:0) = S$, and the only proper ideal $\mathfrak {B}$ of S satisfying $(0^{[p^{n}]}:0) \subseteq (\mathfrak {B}^{[p^{n}]}:\mathfrak {B})$ for all $n\in \mathbb {N}$ is the zero ideal. Thus, the result is true when $\mathfrak {A} = 0$; we therefore assume for the remainder of this proof that $\mathfrak {A} \neq 0$.

Note that $\widehat {R} = \widehat {S}/\mathfrak {A}\widehat {S}$ is again F-pure and that $\widehat {S}$ is an excellent complete regular local ring of characteristic p, with maximal ideal $\mathfrak {M} \widehat {S}$.

We also note that $\mathfrak {B}$ is a fully Φ _R (E)-special ideal of R if and only if $\mathfrak {b} \widehat {R}$ is a fully ${\Phi }_{\widehat {R}}(E)$-special ideal of $\widehat {R}$, by Lemma 2.12. Furthermore, by the faithful flatness of $\widehat {S}$ over S, we have, for $n \in \mathbb {N}$,

$$((\mathfrak{A} \widehat{S})^{[p^{n}]} : \mathfrak{A} \widehat{S}) = (\mathfrak{A}^{[p^{n}]} : \mathfrak{A})\widehat{S} \subseteq (\mathfrak{B}^{[p^{n}]} : \mathfrak{B})\widehat{S} = ((\mathfrak{B} \widehat{S})^{[p^{n}]} : \mathfrak{B} \widehat{S}) $$

if and only if $(\mathfrak {A}^{[p^{n}]} : \mathfrak {A}) \subseteq (\mathfrak {B}^{[p^{n}]} : \mathfrak {B})$. Therefore, we can, and do, assume henceforth in this proof that S is complete.

Let $E_{S} := E_{S}(S/\mathfrak {M})$. Now $(0:_{E_{S}}\mathfrak {A}) = E := E_{R}(R/\mathfrak {m})$ and $(0:_{E_{S}}\mathfrak {B}) = (0:_{E}\mathfrak {b})$. Note that $\mathfrak {B}$ is fully Φ _R (E)-special if and only if, for each $n \in \mathbb {N}$ and each $r \in \mathfrak {b}$, the element r x ⁿ∈R x ⁿ annihilates the R-submodule $(0:_{E}\mathfrak {b})$ of the 0th component E of Φ _R (E).

Let $n \in \mathbb {N}$. There is an exact sequence of (S,S)-bimodules

$$0 \longrightarrow \mathfrak{A} Sx^{n}\stackrel{\subseteq}{\longrightarrow} Sx^{n} \stackrel{\nu}{\longrightarrow} Rx^{n} \longrightarrow 0, $$

where $\nu (sx^{n}) = (s + \mathfrak {A})x^{n}$ for all s∈S. The map

$$Sx^{n} \otimes_{S}(0:_{E_{S}}\mathfrak{A}) \longrightarrow Rx^{n} \otimes_{S}(0:_{E_{S}}\mathfrak{A}) = Rx^{n} \otimes_{R}(0:_{E_{S}}\mathfrak{A}) = Rx^{n} \otimes_{R}E $$

induced by ν therefore has kernel $\mathfrak {A}(Sx^{n} \otimes _{S}(0:_{E_{S}}\mathfrak {A}))$.

It follows that $\mathfrak {B}$ is fully Φ _R (E)-special if and only if, for all $n \in \mathbb {N}$, $s\in \mathfrak {B}$ and $g \in (0:_{E_{S}}\mathfrak {B}) = (0:_{E}\mathfrak {b})$, the element s x ⁿ⊗g of $Sx^{n} \otimes _{S}(0:_{E_{S}}\mathfrak {A})$ lies in

$$\mathfrak{A}(Sx^{n} \otimes_{S}(0:_{E_{S}}\mathfrak{A})). $$

In other words, $\mathfrak {B}$ is fully Φ _R (E)-special if and only if, for all $n \in \mathbb {N}$, we have

$$\mathfrak{B}(Sx^{n} \otimes_{S}(0:_{E_{S}}\mathfrak{B})) \subseteq \mathfrak{A}(Sx^{n} \otimes_{S}(0:_{E_{S}}\mathfrak{A})). $$

(We are here identifying $Sx^{n} \otimes _{S}(0:_{E_{S}}\mathfrak {B})$ and $Sx^{n} \otimes _{S}(0:_{E_{S}}\mathfrak {A})$ with submodules of S x ⁿ⊗ _S E _S in the obvious ways, using the faithful flatness of S ⁽ⁿ⁾ over S.)

By [11, 4.2(iii)], we have S x ⁿ⊗ _S E _S ≅E _S . Since S is complete, each submodule T of E _S satisfies $T = (0:_{E_{S}}(0:_{S}T))$. Set N:=S x ⁿ⊗ _S E _S . Thus

$$\mathfrak{A}(Sx^{n} \otimes_{S}(0:_{E_{S}}\mathfrak{A})) = (0:_{N}(0:_{S}(\mathfrak{A}(Sx^{n} \otimes_{S}(0:_{E_{S}}\mathfrak{A}))))) = (0:_{N}(\mathfrak{A}^{[p^{n}]}:\mathfrak{A})), $$

by Lemma 3.1. Similarly, $\mathfrak {B}(Sx^{n} \otimes _{S}(0:_{E_{S}}\mathfrak {B})) = (0:_{N}(\mathfrak {B}^{[p^{n}]}:\mathfrak {B}))$. It follows that $\mathfrak {B}$ is fully Φ _R (E)-special if and only if

$$(0:_{N}(\mathfrak{B}^{[p^{n}]}:\mathfrak{B})) \subseteq (0:_{N}(\mathfrak{A}^{[p^{n}]}:\mathfrak{A})) \quad \text{for all~} n \in \mathbb{N}, $$

that is (since N≅E _S ), if and only if $(\mathfrak {A}^{[p^{n}]}:\mathfrak {A}) \subseteq (\mathfrak {B}^{[p^{n}]}:\mathfrak {B})$ for all $n \in \mathbb {N}$. □

Theorem 3.3

Suppose that $R = S/\mathfrak {A}$ is a homomorphic image of an excellent regular local ring $(S,\mathfrak {M})$ of characteristic p, modulo a proper ideal $\mathfrak {A}$ . Assume that R is F-pure.

Then each Φ(E)-special ideal of R is fully Φ(E)-special.

Proof

Once again, the claim is easy to prove if $\mathfrak {A} = 0$, and so we assume henceforth in this proof that $\mathfrak {A} \neq 0$

Note that $\widehat {R} = \widehat {S}/\mathfrak {A}\widehat {S}$ is again F-pure and that $\widehat {S}$ is an excellent complete regular local ring of characteristic p, with maximal ideal $\mathfrak {M} \widehat {S}$.

Let $\mathfrak {B}$ be a Φ(E)-special R-ideal with $\mathfrak {b} \neq R$. Then $\mathfrak {b} = \mathfrak {c}\cap R$ for some ${\Phi }_{\widehat {R}}(E)$-special ${\widehat R}$-ideal $\mathfrak {c}$. (We have used Lemma 2.11 here.) Let $\mathfrak {c}$ be the unique ideal of ${\widehat S}$ that contains $\mathfrak {A} {\widehat S}$ and is such that $\mathfrak {C}/\mathfrak {A} {\widehat S} = \mathfrak {c}$. By Proposition 2.7, the ideal $\mathfrak {c}$ of ${\widehat R}$ is fully ${\Phi }_{\widehat {R}}(E)$-special, and so, by Proposition 3.2, we have

$$(\mathfrak{A}^{[p^{n}]}:\mathfrak{A}){\widehat S} = ((\mathfrak{A}{\widehat S})^{[p^{n}]}:\mathfrak{A}{\widehat S}) \subseteq (\mathfrak{C}^{[p^{n}]}:\mathfrak{C}) \quad \text{for all~} n \in \mathbb{N}. $$

Set $\mathfrak {C} \cap S := \mathfrak {B}$, so that $\mathfrak {B}/\mathfrak {A} = \mathfrak {b}$.

Let $n \in \mathbb {N}$ and $s \in (\mathfrak {A}^{[p^{n}]}:\mathfrak {A})$. Therefore, $s \in (\mathfrak {C}^{[p^{n}]}:\mathfrak {C})$. It follows from G. Lyubeznik and K. E. Smith [6, Lemma 6.6] that $\mathfrak {C}^{[p^{n}]}\cap S = (\mathfrak {C}\cap S)^{[p^{n}]}$. (Lyubeznik’s and Smith’s proof of this result uses work of N. Radu [9, Corollary 5], which, in turn, uses D. Popescu’s general Néron desingularization [7, 8].) We can now deduce that

$$s(\mathfrak{C}\cap S) \subseteq s\mathfrak{C} \cap S \subseteq \mathfrak{C}^{[p^{n}]}\cap S = (\mathfrak{C}\cap S)^{[p^{n}]}, $$

so that $s \in ((\mathfrak {C}\cap S)^{[p^{n}]}:\mathfrak {C}\cap S) = (\mathfrak {B}^{[p^{n}]}:\mathfrak {B})$.

We have thus shown that $(\mathfrak {A}^{[p^{n}]}:\mathfrak {A}) \subseteq (\mathfrak {B}^{[p^{n}]}:\mathfrak {B})$ for all $n\in \mathbb {N}$, so that $\mathfrak {b} = \mathfrak {B}/\mathfrak {A}$ is fully Φ(E)-special by Proposition 3.2. □

In the case where R is an F-pure homomorphic image of an excellent regular local ring of characteristic p, the characterization of $\mathcal {I}({\Phi }(E))$ afforded by Proposition 3.2 and Theorem 3.3 enables us to see that set behaves well under localization. As the ideals in $\mathcal {I}({\Phi }(E))$ are precisely those that can be expressed as intersections of finitely many prime members of $\mathcal {I}({\Phi }(E))$, it is of interest to examine the behaviour of $\mathcal {I}({\Phi }(E))\cap \text {Spec} (R)$ under localization. The next proposition, which is an extension of part of [12, Proposition 2.8], is in preparation for this investigation.

Proposition 3.4

Let S be a regular local ring of characteristic p, and let $n \in \mathbb {N}$ . Let $\mathfrak {A},\mathfrak {B}_{1}, \ldots , \mathfrak {B}_{t},\mathfrak {C}$ be ideals of S with $0 \neq \mathfrak {A} \neq S$ , and let $\mathfrak {A} = \mathfrak {Q}_{1} \cap {\ldots } \cap \mathfrak {Q}_{t}$ be a minimal primary decomposition of $\mathfrak {A}$.

• (i) We have $(\mathfrak {B}_{1} \cap {\cdots } \cap \mathfrak {B}_{t})^{[p^{n}]} = \mathfrak {B}_{1}^{[p^{n}]}\cap {\cdots } \cap \mathfrak {B}_{t}^{[p^{n}]}$.

• (ii) If $\mathfrak {Q}$ is a $\mathfrak {P}$ -primary ideal of S, then $\mathfrak {Q}^{[p^{n}]}$ is also $\mathfrak {P}$ -primary.

• (iii) The equation $\mathfrak {A}^{[p^{n}]} = \mathfrak {Q}_{1}^{[p^{n}]} \cap {\cdots } \cap \mathfrak {Q}_{t}^{[p^{n}]}$ provides a minimal primary decomposition of $\mathfrak {A}^{[p^{n}]}$.

• (iv) We have $(\mathfrak {A} :\mathfrak {C})^{[p^{n}]} = (\mathfrak {A}^{[p^{n}]}:\mathfrak {C}^{[p^{n}]})$ and $(\mathfrak {A}^{[p^{n}]} : \mathfrak {A}) \subseteq \left ((\mathfrak {A} :\mathfrak {C})^{[p^{n}]} : (\mathfrak {A} : \mathfrak {C})\right )$.

• (v) If $\mathfrak {P}$ is an associated prime ideal of $\mathfrak {A}$ , then $(\mathfrak {A}^{[p^{n}]} : \mathfrak {A}) \subseteq (\mathfrak {P}^{[p^{n}]} : \mathfrak {P})$.

• (vi) Since $0 \neq \mathfrak {A} \neq S$ , we have $(\mathfrak {A}^{[p^{n}]} : \mathfrak {A}) \neq S$ . If $\mathfrak {P}_{1} := \sqrt {\mathfrak {Q}_{1}}$ is a minimal prime ideal of $\mathfrak {A}$ , then $\mathfrak {P}_{1}$ is a minimal prime ideal of $(\mathfrak {A}^{[p^{n}]} : \mathfrak {A})$ and the unique $\mathfrak {P}_{1}$ -primary component of $(\mathfrak {A}^{[p^{n}]} : \mathfrak {A})$ is $(\mathfrak {Q}_{1}^{[p^{n}]} : \mathfrak {Q}_{1})$.

Proof

Parts (i), (ii) and (iii) were essentially proved in [12, Proposition 2.8], while parts (iv), (v) and (vi) can be proved by obvious modifications of the arguments used to prove the corresponding parts of [12, Proposition 2.8]. □

Corollary 3.5

Suppose that R is F-pure and a homomorphic image of an excellent regular local ring S of characteristic p modulo a proper ideal $\mathfrak {A}$ . Let $\mathfrak {p} \in \text {Spec} (R)$ . Then

$$\mathcal{I}_{R_{\mathfrak{p}}}({\Phi}_{R_{\mathfrak{p}}}(E_{R_{\mathfrak{p}}}(R_{\mathfrak{p}}/\mathfrak{p} R_{\mathfrak{p}}))) \cap \text{Spec} (R_{\mathfrak{p}}) = \left\{ \mathfrak{q} R_{\mathfrak{p}} : \mathfrak{q} \in \mathcal{I}({\Phi}(E))\cap \text{Spec} (R) \text{~and~} \mathfrak{q} \subseteq \mathfrak{p}\right\}. $$

Proof

Note that, by M. Hochster and J. L. Roberts [3, Lemma 6.2], the localization $R_{\mathfrak {p}}$ is again F-pure. The claim is easy to prove when $\mathfrak {A} = 0$, and so we assume that $\mathfrak {A} \neq 0$.

For each lower case fraktur letter that denotes an ideal of R, let the corresponding upper case fraktur letter denote the unique ideal of S that contains $\mathfrak {A}$ and has quotient modulo $\mathfrak {A}$ equal to the specified ideal of R. For example, $\mathfrak {P}$ denotes the unique ideal of S that contains $\mathfrak {A}$ and is such that $\mathfrak {P}/\mathfrak {A} = \mathfrak {p}$.

Note that $R_{\mathfrak {p}} \cong S_{\mathfrak {P}}/\mathfrak {A} S_{\mathfrak {P}}$ is again a homomorphic image of an excellent regular local ring S of characteristic p. Let $\mathfrak {q} \in \text {Spec} (R)$ with $\mathfrak {q} \subseteq \mathfrak {p}$.

Suppose first that $\mathfrak {q} \in \mathcal {I}({\Phi }(E))\cap \text {Spec} (R)$. By Theorem 3.3, we see that $\mathfrak {Q}$ is fully Φ(E)-special; use of Proposition 3.2 shows that $(\mathfrak {A}^{[p^{n}]}:\mathfrak {A}) \subseteq (\mathfrak {Q}^{[p^{n}]}:\mathfrak {Q})$ for all $n \in \mathbb {N}$. Therefore

$$((\mathfrak{A} S_{\mathfrak{P}})^{[p^{n}]}:\mathfrak{A} S_{\mathfrak{P}}) \subseteq ((\mathfrak{Q} S_{\mathfrak{P}})^{[p^{n}]}:\mathfrak{Q} S_{\mathfrak{P}}) \quad \text{for all~} n \in \mathbb{N}. $$

Since the standard isomorphism $S_{\mathfrak {P}}/\mathfrak {A} S_{\mathfrak {P}} \stackrel {\cong }{\longrightarrow } R_{\mathfrak {p}}$ maps $\mathfrak {Q} S_{\mathfrak {P}}/\mathfrak {A} S_{\mathfrak {P}}$ onto $\mathfrak {q} R_{\mathfrak {p}}$, it follows from Proposition 3.2 that $\mathfrak {q} R_{\mathfrak {p}}$ is fully ${\Phi }_{R_{\mathfrak {p}}}(E_{R_{\mathfrak {p}}}(R_{\mathfrak {p}}/\mathfrak {p} R_{\mathfrak {p}}))$-special.

Conversely, suppose that $\mathfrak {q} R_{\mathfrak {p}}$ is ${\Phi }_{R_{\mathfrak {p}}}(E_{R_{\mathfrak {p}}}(R_{\mathfrak {p}}/\mathfrak {p} R_{\mathfrak {p}}))$-special, so that, by Theorem 3.3, it is fully ${\Phi }_{R_{\mathfrak {p}}}(E_{R_{\mathfrak {p}}}(R_{\mathfrak {p}}/\mathfrak {p} R_{\mathfrak {p}}))$-special. By Proposition 3.2, this means that

$$((\mathfrak{A} S_{\mathfrak{P}})^{[p^{n}]}:\mathfrak{A} S_{\mathfrak{P}}) \subseteq ((\mathfrak{Q} S_{\mathfrak{P}})^{[p^{n}]}:\mathfrak{Q} S_{\mathfrak{P}}) \quad \text{for all~} n \in \mathbb{N}. $$

Let ^e and ^c denote extension and contraction of ideals under the natural ring homomorphism $S \longrightarrow S_{\mathfrak {P}}$. Contract the last displayed inclusion relations back to S to see that

$$(\mathfrak{A}^{[p^{n}]}:\mathfrak{A} ) \subseteq (\mathfrak{A}^{[p^{n}]}:\mathfrak{A} )^{ec} \subseteq (\mathfrak{Q}^{[p^{n}]}:\mathfrak{Q} )^{ec} = (\mathfrak{Q}^{[p^{n}]}:\mathfrak{Q} ) \quad \text{for all~} n \in \mathbb{N} $$

because $(\mathfrak {Q}^{[p^{n}]}:\mathfrak {Q})$ is $\mathfrak {Q}$-primary (for all $n\in \mathbb {N}$), by Proposition 3.4(vi). It follows from Proposition 3.2 that $\mathfrak {Q}/\mathfrak {A} = \mathfrak {q}$ is fully Φ(E)-special. □

We can now recover a special case of a result of Lyubeznik and Smith.

Corollary 3.6

(G. Lyubeznik and K. E. Smith [6, Theorem 7.1]) Suppose that R is F-pure and a homomorphic image of an excellent regular local ring S of characteristic p modulo a proper ideal $\mathfrak {A}$ . Let $\mathfrak {p} \in \textnormal {Spec}(R)$ . Then the big test ideal of $R_{\mathfrak {p}}$ is the extension to $R_{\mathfrak {p}}$ of the big test ideal of R. In symbols, $\widetilde {\tau }(R_{\mathfrak {p}}) = \widetilde {\tau }(R)R_{\mathfrak {p}}$.

Proof

The big test ideal $\widetilde {\tau }(R)$ of R is equal to the intersection of the (finitely many) members of $\mathcal {I}({\Phi }(E))\cap \text {Spec} (R)$ of positive height, and a similar statement holds for $R_{\mathfrak {p}}$. The claim therefore follows from Corollary 3.5. □

Some results were obtained in [14, Theorem 3.1] for an F-pure complete local ring of characteristic p. We can now use Theorem 3.3 to establish analogous results for an F-pure homomorphic image of an excellent regular local ring of characteristic p.

Theorem 3.7

Suppose $(R,\mathfrak {m})$ is F-pure and that every Φ(E)-special ideal of R is fully Φ(E)-special. (For example, by Theorem 3.3, this would be the case if R were a homomorphic image of an excellent regular local ring of characteristic p.) Let $\mathfrak {c}$ be a proper ideal of R that is Φ(E)-special. In the light of Theorem 2.6, let $\mathfrak {p}_{1}, \ldots ,\mathfrak {p}_{w}$ be prime ideals of R for which the multiplicatively closed subset $S = R \setminus \bigcup _{i=1}^{w}\mathfrak {p}_{i}$ of R satisfies $\mathfrak {c} = \tau ^{S}(R)$ . Set J:=Δ ^S (Φ(E)), a graded left R[x,f]-module.

• (i) We have $J = 0^{*,S}_{E} \oplus 0^{*,S}_{Rx\otimes _{R}E} \oplus {\cdots } \oplus 0^{*,S}_{Rx^{n}\otimes _{R}E} \oplus {\cdots } = \text {ann}_{\Phi (E)}(\mathfrak {c} R[x,f])$.

• (ii) When we regard J as a graded left $(R/\mathfrak {c})[x,f]$ -module in the natural way, it is x-torsion-free and has $\mathcal {I}_{R/\mathfrak {c}}(J) = \left \{\mathfrak {g}/\mathfrak {c} : \mathfrak {g} \in \mathcal {I}({\Phi }(E)) : \mathfrak {g} \supseteq \mathfrak {c} \right \}$.

• (iii) The 0th component J ₀ of J is $(0:_{E}\mathfrak {c})$ ; as $R/\mathfrak {c}$ -module, this is isomorphic to $E_{R/\mathfrak {c}}((R/\mathfrak {c})/(\mathfrak {m}/\mathfrak {c}))$.

• (iv) The ring $R/\mathfrak {c}$ is F-pure.

• (v) We have $\mathcal {I}({\Phi }_{R/\mathfrak {c}}(J_{0})) \subseteq \mathcal {I}_{R/\mathfrak {c}}(J)$ , so that

  $$\left\{ \mathfrak{d} : \mathfrak{d} \text{~is an ideal of \textit{R} with~} \mathfrak{d} \supseteq \mathfrak{c} \text{~and~} \mathfrak{d}/\mathfrak{c} \in \mathcal{I}({\Phi}_{R/\mathfrak{c}}(J_{0}))\right\} \subseteq \mathcal{I}({\Phi}_{R}(E)). $$

Proof

Since the Φ(E)-special ideal $\mathfrak {c}$ is fully Φ(E)-special, we have $J_{0} = (0:_{E}\mathfrak {c})$. Given this observation, one can now use the arguments employed in the proof of [14, Theorem 3.1] to furnish a proof of this theorem. □

The next corollary follows from Theorem 3.7 just as, in [14], Corollary 3.2 follows from Theorem 3.1.

Corollary 3.8

Suppose that $(R,\mathfrak {m})$ is local, F-pure and that every Φ(E)-special ideal of R is fully Φ(E)-special. (For example, by Theorem 3.3, this would be the case if R were a homomorphic image of an excellent regular local ring of characteristic p.) Let $\mathfrak {c}$ be a proper ideal of R that is Φ(E)-special. Denote $R/\mathfrak {c}$ by $\overline {R}$ , and note that $\overline {R}$ is F-pure, by Theorem 3.7(iv). Let T be a multiplicatively closed subset of $\overline {R}$ which is the complement in $\overline {R}$ of the union of finitely many prime ideals. The finitistic T-test ideal$\tau ^{\text {fg},T} (\overline {R})$ of $\overline {R}$ is defined to be $\bigcap _{L}(0:_{\overline {R}}0^{*,T}_{L})$ , where the intersection is taken over all finitely generated $\overline {R}$ -modules L.

• (i) If $\mathfrak {h}$ denotes the unique ideal of R that contains $\mathfrak {c}$ and is such that $\mathfrak {h}/\mathfrak {c} = \tau ^{\text {fg},T}(\overline {R})$ , the finitistic T-test ideal of $\overline {R}$ , then $\mathfrak {h} \in \mathcal {I}({\Phi }(E))$.

• (ii) In particular, if $\mathfrak {h}^{\prime }$ denotes the unique ideal of R that contains $\mathfrak {c}$ and is such that $\mathfrak {h}^{\prime }/\mathfrak {c} = \tau (\overline {R})$ , the T-test ideal of $\overline {R}$ , then $\mathfrak {h}^{\prime } \in \mathcal {I}({\Phi }(E))$.

• (iii) If $\mathfrak {g}$ denotes the unique ideal of R that contains $\mathfrak {c}$ and is such that $\mathfrak {g}/\mathfrak {c} = \tau ^{T}(\overline {R})$ , the T test ideal of $\overline {R}$ , then $\mathfrak {g} \in \mathcal {I}({\Phi }(E))$.

• (iv) In particular, if $\mathfrak {g}^{\prime }$ denotes the unique ideal of R that contains $\mathfrak {c}$ and is such that $\mathfrak {g}^{\prime }/\mathfrak {c} = \widetilde {\tau }(\overline {R})$ , the big test ideal of $\overline {R}$ , then $\mathfrak {g}^{\prime } \in \mathcal {I}({\Phi }(E))$.

Proof

Straightforward modifications of the arguments given in the proof of [14, Corollary 3.2] will provide a proof for this. □

Lemma 3.9

Assume that $(R,\mathfrak {m})$ is local, F-pure and a homomorphic image of an excellent regular local ring of characteristic p.

• (i) There is a strictly ascending chain $0 = \tau _{0} \subset \tau _{1} \subset {\cdots } \subset \tau _{t} \subset \tau _{t+1} = R$ of radical ideals of R such that, for each i=0,…,t, the reduced local ring R/τ _i is F-pure and its test ideal is τ _i+1 /τ _i . We call this the test ideal chain of R. All of τ ₀ =0,τ ₁ ,⋯ ,τ _t , and all their associated primesm, belong to $\mathcal {I}({\Phi }(E))$.

• (ii) There is a strictly ascending chain $0 = \widetilde {\tau }_{0} \subset \widetilde {\tau }_{1} \subset {\cdots } \subset \widetilde {\tau }_{w} \subset \widetilde {\tau }_{w+1} = R$ of radical ideals in $\mathcal {I}({\Phi }(E))$ such that, for each i=0,…,w, the reduced local ring $R/\widetilde {\tau }_{i}$ is F-pure and its big test ideal is $\widetilde {\tau }_{i+1}/\widetilde {\tau }_{i}$ . We call this the big test ideal chain of R. All of $\widetilde {\tau }_{0} = 0,\widetilde {\tau }_{1}, {\cdots } ,\widetilde {\tau }_{w}$ , and all their associated primes, belong to $\mathcal {I}({\Phi }(E))$.

Note

In the case when R is an (F-pure) homomorphic image of an F-finite regular local ring, part (i) of this result is known and due to Janet Cowden Vassilev [16, §3].

Proof

• (i) Set τ ₁:=τ(R), and note that $\tau (R) \in \mathcal {I}({\Phi }(E))$. If τ ₁≠R, apply Theorem 3.7 with the choice $\mathfrak {c} = \tau (R) = \tau _{1}$. That shows that R/τ ₁ is F-pure. Now argue by induction on $\dim R$, noting that R/τ ₁ is a homomorphic image of an excellent regular local ring of characteristic p. Use Theorem 3.7(v) to show that all of τ ₀,τ ₁,…,τ _t belong to $\mathcal {I}({\Phi }(E))$.

• (ii) This is proved similarly.

□

4 The F-Finite Case

In the F-finite case, the results above have strong connections with work of K. Schwede in [10], and the purpose of this section is to explore some of those connections. The introduction contains a description of certain properties of the set of all uniformly F-compatible ideals in an F-finite, F-pure local ring R, and some of these are similar to properties of the set of all fully Φ(E)-special ideals of R: we shall show in this section that, in this special case, an ideal of R is uniformly F-compatible if and only if it is Φ(E)-special, and that this is the case if and only if it is fully Φ(E)-special.

Definition 4.1

Suppose that R is F-finite, let $\mathfrak {B}$ be an ideal of R. Then $\mathfrak {B}$ is said to be uniformly F-compatible if, for every n>0 and every ϕ∈Hom _R (R ⁽ⁿ⁾,R), we have $\phi (\mathfrak {b}^{(n)}) \subseteq \mathfrak {b}$.

Proposition 4.2

(Schwede [10, Lemma 5.1]) Suppose that $(R,\mathfrak {m})$ is F-finite, let $\mathfrak {B}$ be an ideal of R. Then $\mathfrak {B}$ is uniformly F-compatible if and only if $(0:_{E}\mathfrak {b}) \subseteq (\textnormal {ann}_{\Phi (E)}(\mathfrak {b} R[x,f]))_{0}$.

Thus, when R is F-finite and F-pure, $\mathfrak {B}$ is uniformly F-compatible if and only if it is fully Φ(E)-special.

Proof

Let $n\in \mathbb {N}$ and r∈R. Multiplication by r yields an R-homomorphism of R ⁽ⁿ⁾, which, strictly speaking, we should denote by $r\text {Id}_{R^{(n)}}$. Also f ⁿ:R→R ⁽ⁿ⁾ is an R-homomorphism. Thus we can consider the composition of R-homomorphisms R→f ⁿ R ⁽ⁿ⁾→r R ⁽ⁿ⁾.

Application of the functor ∙ ⊗ _R E yields a composition of R-homomorphisms

$$R \otimes_{R}E \longrightarrow R^{(n)}\otimes_{R}E \stackrel{r}{\longrightarrow} R^{(n)}\otimes_{R}E, $$

where the ‘r’ over the second arrow is an abbreviation for $r\text {Id}_{R^{(n)}}\otimes _{R} E$. But R ⁽ⁿ⁾≅R x ⁿ as (R,R)-bimodules; furthermore, $(0:_{E}\mathfrak {b}) \cong \text {Hom}_{R}(R/\mathfrak {b},E)$. It follows that $(0:_{E}\mathfrak {b}) \subseteq (\text {ann}_{\Phi (E)}(\mathfrak {b} R[x,f]))_{0}$ if and only if, for all $n \in \mathbb {N}$ and all $r \in \mathfrak {b}$, the composition

$$(0:_{E}\mathfrak{b}) \stackrel{\subseteq}{\longrightarrow}E \stackrel{\cong}{\longrightarrow} R \otimes_{R}E \longrightarrow R^{(n)}\otimes_{R}E \stackrel{r}{\longrightarrow} R^{(n)}\otimes_{R}E $$

(in which the second map is the natural isomorphism) is zero.

Let M be an R-module. Recall that there is an R-homomorphism

$$\xi_{M} : M \otimes_{R} E \longrightarrow \text{Hom}_{R}(\text{Hom}_{R}(M,R),E) $$

such that, for m∈M, e∈E and g∈Hom _R (M,R), we have $\left (\xi _{M}(m \otimes e) \right ) (g) = g(m)e$. Furthermore, as M varies, the ξ _M constitute a natural transformation of functors; also ξ _M is an isomorphism whenever M is finitely generated. We shall use D to denote the functor Hom _R (∙ ,E).

Since R ⁽ⁿ⁾ is a finitely generated R-module, $(0:_{E}\mathfrak {b}) \subseteq (\text {ann}_{\Phi (E)}(\mathfrak {b} R[x,f]))_{0}$ if and only if, for all $n \in \mathbb {N}$ and all $r \in \mathfrak {b}$, the composition

$$D(R/\mathfrak{b}) \rightarrow D(R) \stackrel{\cong}{\rightarrow} D(\text{Hom}_{R}(R,R)) \rightarrow D(\text{Hom}_{R}(R^{(n)}, R)) \stackrel{r}{\rightarrow} D(\text{Hom}_{R}(R^{(n)}, R)) $$

is zero. (Here, the first map is induced from the natural epimorphism $R \longrightarrow R/\mathfrak {b}$, the second map is the natural isomorphism, and the sequence from the middle term rightwards is the result of application of the functor Hom _R (Hom _R (∙ ,R),E) to the composition R→f ⁿ R ⁽ⁿ⁾→r R ⁽ⁿ⁾ described at the beginning of the proof.)

Since D is a faithful functor (because E is an injective cogenerator for R), we can deduce that $(0:_{E}\mathfrak {b}) \subseteq (\text {ann}_{\Phi (E)}(\mathfrak {b} R[x,f]))_{0}$ if and only if, for all $n \in \mathbb {N}$ and all $r \in \mathfrak {b}$, the composition

$$\text{Hom}_{R}(R^{(n)}, R)\stackrel{r}{\longrightarrow}\text{Hom}_{R}(R^{(n)}, R) \longrightarrow \text{Hom}_{R}(R,R) \stackrel{\cong}{\longrightarrow} R \longrightarrow R/\mathfrak{b} $$

is zero, that is, if and only if $\mathfrak {B}$ is uniformly F-compatible. □

Proposition 4.3

(Schwede [10]) Suppose that $(R,\mathfrak {m})$ is F-finite, and let $\mathfrak {A}$ be an ideal of R. Note that the completion ${\widehat R}$ of R is again F-finite.

• (i) If $\mathfrak {A}$ is a uniformly F-compatible ideal of R, then $\mathfrak {a}{\widehat R}$ is a uniformly F-compatible ideal of ${\widehat R}$ . See Schwede [10, Lemma 3.9].

• (ii) If $\mathfrak {c}$ is a uniformly F-compatible ideal of ${\widehat R}$ , then $\mathfrak {C} \cap R$ is a uniformly F-compatible ideal of R. See Schwede [10, Lemma 3.8].

Proof

For a finitely generated R-module M, we identify ${\widehat M}$ with $M \otimes _{R}{\widehat R}$ in the usual way, and we note that there is a natural ${\widehat R}$-isomorphism $\psi _{M} : \text {Hom}_{R}(M,R)\otimes _{R}{\widehat R} \stackrel {\cong }{\longrightarrow } \text {Hom}_{{\widehat R}}(M\otimes _{R}{\widehat R},R\otimes _{R}{\widehat R})$ for which $\psi _{M}(g \otimes {\widehat r}) = {\widehat r}(g \otimes \text {Id}_{{\widehat R}})$ for all g∈Hom _R (M,R) and ${\widehat r} \in {\widehat R}$. Let $n \in \mathbb {N}$. Consideration of Cauchy sequences shows that $\widehat {M^{(n)}} = {\widehat M}^{(n)}$. In particular, $\widehat {R^{(n)}} = {\widehat R}^{(n)}$ and $\widehat {\mathfrak {a}^{(n)}} = (\widehat {\mathfrak {a}})^{(n)} = (\mathfrak {a}{\widehat R})^{(n)}$.

There is an ${\widehat R}$-isomorphism $\gamma : R^{(n)}\otimes _{R}{\widehat R} \stackrel {\cong }{\longrightarrow } {\widehat R}^{(n)}$ which maps $\mathfrak {a}^{(n)}\otimes _{R}{\widehat R}$ onto $(\mathfrak {a}{\widehat R})^{(n)}$. Also, the natural ${\widehat R}$-isomorphism $\delta : R\otimes _{R}{\widehat R}\stackrel {\cong }{\longrightarrow } {\widehat R}$ maps $\mathfrak {a} \otimes _{R}{\widehat R}$ onto $\mathfrak {a} {\widehat R}$.

• (i) Let $\theta \in \text {Hom}_{{\widehat R}}(R^{(n)}\otimes _{R}{\widehat R},R\otimes _{R}{\widehat R})$. By the above, there exist ϕ ₁,…,ϕ _t ∈Hom _R (R ⁽ⁿ⁾,R) and ${\widehat r}_{1}, \ldots , {\widehat r}_{t} \in {\widehat R}$ such that $\theta = {\widehat r}_{1}(\phi _{1}\otimes \text {Id}_{{\widehat R}}) + {\cdots } + {\widehat r}_{t}(\phi _{t}\otimes \text {Id}_{{\widehat R}})$. Since $\phi _{i}(\mathfrak {a}^{(n)}) \subseteq \mathfrak {a}$ for all $n \in \mathbb {N}$ and i=1,…,t, we see that $\theta (\mathfrak {a}^{(n)}\otimes _{R}{\widehat R}) \subseteq \mathfrak {a}\otimes _{R}{\widehat R}$ for all $n\in \mathbb {N}$. Use of the above-mentioned isomorphisms γ and δ now enables us to conclude that $\mathfrak {a}{\widehat R}$ is a uniformly F-compatible ideal of ${\widehat R}$.

• (ii) Let ϕ∈Hom _R (R ⁽ⁿ⁾,R), and set $\mathfrak {c} := \mathfrak {C} \cap R$. Then

  $$\phi\otimes\text{Id}_{{\widehat R}} \in \text{Hom}_{{\widehat R}}(R^{(n)}\otimes_{R}{\widehat R},R\otimes_{R}{\widehat R}) $$

  and $\delta \circ (\phi \otimes \text {Id}_{{\widehat R}}) \circ \gamma ^{-1}$ maps $\mathfrak {C}^{(n)}$ into $\mathfrak {c}$, and therefore maps $(\mathfrak {c} {\widehat R})^{(n)}$ into $\mathfrak {c}$. Therefore $\delta \circ (\phi \otimes \text {Id}_{{\widehat R}})$ maps $\mathfrak {c}^{(n)}\otimes _{R}{\widehat R}$ into $\mathfrak {c}$, so that $\phi (a) \in \mathfrak {C}\cap R = \mathfrak {c}$ for all $a \in \mathfrak {c}^{(n)}$. Therefore $\mathfrak {c}$ is a uniformly F-compatible ideal of R.

□

Theorem 4.4

Suppose that $(R,\mathfrak {m})$ is F-pure and F-finite. Then each Φ(E)-special ideal $\mathfrak {A}$ of R is automatically fully Φ(E)-special.

Proof

Note that ${\widehat R}$ is also F-pure, by Hochster and Roberts [3, Corollary 6.13]. Also, ${\widehat R}$ is F-finite, because the completion of the finitely generated R-module R ⁽¹⁾ is ${\widehat R}^{(1)}$.

Thus, by definition, $\mathfrak {A}$ is the R-annihilator of an R[x,f]-submodule of Φ(E). It follows from Lemma 2.11 that $\mathfrak {a} = \mathfrak {A} \cap R$ for some ideal $\mathfrak {A}$ of $\widehat {R}$ that is the $\widehat {R}$-annihilator of an $\widehat {R}[x,f]$-submodule of ${\Phi }_{\widehat {R}}(E)$. Thus $\mathfrak {A}$ is ${\Phi }_{\widehat {R}}(E)$-special. It follows from Proposition 2.7 that $\mathfrak {A}$ is a fully ${\Phi }_{\widehat {R}}(E)$-special ideal of $\widehat {R}$, and so is uniformly F-compatible, by Proposition 4.2. Therefore, by Proposition 4.3(ii), the contraction $\mathfrak {A} \cap R = \mathfrak {a}$ is a uniformly F-compatible ideal of R, and is therefore fully Φ(E)-special, by Proposition 4.2 again. □

Corollary 4.5

Suppose that $(R,\mathfrak {m})$ is F-pure and F-finite; let $\mathfrak {A}$ be an ideal of R. Then the following statements are equivalent:

• (i) $\mathfrak {A}$ is uniformly F-compatible;

• (ii) $\mathfrak {A}$ is Φ(E)-special;

• (iii) $\mathfrak {A}$ is fully Φ(E)-special.

Proof

This is now immediate from Proposition 4.2 and Theorem 4.4. □

Question 4.6

Suppose that $(R,\mathfrak {m})$ is F-pure.

We have seen that each Φ(E)-special ideal of R is fully Φ(E)-special if R is complete (by Proposition 2.7) or if R is a homomorphic image of an excellent regular local ring of characteristic p (by Theorem 3.3) or if R is F-finite (by Theorem 4.4).

Note that each complete local ring is excellent, and that each F-finite local ring of characteristic p is excellent (by E. Kunz [4, Theorem 2.5]). The above results raise the following question. If the F-pure local ring R is excellent, is it the case that every Φ(E)-special ideal of R is fully Φ(E)-special?

5 A Generalization of Aberbach’s and Enescu’s Splitting Prime

Recall from [6, Remark 2.8 and Proposition 2.9] that G. Lyubeznik and K. E. Smith defined $(R,\mathfrak {m})$ to be strongly F-regular (even in the case where R is not F-finite) precisely when the zero submodule of E is tightly closed in E. See M. Hochster and C. Huneke [2, §8].

Theorem 5.1

Suppose that $(R,\mathfrak {m})$ is F-pure and that every Φ(E)-special ideal of R is fully Φ(E)-special. (For example, by Theorem 3.3, this would be the case if R were a homomorphic image of an excellent regular local ring of characteristic p; it would also be the case if R were F-finite, by Theorem 4.4.)

• (i) There exists a unique largest Φ(E)-special proper ideal, $\mathfrak {c}$ say, of R and this is prime. Furthermore, $R/\mathfrak {c}$ is strongly F-regular.

• (ii) Let T be the R[x,f]-submodule of Φ(E) generated by $(0:_{E}\mathfrak {m}) \subseteq R\otimes _{R}E$ . Then $\text {gr-ann}_{R[x,f]}T = \mathfrak {c} R[x,f]$.

Proof

• (i) By Corollary 2.10, there is a unique largest Φ(E)-special proper ideal $\mathfrak {c}$ of R, and this is prime. By Corollary 3.8(iv), the big test ideal of $R/\mathfrak {c}$ is $R/\mathfrak {c}$ itself, so that $1_{R/\mathfrak {c}}$ is a big test element for $R/\mathfrak {c}$. Therefore, the zero submodule of $E_{R/\mathfrak {c}}(R/\mathfrak {m})$ is tightly closed in $E_{R/\mathfrak {c}}(R/\mathfrak {m})$, and so $R/\mathfrak {c}$ is strongly F-regular.

• (ii) Note that T is the image of the R[x,f]-homomorphism

  $$R[x,f] \otimes_{R} (0:_{E}\mathfrak{m}) \longrightarrow R[x,f] \otimes_{R} E = {\Phi}(E) $$

  induced by the inclusion map $(0:_{E}\mathfrak {m}) \stackrel {\subseteq }{\longrightarrow } E$. Let $\mathfrak {d}$ be the Φ(E)-special ideal of R for which $\text {gr-ann}_{R[x,f]}T = \mathfrak {d} R[x,f]$. Since $\mathfrak {d}$ annihilates $(0:_{E}\mathfrak {m})$, we see that $\mathfrak {d}$ is proper. Suppose that there exists $\mathfrak {h} \in \mathcal {I}({\Phi }(E))$ such that $\mathfrak {d} \subset \mathfrak {h} \subseteq \mathfrak {m}$. (The symbol ‘$\subset$’ is reserved to denote strict inclusion.) Thus, we have $(0:_{E}\mathfrak {m}) \subseteq (0:_{E}\mathfrak {h}) \subseteq (0:_{E}\mathfrak {d})$. But we know that every Φ(E)-special ideal of R is fully Φ(E)-special, and therefore $(0:_{E}\mathfrak {h}) \subseteq (\text {ann}_{\Phi (E)}(\mathfrak {h} R[x,f]))_{0}$. Since $\text {ann}_{\Phi (E)}(\mathfrak {h} R[x,f])$ is an R[x,f]-submodule of Φ(E), it follows that

  $$T \subseteq \text{ann}_{\Phi(E)}(\mathfrak{h} R[x,f]) \subseteq \text{ann}_{\Phi(E)}(\mathfrak{d} R[x,f]). $$

  Now take graded annihilators: in view of the bijective correspondence between the sets $\mathcal {I}({\Phi }(E))$ and $\mathcal {A}({\Phi }(E))$ alluded to in the Introduction, we have

  $$\begin{array}{@{}rcl@{}} \mathfrak{d} R[x,f] & =& \text{gr-ann}_{R[x,f]}(\text{ann}_{\Phi(E)}(\mathfrak{d} R[x,f]))\\ &\subseteq& \text{gr-ann}_{R[x,f]}(\text{ann}_{\Phi(E)}(\mathfrak{h} R[x,f])) = \mathfrak{h} R[x,f]\\ & \subseteq& \text{gr-ann}_{R[x,f]}T = \mathfrak{d} R[x,f]. \end{array} $$

  Hence $\mathfrak {h} = \mathfrak {d}$ and we have a contradiction.

Thus $\mathfrak {d}$ is a maximal member of the set of proper Φ(E)-special ideals of R; therefore $\mathfrak {d} = \mathfrak {c}$. □

Definition 5.2

(I. M. Aberbach and F. Enescu [1, Definition 3.2]) Suppose $(R,\mathfrak {m})$ is F-finite and reduced. Let u be a generator for the socle $(0:_{E}\mathfrak {m})$ of E. Aberbach and Enescu defined

$$\mathfrak{P} = \left\{r \in R : r\otimes u = 0 \text{~in~} R^{(n)}\otimes_{R}E \text{~for all~}n \gg 0\right\}, $$

an ideal of R.

In [1, §3], Aberbach and Enescu showed that in the case where $(R,\mathfrak {m})$ is F-finite and F-pure, and with the notation of 5.2, the ideal $\mathfrak {P}$ is prime and is equal to the set of elements c∈R for which, for all $e \in \mathbb {N}$, the R-homomorphism $\phi _{c,e} : R \longrightarrow R^{1/p^{e}}$ for which $\phi _{c,e}(1) = c^{1/p^{e}}$ does not split over R. Aberbach and Enescu call this $\mathfrak {P}$ the splitting prime for R. By [1, $R/\mathfrak {P}$ is strongly F-regular.

Proposition 5.3

Suppose that $(R,\mathfrak {m})$ is F-finite and F-pure. Let $\mathfrak {P}$ be Aberbach’s and Enescu’s splitting prime, as in 5.2. Let $\mathfrak {Q}$ be the unique largest Φ(E)-special proper ideal of R, as in Theorem 5.1. Then $\mathfrak {P} = \mathfrak {q}$.

Proof

Let u be a generator for the socle $(0:_{E}\mathfrak {m})$ of E. We can write

$$\mathfrak{P} = \left\{r \in R : rx^{n}\otimes u = 0 \text{~in~} Rx^{n}\otimes_{R}E \text{~for all~}n \gg 0\right\}. $$

Now for a positive integer j and r∈R, if r x ^j⊗u=0 in Φ(E), then

x(r x ^j−1⊗u)=r ^p x ^j⊗u=0,

so that r x ^j−1⊗u=0 because the left R[x,f]-module Φ(E) is x-torsion-free. Therefore

$$\mathfrak{P} = \left\{r \in R : rx^{n}\otimes u = 0 \text{~in~} Rx^{n}\otimes_{R}E \text{~for all~}n \geq 0\right\}.$$

Let T be the R[x,f]-submodule of Φ(E) generated by $(0:_{E}\mathfrak {m}) \subseteq R\otimes _{R}E$. We thus see that $\mathfrak {P} R[x,f] = \text {gr-ann}_{R[x,f]}T$, and this is $\mathfrak {q} R[x,f]$ by Theorem 5.1. Hence $\mathfrak {P} = \mathfrak {q}$. □

Remark 5.4

Suppose that $(R,\mathfrak {m})$ is F-pure and a homomorphic image of an excellent regular local ring S of characteristic p modulo an ideal $\mathfrak {A}$. By Theorem 5.1(i), there exists a unique largest Φ(E)-special proper ideal, $\mathfrak {Q}$ say, of R and this is prime. Let $\mathfrak {Q}$ be the unique ideal of S containing $\mathfrak {A}$ for which $\mathfrak {Q}/\mathfrak {A} = \mathfrak {q}$.

• (i) The results of this section suggest that $\mathfrak {Q}$ can be viewed as a generalization of Aberbach’s and Enescu’s splitting prime: for example, Proposition 5.3 shows that $\mathfrak {Q}$ is that splitting prime in the case where R is, in addition, F-finite.

• (ii) Note that $R/\mathfrak {q}$ is strongly F-regular (in the sense of Lyubeznik and Smith mentioned at the beginning of the section).

• (iii) By Proposition 3.2, we have $(\mathfrak {A}^{[p^{n}]} : \mathfrak {A}) \subseteq (\mathfrak {Q}^{[p^{n}]} : \mathfrak {Q})$ for all $n\in \mathbb {N}$. In the special case in which S is F-finite, this result was obtained by Aberbach and Enescu [1, Proposition 4.4].