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.
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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
The inverse bijection, \({\Delta }^{-1} : \mathcal {I}(G) \longrightarrow \mathcal {A}(G),\) also order-reversing, is given by
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
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 ΔS(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 ΔS(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
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
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),
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
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
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
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
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,
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
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)
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 \(n \in \mathbb {N}\),
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
where \(\nu (sx^{n}) = (s + \mathfrak {A})x^{n}\) for all s∈S. The map
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
In other words, \(\mathfrak {B}\) is fully Φ R (E)-special if and only if, for all \(n \in \mathbb {N}\), we have
(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
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
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
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
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
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
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
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
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
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
(in which the second map is the natural isomorphism) is zero.
Let M be an R-module. Recall that there is an R-homomorphism
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
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
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
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
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
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].
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Dedicated to Ngo Viet Trung, on the occasion of his sixtieth birthday
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Sharp, R.Y. Graded Annihilators and Uniformly F-Compatible Ideals. Acta Math Vietnam 40, 179–195 (2015). https://doi.org/10.1007/s40306-015-0114-3
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DOI: https://doi.org/10.1007/s40306-015-0114-3
Keywords
- Commutative Noetherian local ring
- Prime characteristic
- Frobenius homomorphism
- Tight closure
- Test element
- Excellent ring
- Frobenius skew polynomial ring
- Graded annihilator
- F-pure ring
- Uniformly F-compatible ideal