Characteristic-free test ideals

Tight closure test ideals have been central to the classification of singularities in rings of characteristic $p>0$, and via reduction to characteristic $p$, in equal characteristic zero as well. A summary of their properties and applications can be found in"A survey of test ideals"by Karl Schwede and Kevin Tucker. In this paper, we extend the notion of a test ideal to arbitrary closure operations, particularly those coming from big Cohen-Macaulay modules and algebras, and prove that it shares key properties of tight closure test ideals. Our main results show how these test ideals can be used to give a characteristic-free classification of singularities, including a few specific results on the mixed characteristic case. We also compute examples of these test ideals.


Introduction
The test ideal originated in the study of tight closure [HH90]. Since then, it has been used to define a classification of singularities in rings of characteristic p > 0 [HH90, HH94,HH89], which aligns well with the classification of singularities in equal characteristic 0 [Smi00,Har01]. The general idea is that the larger the test ideal, the closer the ring is to being regular, and the smaller the test ideal, the singular the ring is. The gap in the literature on test ideals is the mixed characteristic case. Recent work of Ma and Schwede [MS18a,MS18b] has begun to fill in this gap, from the perspective of test ideals of pairs. However, most existing results are heavily dependent on the characteristic of the ring, and it is not always known whether corresponding definitions actually agree. In this paper, we study a generalization of the test ideal in a characteristic-free setting. We study test ideals from the perspective of closure operations, mimicking the approach of Hochster and Huneke [Hoc07] with regard to the tight closure test ideal but broadening our definition to include test ideals coming from arbitrary closure operations.
We are motivated by work of the second named author on the connections between closure operations given by big Cohen-Macaulay modules and algebras, and the singularities of the ring [R.G16b, RG18], and encouraged by the fact that these connections hold in all characteristics. More precisely, in [R.G16b], the second named author proves that a ring is regular if and only if all closure operations satisfying certain axioms (Dietz closures) act trivially on modules over the ring. Since big Cohen-Macaulay modules give Dietz closures, we expect further connections to hold between the singularities of the ring and the big Cohen-Macaulay module closures over the ring, and we give some of those connections in this paper. In order to do this, we define and study the test ideals given by closures coming from big Cohen-Macaulay modules and algebras. See Section 3 for details.
We prove that the test ideal of a module closure has multiple equivalent definitions, which we use to get our main results connecting singularities to big Cohen-Macaulay module test ideals.
Theorem 1.1. Let (R, m, k) be a local ring and E = E R (k) the injective hull of the residue field.
(1) Let cl be a residual closure operation. Then the test ideal τ cl (R) = Ann 0 cl E . (Proposition 3.9) (2) Let cl = cl B be a module closure. If R is complete or B is finitely-presented, then τ cl (R) = f ∈Hom R (B,R) f (B). (Theorem 3.12) In particular, the second result is similar to the result that the tight closure test ideal τ * (R) = e≥0 φ∈Hom R (R 1/p e ,R) φ((cR) 1/p e ) for particular elements c [HT04]. This perspective on the tight closure test ideal is one of the major tools used to study it, as described in [ST12]. Our second definition also coincides with the trace ideal of the module B, as studied in [Lam99,Lin17]. By drawing this connection, we open the door for future results on test ideals using the theory of trace ideals, and vice versa. In an upcoming paper with Neil Epstein, the second named author has generalized this to a duality between closure operations and interior operations on finitely-generated and Artinian modules over complete local rings. One important consequence of these results is that when the ring is complete and cl is a big Cohen-Macaulay module closure, τ cl (R) is nonzero (Corollary 3.15).
We also define a finitistic test ideal and discuss cases where it is equal to the (big) test ideal. In the Gorenstein case, the test ideal of an algebra closure is the whole ring if and only if the corresponding finitistic test ideal is also the whole ring (Proposition 3.10).
One advantage to working with test ideals of module closures is that, as a consequence of Theorem 3.12, when the module is finitely-generated, we can compute its test ideal in Macaulay2. This is in contrast to the tight closure test ideal, which is difficult to compute in general. In Section 5, we compute examples of test ideals of finitely-generated Cohen-Macaulay modules, and in some cases are able to compute or approximate the "smallest" Cohen-Macaulay test ideal.
In summary, our results on the classification of singularities via test ideals are: Theorem 1.2. Assume that R is a complete local domain.
(1) R is regular if and only if τ cl B (R) = R for all big Cohen-Macaulay R-modules B. (Corollary 3.5) (2) If R has characteristic p > 0, then R is weakly F-regular if and only if the finitistic test ideal τ f g cl B (R) = R for all big Cohen-Macaulay algebras B. (Corollary 4.22) (3) If τ cl B (R) = R for some big Cohen-Macaulay module B, then R is Cohen-Macaulay.
(Corollary 3.6) (4) If R is a Cohen-Macaulay ring with a canonical module ω, R is Gorenstein if and only if τ cl ω (R) = R. (Corollary 3.18) (5) If B is a finitely-generated Cohen-Macaulay module, then V (τ cl B (R)) ⊆ Sing(R). (Corollary 4.10) (6) If R has finite Cohen-Macaulay type, then τ cl B (R) is m-primary for all finitely-generated Cohen-Macaulay modules B. (Proposition 4.13) (7) If R has countable Cohen-Macaulay type, then τ cl B (R) may not be m-primary, even if B is a finitely-generated Cohen-Macaulay module. (Example 5.5) We apply our techniques to the case of mixed characteristic rings in Section 6. We propose a mixed characteristic closure operation that satisfies Dietz's axioms (these guarantee that it acts like a big Cohen-Macaulay module closure-see [Die10, R.G16b] for details), and prove that its test ideal can be viewed in three different ways similar to those we gave for module closures earlier. In addition to demonstrating how our results can be used in mixed characteristic, this section shows how our proof techniques can be applied to a broader group of closures than module closures.

Preliminaries
In this section we recall the concepts of closure operations and trace ideals. We record their basic properties for later use and give the appropriate references for their proofs.
2.1. Closure Operations. Given a submodule N of a module M , we would like to find a submodule of M containing N that also satisfies some desired properties. This idea is encoded in the following definition.
Definition 2.1. A closure operation cl on a ring R is a map, which to each pair of modules ( (4) (Generalized Colon-Capturing) Let x 1 , . . . , x k+1 be a partial system of parameters for R, and let J = (x 1 , . . . , x k ). Suppose that there exists a surjective homomorphism Note that these axioms are independent of each other, and an arbitrary closure operation on any ring R can satisfy some subset of them.
Remark 2.3. The careful reader will note that the axioms, as expressed here, are set in a more general setting than in [Die10]. In [Die10] the axioms were defined only for complete rings, but this hypothesis was not needed. They were also defined only for finitely-generated modules in [Die10], but the definitions were later used for arbitrary modules in [Die18].
Associated to any R-module B we define a closure operation as follows.
Definition 2.4. Given an R-module B (not necessarily finitely-generated), we define a closure operation cl B on R by for any pair of R-modules N ⊆ M and u ∈ M . This is called a module closure.
When B is an R-algebra, the previous definition can be simplified to u ∈ N cl B M if and only if Remark 2.5. We can extend this closure operation to families of modules in certain circumstances. Let B = {B i } i∈I be a collection of R-modules. We define cl B = cl B i . This is not in general a closure operation (it is not necessarily idempotent), but if the ring is Noetherian, it can be extended to one by iteration as in [Eps12, Construction 3.1.5]. Alternatively, if the family is directed under generation (see Definition 2.18), then cl B does form a closure operation. In particular, if the B i are R-algebras that form a directed family, then cl B is a closure operation.
Definition 2.6. Let (R, m) be a local ring. We say that an R-module B (not necessarily finitelygenerated) is a big Cohen-Macaulay R-module if mB = B and every system of parameters on R is a regular sequence on B. Note that these modules are sometimes referred to as balanced big Cohen-Macaulay R-modules.
Theorem 2.7 ( [Die10]). If B is a big Cohen-Macaulay module, then cl B is a Dietz closure.
Lemma 2.8 ([R.G16b, Lemma 3.2]). Let R be any ring and B any R-module (not necessarily finitely-generated). Then cl B satisfies the first two axioms of a Dietz closure, i.e., cl B is functorial and semi-residual.
Remark 2.9. Note that when M = R and N = I = (f 1 , . . . , f n ) ⊆ R is an ideal we have u ∈ I cl B R if and only if uB ⊆ IB. That is, the closure of of an ideal is the collection of all elements that get multiplied inside I by B, or equivalently Alternatively, we can write I cl B R as the set of elements u of R for which the equation ub = f 1 X 1 + . . . + f n X n has a solution (X 1 , . . . , X n ) in B ⊕n for every b ∈ B. Or in the case that B is an R-algebra, it is enough to check that has a solution.
We will sometimes write I cl B when R is clear from context.
The following examples show that familiar ideals and closure operations are particular examples of module closures.
Example 2.10. Suppose that B = R/J, then we have that I cl B = I + J.
Example 2.12. If R is a domain of characteristic p > 0 and B = R 1/p e for some e > 0, then for an ideal I ⊆ R, u ∈ I cl B if uR 1/p e ⊆ IR 1/p e or equivalently u p e ∈ I [p e ] .
If instead B = R 1/p ∞ , then for an ideal I ⊆ R, u ∈ I cl B if uR 1/p ∞ ⊆ IR 1/p ∞ which in turn is equivalent to uR 1/p e ⊆ IR 1/p e for some e > 0, that is u p e ∈ I [p e ] for some e > 0. This is known as the Frobenius Closure. Proposition 2.14. Let R be a ring possessing a closure operation cl. In the following, N and N ′ are R-submodules of the R-module M , I is a set, and N i ⊆ M i for i ∈ I are R-modules.
(a) Suppose that cl satisfies the Functoriality Axiom and the Semi-residuality Axiom. Let Suppose that R is a domain, cl satisfies the Functoriality Axiom, 0 cl R = 0, and M is a torsion-free finitely-generated R-module. Then 0 cl M = 0.
(h) Suppose that (R, m) is local and cl satisfies the Functoriality Axiom, the Semi-residuality Axiom, and the Faithfulness Axiom. Then, for M a finitely-generated R-module, and N ⊂ M , N cl M ⊆ N + mM . When the closure operation satisfies the functoriality and semi-residual axioms, the elements of the ring multiplying the closure inside the original module can be seen as an annihilator, more precisely: Lemma 2.15. Let cl be a closure operation that is functorial and semi-residual. Then for any R-module M and any R-submodule N of M , we have that N : R N cl M = Ann R 0 cl M/N . In particular, this holds for module closures.
Proof. It is enough to prove that N : Proposition 2.16. Let B, N and M be R-modules, such that N ⊆ M . If R → S is a ring morphism, then for all b ∈ B and all s ∈ S. But we can rewrite the previous expression as for all b ∈ B and all s, S⊗ R M . Corollary 2.17. Let B be an R-module and cl B the associated module closure. For any ideal I in R and any prime ideal P , Similarly, if R is a local ring andR is its completion at the maximal ideal, then Definition 2.18. Recall that a module B is said to generate a module D if some direct sum of copies of B maps onto D.
The generation property enables us to compare the closures given by B and D. Before we give the precise result we need a lemma. Let b 1 , . . . , b r be a generating set for B and E be the injective hull of the residue field of R. We have a map h : given by h(f ) = (b 1 ⊕. . . ⊕b r )⊗f . The kernel of this map is the set of elements f of This is equal to 0 cl B Hom R (D,E) . Hence, by our assumption, 0 cl B Hom R (D,E) ⊆ 0 cl D Hom R (D,E) , but the latter is 0 by the preceding lemma. This implies that h is injective.
Since h is injective, its Matlis dual By Hom-tensor adjointness, we have Under this isomorphism, a map ψ :

Put together, this gives us a surjective map
Hom(B ⊕r , Hom R (Hom R (D, E), E))) → Hom R (Hom R (D, E), E)) that sends ψ : Since R is complete and D is finitely-generated, D ∼ = Hom R (Hom R (D, E), E), and therefore the map Hom(B ⊕r , D) ։ D.
given by (ψ : B ⊕r → D) → ψ(b 1 ⊕ . . . ⊕ b r ) is surjective. Hence for every d ∈ D, there is a map B ⊕r → D whose image contains d. Therefore, B generates D.
The following proposition characterizes regular rings in terms of the behaviour of Dietz closures. This result describes an important connection between the behavior of big Cohen-Macaulay module closure operations and the singularities of the ring. Note that this result holds regardless of the characteristic of R, as by [HH92,And18], we know that big Cohen-Macaulay algebras (and in particular big Cohen-Macaulay modules) exist over complete local domains of any characteristic.
In fact, the proof of this statement in [R.G16b] uses the fact that big Cohen-Macaulay modules over regular rings are faithfully flat [HH92], and we get the following corollary to Theorem 2.21 and its proof in [R.G16b]: Corollary 2.22. Suppose that (R, m) is a local domain with a big Cohen-Macaulay module B (in particular, R may be any complete local domain). Then R is regular if and only if all big Cohen-Macaulay module closures on R are trivial (on submodules of all R-modules).
Remark 2.23. Let (R, m, k) be a Cohen-Macaulay local domain of dimension d. If R is approximately Gorenstein (for example if dim(R) = 1), then for all n ≥ d, the R-modules syz n (k) induce Dietz closures that are trivial if and only if R is regular [R.G16b]. So when R is not regular, syz n (k) gives an example of a nontrivial Dietz closure on R.
We also have the following: Lemma 2.24. Let R be a local domain with a big Cohen-Macaulay R-module B such that cl B is trivial on ideals of R. Then R is Cohen-Macaulay.
Proof. The closure cl B captures colons, so for all partial systems of parameters x 1 , . . . , x k+1 on R, we must have

Trace ideals and Modules.
Definition 2.25. Let R be a ring and A, B R-modules. The trace of A with respect to B is defined as where the sum runs over all R-linear maps from B to A.
That is, the trace of a module A with respect to another module B is the submodule generated by the images of all possible maps from B to A.
(1) B generates A if and only if tr B (A) = A. One example where B generates A is when there is a surjective map from B to A, or if B = R.
(2) When A = R, this is also referred to as the trace ideal, tr B (R) [Lam99].
We collect some basic properties of the trace in the next proposition.
. Let R be a ring, and A, B, C R-modules. The following holds.
(1) We have (2) The behavior with respect to direct sums is given by (3) More generally, if {B i } i∈I is an arbitrary family of R-modules, then (4) For tensor products, we have Furthermore, if B generates Hom R (C, A) or C generates Hom R (B, A), then the equality holds.  Proof.
(1) This is clear from the definition.
(2) From the definition we see that (3) We proceed as in the previous case and the result follows.
To get the equality, assume that B generates Hom R (C, A). Then for a ∈ tr B (A) ∩ tr C (A) there exists φ : C → A such that φ(c) = a for some c ∈ C. Now as B generates Hom R (C, A), there exists a map Ψ : B → Hom R (C, A) and an element b ∈ B such that Ψ(b) = φ. Consider the map B ⊗ C → A given by y ⊗ z → Ψ(y)(z). This map is well defined and b ⊗ c → a. The result follows. The case where C generates Hom R (B, A) works the same way. (5) This follows from the fact that every element a in tr C (A) can be obtained via a map C → A and an element c ∈ C. This element c will be in the image of some map B → C, and so its image a in A can be obtained via the composition B → C → A. Hence a ∈ tr B (A). (6) Follows as in the references, where the hypothesis that A is finitely-generated used by Lindo is not needed.
The last assertion is trivial after noting that The result below relates traces of modules in an exact sequence.
Proposition 2.28. Let 0 → B α − → C → D → 0 be a short exact sequence of R-modules, and A to be any other R-module.
Proof. By Proposition 2.27 part 5 we have that tr D (A) ⊆ tr C (A). Let a ∈ tr B (A). Then there exist φ ∈ Hom R (B, A) and b ∈ B such that φ(b) = a. From the exact sequence we can conclude that for any r ∈ Ann R (Ext 1 . The result follows.

Test ideals and Trace Ideals
In this section we define the test ideal of an arbitrary closure operation, give some of its basic properties, and prove that the test ideal of a module closure is a trace ideal.
Definition 3.1. Let R be a ring and cl be a closure operation on R-modules. The big test ideal of R associated to cl is defined as where the intersection runs over any (not necessarily finitely-generated) R-modules N, M . In the case that cl is generated from a R-module B, (resp. a family B) that is cl = cl B we also denote this ideal by τ B (R) (resp. τ B (R)). We sometimes refer to the big test ideal as the test ideal.
Similarly, we define the finitistic test ideal of R associated to cl as In the case where cl = cl B for some R-module B, we denote this ideal by τ fg B (R). Note that the big test ideal is always contained in the finitistic test ideal.
When cl is tight closure, these definitions agree with the the tight closure test ideal as given in [HH90,Definition 8.22]. As an immediate consequence we get: Additionally, Proof. Lemma 2.15 implies that for any R-modules N ⊆ M , N : N cl M = Ann R 0 cl M/N , and so The second result follows as the intersection will be over all M/N finitely-generated.
Remark 3.4. The finitistic test ideal is sometimes taken as the intersection over all R-modules N ⊆ M where M is finitely-generated. If cl is functorial and semi-residual, then by the proof of Lemma 3.3, this is equal to and so it is equal to our definition of the finitistic test ideal. In particular, this holds for module closures.
Corollary 3.5. If R is a regular local ring, and cl is a Dietz closure on R, then τ fg cl (R) = R. If cl = cl B for some big Cohen-Macaulay module B, then τ cl (R) = R as well. In fact, if R is a complete local domain, R is regular if and only if τ B (R) = R for all big Cohen-Macaulay modules B.
Proof. The first claim follows from the definition of a test ideal, Proposition 2.21, and Lemma 3.3: if R is regular and cl is a Dietz closure, cl is trivial on finitely-generated R-modules, so τ fg cl (R) = R. It follows from the definition that τ cl (R) ⊆ τ fg cl (R), leading to the following question that is still open in most cases for the tight closure test ideal.
Question 3.7. Do the big test ideal and the finitistic test ideal coincide? More specifically, what are the conditions needed on a ring R or on a closure operation cl so that τ cl (R) = τ fg cl (R)? The following result answers this question in one special case. We will be able to say more once we prove Proposition 3.9, our first result giving an alternate definition of the test ideal.
Proposition 3.8. Let B be a directed family of flat R-algebras, or a single flat R-module B. Then τ B (R) = τ fg B (R). Proof. Let cl = cl B , and clfg denote the closure given by: We claim that clfg = cl. To see that clfg ⊆ cl, note that by part (f) of Proposition 2.14, for Note that M/N is the union of its finitely-generated submodules, so M can be written as the union of the M 0 above. Hence the final step is equal to The forward inclusion always holds. For the reverse inclusion, suppose that u ∈ τ fg B (R). We would like to show that for arbitrary R-modules N ⊆ M , Hence uv ∈ N . This implies that u ∈ τ B (R), which gives us the result.
Proposition 3.9. Let cl be a closure on a local ring (R, m, k) satisfying the first two Dietz axioms, functoriality and semi-residuality, and E = E R (k) be the injective hull of the residue field k. Let τ cl (R) denote the big test ideal associated to cl. To see that this is contained in Ann R 0 clfg E , notice that every element v ∈ 0 clfg E is contained in 0 cl E ′ for some finitely-generated E ′ ⊆ E. So an element u that kills 0 cl M for every finitely-generated R-module M will kill v. Hence τ fg cl (R) ⊆ Ann R 0 clfg E . For the reverse inclusion, let u ∈ R − {0} such that u0 clfg E = 0, and let M be a finitely-generated R-module such that u0 cl M = 0. The rest of the argument follows as for the non-finitely-generated case, with the addition to the last line that since M is finitely-generated, u0 cl M ⊆ u0 clfg E = 0. Using this alternative description of the test ideal, we give an additional partial answer to Question 3.7. This result is the module-closure version of Theorem 3.1 of [HH89] or the notes of October 22nd and 24th of [Hoc07].
Proposition 3.10. Let R be a Gorenstein local ring, and B any R-algebra or finitely-generated R-module. Then τ fg B (R) = R if and only if τ B (R) = R. Proof. We always have τ B (R) ⊆ τ fg B (R), so the reverse direction holds without the Gorenstein assumption on R. For the forward statement, denote cl B by cl, and suppose that τ fg B (R) = R. Then, I cl R = I for all ideals I of R. Let x 1 , . . . , x d be a system of parameters on R, and I t = (x t 1 , x t 2 , . . . , x t d ). Since R is Gorenstein local, we have E R (k) = lim − →t R/I t , where the maps R/I t → R/I t+1 are given by multiplication by y = x 1 · · · x d . Using the notation of [Hoc07, Lecture of October 24th], let us denote the equivalence class of an element of R under the composition R ։ R/I t ֒→ E by (u; I t ). So (u; I t ) = (uy r ; I t+r ). Suppose that some element v = (u; This holds if and only if for each i, there is some r i such that b i ⊗ uy r i +s = 0 in B ⊗ R/I t+r i +s for all s ≥ 0. Set r = max i {r i }. Identifying B ⊗ R/I t with B/I t B, this implies that uy r b i ∈ I t+r B for each i. But this is exactly equivalent to uy r ∈ (I t+r ) cl R . Since by assumption (I t+r ) cl R = I t+r , we have uy r ∈ I t+r for sufficiently large values of r. This implies that v = 0 in E. Hence 0 cl E = 0, and thus τ B (R) = R. We can use the previous result to give a similar result for families. Proof. Let cl = cl B . The piece we need to prove is that if τ fg By Proposition 3.10, this implies that τ cl B (R) = R. Hence by Proposition 3.9 0 cl B E = 0, whch implies that v = 0. Therefore, 0 cl E = 0, and so τ cl (R) = R.
The following theorem connects test ideals with trace ideals, and is the key component of many of our results. This connects the idea of the test ideal with representation theoretic ideas.
Theorem 3.12. Let R be local and cl = cl B for some R-module B. If B is a finitely presented R-module or R is complete then τ cl (R) = tr cl (R) Proof. Let E = E R (k) be the injective hull of the residue field k of R. By Proposition 3.9, τ cl (R) = Ann R (0 cl E ) = (0 : 0 cl E ); hence c ∈ τ cl (R) if and only if c · 0 cl where the map E → B ⊗ E corresponding to b ∈ B is given by e → b ⊗ e. Since E is Artinian, there are elements b 1 , . . . , b n ∈ B such that this is equal to b∈{b 1 ,...,bn} We can rewrite this as ker(φ), where φ = (φ 1 , . . . , φ n ) : First, suppose that c ∈ τ cl (R), so that c · 0 cl E = c ker(φ) = 0. Then 0 cl E ⊆ Ann E (c), and by Matlis duality the map is surjective. But applying Hom R (_, E) to the exact sequence gives where ith map HomR(B ⊗R,R) →R is given by φ → φ(b i ). From the surjectionR/cR → HomR(0 cl E , E) we can now conclude that Im(HomR(B ⊗R,R) →R).
In the complete case, the denominator is contained in tr cl (R), so this implies that cR ⊆ tr cl (R). In the case that B is finitely presented, since Hom commutes with flat base change, the last expression is equal to It then follows by the faithful flatness of completion that For the reverse containment, suppose that c ∈ tr cl (R). Then there are Im(Hom R (B, R) → R).

Hence we have a surjection
.
Applying Hom R (_, E), we get an injection But the module on the left is 0 cl E . Hence c ∈ Ann R 0 cl E , which is equal to τ cl (R).
Remark 3.13. The second direction of the previous theorem works in greater generality; in particular it shows that for any local ring R (not necessarily complete) and any R-module B (not necessarily finitely-generated) we have Remark 3.14. The following example shows that when R is not complete and B is not finitely presented the trace ideal may differ from the test ideal. We start with [DS16, Example 4.5.1] which allows us to build a DVR V whose fraction field is F p (x, y). In this case V is a Noetherian, regular ring of dimension 1, which is not F -finite. By [DS16, Lemma 2.4.2] this implies that Hom V (V 1/p , V ) = 0, hence we have tr V 1/p (V ) = 0.
On the other hand, as V is a regular ring of dimension one, it is a domain. Hence V 1/p is torsion-free. Additionally, mV 1/p = V 1/p , so V 1/p is a Cohen-Macaulay module. This implies that τ V 1/p (V ) = V = 0. [Note: The paper as originally published has an error, which the authors corrected in an erratum, but the example and the lemma we are using are correct.] The following results use Theorem 3.12 to extend our knowledge of test ideals and closure operations, and in particular give an important case when the test ideal is nonzero.
Corollary 3.15. If R is local, cl = cl B for some solid R-module B, and either R is complete or B is finitely-generated, then we have τ cl (R) = 0. Consequently, τ fg cl (R) = 0 as well. In particular, if R is a complete local domain and B is a big Cohen-Macaulay R-module, then τ B (R) = 0.
Proof. Assume that cl = cl B for some solid R-module B. Since τ cl (R) = tr cl (R), and there is a nonzero map B → R, τ cl (R) = 0.
If R is a complete local domain, then B is solid [Hoc07, Lecture of September 7th], and the last statement follows.
is nonzero as well. Corollary 3.17. Let R be local, S an R-module, and either R is complete or S is finitelygenerated. Then τ S (R) = R if and only if S has a free summand, and consequently, cl S is trivial if and only if S has a free summand.
Proof. By part vii of Proposition 2.27, tr S (R) = R if and only if S has a free summand. Additionally, by Theorem 3.12, τ S (R) = tr S (R), and by Corollary 3.2, τ S (R) = R if and only if cl S is trivial.
The following corollary is generally known for trace ideals. We give a test ideal interpretation of the result. Proof. R is Gorenstein if and only if ω is free. Since ω is a rank 1 R-module, it has a free summand if and only if it is free. The result now follows from Corollary 3.17.
Corollary 3.19. Let A and B be R-modules satisfying the conditions of the theorem. If cl A and cl B are the closure operations associated to A and B, then Proof. This follows from the previous Theorem and Proposition 2.27 part (2).

Test ideals of Families
We extend the concept of test ideal introduced in the previous setting to that of families of modules. We can make this definition even when the family of modules does not give an idempotent closure operation, which is one way to deal with the question of how large the sum of the corresponding module closure operations can be (discussed in [R.G16b, Section 9.2]). We will then discuss the test ideals of specific families of big Cohen-Macaulay modules and algebras and connect them to the singularities of the ring. We list an immediate set of properties Lemma 4.2. Let R be a commutative ring and B, C families of R-modules, then Note that if B is a directed family of R-algebras or of R-modules directed under generation, so that it defines a closure operation, then this definition of the test ideal agrees with our prior definition:  Corollary 4.5. Let R be a complete local domain. If S is a directed family of R-algebras or a family of R-modules directed under generation (so that cl S is a closure operation), then cl S is trivial if and only if for every S ∈ S, S has a free summand.

Proof. By Proposition 4.3 and Corollary 4.4,
We know that cl S is trivial if and only if τ cl S = R. The right hand side is equal to R if and only if tr S (R) = R for all S ∈ S, which holds if and only if each S has a free summand (Lemma 2.27, part (6)).
Alternatively, this follows from Definition 4.1 and Corollary 3.17.
Ideally, we want to consider the test ideal coming from the family of all Cohen-Macaulay modules, since a ring is regular if and only if the test ideals of these modules are equal to the whole ring by Corollary 3.5. The collection of Cohen-Macaulay modules is not generally a set, so we work with the following family instead: Remark 4.6. Let R be any Cohen-Macaulay ring and consider the full subcategory of M od(R) consisting of big Cohen-Macaulay modules over R. For any set S the module R S is in this subcategory, hence there is an embedding of the category of sets to the category of Cohen-Macaulay modules over R. The former is not a small category, so the latter is not a small category either.
To avoid this complications we bound the size of the modules we consider. Let R be a local ring and Bas be a fixed infinite set. Let CM (R) be the full subcategory of big Cohen-Macaulay R-modules that are quotients of free R-modules R S with S ⊆ Bas. This is a small category, and therefore we can consider the set of objects in this category. For the purposes of this paper, it is enough for Bas to have countable order, and we denote the set of objects by CM .
Definition 4.7. Let R be a complete local domain. We define the singular test ideal to be where CM is defined as in Remark 4.6. The following results connect the test ideals of big Cohen-Macaulay modules to the singular locus of the ring, and are used to get more specific results on test ideals of big Cohen-Macaulay modules over rings with finite Cohen-Macaulay type.
Theorem 4.9. Let B be a finitely-generated Cohen-Macaulay module over a local domain R. Then V (tr B (R)) is contained in the singular locus of R.
Proof. Suppose otherwise, then there exists P ∈ Spec(R) such that R P is a regular ring and tr B (R) ⊆ P . After localizing at P this implies tr B P (R P ) ⊆ P R P . Since B is faithful over R, B P is nonzero. It is also finitely-generated, so by Nakayama's lemma P B P = B P . Now, B P is a Cohen-Macaulay module over the regular local ring R P , hence faithfully flat over R P [HH92, Pag. 77], a local ring, and hence τ B P (R P ) = R P (B P gives the trivial closure, so it gives the whole ring as the test ideal). This implies that tr B P (R P ) = R P , a contradiction. This leads to a statement for test ideals. Proof. This follows immediately from the previous result and Theorem 3.12.
Remark 4.11. We denote by M CM (R) the set of all finitely-generated Cohen-Macaulay modules over R, commonly known as the maximal Cohen-Macaulay R-modules to distinguish them from Cohen-Macaulay modules of non-maximal depth, which are not discussed in this paper. We will write just M CM if R is understood from the context. Definition 4.12. Let R be a local ring. R has finite Cohen-Macaulay type if R has finitely many indecomposable finitely-generated Cohen-Macaulay modules.
If R is a local ring of finite Cohen-Macaulay type, we know the following: • (Auslander [LW12, Theorem 7.12]) R has isolated singularities.
• If R is not regular then the top dimensional syzygy S of the residue field k is a finitely-  Proof. Let M be a finitely-generated Cohen-Macaulay module over R. Then by Proposition 4.9, since R has an isolated singularity, tr M (R) is either m-primary or R. From the facts above there is at least one MCM module (say the top dimensional syzygy) that gives an m-primary trace ideal. Since the finite intersection of m-primary ideals is m-primary, the result follows.
The following results connect the trace ideal, and hence the test ideal, to the socle of the ring (the set of elements annihilated by the maximal ideal m). Rings with nonzero socle are not reduced.
Lemma 4.14. Let (R, m) be a local ring and B an R-module such that B/mB is nonzero (for example, B could be a nonzero finitely-generated module). Then soc(R) ⊆ tr B (R).
Proof. Since B = mB, B/mB is a nontrivial R/m-vector space, so we can find a surjective morphism from B/mB to R/m. In particular we have a surjection B → R/m. If x is an element of the socle of R, then there is a map from B → R/m → R that first sends B onto R/m and then to R/(xm) ∼ = R via multiplication by x. Some element of B maps to 1 in R/m, and this maps to x in R. From this we see that soc(R) ⊆ int B (R).
Corollary 4.15. Let R be a local ring and B an R-module such that B/mB is nonzero. If B is finitely-presented or R is complete then soc(R) ⊆ τ B (R).
Proof. This follows from Theorem 3.12 and Lemma 4.14.
As a consequence of these results, when R is zero-dimensional, we can say exactly what the singular test ideal is. If in addition R is complete, then τ sing (R) = soc(R).
Proof. By Lemma 4.14, we know that for each B ∈ CM , tr B (R) ⊇ soc(R). Hence For the other inclusion, note that since R is zero dimensional, k = R/m is a Cohen-Macaulay module. The image of any map from k to R lives in soc(R). So tr k (R) ⊆ soc(R). Hence The last line follows from Corollary 4.4.
In the one-dimensional case, we prove that τ M CM (R) = tr M CM (R) = 0 without the hypothesis that R is complete. We use several definitions from [LW12, Chapter 4] .
Definition 4.17. Let R be a domain of dimension one (so R is Cohen-Macaulay), let K be the fraction field of R, and letR be the integral closure of R in K. The conductor c = (R : RR ) is the largest common ideal of R andR, and is nonzero.
If M is a finitely-generated Cohen-Macaulay R-module, then M is torsion-free. We useRM to denote theR-submodule of K ⊗ R M generated by Im(M → K ⊗ R M ). This module is R-projective [LW12, Chapter 4]. Proof. Let M be a finitely-generated Cohen-Macaulay module over R. ThenRM is a projective module over the regular ringR. It follows that there is a map φ :RM →R that sends rm → 1. Therefore after multiplying by an element c ∈ (R : R) we have that the map cφ sends (cr)m → c and has image in R, so we conclude that (R : R) ⊆ tr M (R). The result follows.
We now discuss the test ideal given by the family of big Cohen-Macaulay R-algebras. The following result of Hochster indicates that tight closure on finitely-generated R-modules comes from big Cohen-Macaulay R-algebras. Our study of the test ideal coming from the family of big Cohen-Macaulay algebras is motivated by the view that big Cohen-Macaulay algebras are a useful tight closure replacement in all characteristics. We can also define the finitistic version, The following result indicates why big Cohen-Macaulay algebra test ideals are a good tight closure replacement.
Theorem 4.21. Let R be a complete local domain of characteristic p > 0. Then τ f g CM A (R) as defined above is equal to the finitistic tight closure test ideal.
Proof. By Theorem 4.19, for finitely-generated R-modules N ⊆ M , N cl B M ⊆ N * M for every big Cohen-Macaulay algebra B. Hence for each B ∈ CM A, τ f g cl B (R) ⊇ τ f g * (R). This implies that For the other direction, note that for each finitely-generated R-module M , there exist . Hence This implies that But the left hand side contains and the right hand side is equal to τ f g * (R). Hence which gives us equality.
This result only concerns the finitistic test ideal because it is unknown whether tight closure and big Cohen-Macaulay algebras give the same closure operation on all R-modules, or even the same big test ideal. We are still able to get the following consequence: If R is a complete local domain of equal characteristic, Dietz and R.G. [Die07,DR17] construct a directed family of big Cohen-Macaulay algebras, i.e., a family of big Cohen-Macaulay Ralgebras such that given big Cohen-Macaulay algebras B and B ′ , there is a big Cohen-Macaulay algebra C and R-algebra maps B, B ′ → C that give rise to the following commutative diagram, where the maps R → B and R → B ′ send 1 → 1: In characteristic p > 0, this includes all big Cohen-Macaulay R-algebras; in equal characteristic 0, this includes all big Cohen-Macaulay R-algebras that are ultrarings. In these cases, we use the closure operation given by the family of big Cohen-Macaulay R-algebras to define the test ideal.
Definition 4.23. Let W be an infinite set with a non-principal ultrafilter W. For each w ∈ W , take a ring A w . The ultraproduct A ♮ of the A w (with respect to W) is the quotient (Π w A w )/I null , where I null is the ideal of elements (x w ) w∈W of Π w A w where x w = 0 for all w in some subset V of W contained in W. Any such ring A ♮ is called an ultraring.
For our purposes, we will be dealing with rings of equal characteristic 0 that are ultraproducts of rings of characteristic p > 0, as in [DR17].  In either case, we can define the test ideal of the directed family as in Definition 4.1.
Corollary 4.26. Let R be a complete local domain of equal characteristic and let B be either the set of all big Cohen-Macaulay R-algebras (if R has characteristic p > 0) or the set of big Cohen-Macaulay R-algebras that are also ultrarings (if R has equal characteristic 0), in both cases following the setup of Remark 4.6 to ensure we get a set. Then τ B (R) is equal to the test ideal of the closure cl B .

Examples
In this section we compute test ideals and trace ideals. In these examples, we compute Hom R (B, R) for various Cohen-Macaulay modules B, and look at the images of these maps in R. In the situation of Theorem 3.12, this gives us the test ideal τ B (R), and in general it gives us the trace ideal tr B (R).
Example 5.1. Let R be a complete PID. Then for any family of R-modules F we either have tr F (R) = 0 or tr F (R) = R. Indeed, if tr F (R) = 0 then it is a principal ideal I. Let I → R be an isomorphism. Composing this isomorphism with the elements of Hom R (F, R), whose images add up to all of I, we have for each element of R a map from F → R whose image includes that element. Hence tr F (R) = R.
If R is also local and B is any big Cohen-Macaulay R-module, B is solid (i.e. Hom R (B, R) = 0), so tr B (R) = R. Hence τ sing (R) = R.
But this is not always true in the general one-dimensional case, as the following example shows.
This is a finitely-generated Cohen-Macaulay R-module. There is no surjective map B → R. Indeed, if there were then there would be a, b ∈ R such that (at 4 + bt 3 , at 3 + bt 2 ) → 1. But note that if e = at 2 (t 3 , t 2 ) + b(t 4 , t 3 ) = (at 5 + bt 4 , at 4 + bt 3 ) ∈ B maps to x ∈ R, we also have t 3 (at 4 + bt 3 , at 3 + bt 2 ) → t 3 , but t 3 (at 4 + bt 3 , at 3 + bt 2 ) = t 2 (at 5 + bt 4 , at 4 + bt 3 ) = t 2 e → t 2 x. This implies that t 2 x = t 3 . However there is no element of R that satisfies this equation. Now consider the map B → R given by (c, d) → d. The image of this map is the ideal m = (t 2 , t 3 ). Hence we can conclude that tr B (R) = m.
, where k is a field. By [Yos90, Proposition 1.16] high syzygies (dim(R) or higher) K of the residue field k are Cohen-Macaulay modules if they are nonzero, and by Remark 2.23 cl = cl K is non-trivial. Hence τ K (R) = R. Using Macaulay2 we find that the free resolution for the residue field has the form where the map R 4 → R 3 is given by the matrix Hence K = syz 2 (k) is the R-submodule of R 3 generated by the columns of this matrix. Let I = (x 2 , y 2 ). Then rad(I) = m. We claim that xy ∈ I cl K . Since I is an ideal, I cl K = (IK : K).
Hence it is enough to show that xyK ⊆ IK. Multiplying xy by each of the columns of the matrix above, we have which implies that xyK ⊆ IK. Hence I cl K = m, and so I : . Since M ∼ = (x 2 , xy)R ∼ = (xy, y 2 )R, τ sing (R) = τ M (R) must contain m = (x 2 , xy, y 2 )R. However, since R is not regular, τ sing (R) = R. Therefore, τ sing (R) = m.
The following example is of a ring with countable Cohen-Macaulay type whose singular test ideal is not primary to the maximal ideal. This indicates that Proposition 4.13 does not hold even for fairly nice rings with infinite Cohen-Macaulay type.
Example 5.5. Let R = k[[x, y, z]]/(x 2 y + z 2 ), where k is a field of arbitrary characteristic. This ring is known as the D ∞ hypersurface singularity and as the Whitney Umbrella. By [LW12,Proposition 14.19], this ring has countable Cohen-Macaulay type and the idecomposable, non-free finitely-generated Cohen-Macaulay modules are obtained as the cokernel of one of the following matrices.
• M = coker z y −x 2 z , Let's compute the corresponding test ideals. As the ring R is a complete local domain, by Theorem 3.12 we only need to compute the trace ideal of R with respect to these modules.
• M : A map φ from M to R is the same as a map from R 2 → R whose kernel contains < (z, x 2 ), (−y, z) >. That is, we must have that zφ(e 1 )+x 2 φ(e 2 ) = −yφ(e 1 )+zφ(e 2 ) = 0, or in an equivalent way, we want solutions for with a, b ∈ R. We first find the solutions in the fraction field and then determine when they are in R. To do this, we row reduce this matrix by multiplying the second row by x 2 and then adding the z times the first row, which gives us This means that we need az = bx 2 . As we want a, b ∈ R, this is equivalent to saying a ∈ x 2 : z and b ∈ z : x 2 . It follows that τ M (R) = tr M (R) = (x 2 : z) + (z : x 2 ). As both ideals are proper, tr M (R) = R. Now, note that from the equation z 2 = −x 2 y we have that (x 2 , y, z) = tr M (R). • N: A similar procedure implies τ N (R) = tr N (R) = (z : x) + (x : z), which is equal to (xy, z) + (z, x) = (x, z). • M j : After transposing and row reducing we obtain the system Some possibilities that satisfy this equation are (found in Macaulay2 for particular values of j, but easy to check that they are correct for any j): Hence (x, y j , z) ⊆ τ M j (R). (In fact, computations in Macaulay2 confirm that these choices generate all maps M j → R, so the two ideals are equal.) • N j : As in the previous case, transposing and row reducing we obtain the system az − cx − dy j = 0 bz + dxy = 0.
In particular the following are solutions to this set of equations so (x, y j , z) ⊆ τ N j (R). (As with M j , Macaulay2 computations confirm that they are actually equal.) From this we can conclude that the intersection of τ B (R) over all finitely-generated Cohen-Macaulay R-modules B is (x 2 , y, z) ∩ (x, z) ∩ (x, y j , z) ∩ (x, y j , z) = (x 2 , z).
Notice that this is not primary to the maximal ideal, and so the singular test ideal, which is contained in this ideal, is also not m-primary.
Even though we have only defined test ideals for domains, we can compute trace ideals without this hypothesis. In the next example we compute the trace ideal of a non-domain ring with respect to its finitely-generated Cohen-Macaulay modules.
Example 5.6. Let R = k[x, y, z]/(xz), where k is an algebraically closed field of characteristic not equal to 2. We will use i to denote √ −1. In this case R has countably infinite Cohen-Macaulay type, that is, up to isomorphism, there are countably many indecomposable finitelygenerated Cohen-Macaulay R-modules. By isomorphism with k[x, y, z]/(x 2 + z 2 ) via we see that this is the same as the example in [LW12,Proposition 14.17]. Hence the indecomposable finitely-generated Cohen-Macaulay R-modules are given as the cokernels M j of We claim that tr M j (R) = tr M ′ j (R) = (x, y j , z) and that tr M (R) = tr M ′ (R) = (x, z). A map from M j → R must send its natural generators to elements a, b satisfying the relations az = 0 −ay j + bx = 0.
The first implies that a = f x for some f ∈ R, and so x(b − f y j ) = 0. This, in turn, implies b = f y j + gz for some g ∈ R. This implies that a, b ∈ (x, z, y j ). Now, choosing f = 1 and g = 0 gives the solution a = x and b = y j . This implies that tr M j (R) ⊇ (x, y j ). Similarly, choosing f = 0 and g = 1 gives tr M j (R) ⊇ (z), hence tr M j (R) = (x, z, y j ). The other cases are similar.
This implies that where the intersection is taken over all finitely-generated indecomposable Cohen-Macaulay Rmodules.
Remark 5.7. This example shows an interesting behavior, that is we have: • There is a closed property kind of behaviour, that is " lim "M j = ∩M j = M .
• There are surjective maps M j → M , which implies tr M (R) ⊆ tr M j (R) for all j, as we saw in the example.
One more example of modules for which we can say something is the following Example 5.8. Let R = k[x, y, z]/(z 2 ) localized at (x, y, z) and set We make M n an R-module via x y 0 0 · · · 0 0 x y 0 · · · 0 0 0 x y · · · 0 . . .
By Proposition 3.4 of [LW12], M n is an indecomposable Cohen-Macaulay module over R for all n ≥ 2. We compute int Mn (R). Let e 1 , . . . , e 2n be the obvious set of generators for M n . For any map ψ : M n → R we have that ψ is determined by ψ(e i ). Notice that z has the following action on the e i : We have a map ψ : M n → R sending e 1 → z e n+1 → x e n+2 → y e i → 0 for all other i.
To see that this is an R-linear map, we check that the action of z is compatible with the map. We have zψ(e n+1 ) = zx and ψ(ze n+1 ) = ψ(xe 1 ) = xψ(e 1 ) = xz.
The existence of this map shows that (x, y, z) ⊆ tr Mn (R) for n ≥ 2. Hence To see that these are in fact equal, suppose there is a map ψ : M n → R sending e i → 1 for some i. If i ≤ n, we have z = zψ(e i ) = ψ(ze i ) = 0, which is a contradiction. If i = n + 1, we have z = zψ(e i ) = ψ(ze i ) = ψ(xe 1 ) = xψ(e 1 ), which is also a contradiction as z ∈ (x)R. If i > n + 1, we have z = zψ(e i ) = ψ(ye n−i−1 + xe n−i ) = yψ(e n−i−1 ) + xψ(e n−i ), which is a contradiction since z ∈ (x, y)R. Hence 1 ∈ tr Mn (R), which implies that tr Mn (R) = (x, y, z).
Remark 5.9. Given Proposition 2.20 it is natural to ask whether τ T (R) ⊆ τ S (R) if and only if S generates T . Note that the "if" part follows from Proposition 2.20. But as Example 5.10 shows the other direction is false, even in the case of finitely-generated Cohen-Macaulay modules. Similarly, Hom(M 3 , R) = Im x z 3 z y , and the homomorphisms send s 3 → x, t → z, or s 3 → z 3 , t → y. So tr M 1 (R) = tr M 3 (R) = m. But M 1 and M 3 are distinct indecomposable Cohen-Macaulay R-modules, so neither generates the other. As M 1 and M 3 are finitely-generated R-modules, tr M i (R) = τ M i (R) for i = 1, 3, so M 1 and M 3 are two R-modules that give the same test ideal, but neither one generates the other.

Mixed Characteristic
Recently, André proved the existence of big Cohen-Macaulay algebras in mixed characteristic [And18]. We take advantage of this result and of almost big Cohen-Macaulay algebras as defined by Roberts [Rob10] and used by André to define a closure operation in mixed characteristic, and to prove that the corresponding test ideal can be written as a variant on a trace ideal, paralleling our results in previous sections. This demonstrates that the arguments used in earlier sections can be adapted to apply to closures that are variations on module closures.
Our closure is similar to dagger closure as defined by Hochster and Huneke [HH91]. The key difference is that we have replaced R + , the absolute integral closure of R, with an arbitrary almost big Cohen-Macaulay algebra. We are also using small powers of a particular element as our "test elements", as is usual in working with perfectoid algebras, rather than using arbitrary elements of small order as in [HH91].
In this section, let (R, m) be a complete local domain of dimension d > 0 and mixed characteristic (0, p), T a p-torsion free algebra, and π ∈ T a non-zerodivisor such that T contains a compatible system of p-power roots of π, i.e. a set of elements {π 1/p n } n≥1 such that (π 1/p n ) p m = π 1/p n−m for all m ≤ n. We will denote this system of p-power roots of π by π 1/p ∞ .
for all n, n ′ > 0. As in the proof of idempotence, this implies that π 1/p n ⊗u ∈ Im(T ⊗Iv → T ⊗M ) for all n > 0. Therefore u ∈ (Iv) cl M , which completes the proof of generalized colon-capturing. As a corollary to Proposition 3.9, since cl is residual, τ cl (R) = Ann R 0 cl E R (k) . Definition 6.4. Let cl be the closure from Definition 6.2. We define tr cl (R) = n>0 ψ:T →R ψ(π 1/p n T ) = n>0 Im(T * ⊗ π 1/p n T → R), where T * = Hom R (T, R) and the map sends h ⊗ x → h(x).
Theorem 6.5. Let R be a complete local domain and let cl be the closure defined above. Then τ cl (R) = tr cl (R).