A ZariskiNagata theorem for smooth algebras
Abstract.
In a polynomial ring over a perfect field, the symbolic powers of a prime ideal can be described via differential operators: a classical result by Zariski and Nagata says that the th symbolic power of a given prime ideal consists of the elements that vanish up to order on the corresponding variety. However, this description fails in mixed characteristic. In this paper, we use derivations, a notion due to Buium and Joyal, to define a new kind of differential powers in mixed characteristic, and prove that this new object does coincide with the symbolic powers of prime ideals. This seems to be the first application of derivations to commutative algebra.
1. Introduction
The subject of symbolic powers is both a classical commutative algebra topic and an active area of current research. While there are many open problems in the setting of algebras containing a field, even the results that are wellunderstood for algebras over fields are mostly open for algebras and local rings of mixed characteristic. Thanks to the perfectoid spaces techniques of Schölze [Scholze] as applied to commutative algebra by André and Bhatt, a major advance has happened recently [AndreDirectSummand, BhattDirectSummand]. Ma and Schwede have shown that a theorem of EinLazersfeldSmith [ELS] and HochsterHuneke [HH] on the uniform containment of symbolic and ordinary powers of ideals holds for regular rings of mixed characteristic [MaSchwedeSymbPowers].
In this paper, we are interested in generalizing another classical result on symbolic powers to the case of mixed characteristic and smooth algebras: the ZariskiNagata theorem, which establishes that symbolic powers of prime ideals can be described using differential operators.
Zariski’s main lemma on holomorphic functions [Zariski], together with work by Nagata [Nagata, p. 143], states that if is a prime ideal in a polynomial ring over a field, then
This result was later refined by Eisenbud and Hochster [EisenbudHochster], and can be rephrased using differential powers of ideals, a fact which was wellknown in characteristic and extended to perfect fields; see [SurveySP]. More precisely, if is a smooth algebra over a perfect field , and is a prime ideal in , the ZariskiNagata theorem states that the th symbolic power of consists of the elements in that are taken inside by every linear differential operator of order at most .
Rather than using perfectoid techniques, our generalization of ZariskiNagata makes use of a different arithmetic notion of derivative, the notion of a derivation, defined by Joyal [Joyal] and Buium [Buium1995] independently. From a commutative algebra point of view, derivations are rather exotic maps from a ring to itself — in particular, they are not even additive — but they do have many applications to arithmetic geometry, such as in [BuiumADE, BuiumDCI, Borger]. To the best of our knowledge, this is the first application of derivations to commutative algebra.
While our results cover a more general setting, let us describe the case where , where denotes the integers or the adic integers . Given a prime ideal in , we study two different types of differential powers associated to . The first one is defined just in terms of differential operators, as in the statement of the ZariskiNagata theorem. More precisely, given an integer , the th (linear) differential power of is defined as
where is the set of linear differential operators on of order at most (see Definition 2.3). If does not contain any prime integer, then coincides with the th symbolic power of .
Theorem A.
(see Theorem 3.9) Let , where or , and be a prime ideal of such that . Then for all .
More generally, the previous result holds if is an essentially smooth algebra over , where is either or a DVR of mixed characteristic.
If the prime ideal contains a prime integer , then differential powers are not sufficient to characterize symbolic powers, as one can see in Remark 3.11. To overcome this issue, we combine differential operators and derivations to define the mixed differential powers of an ideal. Given a fixed derivation , the th mixed differential power of is the ideal given by
In principle, the mixed differential powers of an ideal depend on the choice of a derivation . However, in our setting is independent of the choice of the derivation (see Corollary LABEL:mixed_powers_independent_of_delta). This new notion of mixed differential powers allows us to characterize symbolic powers of prime ideals that contain a given integer .
Theorem B.
(see Theorem LABEL:thm_mdiff=symb) Let , where or , and be a prime ideal of such that , for a prime . Then for all .
More generally, we show this holds for an essentially smooth algebra over , where is either or a DVR with uniformizer , as long as has a derivation and satisfies some additional assumptions (e.g., is perfect).
2. Background
2.1. Essentially smooth algebras
Throughout, we say that a ring is smooth over a subring if the inclusion map is formally smooth and is a finitely generated algebra. We say that is essentially smooth over if the inclusion map is formally smooth and is a localization of a finitely generated algebra (i.e., is essentially of finite type over ). Polynomial extensions of rings are smooth. The following structure lemma asserts that in our setting, every smooth map is locally of this form.
Lemma 2.1.
Proof.
By assumption, we can write as , where , for some prime ideal of and an ideal of . We prove the claim that by induction on .
Case (1): If , there is nothing to show, since our assumptions imply that . We now assume . If is a generating set of , we may assume that every belongs to , otherwise is isomorphic to the localization of a polynomial ring over in one fewer variable, and we are done by induction. Then, , but if , then , and is not regular, so that is not smooth over . Thus, .
Case (2): If , there is nothing to show, since our assumptions imply that . We now assume . Observe that, if we set , then . If is a generating set of , we can write , where and . We may assume that every belongs to , otherwise is isomorphic to the localization of a polynomial ring over in one fewer variable, and we are done by induction. Tensoring with we obtain
Since is formally smooth by assumption, the fiber must be regular.
As we previously reduced to the case for all , we must have . To see this, note that if the image of inside the regular local ring contains a nonzero element, then , but and have the same embedding dimension. So cannot be regular, unless maps to zero in .
In particular, we have for all , that is, for some ideal . Unless , this contradicts the fact that is formally smooth, hence flat [EGAIV, 17.5.1]. It follows that , and we have proved that for some polynomial algebra and some prime ideal . ∎
Lemma 2.2.
Let either

be a field, or

be or a DVR with uniformizer , and .
Let be a local ring that is essentially smooth over , and suppose that . Assume that the field extension is separable. If is a minimal generating set of in Case (1) or is a minimal generating set of in Case (2), then there is a free basis for that contains .
Proof.
We can write by Lemma 2.1. The module is free of rank . It suffices to show that the images of are linearly independent in . We apply the second fundamental sequence of Kähler differentials [Matsumura, Theorem 25.2] to the maps to obtain a rightexact sequence
where .
The transcendence degree of over is . By [Hochster615, p. 24] (or see [Matsumura, §26]), since is a separable field extension, is a vector space of dimension at most . In Case (1), , so must be injective.
In Case (2), , so the kernel of must be at most onedimensional. The class of modulo is in the kernel of , so the images of the other basis elements must be linearly independent. This concludes the proof. ∎
2.2. Differential operators
We now review some results regarding differential operators that we use in the rest of the paper. A general reference for differential operators is [EGAIV, Chapter 16]; specific references to the facts we need are given below.
Definition 2.3.
[EGAIV, Section 16.8] Let be a map of rings. The linear differential operators on of order are defined inductively as follows:

.

.
Lemma 2.4.
Let be a formally smooth map of rings. Suppose that is free, e.g., is local, and let be a free basis for . Then there exists a family of differential operators such that

for all with for all , and

for all with for some .
The module is a free module generated by for each .
Proof.
By [EGAIV, Theorem 16.10.2], is differentially smooth over . Then, the statement above is the content of [EGAIV, Theorem 16.11.2]. ∎
2.3. Derivations
Fix a prime , and let be a ring on which is a nonzerodivisor. The following operators were introduced independently in [Joyal] and [Buium1995]:
Definition 2.5.
We say that a settheoretic map is a derivation if is a ring homomorphism. Equivalently, is a derivation if and satisfies the following identities for all :
(2.3.1) 
and
(2.3.2) 
where . If is a derivation, we set to be the fold selfcomposition of ; in particular, is the identity. We set to be the set of derivations on . For all positive integers , we let
For a thorough development of the theory of derivations, see [BuiumADE].
Note that having a derivation on is equivalent to having a lift of the Frobenius map . Indeed, it follows from the definition that if is a map such that the induced map is the Frobenius map, then is a derivation. For example, if , then the map that sends a polynomial to
is a derivation.
However, not every ring admits a derivation. See [Dupuy], or consider the following example:
Example 2.6.
However, we do have the following:
Proposition 2.7.
A ring admits a derivation in each of the following cases:

,

is the Witt vectors over for some perfect ring of positive characteristic,

is a polynomial ring over a ring that admits a derivation, or

is adically complete and formally smooth over a ring that admits a derivation.
Suppose also that is a derivation on and is a prime of containing . Then there exists a derivation on such that for all when:
Proof.
As we have noted before, showing that a derivation exists is equivalent to proving that there exists a lift of Frobenius. Moreover, to verify that a derivation extends , it suffices to check that the associated lift of Frobenius extends the other. We verify for (1)–(4) that there is a lift of Frobenius.
(1): The identity on is a lift of Frobenius.
(2): The Witt vectors admit a functorially induced Frobenius.
(3): Extend a lift of Frobenius on by sending each variable to its th power.
(4): Let be the composition of the quotient map with the Frobenius map on . Since is formally smooth, there is a map such that , where is the natural surjection of . Inductively, by formal smoothness one obtains a family of maps such that . This compatible system of maps induces a map that is a lift of the Frobenius.
(a): Let be a lift of the Frobenius. We note that if , then, since and , we also have , and hence . It follows that induces a map . Now, we claim that is a lift of the Frobenius as well. In fact, if , observe that . To see this, note that
where the numerator is a multiple of , and the denominator is not.
(b): Given a lift of the Frobenius on , to see that it extends to it suffices to check that . In fact, in this case, we have for all positive integers , since is a ring homomorphism, and it follows that is adically continuous. To see that , observe that, for , , so , because by assumption. ∎
Remark 2.8.
Repeated application of Equation (2.3.2) shows that a derivation sends the prime ring of (i.e., the canonical image of ) to itself. If has characteristic zero, so that its prime ring is , any derivation on restricts to a derivation on . On the other hand, there is a unique derivation on given by the Fermat difference operator: . In particular, when has characteristic zero, every derivation satisfies for all .
Remark 2.9.
Let be a ring, and be ideals. Let be a derivation. If and , then modulo . In fact, we have that
because for all . In particular, we have that . With similar considerations, one can show that if , then modulo .
3. Results
3.1. Primes not containing
In this subsection, we focus on differential and symbolic powers of prime ideals that do not contain any prime integer. To study symbolic powers of such ideals, we use differential operators.
Definition 3.1.
[SurveySP] Let be a ring, be a subring of , and be an ideal of . The th (linear) differential power of is
The following proposition is a generalization of [SurveySP, Proposition 2.4].
Proposition 3.2.
Let be a ring, be a subring of , and an ideal of . The following properties hold:

is an ideal, and .

for all .

For any , and any , we have . In particular, .

If is a prime ideal of , and is primary, then is primary.

If is a prime ideal of , and is primary, then .
Proof.
The proof of (1) and (2) is analogous to that of [SurveySP, Proposition 2.4], where the same claim is made for the case when is a field.
For part (3), we first proceed by induction on . If , then is just multiplication by an element of , and the statement is clear. If , we proceed by induction on . If , then is trivial. By induction, assume that . To conclude the proof that , it suffices to show that for all and . To see this, observe that , with . By the inductive hypothesis on , we have that . Since and , we also have , and the claim follows. In particular, this shows that for all of order up to , so that .
To show (4), we first observe that by parts (1) and (3), so that . To prove that is primary, we proceed by induction on . The case is true by assumption, since . Let , with . Observe that by part (2), and by the inductive hypothesis the latter ideal is primary. Since , it follows that . Let , so that . It follows that , by definition of . On the other hand, we also have , and thus . Using again that , and that is primary, it follows that . Since was arbitrary, we conclude that , and thus is primary. ∎
Corollary 3.3.
In the context of Lemma 2.4, fix such that , for some . Consider the ideal , and let . Then:

, and

For all with , .
Proof.
We will need the following lemma on the behavior of differential powers under localization. The following is from forthcoming work by Brenner, NúñezBetancourt, and the third author [DiffSig]. We include a proof here for completeness, while we refer the reader to [DiffSig] for a thorough treatment and other applications of differential powers. We thank Holger Brenner and Luis NúñezBetancourt for allowing us to share this result here.
Lemma 3.4 ([DiffSig]).
Let be a ring, be a subring of , be a multiplicatively closed subset of , , and be an ideal of such that . Suppose that is essentially of finite type over . Then

,

, and

.
We first record the following lemma, which is wellknown in the case is a field.
Lemma 3.5.
[DiffSig] With notation as above, there are isomorphisms
In particular, every extends to an element .
Proof.
By [EGAIV, 16.8.1] and [EGAIV, 16.8.8], there are isomorphisms for all algebras , where denotes the module of principal parts. By [EGAIV, 16.4.22], each is a finitely generated module. By [EGAIV, 16.4.15.1], there are isomorphisms , and by [EGAIV, 16.4.14.1] these modules are isomorphic to . We caution the reader that the proof of [EGAIV, 16.4.14] contains an error, but the statements are correct. The stated isomorphisms now follow. ∎
Proof of Lemma 3.4.
Part (1) is immediate from the previous lemma.
We prove part (2). In order to show that , it suffices to prove that if for some , then . For any , by Lemma 3.5 there exists and such that for all . The claim is then clear. For the other containment, suppose that , and . If , then it extends to a differential operator such that . By hypothesis, this element is in . Thus, lies in .
We now prove part (3). To show that , we proceed by induction on . The case is trivial. Let , , and . Then, . By the induction hypothesis, . By Lemma 3.5, there exists and such that . Since by hypothesis, we also have that . We now prove the containment . Since elements of extend to elements of , we know that for all . But then
since multiplication by an element does not increase the order of a differential operator. By hypothesis, , so , as required. ∎
As noted in the introduction, the ZariskiNagata theorem can be stated in terms of differential powers of ideals [SurveySP]. Namely, if , is a perfect field, and is a prime ideal of , then for all . We can give a concise proof of the ZariskiNagata theorem for algebras over fields by combining the previous two results.
Theorem 3.6 (ZariskiNagata).
Let be a field, be essentially smooth over , and be a prime ideal of . If is separable, then for all . In particular, if is perfect, then for every prime and all .
Proof.
It suffices to check the equality . By Lemma 3.4, , and since is maximal in , we have that . Thus, it suffices to show .
Let be a minimal generating set for . Suppose that there exists an element of order , meaning that , . Then, we can write for some units , some with , and some . Fix some multiindex with and some unit such that appears in the expression of as above. By Lemma 2.2, form part of a free basis for . Thus, by Theorem 2.4 and Corollary 3.3, there exists a differential operator such that and for each other term in the expression for . Additionally, since and , we have by Proposition 3.2 (3). It follows that , contradicting the assumption . ∎
We note that the essential smoothness hypothesis is necessary.
Example 3.7.
The following example shows that the conclusion of the ZariskiNagata theorem may fail if the field extension is not separable.
Example 3.8.
We are now ready to state our first main result: a version of the ZariskiNagata theorem for prime ideals that do not contain any prime integer.
Theorem 3.9.
Let be either or a DVR of mixed characteristic. Let be an essentially smooth algebra. If is such that , then .
Proof.
Let and . We note that is a field of characteristic zero, and is formally smooth and essentially of finite type over . Observe that is a prime ideal in . We claim that . Indeed, , so , and thus
As an application of Theorem 3.9, we obtain a generalization of Zariski’s main lemma on holomorphic functions [Zariski, EisenbudHochster].
Corollary 3.10.
Let be as in Theorem 3.9, and assume that is smooth over . Let , and . For a prime not containing any prime integer, set . We have .
Proof.
By Theorem 3.9, it suffices to show that . Since for all , it follows that . For the converse, we claim that . Let , and let be a prime integer. Note that , where the equality follows from the fact that is a HilbertJacobson ring, and that . Observe that is a nonzerodivisor on , because for all . Let . It follows from the containments proved above that we can write , where