The Braun-Kemer-Razmyslov Theorem for affine PI-algebras

A self-contained, combinatoric exposition is given for the Braun-Kemer-Razmyslov Theorem over an arbitrary commutative Noetherian ring.


Introduction
One of the major theorems in the theory of PI algebras is the Braun-Kemer-Razmyslov Theorem (Theorem 1.1 below). We preface its statement with some basic definitions. Definition 1. 1. An algebra is affine over a commutative ring if is generated as an algebra over by a finite number of elements 1 , . . . , ℓ ; in this case we write = { 1 , . . . , ℓ }.
2. We say the algebra is finite if is spanned as a -module by finitely many elements.
The aim of this article is to present a readable combinatoric proof (essentially self-contained in characteristic 0).
Let us put the BKR Theorem into its broader context in PI theory. We say a ring is Jacobson if the Jacobson radical of every prime homomorphic image is 0. For PI-rings, this means every prime ideal is the intersection of maximal ideals. Obviously any field is Jacobson, since its only prime ideal 0 is maximal. Furthermore, any commutative affine algebra over a field is Noetherian by the Hilbert Basis Theorem and is Jacobson, in view of [28,Proposition 6.37], often called the "weak Nullstellensatz," implying the following two results: (cf. Proposition 1.11) If a commutative algebra is affine over a field, then Jac( ) is nilpotent.
(Special case of Theorem 2) If is a finite algebra over an affine central subalgebra over a field, then Jac( ) is nilpotent. (Sketch of proof: Passing to homomorphic images modulo prime ideals, we may assume that is prime PI, and is an affine domain over which is torsion-free. The maximal ideals of lift up to maximal ideals of , in view of Nakayama's lemma, implying ∩ Jac( ) ⊆ Jac( ) = 0. If 0 ̸ = ∈ Jac( ), then writing as integral over , we have the nonzero constant term in ∩ Jac( ) = 0, a contradiction.) Since either of these hypotheses implies that is a PI-algebra, it is natural to try to find an umbrella result for affine PI-algebras, which is precisely the Braun-Kemer-Razmyslov Theorem. This theorem was proved in several stages. Amitsur [1,Theorem 5], generalizing the weak Nullstellensatz, proved that if is affine over a commutative Jacobson ring, then Jac( ) is nil. In particular, is a Jacobson ring. (Later, Amitsur and Procesi [3,Corollary 1.3] proved that Jac( ) is locally nilpotent.) Thus, it remained to prove that every nil ideal of is nilpotent.
It was soon proved that this does hold for an affine algebra which can be embedded into a matrix algebra, see Theorem 1 below. However, examples of Small [33] showed the existence of affine PI algebras which can not be embedded into any matrix algebra. Thus, the following theorem by Razmyslov [22] was a major breakthrough in this area. Following Razmyslov's theorem, Kemer [15] then proved Theorem 1.4. [15] In characteristic zero, any affine PI algebra satisfies some Capelli identity (see Theorem 3.3).
Thus, Kemer completed the proof of the following theorem: Theorem 1.5 (Kemer-Razmyslov). If is an affine PI-algebra over a field of characteristic zero, then its Jacobson radical Jac( ) is nilpotent.
Then, using different methods relying on the structure of Azumaya algebras, Braun proved the following result, which together with the Amitsur-Procesi Theorem immediately yields Theorem 1.1: Theorem 1.6. Any nil ideal of an affine PI-algebra over an arbitrary commutative Noetherian ring is nilpotent.
Note that to prove directly that Jac( ) is nilpotent it is enough to prove Theorem 1.6 and show that Jac( ) is nil, which is the case case when is Jacobson, and is called the "weak Nullstellensatz." But the weak Nullstellensatz requires some assumption on the base ring . It can be proved without undue difficulty that is Jacobson when is Jacobson, cf. [26,Theorem 4.4.5]. Thus, in this case the proper general formulation of the nilpotence of Jac( ) is: Theorem 1.7 (Braun). If is an affine PI-algebra over a Jacobson Noetherian base ring, then Jac( ) is nilpotent.
Small has pointed out that Theorems 1.6 and 1.7 actually are equivalent, in view of a trick of [25]. Indeed, as just pointed out, Theorem 1.6 implies Theorem 1.7. Conversely, assuming Theorem 1.7, one needs to show that Jac( ) is nil. Modding out the nilradical, and then passing to prime images, one may assume that is prime. Then one embeds into the polynomial algebra [ ] over the Noetherian ring [ ], and localizes at the monic polynomials over [ ], yielding a Jacobson base ring by [25,Theorem 2.8].
Braun's qualitative proof was also presented in [27,Theorem 6.3.39], and a detailed exposition, by L'vov [19], is available in Russian. A sketch of Braun's proof is also given in [5, Theorem 3.1.1].
Together with a characteristic-free proof of Razmyslov's theorem 1.2 due to Zubrilin [34], Kemer's Theorems 1.4 and 1.8 yield another proof of the Braun-Kemer-Razmyslov Theorem 1.1. The paper [34] is given in rather general circumstances, with some non-standard terminology. Zubrilin's method was given in [7], although [7,Remark 2.50] glosses over a key point (given here as Lemma 2.13), so a complete combinatoric proof had not yet appeared in print with all the details. Furthermore, full combinatoric details were provided in [7] only in characteristic 0 because the conclusion of the proof required Kemer's difficult Theorem 1.8. We need the special case, which we call "Kemer's Capelli Theorem," that every affine PI-algebra over an arbitrary field satisfies some Capelli identity. This can be proved in two steps: First, that satisfies a "sparse" identity, and then a formal argument that every sparse identity implies a Capelli identity. The version given here (Theorem 4.4) uses the representation theory of the symmetric group , and provides a reasonable quartic bound (( − 1) ) for the degree of the sparse identity of in terms of the degree of the given PI of .
It should be noted that every proof that we have cited of the Braun-Kemer-Razmyslov Theorem ultimately utilizes an idea of Razmyslov defining a module structure on generalized polynomials with coefficients in the base ring, but we cannot locate full details of its implementation anywhere in the literature. One of the objectives of this paper is to provide these details, in §2.5 and §2.6.1. Although the proof is rather intricate for a general expository paper, we feel that the community deserves the opportunity to see the complete argument in print.
We emphasize the combinatoric approach here. Aside from the intrinsic interest in having such a proof available of this important theorem (and characteristic-free), these methods generalize easily to nonassociative algebras, and we plan to use this approach as a framework for the nonassociative PI-theory, as initiated by Zubrilin. (The proofs are nearly the same, but the statements are somewhat more complicated. See [6] for a clarification of Zubrilin's work in the nonassociative case.) To keep this exposition as readable as we can, we emphasize the case where the base ring is a field and prove Theorem 1.1 directly by an induction argument without subdividing it into Theorem 1.6 and the weak Nullstellensatz, although we also treat these general cases.
To complete the proof of the BKR Theorem, it remains to prove Kemer's Capelli Theorem. In §3 we provide the proof in characteristic 0, by means of Young diagrams, and §4 contains the characteristic analog (for affine algebras). An alternative proof could be had by taking the second author's "pumping procedure" which he developed to answer Specht's question in characteristic , and applying it to the "identity of algebraicity" [7, Proposition 1.59]. We chose the representationtheoretic approach since it might be more familiar to a wider audience. The proof of Theorem 1.6, over arbitrary commutative Noetherian rings, is given in §5. Remark 1.9. An early version of Theorem 3.16 was written by Amitai Regev, to whom we are indebted for suggesting this project and providing helpful suggestions all along the way. Belov belongs lemma 3.8

Structure of the proof
We assume that is an affine -algebra and satisfies the + 1 Capelli identity Cap +1 (but not necessarily the Capelli identity Cap ), and we induct on : if such satisfies Cap then we assume that Jac( ) is nilpotent, and we prove this for Cap +1 . For the purposes of this sketch, in Steps 1 through 3 and Step 7 we assume that is a field .
The same argument shows that any nil ideal of an affine algebra over a Noetherian ring is nilpotent, yielding Theorem 1.3. For this result we would replace Jac( ) by throughout our sketch.
We write { , , } for the free (associative) algebra over the base ring , with indeterminates , , , , containing one extra indeterminate for further use. This is a free module over , whose basis is the set of words, i.e., formal strings of the letters , , , . The and indeterminates play a special role and need to be treated separately. We write { } for the free subalgebra generated by the and , omitting the and indeterminates.
1. The induction starts with = 1. Then + 1 = 2, and any algebra satisfying Cap 2 is commutative. We therefore need to show that if is commutative affine over a field , then Jac( ) is nilpotent. This classical case is reviewed in §1.3.1.
2. Next is the finite case: If is affine over a field and a finite module over an affine central subalgebra, then Jac( ) is nilpotent. This case was known well before Razmyslov's Theorem, and is reviewed in §1.3.2. Theorem 1.3 follows whenever is a subring of a finite dimensional algebra over a field.
6. We prove that Obst ( ) · ( ( )) 2 = 0 over an arbitrary ring . This is obtained from Step 4 via a sophisticated specialization argument involving free products. 7. We put the pieces together. When is a field, Step 3 shows that Jac( ) ⊆ ( ) for some , and Step 5 shows that Jac( ) ⊆ Obst ( ) for some . Hence which completes the proof of Theorem 1.2. When is Noetherian, any nil ideal of is nilpotent, so the analogous argument shows that ⊆ Obst ( ) for some . Hence

The commutative case
Our main objective is to prove that the Jacobson radical Jac( ) of an affine PI-algebra (over a field) is nilpotent. We start with the classical case for which is commutative. Remark 1.10. Any commutative affine algebra over a Noetherian base ring is Noetherian, by Hilbert's Basis Theorem, and hence the intersection of its prime ideals is nilpotent, cf. [29,Theorem 16.24].
But for any ideal ▷ , the algebra / is also Noetherian, so the intersection of the prime ideals of containing is nilpotent modulo . Proposition 1.11. If is a commutative affine algebra over a field, then Jac( ) is nilpotent.
Proof. The "weak Nullstellensatz" [28, Proposition 6.37] says that is Jacobson, and thus the Jacobson radical Jac( ) is contained in the intersection of the prime ideals of . But any commutative affine algebra is Noetherian, so we conclude with Remark 1.10. 2

The finite case
To extend this to noncommutative algebras, we start with some other classical results: 2. If is a field, then Jac( ) is nilpotent.
Proof. For each 1 ℓ, write each as an × matrix ( ( ) ), for ( ) ∈ , and let be the commutative -subalgebra of generated by these finitely many ( ) ; then is -affine. We can view each in ( ), so ⊆ ( ).
For any maximal ideal of , we see that / is an affine field extension of , and thus is finite dimensional over , by [28,Theorem 5.11]. But then the image of in ( / ) is finite dimensional over , so the image¯of is nilpotent, implying¯= 0. Hence is contained in ∩ ( ) = (∩ ) = 0, where runs over the maximal ideals of . 2 Theorem 2. Suppose is an algebra that is finite over , itself an affine algebra over a field. Then Jac( ) is nilpotent.
So we may assume that is prime. But localizing over the center, we may assume that is a field. Let = dim . Then is embedded via the regular representation into × matrices over a field, and we are done by Theorem 1. 2 Since not every affine PI-algebra might satisfy the hypotheses of Theorem 1, cf. [33] and [18], we must proceed further.
Proof. We get the same result in Definition 2.1 by specializing to ℎ and then to ⃗, as we get by specializing to ℎ ′ , and then to ⃗. 2 This observation is needed in our later specialization arguments.
It suffices to show that ≡ 0 for each . Note that in calculating is unchanged (since it is the last indeterminate), while for all other 's (in particular -for +1 ) we substitute ↦ → ( + 1) , cf. Remark 1.1. Therefore is the sum of the monomials of ( , ) , It is not difficult to see that for > 0, It also follows from the definitions that Summing in (4) we get We need a special sort of alternating polynomials. Our main example is the double Capelli polynomial Here ⃗ and ⃗′ are arbitrary sets of extra indeterminates. We suppress the indeterminates ⃗ , ⃗′ , and 1 , 2 , 3 from the notation, since we do not alter them.

Commutativity of the operators
We usê︁ ℳ instead of ℳ because of the following lemma.
The middle equivalence (ii) is an obvious equality. The first and third equivalences are similar, and we prove the first. By Proposition 2.10, by (ii), and again by Proposition 2.10, modulo +1 we can write ,ℎ 1 ( ).
Note that in the last step, Lemma 2.3 was applied (to ( , ) ,ℎ 1 ( )). [ , ] by the elements Working with the relatively free algebra, our next goal is to prove that , { } ·̂︁ ℳ = 0. For that we shall need the next result.
Using Proposition 2.7, we have that The proof now follows by substituting +1 ↦ → 1 and → ℎ ∈ { }. Proof. We prove that , { } ·̂︁ ℳ = 0, by showing for any doubly alternating polynomial It follows from the action ,ℎ = ( ) ,ℎ ( ) and from Proposition 2.16 that modulo +1 , In order to utilize these results about integrality, we need another concept. We define
Extend to * : [ , ] → as follows: be one of the generators of , .
Proof. The assumption implies that in the above, the identity map = : → satisfies the condition of Lemma 2.18. Hence 0 = ker( ) ⊇ Obst ( ), and the proof follows. 2 This corollary explains the notation Obst ( ): it is the obstruction for each ∈ to be -integral. The next result technically is not needed, but helps to show how Obst behaves. and with the respective conditions of − 1 integrality and of integrality. Take ∈ and ℎ : → with ℎ( ) being − 1 integral over the center of . Then ℎ( ) is also integral over the center of . Hence every ker(ℎ) in Obst −1 ( ) also appears in the intersection Obst ( ) = ∩ ker( ), and the assertion follows. 2

Reduction to finite modules
The reduction to finite modules is done using Shirshov's theorem. Proposition 2.21. Let = { 1 , . . . , ℓ } have PI degree over the base ring . Then the affine algebra / Obst ( ) can be embedded in an algebra which is finite over a central affine subalgebra. Proof.

Let
⊆ be the subset of the words in the alphabet 1 , . . . , ℓ of length . By Shirshov's Height Theorem there exists an integer ℎ such that the set and let , be the ideal We show that ′ is finite over an affine central subalgebra and thus is Noetherian.
Hence˜embeds˜= / Obst ( ) into * . Note that * is a finite algebra over the affine central subalgebra ⊆ * generated by the finitely many central elements ,˜+˜,˜.
Denote * =˜+ ,˜. Then, as in (11), the finite subset In this section we show how Proposition 2.17 implies that Obst ( ) · ( ( )) 2 = 0, thereby completing the proof of Razmyslov's Theorem. For this, we need to specialize down to given algebra , requiring a new construction, the relatively free product, which enables us to handle together with polynomials. Since, to our knowledge, this crucial step, which is needed one way or another in every published proof of the BKR theorem, has not yet appeared in print in full detail, we present two proofs, one faster but more ad hoc (since we intersect with and bypass certain difficulties), and the second more structural.
This construction is universal in the following sense: Any homomorphic image of ⟨ ; ; ⟩ satisfying these identities (from ℐ) is naturally a homomorphic image of ⟨ ; ; ⟩ ℐ . Thus, we have: Although difficult to describe explicitly, the relatively free product is needed implicitly in all known proofs of the Braun-Kemer-Razmyslov Theorem in the literature. From now on, we assume that ℐ contains +1 , so that we can work witĥ︁. Let̂︁ ℳ denote the image of ℳ under substitutions to , i.e., the -submodule of { , , } consisting of the images of all doubly alternating polynomials (in 1 . . . , , and in 1 , . . . , ). In view of Lemma 2.13, the natural action of Obst ( ) on̂︁ ℳ respects multiplication by the Proof. If ∈ Obst ( ), then ℳ ∈ ℐ, in view of Lemmas 2.5 and 2.12 and Proposition 2.17, so is 0 modulo ℐ. 2 Corollary 2.26. If ∈ belongs to the -ideal generated by doubly alternating polynomials, then Obst ( ) = 0.
Proof. The element belongs to the linear combinations images of̂︁ ℳ under specializations ↦ → . 2 By Step 7 of Section 1.2, this will complete the proof of the nilpotence of Jac( ) when is a field, or more generally of any nil ideal when is Noetherian, once we complete the proof of Proposition 2.10.

A more formal approach to Zubrilin's argument
Rather than push immediately into , one can perform these computations first at the level of polynomials and then specialize. This requires a bit more machinery, since it requires adjoining the commuting indeterminates , to the free product, but might be clearer conceptually. where ∈ and ℎ are distinct monomials. Proof.

The connection to the group algebra of
We begin with the basic correspondence between multilinear identities and elements of the group algebra over .

Proof of Proposition 2.10
We may assume that ℎ is a new indeterminate . Recall that

Proof of Kemer's "Capelli Theorem,"
To complete the proof of Theorem 1.1, it remains to present an exposition of Kemer's "Capelli Theorem," that any affine PI algebra over a field satisfies a Capelli identity Cap for large enough . This is done by abstracting a key property of Cap , called spareseness. See [7, §2.5.2] for more detail. The major example of a sparse identity is the Capelli identity. One proves rather quickly that any sparse identity implies a Capelli identity, so it remains to show that any affine PI algebra over a field satisfies a sparse identity. There are two possible approaches, both using the classical representation theory of . One proof relies on "the branching theorem," which requires characteristic 0, and the other relies more on the structure of the group algebra [ ], also with the technique of "pumping" polynomial identities, and works in arbitrary characteristic.
Claim: If | 1 |, . . . , | | , then Δ = ( 1 , . . . , ; ⃗ ) is a linear combination of terms The above Claim implies the existence of descending sequences of monomials, under the left lexicographic order. Such a descending sequence must stop. When it stops we have a corresponding monomial having strictly fewer words ′ for which | ′ | . Therefore proving the above Claim will prove the lemma. We now prove the Claim.

Actions of the group algebra
It remains to prove the existence of sparse identities for affine PI-algebras. For this, we turn to the representation theory of . After a brief review of actions of on Young diagrams, we treat the characteristic 0 case, cf. Kemer [15]. The characteristic > 0 proof, which requires some results about modular representations but bypassing branching, is done in §3.2 and §4. Given , ∈ , by convention we take ( ) = ( ( )). The product corresponding (by Definition 2.33) to the monomial can be interpreted in two ways, according to left and right actions of on , described respectively as follows: Let , ∈ . Let = ( ) . Then where (resp. ℛ ) denotes the set of column (resp. row) permutations of the tableau . 2 = for some in the base field . When ̸ = 0, which by [29,Lemma 19.59(i)] is always the case when char( ) does not divide , in particular, when char( ) = 0, we will call the idempotent := −1 the Young symmetrizer of the tableau . itself must correspond to a PI of . := dim is given by the "hook"formula, see for example [30] or [14], where we recall that each "hook"number ℎ for a box is the number of boxes in "hook"formed by taking all boxes to the right of and beneath . (In the literature, one writes instead of , but here we have used throughout for polynomials.) Lemma 3.6. Suppose is a minimal left ideal of a ring . Then the minimal two-sided ideal of containing is a sum of minimal left ideals of isomorphic to as modules.
We let denote the minimal two-sided ideal of [ ] containing .
We define the codimension ( ) = dim The characteristic 0 version of the next result is in [24]. Lemma 3.7. Let be an -algebra, and let be a partition of . If dim > ( ), then ⊆ Id( ) ∩ .
Proof. By Lemma 3.6, is a sum of minimal left ideals, with each such minimal left ideal isomorphic to .
a contradiction. Therefore each ⊆ Id( ) ∩ . ⊆ Id( ) ∩ since equals the sum of these minimal left ideals. 2 3.3. The characteristic 0 case [15] The characteristic 0 case is treated separately here, since it can be handled via the classical representation theory of the symmetric group. By Maschke's Theorem, the group algebra now is a finite direct product of matrix algebras over . We have the decomposition = ⨁︀ ⊢ . Thus, Lemma 3.7 yields at once: Let us review the proof, for further reference. For any box in the (1, ) position, the hook has length + − , so the sum of all hook numbers in the first row is Summing this over all rows yields as desired.
3.3.1. Strong identities Definition 3.9. Let be a PI algebra. The multilinear polynomial ∈ is a strong identity of if for every we have · · ⊆ Id( ).
Канель-Белов А. Я., Роуэн Луис Халли Proof. By "branching," the two-sided ideal generated in by is Hence, ( ) ( ) ⊆ Id( ) for any 2 − 1, and in particular, if ∈ and ∈ then ∈ Id( ). (26) concludes the proof. 2 By Proposition 3.13 and Lemma 3.10 we have just proved Proposition 3.15. Every PI algebra in characteristic 0 satisfies non-trivial strong identities. Explicitly, let char( ) = 0 and let satisfy an identity of degree . Let , be natural numbers such that + · 2 ( − 1) 4 , and let = ( ) be the × rectangle. Then every ∈ is a strong identity of . The degree of such a strong identity is . We can choose for example We summarize: Theorem 3.16. Every affine PI algebra over a field of characteristic 0 satisfies some Capelli identity. Explicitly, we have the following: (a) Suppose the -algebra satisfies an identity of degree . Then satisfies a strong identity of degree

Actions of the group algebra on sparse identities
Although the method of §3.2 is the one customarily used in the literature, it does rely on branching and thus only is effective in characteristic 0. A slight modification enables us to avoid branching. The main idea is that any sparse identity follows from an identity of the form since we could then specialize +1 , . . . , 2 to whatever we want. Thus, letting ′ denote the subspace of 2 generated by the words (1) +1 · · · ( ) 2 , we can identify the sparse identities with [ ]-subbimodules of ′ inside 2 . But there is an as [ ]-bimodule isomorphism : → ′ , given by (1) · · · ( ) → (1) +1 · · · ( ) 2 . In particular ′ has the same simple [ ]-subbimodules structure as and can be studied with the same Young representation theory, although now we only utilize the left action of permutations.
One such example is when is the staircase, which we define to be the Young tableau whose rows have length , − 1, . . . , 1. This gave rise to the James-Mathas conjecture [21] of conditions on characterizing when is simple in characteristic > 2, which was solved by Fayers [9].

Kemer's Capelli Theorem for all characteristics
In this section we give a proof of Kemer's "Capelli Theorem" over a field of any characteristic. In fact in characteristic Kemer proved a stronger result, even for non-affine algebras. Theorem 4.1. [17] Any PI algebra over a field of characteristic > 0 satisfies a Capelli identity Cap for large enough .
This fails in characteristic 0, since the Grassmann algebra does not satisfy a Capelli identity. The proof of Theorem 4.1 given in [17] is quite complicated; an elementary proof using the "identity of algebraicity" is given in [7, §2.5.1], but still requires some computations. In the spirit of providing a full exposition which is as direct as possible, we treat only the affine case via representation theory, in which case characteristic > 0 works analogously to characteristic 0. This produces a much better estimate of the degree of the sparse identity, which we obtain in Theorem 4.4.
In view of Theorem 3.3, it suffices to show that any affine PI algebra satisfies a sparse identity. Although we cannot achieve this through branching, the ideas of the previous section still apply, using [9].

Simple Specht modules in characteristic > 0
In order to obtain a -version of Proposition 3.13 in characteristic > 2, first we need to find a class of partitions satisfying Fayer's criterion.
For a positive integer , define to be the -adic valuation, i.e., ( ) is the largest power of dividing . Also, temporarily write ℎ ( ; ) for ℎ where is the box in the , position. The Канель-Белов А. Я., Роуэн Луис Халли James-Mathas conjecture for ̸ = 2, proved in [9], is that is simple if and only if there do not exist , , ′ , ′ for which (ℎ ( ; ) ) > 0 with (ℎ ( ; ) ), (ℎ ( ′ ; ) ), (ℎ ( ; ′ ) ) all distinct. Of course this is automatic when each hook number is prime to , since then every (ℎ ( ; ) ) = 0. Example 4. A wide staircase is a Young tableau whose rows have all have lengths different multiples of − 1, the first row of length ( − 1) , the second of length ( − 1)( − 1), and so forth until the last of length − 1. The number of boxes is When = 2, the wide staircase just becomes the staircase described earlier.
In analogy to Example 3, the dimension of the "wide staircase" can be estimated as follows: We write = ( − 1) ′ + ′′ for 1 ′′ − 1. The hook of a box in the ( , ) position has length ( + 1 − )( − 1) + 1 − , and depth + 1 − ′ − , so the hook number is which is prime to . Thus each wide staircase satisfies a stronger condition than Fayer's criterion.
The dimension can again be calculated by means of the hook formula. The first − 1 boxes in the first row have hook numbers  In particular, if 3 (2 +1) , then .
Proof. We imitate the proof of Lemma 3.11. Since the geometric mean is bounded by the arithmetic mean,
Let ⊢ be any partition of corresponding to the "wide staircase" . Then the elements of the corresponding [ ]-bimodule ⊆ ′ are sparse identities of .
Proof. By Remark 3.12, and we conclude from Lemma 3.18. . This concludes the proof of Theorem 4.1 in the affine case.

Results and proofs over Noetherian base rings
We turn to the case where is a commutative Noetherian ring. In general, we say a -algebra is PI if it satisfies a polynomial identity having at least one coefficient equal to 1. Let us indicate the modifications that need to be made in order to obtain proofs of Theorems 1.6 and 1.7.
The method of proof of Theorem 1(2) (for the case in which the base ring is a field) was to verify the "weak Nullstellensatz", and a similar proof works for commutative when is Jacobson, cf. [26,Proposition 4.4.1]. Thus we have Theorems 1.6 and 1.7 in the commutative case, which provide the base for our induction to prove Theorem 1.3. The argument is carried out using Zubrilin's methods (which were given over an arbitrary commutative base ring.) It remains to find a way of proving Kemer's Capelli Theorem over arbitrary Noetherian base rings. One could do this directly using Young diagrams, but there also is a ring-theoretic reduction.
Proof. (i) Viewing the symmetric group 1 × · · · × ˓→ , we partition into orbits under the subgroup 1 × · · · × and match the permutations in Cap . (ii) This time we note that any interchange of two odd-order sets of letters has negative sign, so we partition into parts each with letters. (iii) Any algebra satisfying Cap for even, also satisfies Cap +1 , and + 1 is odd. 2 Thus, it suffices to prove that satisfies a product of Capelli identities. Theorem 5.2. Any affine PI algebra over a commutative Noetherian base ring satisfies some Capelli identity.
Proof. By Noetherian induction, we may assume that the theorem holds for every affine PI-algebra over a proper homomorphic image of .
First we do do the case where is an integral domain, and = { 1 , . . . , ℓ } satisfies some multilinear PI . It is enough to assume that is the relatively free algebra { 1 , . . . , }/ (where is the T-ideal generated by ). Let be the field of fractions of . Then := ⊗ is also a PI-algebra, and thus, by Theorem 4.1 satisfies some Capelli identity 1 = Cap . Thus the imagē 1 of 1 in becomes 0 when we tensor by , which means that there is some ∈ for which 1 = 0. Letting ′ denote the T-ideal of generated by the image of 1 , we see that ′ = 0. If = 1 then we are done, so we may assume that ∈ is not invertible. Then / is an affine PI-algebra over the proper homomorphic image / of , and by Noetherian induction, satisfies some Capelli identity Cap , so /( ∩ ′ ) satisfies Cap max{ , } . But ∩ ′ is nilpotent modulo ′ = ′ = 0, implying by Lemma 5.1 that satisfies some Capelli identity. For the general case, the nilpotent radical of is a finite intersection 1 ∩ · · · ∩ of prime ideals. By the previous paragraph, / , being an affine PI-algebra over the integral domain / , satisfies a suitable Capelli identity Cap , for 1 , so / ∩ ( ) satisfies Cap , where = max{ 1 , . . . , }. But ∩( ) is nilpotent modulo , so, by Lemma 5.1, / satisfies a suitable Capelli identity Cap . Furthermore, = 0 for some , implying again by Lemma 5.1 that satisfies . 2