The finite basis problem for varieties modulo x^3=0

Introduction

Let $K$ be a commutative associative ring with unity and $A$ be an associative algebra without unity over $K$ freely generated by (countably many) generators $x_1, x_2, \ldots$ We will often denote these generators by $x$ and $y$. An element $v=v(x_1,\ldots,x_n)\in A$ is called a (polynomial) identity of an associative $K$-algebra $M$ if $v(f_1,\ldots,f_n)=0$ for any $f_1,\ldots,f_n\in M$. In that case, we also say that $M$ satisfies the identity $v\equiv0$. More generally, we say that $M$ satisfies an identity $v\equiv u$ if $M$ satisfies the identity $v-u\equiv0$.

For example, any commutative algebra satisfies the identity $[x,y]\equiv0$, where $[x,y]=xy-yx$ is the commutator, a nil-algebra satisfies the identity $x^n\equiv0$ for some positive integer $n$, a nilpotent algebra of degree at most $s$ satisfies the identity $x_1x_2\cdots x_s\equiv0$. If $\{v_t\mid t\in\Omega\}$ is a set of identities, then the class of all $K$-algebras satisfying all identities $v_t\equiv0$ for $t\in\Omega$ is called a variety.

A variety is finitely based if it can be defined by a finite set of identities. It was proved by A. Kemer [8] that all varieties of associative algebras over fields of characteristic 0 are finitely based. On the other hand, A. Belov [3], A. Grishin [5], and V. Shchigolev [14] constructed examples of non-finitely generated varieties of associative algebras over fields of positive characteristic. All these examples remain non-finitely based modulo $x^n\equiv0$ for $n$ big enough.

It is easy to see that any variety of associative algebras modulo $x^2\equiv0$ over a Noetherian ring satisfies the identity $xy\equiv-yx$ and thus is finitely-based [9]. Moreover, according to the Dubnov-Ivanov-Nagata-Higman theorem [4, 12, 7], any variety of associative algebras modulo $x^3\equiv0$ over a field of characteristic $p$ with $p>3$ satisfies an identity of nilpotency. It is well known that modulo such an identity any variety of associative algebras is finitely based [2]. On the other hand, C. Gupta and A. Krasilnikov [6] for algebras over fields of characteristic 2 and E. Aladova and A. Krasilnikov [1] for algebras over fields of characteristic 3 constructed examples of non-finitely based varieties of associative algebras modulo $x^6\equiv0$.

Our main result is as follows.

An ideal $V$ of $A$ is called a $T$-ideal, if $V$ is fully characteristic in $A$, that is, $\alpha(V) \subset V$ for any endomorphism $\alpha$ of $A$. It is well known that there exists a one-to-one correspondence between the set of all varieties of associative $K$-algebras and the set of all $T$-ideals of $A$. This correspondence takes a variety $\mathcal{V}$ of associative algebras to the $T$-ideal of all polynomial identities satisfied in all algebras of $\mathcal{V}$. A variety of associative $K$-algebras is finitely based if and only if the corresponding $T$-ideal is finitely generated as a $T$-ideal.

If $\mathcal{V}$ is a variety of associative $K$-algebras and $V$ is the corresponding $T$-ideal of $A$, then the quotient algebra $A/V$ is called the free algebra of $\mathcal{V}$ or a relatively free algebra freely generated by $x_1+V,x_2+V,\ldots$ A fully characteristic ideal in a relatively free algebra is also called a $T$-ideal. There is a similar one-to-one correspondence between the set of all subvarieties of $\mathcal{V}$ and the set of all $T$-ideals in $A/V$.

A submodule $U$ in a relatively free algebra $A/V$ is called a $T$-space if $U$ is fully characteristic in $A/V$. It was proved by V. Shchigolev [16] for algebras over fields of characteristic 0 and by E. Kireeva [9] for commutative algebras over Noetherian commutative associative rings with unity that all $T$-spaces in the corresponding free algebras are finitely generated. Also it is easy to see that all $T$-spaces in a nilpotent relatively free algebra are finitely generated. On the other hand, V. Grishin [5] for algebras over fields of characteristic 2 and V. Shchigolev [15] for algebras over fields of characteristic $p>2$ constructed examples of non-finitely generated $T$-spaces. All these examples remain non-finitely generated modulo $x^n \equiv0$ for some $n$. In particular, in [15] examples of non-finitely generated $T$-spaces modulo $x^3 \equiv0$ in algebras over fields of characteristic 3 were constructed. These $T$-spaces played an important role in the negative solution of the finite basis problem in positive characteristic.

We can note that if all $T$-spaces in a relatively free algebra $A/V$ are finitely generated, then all $T$-ideals in $A/V$ are finitely generated as well, and hence all subvarieties of the variety $\mathcal{V}$ are finitely based.

Let $\mathcal{V}_3$ be the variety of associative $K$-algebras satisfying identity (1) and $V_3$ be the corresponding $T$-ideal in $A$. We will prove the following theorem.

Auxiliary results

This section serves as a preparation for the proofs of the theorems stated in the introduction. It contains various calculations, some of them quite lengthy. So we encourage the reader to use system of computer algebra (for example, Maxima) to verify them. We abbreviate $\widetilde{A}=A/V_3$.

Proof. By definition, $\widetilde{A}$ satisfies the identity

In what follows, we will denote the left-hand side of (2) by $S_3(x,y,z)$.

Proof. Applying Lemma 1 and a routine (computer assisted) calculation, we get that $\widetilde{A}$ satisfies the identity

We will use left normed commutators inductively defined by the rules: $[x_1,x_2]=x_1x_2-x_2x_1$ (as in the introduction); $[x_1,x_2,\ldots,x_n]=[[x_1,x_2,\ldots,x_{n-1}],x_n]$ for $n>2$.

Proof. Denoting $U(x_1,x_2,x_3)=[x_2,x_1,x_3]+[x_3,x_1,x_2]$, we get by Lemma 1 that $\widetilde{A}$ satisfies the identity

Hence this algebra satisfies the identity

and thus also the identity

Finally, as $U(x_3,[x_1,x_2],x_4) \equiv0$, we get

Proof. For $k=0,1,2,3$ we have

“Solving” this system by Cramer’s rule, we get

We denote $g(x,y)=x^2y+xyx+yx^2$. Following [11], we note that

Hence

Denoting $h(y,t)=yt+ty$, we get

whence

We conclude by

We also will use the following easy lemma, whose proof we leave to the reader.

Main proofs

Proof of theorem 2. Part 1. Let the characteristic of $K$ be equal to $p$, where $p$ is a prime number distinct from 3. If $p=2$ then by Lemma 2 the algebra $\widetilde{A}$ satisfies identity (3). Substituting $[x_{6},x_{7}]$ for $x_{2}$, we see that $\widetilde{A}$ also satisfies the identity $[x_{1},x_{2}][x_{6},x_{7}][x_{4},x_{5}]\equiv0$. Thus by [10 Theorem 4], all T-spaces in $\widetilde{A}$ are finitely generated. If $p>3$ then $\widetilde{A}$ satisfies identity (5). Since $2\cdot1_{K}$ and $3\cdot1_{K}$ are invertible in this case, the algebra $\widetilde{A}$ is nilpotent and all T-spaces in $\widetilde{A}$ are finitely generated as well.

Part 2. Now let the characteristic of $K$ be equal to $p^{s}$, where $p$ is a prime number distinct from 3 and $s$ is a positive integer greater than 1. Let

be an increasing chain of T-spaces in $\widetilde{A}$. We are going to prove by decreasing induction on $r=0,\ldots,s$ that this chain stabilizes if all its elements are contained in $p^{r}\widetilde{A}$. The base case $r=s$ is obvious, as $p^{s}\widetilde{A}=0$.

Suppose that $r<s$ and $U_{i}\subseteq p^{r}\widetilde{A}$ for any $i$. We consider two chains

and

By Lemma 5, it suffices to prove that both chains stabilize. The latter chain stabilizes by the inductive hypothesis. To prove that the former chain also stabilizes, we consider the homomorphism

given by $f+p\widetilde{A}\mapsto p^{r}f+p^{r+1}\widetilde{A}$. It is clearly surjective. As the modules $(U_{i}+p^{r}\widetilde{A})/p^{r+1}\widetilde{A}$ are T-spaces in $\widetilde{A}/p^{r+1}\widetilde{A}$, their inverse images under $\chi$ are T-spaces in $\widetilde{A}/p\widetilde{A}$, which is a relatively free algebra over $\mathbb{Z}/p\mathbb{Z}$. These inverse images form an increasing chain, which stabilizes by Part 1. Applying $\chi$, we get that chain (7) also stabilizes.

Part 3. Now let the characteristic of $K$ be equal to a positive integer $n$ not divisible by 3. We consider the factorization

where $p_{1},\ldots,p_{r}$ are pairwise distinct prime numbers not equal to 3. We denote $n_{i}=n/p_{i}^{s_{i}}$ for all $i$. Since $n_{1},\ldots,n_{r}$ are relatively prime, there exist integers $u_{1},\ldots,u_{r}$ such that $n_{1}u_{1}+\cdots+n_{r}u_{r}=1$. We denote $e_{i}=n_{i}u_{i}\cdot1_{K}$. It is easy to check that $e_{i}e_{j}=0$ for $i\ne j$ and $e_{i}^{2}=e_{i}$. Hence we get the following decomposition:

where

The algebra $\widetilde{A}/p_{i}^{s_{i}}\widetilde{A}$ is an algebra over the ring $\mathbb{Z}/p_{i}^{s_{i}}\mathbb{Z}$.

Let

be an increasing chain of T-spaces in $\widetilde{A}$. Then we get the chain

for each $i=1,\ldots,r$. Every such chain is a chain of T-spaces in $\widetilde{e_iA}$ and thus stabilizes by Part 2. Hence it follows that chain (8) stabilizes as well. □

Proof of Theorem 1. If $K$ is a ring of characteristic 3, then by Lemma 3 the algebra $\widetilde{A}$ satisfies identity (4) and hence all T-ideals in $\widetilde{A}$ are finitely generated by [13]. Therefore, arguing similarly to the proof of Theorem 2, we get that all T-ideals in $\widetilde{A}$ are finitely generated if $K$ is a ring of characteristic $n\ne0$.

Let now $K$ be a ring of characteristic 0. We denote by $I$ the T-ideal in $\widetilde{A}$ generated by $x_1\cdots x_7+V_3$. Let

be an increasing chain of T-ideals in $\widetilde{A}$. Then we get two chains

By Lemma 5, it suffices to prove that both chains stabilize.

The ideals $(I_j+I)/I$ are T-ideals in the nilpotent algebra $\widetilde{A}/I$ and hence the former chain stabilizes.

To prove that the latter chain stabilizes, we consider the T-ideal $N=2^4 3^2\widetilde{A}$ in $\widetilde{A}$. Again it suffices to prove that both chains

stabilize.

The ideals $((I_j\cap I)+N)/N$ are T-ideals in the algebra $\widetilde{A}/N$, which is an algebra over $\mathbb{Z}/2^4 3^2\mathbb{Z}$ and hence chain (10) stabilizes. Chain (11) stabilizes since $I\cap N$ is the T-ideal in $\widetilde{A}$ generated by $2^4 3^2x_1\cdots x_7+V_3$ and hence $I\cap N=\{0\}$ by Lemma 4. □

References

[(E. Kireeva)] Bauman Moscow State Technical University, ul. Baumanskaya 2-ya, 5/1, Moscow, 105005, Russia Email address: [email protected]

[(A. Krasilnikov)] Departamento de Matemática, Universidade de Brasília, 70910-900 Brasília, DF, Brazil Email address: [email protected]

[(V. Shchigolev)] Financial University under the Government of the Russian Federation, Leningradsky Prospekt 49, Moscow, 125993, Russia Email address: [email protected]