The Kurosh Problems for PI algebras via Combinatorics

This is a continuing series of posts on some of the highlights of a class at UCSD being given by Efim Zelmanov.

For a field $F$, recall the Kurosh problem as I stated it for algebras.

Kurosh Problem for nilalgebras: Suppose $A$ is a finitely generated $F$-algebra which is nil. That is $\forall a\in A\exists N\in\mathbb Z a^N=0$. Is $A$ finite dimensional (equivalently, nilpotent)?

We saw that the Golod-Shafarevich construction gives a counterexample. However, it turns out that the answer is positive for PI algebras, which we will show. There is another version of the Kurosh problem which I didn't state since it was not relevant for the Burnside problem. It replaces the concept of "nil" with "algebraic."

Definition: An element $a\in A$ is said to be algebraic if there is a polynomial $f(t)$ with $F$-coefficients such that $f(a)=0\in A$. An algebra is said to be algebraic if every element is algebraic.

Kurosh Problem for algebraic algebras: Suppose $A$ is algebraic and finitely generated. Is $A$ finite dimensional?

Since nil implies algebraic, the Golod-Shafarevich construction shows that this is false in general, but we will show that it is indeed true for PI algebras.

Everything follows from a combinatorial lemma about words. But first we need a definition. Let $X$ be a finite alphabet with a fixed order, say $x_1<x_2\ldots x_m$. Order the set of words $X^*$ lexicographically, with the convention that if $v$ is an initial word for $w$, then $v>w$. (This is the opposite of the usual convention.) So for example $x_1x_2<x_1^2$ and $x_1x_2>x_1x_2x_3$.

Definition: A word $w\in X^*$ is $n$-divisible if $w=w'u_1\cdots u_n w''$ for some subwords $u_i$ and every nontrivial permutation $u_{\sigma(1)}\ldots u_{\sigma(n)}$ is a smaller word than $u_1\ldots u_n$. If a word is not $n$-divisible, we say it is $n$-indivisible.

If $A$ satisfies a polynomial identity of degree $n$, then by linearizing we can assume it is of the form $x_1\ldots x_n+\sum_{1\neq\sigma\in \mathcal S_n}\alpha_\sigma x_{\sigma(1)}\cdots x_{\sigma(n)}$. Plugging in $x_i=u_i$, we can rewrite an $n$-divisible word as a linear combination of smaller words. Hence $A$ is spanned by words in its generators which are $n$-indivisible.

Shirshov $\mathcal N(m,k,n)$ lemma: There exists a function $\mathcal N\colon \mathbb N^3\to \mathbb N$ such that any word of length $\geq \mathcal N(m,k,n)$ in the alphabet $X=\{x_1,\ldots, x_m\}$ either has a subword of the form $v^k$ or is $n$-divisible.

Proof of Shirshov lemma:
We proceed by induction on the pair $(n,m)$ ordered lexicographically. For the base cases $m=1$, we can take $\mathcal N(1,k,n)=k$.  So now suppose we know how to define it for all pairs less than $(n,m)$.

Define a set of words
$$T=\{x_m^sx_{i_1}\cdots x_{i_l}\,|\, s\geq 1, \ell\geq 1, \forall j\,i_j<m, s<k, \ell<\mathcal N(m-1,k,n)\}$$
Note that $T$ is finite. We say a word is a $T$-word if it is a product of elements of $T$. Any word that starts with $x_m$ and ends with some $x_i\neq x_m$, and is both $n$-indivisible and doesn't contain any $k$th powers must be a $T$ word. This is because any such word can be written as a product of $T$-type words except that $s$ and $\ell$ could be large. However, the size of $s$ is bounded by the hypothesis of no $k$th powers and each $\ell$ is inductively bounded by $\mathcal N(m-1,k,n)$ since it does not contain any powers and is $n$-indivisible.

Now we want to show that a long word $v$ in $X$ must contain $k$th powers or be $n$-divisible. So suppose $v$ does not contain $k$th powers and is $n$-indivisible. Any such word $v$ can be written as $v'(x_1,\ldots,x_{m-1})v''x_m^\mu$, where $v''$ is a $T$-word, $\ell(v')<\mathcal N(m-1,k,n)$ and $\mu<k$. So we need to control the length of the $T$-word $v''$.

Now $T$-words can be ordered by looking at the order with respect to $X$ or with respect to $T$. Luckily, or actually by design, these give the same order on $T$-words of the same composition.

Lemma: Given two $T$-words $u,v$ of the same composition (that is they use the same letters with the same multiplicity from the alphabet $X$, or equivalently $T$), then
$$u<_T v\Leftrightarrow u<_X v.$$
Proof of lemma:
If $u$ and $v$ begin with the same letter of $T$, then proceed by induction. Otherwise, suppose that the first $T$ word of $v$ is an initial word of the first $T$ word of $u$. Then, $u<v$ with respect to both orders. $\Box$

Now suppose that $\ell_T(v'')>\mathcal N(|T|,k,n-1)+1$. Then $v''=\underline{\hspace{2em}}t$, where there is an $(n-1)$-division in the initial space. So we need such an $n-1$ division to promote to an $n$-division. Suppose $u_1\cdots u_{n-1}$ is an $n-1$ division, where the $u_i$ are $T$-words.

Lemma: $u_1\cdots u_{n-1} x_m$ is $n$-divisible.
Proof of lemma:
Write $u_i=x_mu'_i$. Divide the word as $$\underbrace{x_m}\underbrace{u_1'x_m}\underbrace{u_2'x_m}\cdots\underbrace{u'_{m-1}x_m}.$$
 Every permutation of these words decreases the order, given our assumption that this is true for the $u_i$'s. $\Box$

So, to finish the proof, we can define $\begin{multline*} \mathcal N(m,k,n)=[\mathcal N(m-1,k,n)+k] +\\ [\mathcal N(m-1,k,n)+k]\cdot\mathcal N(m^{\mathcal N(m-1,k,n)},k,n-1) \end{multline*}$
Here the first $[\mathcal N(m-1,k,n)+k]$ controls the length of $v'$ and $x_m^\mu$. The second $[\mathcal N(m-1,k,n)+k]$ bounds the length of an element of $T$, while $m^{\mathcal N(m-1,k,n)}$ is an upper bound on the size of $T$.

This completes the proof of the Shirshov Lemma. $\Box$

We actually need a slightly strengthened version of the Shirshov Lemma, which itself requires an elementary lemma.

Lemma: Let $v,w\in X^*$ be words in the alphabet $X$. Then $v,w$ commute iff they are powers of the same word.
Proof of lemma: 
We use induction on the sum of lengths $\ell(v)+\ell(w)$. Assume that $vw=wv$. Then if $\ell(v)=\ell(w)$, it is clear that $v=w$, since they are the initial words on both sides of the equality. On the other hand if $\ell(v)<\ell(w)$, say, then $v$ is an initial word of $w$: $w=vw'$. So $vvw'=vw'v$, and hence $vw'=w'v$. By induction $v=u^k$ and $w'=u^\ell$, and $w=u^{k+\ell}$. $\Box$

Lemma: Let $v$ be a word that is not a nontrivial power. If $\ell(v)>\geq n$, then $v^{2n}$ is $n$-divisible.
Proof of lemma:
Write $v=x_{i_1}\cdots x_{i_\ell}$ for $\ell\geq n$. Consider all cyclic permutations of the word. If two cyclic permutations are equal, then $v'v''=v''v'$ for some decomposition, meaning that $v$ is a proper power by the previous lemma. Now write
$$v=v_1'v_1''=v_2'v_2''=\cdots=v_\ell'v_\ell',$$
which are all the ways of decomposing $v$ into two words. Order them in such a way that
$$v_1''v_1'>v_2''v_2'>\cdots>v_\ell'' v_\ell'.$$
Note that $v^2$ contains all of these as subwords: $v^2=v_i'\underbrace{v_i''v_i'}v_i''$. Consider
$$v^{2n}=v^2\cdots v^2=(v_1'\underbrace{v_1''v_1'v_1'')(v_2'}\underbrace{v_2''v_2'v_2'')(v_3'}v_3''v_3'v_3'')\cdots$$
The bracketed subwords give an $n$-division. $\Box$

This leads to the following strengthening of the Shirshov Lemma:
Theorem: Let $k\geq 2n$. $\exists \mathcal N(m,k,n)$ such that a word $w$ of length $\geq \mathcal N$ is $n$-divisible or contains a subword $v^k$ with $\ell(v)<n$.

We can now deduce some interesting consequences.

Corollary: Let $A=\langle a_1,\cdots, a_m\rangle$ be a finitely generated PI $F$-algebra such that every product $a_{i_1}\ldots a_{i_s}$ for $s<n$ is nilpotent. Then $A$ is a nilpotent algebra, and hence finite dimensional.
Proof of corollary:
Let $k$ be the maximum of all the degrees of nilpotency and $2n$. Then $A^{\mathcal N(m,k,n)}=(0)$. $\Box$
This implies the Kurosh problem for nilalgebras.

Corollary: Suppose that all words $a_{i_1}\cdots a_{i_s}$ $s<n$ are algebraic. Then $A$ is finite dimensional.
Proof of corollary:
$A$ is generated by products of length $<\mathcal N(m,k,n)$. $\Box$

Corollary: Suppose $A$ is a finitely generated nilalgebra where the degrees of nilpotency are bounded. Then $A$ is finite dimensional.
Proof of corollary:
We have $x^n=0$ for all $x\in A$, so it satisfies a polynomial identity. $\Box$

Corollary: Suppose $A$ is a finitely generated algebraic algebra of bounded degree. Then $A$ is finite dimensional.
Proof of corollary:
Say $n$ bounds the degree. Then $S_n([y,x^n],[y,x^{n-1}],\cdots,[y,x])=0$. Why? If $x$ is algebraic, of degree $n$, then $\{x,\ldots, x^n\}$ are linearly dependent, so that the standard polynomial will be $0$. We put the $y'$s in there so the identity is not itself $0$ as a noncomuttatuve polynomial. $\Box$