a polynomial identity $f(x_1,\ldots,x_n)=0$. It is possible to prove that then it satisfies an admissible polynomial identity (that is, a polynomial identity with coefficient $1$ at some term of the highest degree): $$x_1 x_2\cdots x_n = \sum_{\sigma}\alpha_{\sigma}x_{\sigma(1)}
x_{\sigma(2)}\cdots x_{\sigma(n)},$$ where $\alpha_{\sigma}$ belong to the ground field. In this case, if $W = v u_1 u_2 \cdots u_n$ is $n$-divisible then for any permutation $\sigma$ the word $W_{\sigma} = vu_{\sigma(1)}u_{\sigma(2)}\cdots u_{\sigma(n)}$ is lexicographically smaller than $W$, and thus an $n$-divisible word can be represented as a linear combination of lexicographically smaller words. Hence a $\PI$-algebra has a basis consisting of non-$n$-divisible words. By Shirshov theorem on height, a $\PI$-algebra has bounded height. In particular, if a $\PI$-algebra satisfies $x^n = 0$ then it is nilpotent, that is, any its word of length exceeding some $N$ is identically zero. Surveys on height theorem can be found in [@BBL97; @Kem09; @BelovRowenShirshov; @Ufn90; @Dr04].

This theorem implies the positive solution of Kurosh problem and of other Burnside-type problems for $\PI$-колец. Indeed, if $Y$ is a Shirshov basis and all its elements are algebraic then the algebra $A$ is finite-dimensional. Thus Shirshov theorem explicitly indicates a set of elements whose algebraicity makes the whole algebra 