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Makar-Limanov's problem on values of polynomials on matrices

Louis H. Rowen, Uzi Vishne

Abstract

Suppose $F$ is a field and $ f(X_1, \dots,X_m)$ is a noncommutative polynomial. Makar-Limanov asked whether $f$ evaluated on $M_n(F)$ necessarily has some evaluation of bounded rank independent of $n.$ Answering this query, we show moreover that there are numbers $d\le °f$ and $m'\le m$ such that, under suitable restrictions on $F,$ for any $n \ge m'$ and any $β_i$ in $F$, there are matrices $A_1,\dots,A_m$ in~$M_n(F)$ such that $f(A_1,\dots,A_m)$ is diagonal, and the upper left $(n-m')\times (n-m')$ piece of $f(A_1,\dots,A_m)$ can be taken to be $\diag{β_1,\dots, β_{n-m'}}$, for indeterminates~$β_i$. When f is multilinear, $F$ can be an arbitrary field, and we can take $m' = m-1$. When $f$ is completely homogeneous, $F$ must be closed under $d$ roots. In general, $F$ can be any field closed under roots of polynomials of degree $\le d$. As Makar-Limanov observed, this leads to an immediate, characteristic-free proof of the Freiheitsatz for associative algebras over an algebraically closed field. Also, we show that if $f$ is not a polynomial identity of $ k \times k $ matrices, then for generic matrices $Y_1,\dots,Y_m$, at least $ n - k $ characteristic values of $ f(Y_1,\dots,Y_m) $ are algebraically independent.

Makar-Limanov's problem on values of polynomials on matrices

Abstract

Suppose is a field and is a noncommutative polynomial. Makar-Limanov asked whether evaluated on necessarily has some evaluation of bounded rank independent of Answering this query, we show moreover that there are numbers and such that, under suitable restrictions on for any and any in , there are matrices in~ such that is diagonal, and the upper left piece of can be taken to be , for indeterminates~. When f is multilinear, can be an arbitrary field, and we can take . When is completely homogeneous, must be closed under roots. In general, can be any field closed under roots of polynomials of degree . As Makar-Limanov observed, this leads to an immediate, characteristic-free proof of the Freiheitsatz for associative algebras over an algebraically closed field. Also, we show that if is not a polynomial identity of matrices, then for generic matrices , at least characteristic values of are algebraically independent.
Paper Structure (3 sections, 5 theorems, 13 equations, 1 figure)

This paper contains 3 sections, 5 theorems, 13 equations, 1 figure.

Table of Contents

  1. Introduction
  2. Proofs of the theorems
  3. Proof of Proposition \ref{['PropA']}

Key Result

Theorem 1

If $F$ is algebraically closed, then every polynomial $f(X_1,\dots,X_m)$ is $m'$-rich for a suitable $m' < m \deg f$.

Figures (1)

  • Figure 1: The matrices $x_1,\dots,x_n$ for $n=9$ and $m=4$

Theorems & Definitions (12)

  • Theorem 1
  • Theorem 2
  • Theorem 3
  • Corollary 4
  • proof : Proof of Corollary \ref{['CorA']}, due to Makar-Limanov
  • Proposition A
  • proof : Proof of Theorem \ref{['T3.multi']}
  • Remark 5
  • proof : Proof of Theorem \ref{['T3']}(1)
  • proof : Proof of Theorem \ref{['T3']}(2)
  • ...and 2 more