Corrigendum: The Gerstenhaber problem for commuting triples of matrices is “decidable”

Abstract

I correct a slip in an argument in the above paper. This does not affect the main result: the Gerstenhaber Problem is Turing computable for all fields.

Keywords:

• Correction
• finite fields
• Gerstenhaber problem

2020 Mathematics Subject Classification:

• Primary: 13E15
• 15A27; Secondary: 13E10
• 15-04
• 15A21
• 03D99

A very diligent reviewer of my paper [1] for Mathematics Reviews has pointed out a slip in an argument I gave. This doesn’t affect the main result (the Gerstenhaber Problem (GP) is indeed Turing computable for all fields F) but two results in Section 2: Theorem 2.1 and Corollary 2.2. In 2.1 “prime subfield” needs to be replaced by “subfields which are finite extensions of the prime subfield”. In 2.2 the field $\mathbb{Z}/p$ of integers mod p needs to be replaced by “finite subfields of characteristic p”. Here are the amended Theorem and Corollary.

Theorem 2.1.

The GP has a positive answer for a given field F if and only if it has a positive answer for subfields K which are finite extensions of the prime subfield F₀ of F. (In characteristic 0 this means K is an algebraic number field, and in nonzero characteristic K is a finite field.)

Corollary 2.2.

1. The GP has a positive answer for fields F of characteristic p if and only if it is positive for finite fields of characteristic p.

2. The GP has a positive answer for fields F of characteristic 0 if the GP is positive for finite fields.

The amended theorem has already been established in the original proof. The incorrect part of the original proof of 2.1 is the mistaken claim in the last sentence that $${\dim }_P P[\overline{A},\overline{B},\overline{C}]=[K:P]{\dim }_K K[\overline{A},\overline{B},\overline{C}].$$

This is not always so (the right hand side is instead the P-dimension of $K[\overline{A},\overline{B},\overline{C}]$).

Proof

of 2.2. (1) is clear. The original argument of 2.2 showing (correctly) that a negative answer for the GP for the prime subfield $F=\mathbb{Q}$ would lead to a negative answer over some $\mathbb{Z}/p$ used reduction from matrices over F to matrices over the principal ideal domain $R=\mathbb{Z}$ (by clearing fractions), then to matrices over $\overline{R}=\mathbb{Z}/p$ for some prime p of R. The new reductions require a little more subtlety but the general argument is the same: reduce from matrices over an algebraic number field F to matrices over the Dedekind domain R of algebraic integers of F (those elements of F satisfying a monic polynomial with integer coefficients), then to matrices over $\overline{R}=R/P$ for some maximal ideal P of R. The field $\overline{R}$ can be infinite for general Dedekind domains but what is special here is that $\overline{R}$ is actually a finite field (see Serge Lang’s book “Algebraic Number Theory”). Also, R is a Jacobson ring because it is finitely generated over the Jacobson subring $\mathbb{Z}.$ Inasmuch as R is a prime ring, the prime ideal 0 must now be an intersection of maximal ideals (so the Jacobson radical of R is 0). The upshot is if $d\ne 0$ in R, then $\overline{d}\ne 0$ in some $\overline{R}.$ The proof can now be readily completed. [Note that F is the field of fractions of R so we can clear fractions as before.].□

Despite the changes in Section 2, the argument in Section 3 for establishing the GP is Turing computable for all fields F carries over. There is no problem in nonzero characteristic by 2.2(1). In characteristic 0, we can reduce to algebraic number fields by 2.1. These take the form $F=\mathbb{Q}[x]/(q(x))$ where q(x) is an irreducible rational polynomial. However, the set of irreducible rational polynomials is formally enumerable because a given rational polynomial can be determined to be irreducible or not in a finite number of steps, using Kronecker’s method or the Zassenhaus algorithm. Matrix calculations (computing rank etc.) over the algebraic integers of F can still be done exactly and algorithmically. Thus our earlier argument for $\mathbb{Q}$ carries over.

The good news, computationally, is that computer searches for an example where the GP has a negative answer still come down to matrices over $\mathbb{Z}/p$ using mod p arithmetic for various primes p of $\mathbb{Z}.$ This is because we can represent the elements of a finite field K of order p^t as polynomials f(U) in a t × t companion matrix U of an irreducible polynomial q(x) over $\mathbb{Z}/p$ of degree t, and then commuting triples A, B, C of n × n matrices over K as commuting triples $A\prime ,B\prime ,C\prime$ of tn × tn matrices over ${\mathbb{Z}}_p=\mathbb{Z}/p.$ Now $$(\ast ){\dim }_{K}K[A,B,C]=(1/t)({\dim }_{{\mathbb{Z}}_p}{\mathbb{Z}}_p[A\prime ,B\prime ,C\prime ,V])$$ where V is the n × n block diagonal matrix $\text{diag}(U,U,\dots ,U).$ This is because, as ${\mathbb{Z}}_p$ spaces, $K[A,B,C] \cong {\mathbb{Z}}_p[A\prime ,B\prime ,C\prime ,V].$ Inasmuch as the ${\mathbb{Z}}_p$ dimension of $K[A,B,C] = [K:{\mathbb{Z}}_p]({\dim }_{K}K[A,B,C]) = t({\dim }_{K}K[A,B,C]),$ the equation (*) follows.