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Can there be a polynomial over the field $F_p$ of $p$ elements ($p$ prime) in non-commuting variables $X_1,..., X_r$ such that:

1) $f(A_1,...,A_r)=0$ for every $n \times n$ matrices $A_1,...,A_r$ over $F_p$; and

2) $f$ has a non-trivial monomial of total degree less than $n$ ?

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The answer may be "no". Try looking at the proof of L'vov-Kruse theorem (that every finite ring has finite basis of identities). The proof is, in particular, in my book (referred to in the answer below). – Mark Sapir Jul 13 '13 at 6:08
There are three people you can ask (none of them is active on MO as far as I know): Alexei Belov-Kanel from Bar Ilan, Israel, Alexei Krasilnikov from the University of Brasilia, and Mikhail Volkov from the Ural Federal University, Russia. Their email addresses can be found easily. – Mark Sapir Jul 13 '13 at 6:17

2 Answers 2

The answer to 1 is yes. There is the so called standard identity. It can be found in Mark Sapir's book, page 147 in the PDF numbering. I am not sure about 2, maybe some expert will pipe in.

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The smallest degree of a homogeneous polynomial identity on $n\times n$ matrices is $2n$ if $n>2$ or $p$ is odd, according to Amitsur and Levitzky. – Mariano Suárez-Alvarez Jul 12 '13 at 19:32
This standard polynomial does not use the field. It holds for any $2n$ $n \times n$ matrices with symbolic entries. – Aaron Meyerowitz Jul 12 '13 at 20:45
The answer is CW so please edit and fix. – Benjamin Steinberg Jul 12 '13 at 22:04
Thank you for your answer, David. My question is however if there exists a polynomial satisfying both 1) and 2). – Idanps Jul 13 '13 at 5:39
@AaronMeyerowitz, I actually did not write exactly what I meant: the theorem A-L prove is that a non-zero identity of minimal degree is homogeneous and equal (up to a scalar) to the standard one if either $n>2$ or the field has more than two elements (they show that there exist identitie of minimal degree which are not homogeneous in the two exceptional cases: in $M_1(\mathbb F_2)$ one can take $x^2-x$, for example) – Mariano Suárez-Alvarez Jul 15 '13 at 6:56

I tried to just comment, but it seems that I do not have enough "reputation" for that. So I am commenting here.

The answer would be "no" if your field were infinite. Since in that case, each homogeneous component of a polynomial identity would also be an identity (see [1], prop. 4.2.3). And that would imply that you would have a multilinear identity of the same degree of such monomial (use the multilinearization process (see [1])). Finally, it is easy to show that there does not exist polynomial identities of degree less than 2n for M_n(F) (staircase argument see [1] Exercise 7.1.2 - this is true for any field).

Ok, but it was not asked about infinite fields.

For matrices over finite fields, there are descriptions for bases of identities for matrices of order up to 4. (see [2,3,4]) I checked the papers for 2 and 3 (unfortunately the case $n=4$ is in russian), and some of the generators of the ideal of identities of such algebras are polynomials which contains monomials of degree exactly n (2 or 3, in this case), but not less than n. So, it does not answer your question.

At first sight I thought the answer to your question would be "no" even if you were asking for the degree of your monomial to have degree less than 2n. For 2n, the papers [2] and [3] show I was wrong.


[1] V. Drensky, Free Algebras and PI-algebras: Graduate Course in Algebra, Springer, Singapore, 1999.

[2] Yu. N. Mal'tsev, E. N. Kuz'min, A basis for the identities of the algebra of second-order matrices over a finite field, Algebra Log. 17 (1978) 18–21

[3] G. Genov, Basis for identities of a third order matrix algebra over a finite field, Algebra Log. 20 (1981) 241–257.

[4] G. Genov, P. Siderov, A basis for identities of the algebra of fourth-order matrices over a finite field. I, II, Serdica 8 (1982) 313–323, 351–366 (in Russian).

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