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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
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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

1 Answer 1

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
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@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

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