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Let $K$ be an inifinite field of characteristic different from 2. The well-known Amitsur-Levitzki theorem states that the algebra $M_n(K)$ satisfy the standard polynomial identity of degree $2n$, $$s_{2n}(x_1,\dots,x_{2n})=\sum_{\sigma\in S_{2n}}(-1)^{\sigma}x_{\sigma(1)}\cdots x_{\sigma(2n)}$$ Moreover, it does not satisfy any other identity of degree less than $2n$. In particular, if $m < n$, $s_{2m}$ is an identity for $M_m(K)$ and is not an identity for $M_n(K)$.

My question is the following:

If $R$ is a unitary associative noncommutative $K$-algebra that satisfy a polynomial identity, is it true that if $m < n$ then there is an identity of $M_m(R)$ which is not an identity for $M_n(R)$?

In the language of T-ideals, is the inclusion $T(M_n(R))\subset T(M_m(R))$ a proper one?

Equivalently, the PI-equivalence of $M_n(R)$ and $M_m(R)$ imply that $m = n$?

Of course, if the condition that $R$ is a unitary algebra is removed, nilpotent algebras can give counter-examples.

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    $\begingroup$ The A-L theorem actually works for any commutative ring $K$. The very best theorems do not have hypotheses :-) $\endgroup$ – Mariano Suárez-Álvarez Dec 18 '12 at 15:31
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    $\begingroup$ A central polynomial for $M_n(R)$ would fit... $\endgroup$ – Ilya Bogdanov Dec 18 '12 at 16:31
  • $\begingroup$ Ilya Bogdanov, the problem is that we do not know if $M_n(R)$ has a central polynomial. If $R$ is commutative, it is ok, but this is not the case here. $\endgroup$ – Thiago Dec 18 '12 at 23:45
  • $\begingroup$ Related question: Does the truth of any statement of real matrix algebra stabilize in sufficiently high dimension? mathoverflow.net/questions/34186/… $\endgroup$ – Joel David Hamkins May 13 '13 at 1:48
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This is not true in general. For example, if $R$ is a free associative algebra of rank $>1$, then $M_n(R)$ does not have nontrivial identities for any $n$.

Moreover, I believe there are examples of algebras $R$ such that much stronger condition holds: $M_n(R)$ is isomorphic to $M_m(R)$ for some $m \ne n$. This reminds cancellation problems in the commutative setting, though I cannot provide examples of such algebras.

This is true if, for example, if $R$ is finite-dimensional and prime (even not necessary associative). Then we can note that $M_n(R)$ is prime too, pass to the algebraic closure of the base field, and invoke theorem of Razmyslov that finite-dimensional prime algebras over algebraically closed fields are determined by their identities (Yu.P. Razmyslov, Identities of Algebras and Their Representations, AMS, 1994 (translation from Russian), around p. 30).

EDIT:

I was not careful enough when reading the question, sorry. The question explicitly asks for situation when $R$ is a PI algebra, so the example with a free algebra obviously does not qualify.

I still think this is not true in the whole generality: probably just the condition of being unitary is to weak, one should demand something like (semi)primeness.

I can think of two approaches. First, there is a lot of works about identities of tensor product of algebras, and of $M_n(R)$ in particular (typical results: if $R$ satisfies the standard identity of degree $k$, then $M_n(R)$ satisfies the standard identity of some given degree in terms of $n$ and $k$, see e.g. M. Domokos, Eulerian polynomial identities and algebras satisfying a standard identity, J. Algebra 169 (1994), N3, 913-928 DOI: 10.1006/jabr.1994.1317). Second, perhaps one can do something along the lines of Sections 4 and 5 of arXiv:0911.5414.

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    $\begingroup$ If $R$ is a ring of row-finite column finite infinite matrices, then $R\cong M_2(R)$. $\endgroup$ – Mariano Suárez-Álvarez May 11 '13 at 10:00
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    $\begingroup$ Thiago has the assumption that $R$ satisfies a nontrivial polynomial identity. Associativity is, of course, an identity, but, I suspect, he just forgot to assume that $R$ is an associative ring. $\endgroup$ – Misha May 12 '13 at 23:26
  • $\begingroup$ Right, Misha! I assume R to be an associative ring. In particular, Pasha, the free associative algebra of rank >1 is not PI, as well. I will fix this. Thanks for the answers. $\endgroup$ – Thiago May 13 '13 at 0:53
  • $\begingroup$ Mariano, also in this case, $R$ is not PI. $\endgroup$ – Thiago May 13 '13 at 0:56
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    $\begingroup$ @Thiago, I was just providing an example to Pasha's second paragraph, quite independently of what the question asked for. $\endgroup$ – Mariano Suárez-Álvarez May 13 '13 at 5:08

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