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 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.

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.