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Do there exist (unital, associative, noncommutative) rings $R$ and $S$, where $\mathbb{M}_2(R)\cong \mathbb{M}_3(S)$, but these matrix rings are not isomorphic to $\mathbb{M}_6(T)$ for any ring $T$?

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    $\begingroup$ I don’t have time just now to write a full answer, but doesn’t this follow from Bergman’s theorem that every monoid is the monoid of finitely generated projective modules for some ring unless it obviously isn’t? $\endgroup$ Commented Feb 23 at 15:16
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    $\begingroup$ @JeremyRickard Ah, so we take the commutative monoid generated by two elements $a,b$, subject to the single relation $a+a=b+b+b$. Taking $R$ to be the endomorphism ring of the projective module corresponding to $a$, and taking $S$ to be the endomorphism ring of the projective module corresponding to $b$, we are done (because, clearly, no element $c$ exists with $6c=2a=3b$). $\endgroup$ Commented Feb 23 at 15:32
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    $\begingroup$ Or $(\mathbb{N}\setminus\{1\},+)$, as some people prefer to call that monoid. $\endgroup$ Commented Feb 23 at 15:43

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If $\Lambda$ is a ring, then the isomorphism classes of finitely generated projective $\Lambda$-modules form a commutative monoid $(A,+)$, with $[P]+[Q]=[P\oplus Q]$. This monoid contains a distinguished element $I$: namely, the class $[\Lambda]$ of the free module of rank one.

This monoid $(A,I)$ with distinguished element satisfies two obvious properties: that $$(\forall x,y\in A),\,\, x+y=0\Rightarrow x=y=0,$$ and that $$(\forall x\in A)(\exists y\in A)(\exists n\in\mathbb{N}),\,\,x+y=nI.$$

$\Lambda$ is a $d\times d$ matrix ring if and only if $I$ is divisible by $d$ in the monoid $A$: if $\Lambda\cong P^{d}$ as right modules, then $\Lambda\cong\mathbb{M}_{d}\left(\operatorname{End}_{\Lambda}(P)\right)$, and if $\Lambda\cong\mathbb{M}_{d}(\Gamma)$ with matrix units $e_{ij}$, then $\Lambda\cong e_{11}(\Lambda)\oplus\cdots\oplus e_{dd}(\Lambda)$ as right modules, where $e_{ij}$ induces an isomorphism $e_{jj}(\Lambda)\cong e_{ii}(\Lambda)$.

In Theorems 6.2 and 6.4 of

Bergman, George M., Coproducts and some universal ring constructions, Trans. Am. Math. Soc. 200, 33-88 (1974),

George Bergman proved that every $(A,I)$ satisfying the conditions above is isomorphic to that coming from some ring $\Lambda$.

Taking $A=(\mathbb{N}\setminus\{1\},+)$ and $I=6$, the distinguished element $I$ is divisible by $2$ and by $3$, but not by $6$, so the ring $\Lambda$ is a $2\times2$ matrix ring and a $3\times3$ matrix ring, but not a $6\times6$ matrix ring.

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    $\begingroup$ This is a really nice answer and shows why full answers are better than comments. I didn’t see from the comment how to tell that it is not a 6x6 matrix algebra for any ring but now it is clear. $\endgroup$ Commented Feb 24 at 15:02

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