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Tom Goodwillie
Tom Goodwillie
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Yes and yes. Let $T=M_m(R)=M_n(S)$.

The center of $T$ is isomorphic to both $R$ and $S$.

The $1\times m$ matrices over $R$ form an $(R,T)$-bimodule and the $n\times 1$ matrices over $S$ form a $(T,S)$-bimodule. Tensor these over $T$ to get an $(R,S)$-bimodule. As an $R$$S$-module the direct sum of $m$ copies of this is free of rank $n$. For a nonzero commutative ring this implies that $m$ divides $n$. (Tensor with a residue field to get a vector space of dimension $\frac{n}{m}$.) Likewise, looking at it as an $S$$R$-module, $n$ divides $m$.

Yes and yes. Let $T=M_m(R)=M_n(S)$.

The center of $T$ is isomorphic to both $R$ and $S$.

The $1\times m$ matrices over $R$ form an $(R,T)$-bimodule and the $n\times 1$ matrices over $S$ form a $(T,S)$-bimodule. Tensor these over $T$ to get an $(R,S)$-bimodule. As an $R$-module the direct sum of $m$ copies of this is free of rank $n$. For a nonzero commutative ring this implies that $m$ divides $n$. (Tensor with a residue field to get a vector space of dimension $\frac{n}{m}$.) Likewise, looking at it as an $S$-module, $n$ divides $m$.

Yes and yes. Let $T=M_m(R)=M_n(S)$.

The center of $T$ is isomorphic to both $R$ and $S$.

The $1\times m$ matrices over $R$ form an $(R,T)$-bimodule and the $n\times 1$ matrices over $S$ form a $(T,S)$-bimodule. Tensor these over $T$ to get an $(R,S)$-bimodule. As an $S$-module the direct sum of $m$ copies of this is free of rank $n$. For a nonzero commutative ring this implies that $m$ divides $n$. (Tensor with a residue field to get a vector space of dimension $\frac{n}{m}$.) Likewise, looking at it as an $R$-module, $n$ divides $m$.

Source Link
Tom Goodwillie
Tom Goodwillie
  • 55.9k
  • 7
  • 151
  • 240

Yes and yes. Let $T=M_m(R)=M_n(S)$.

The center of $T$ is isomorphic to both $R$ and $S$.

The $1\times m$ matrices over $R$ form an $(R,T)$-bimodule and the $n\times 1$ matrices over $S$ form a $(T,S)$-bimodule. Tensor these over $T$ to get an $(R,S)$-bimodule. As an $R$-module the direct sum of $m$ copies of this is free of rank $n$. For a nonzero commutative ring this implies that $m$ divides $n$. (Tensor with a residue field to get a vector space of dimension $\frac{n}{m}$.) Likewise, looking at it as an $S$-module, $n$ divides $m$.