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If $R$ is a commutative ring, and $A \in M_{n \times m}(R)$, $B \in M_{m \times n}(R)$ with the property that $AB = I_n$ and $BA = I_m$, prove that $n = m$.

I’m aware of a proof involving traces in the case that $R$ is a field of characteristic zero, but cannot think of / know of a proof for this more general case.

Cookie Monster
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1 Answers1

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Let $\mathfrak{m}$ be a maximal ideal of $R$, and let $k=R/\mathfrak{m}$. If $C=(c_{ij})\in M_{p\times q}(A)$, define $\bar{C}=(\overline{c_{ij}})_{ij}\in M_{p\times q}(k)$.

It is not difficult to see that $\overline{C_1C_2}=\overline{C}_1 \overline{C}_2$, with obvious notation.

Now if $AB=I_n$ and $BA=I_m$, then $\bar{A}\bar{B}=I_n\in M_n(k)$ and $\bar{B}\bar{A}=I_m\in M_m(k)$. Since $k$ is a field, a classical result of linear algebra tells you that $n=m$.

GreginGre
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  • Oh my goodness, whenever I saw results such as the existence of maximal subgroups or ideals I would kinda tune out, thinking ‘why is this useful’... the fact that any non-trivial ring can be quotiented to give a field now feels extremely useful. – Cookie Monster Nov 26 '19 at 23:46