I have the following question:

Let $S$ be a commutative regular local ring and $\mathfrak{n}$ be its maximal ideal. Let $f\in\mathfrak{n}$ be a non zero-divisor in $S$ and let $m\geq 1$ ne a natural number.

Let $\varphi$ and $\psi$ be two $m\times m$ - matrices with entries in $S$ such that $\varphi\cdot \psi=\psi\cdot \varphi=f\cdot 1_{S^m}$.

We call the ordered pair $(\varphi,\psi)$ reduced, if all entries in $\varphi$ and all entries in $\psi$ are non-units in $S$.

$\mathbf{Question:}$ If $(\varphi,\psi)$ is not reduced, is it then true (and why?), that always one of the following two cases must occur:

1) There are invertible matrices $\alpha$ and $\beta$ such that

$$\begin{array} 0S^m &\stackrel{id_{S^m}}{\longrightarrow}&S^m\\ \downarrow{\beta}&&\downarrow{\alpha}\\ S^m&\stackrel{\varphi}{\longrightarrow}&S^m& \end{array} $$ is commutative

2) There are invertible matrices $\alpha$ and $\beta$ such that

$$\begin{array} 0S^m &\stackrel{f\cdot id_{S^m}}{\longrightarrow}&S^m\\ \downarrow{\beta}&&\downarrow{\alpha}\\ S^m&\stackrel{\varphi}{\longrightarrow}&S^m& \end{array} $$ is commutative ?

Thanks for the help.


1 Answer 1


I don't understand very well the question, or better say why do you think such situation should occur.

If $(\varphi',\psi')$ is an $f$-reduced pair and take $\varphi=\left(\begin{array}{cc}I_r & 0\\ 0 &\varphi'\end{array} \right)$, respectively $\psi=\left(\begin{array}{cc}fI_r & 0\\ 0 &\psi'\end{array} \right)$ you obtain a non-reduced pair and $\varphi$ is not of the form you suggested.

  • $\begingroup$ Thank you very much for your answer. But why it is impossible then that one of the two above cases does occur? $\endgroup$ Feb 3, 2014 at 18:34

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