# Non-zero diagonal through permutation of rows

I'm looking for a reference of the following statement (which can easily be proved by Laplace's formula and induction):

Let $R$ be a commutative ring with identity and let $A$ be an invertible matrix over $R$. Then there is a permutation matrix $P$ such that the diagonal of $PA$ has no zero.

Edit: Of course, $R$ has to be a domain to make the arguments (formula of Leibniz or Laplace) work. Futhermore, it's sufficient to require $det(A) \neq 0$ (instead of $A$ being invertible).

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I'm hoping you won't find a reference of that statement, since it is false. (Consider the 1x1 matrix over the zero ring.) – Ricky Demer May 5 '11 at 21:13
Doesn't a ring with identity have at least two elements? – Tony Huynh May 5 '11 at 21:19
No, Tony. Otherwise algebra would be swamped in statements like "if $I$ is an ideal of $R$, then $R/I$ is a ring or zero". – darij grinberg May 5 '11 at 21:21
Makes sense. Thanks darji. – Tony Huynh May 5 '11 at 21:26
I don't know of a reference, but you can also appeal to the determinant formula as a sum of (permuted) products for a shorter proof. Then you may not need a reference. Gerhard "Ask Me About System Design" Paseman, 2011.05.05 – Gerhard Paseman May 5 '11 at 22:08

Given $A\in{\bf M}_n(R)$, the following statements are equivalent.
• There exists a permutation matrix $P$ such that the diagonal of $PA$ has no zero.
• If an $m\times p$-block of $A$ is mades of zeroes, then $m+p\le n$.