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Let $R$ be a ring with the following property: "Every unimodular matrix (ie. with a determinant that is a unit in $R$) contains at least one entry that is a unit in $R$."

It looks like a very natural property that must have been studied before. What is known about such rings?

1 Answers1

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Not a complete answer but a bit too long for a comment. Since we consider determinants I assume $R$ is commutative. Let $x \in R$ and consider the matrix

$$X = \left[ \begin{array}{cc} 1 + x & -x \\ x & 1 - x \end{array} \right]$$

which by construction has determinant $1$. By hypothesis, at least one entry of this matrix must be a unit, hence either $x$ is a unit, $1 + x$ is a unit, or $1 - x$ is a unit.

This is very close to one of the characterizations of local rings, that for every $x \in R$ either $x$ or $1 - x$ is a unit. It's not hard to show that this characterization is equivalent to the following: if $\sum x_i = 1$ then at least one of the $x_i$ is a unit. Using this characterization it's not hard to see that the desired property holds for all local rings, since if $X$ is a matrix over a local ring such that $\det X = u$ is a unit then, after multiplying the Laplace expansion of $\det X$ by $u^{-1}$, we see that at least one term in the Laplace expansion must be a unit. So we even conclude that at least $n$ entries of $X$, in distinct rows and columns, must be units.

I'd guess that there ought to be an argument in the other direction showing that any ring with this property is local but I'm not quite familiar enough with local rings to see it.

Qiaochu Yuan
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  • @Quiriacus How'd you make that determination? (Why the dual numbers?) – rschwieb Jun 11 '24 at 15:32
  • I think perhaps because you can phrase the determinant condition in terms of elementary sentences, it might pass to products. If so, you'd have it for semiperfect rings, i think. – rschwieb Jun 11 '24 at 16:10
  • @Quiriacus Yes, sorry, along the way I forgot we wanted one that has the property. I'm kind of surprised the dual numbers would do it. Is it simpler to check than it looks? – rschwieb Jun 11 '24 at 17:38
  • @rschwieb My previous reasoning was incorrect, so I thought it best to delete it. Thank you. Now I think that better formulation of the problem is "Every invertible matrix contains at least one entry that is a unit in $R$." Then a non-commutative $R$ can be also considered. – Quiriacus Jun 12 '24 at 04:06
  • @Qiaochu Yuan Thank you very much. – Quiriacus Jun 12 '24 at 04:10