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fix a field $F$ and consider an $n \times n$ matrix $M(X)$ with entries in $F[X]$. the determinant $\det(M(X))$ is itself a polynomial, say $D(X)$.

clearly, if $x \in F$ is such that the specialized $\text{rank}(M(x)) < n$, then $D(x) = 0$. i want to say more:

Claim. If $x \in F$ is such that $\text{rank}(M(x)) \leq n - d$, then $D(x)$ vanishes to order $\geq d$; that is, $(X - x)^d \mid D(X)$ in $F[X]$.

BD107
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    i think this will be harder to prove in characteristic 2 (which is actually what i care about), but I don't even have a proof in any characteristic. – BD107 Mar 04 '24 at 20:30

1 Answers1

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A matrix over a commutative ring is invertible whenever its determinant is, so multiplying $M$ by invertible matrices would change neither quantity. $F[X]$ is a PID, so we have the Smith normal form, and for diagonal matrices the claim is rather evident: at least $d$ diagonal entries are zero in the specialization, and $X-x$ divides each of these.

Amateur_Algebraist
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    thank you; this settles the original question. i hate to move the goalposts; but any leads for the same problem, but for $F$ replaced by a reduced commutative ring with unit (but necessarily an integral domain)? thanks in advance. – BD107 Mar 04 '24 at 21:51
  • @BD107 Do you mean "not necessarily an integral domain"? What kind of rank do you want to use? – Amateur_Algebraist Mar 05 '24 at 10:07
  • yes, and good question—let me break this out into a new question. – BD107 Mar 05 '24 at 13:48
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    done https://math.stackexchange.com/q/4875490/308170 – BD107 Mar 05 '24 at 13:56