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a special case of this question with coefficients in a field was recently asked and answered.

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

suppose that an element $x \in R$ is such that the specialized matrix $M(x)$ has $\text{rank}(M(x)) \leq n - d$, in the sense that there is a subset of $M$'s columns, of size $d$, such that each column in that subset is a linear combination (over $R[X]$) of $M$'s other $n - d$ columns. (this straightforwardly implies that each of $M$'s $n - d + 1 \times n - d + 1$ minors vanish in $R[X]$, by the properties of the determinant.)

Claim. In this case, the determinant $D(X) := \det(M(X))$ vanishes at $x$ to order $\geq d$; that is, $(X - x)^d \mid D(X)$ in $R[X]$.

BD107
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i was able to solve this, in the following way:

let $x \in R$ and $d$ be as in the hypothesis; i.e., $\text{rank}(M(x)) \leq n - d$.

fix any entry $(i, j) \in \{0, \ldots , n - 1\} \times \{0, \ldots , n - 1\}$. by Euclidean division, we can re-express the polynomial entry $M_{i, j}(X) = (X - x) \cdot M'_{i, j}(X) + M_{i, j}(x)$, where $M'_{i, j}(X) \in R[X]$ is some arbitrary other polynomial.

now consider the determinant of the resulting matrix $M(X)$, where we write the variant $(X - x) \cdot M'_{i, j}(X) + M_{i, j}(x)$ in each cell $(i, j)$. we get a complicated determinant of binomials. as in the usual binomial theorem, we can "stratify" the result according to how many powers of $(X - x)$ are present. our goal is to show that the entire resulting expression is divisible by $(X - x)^d$. of course we can simply ignore each stratum containing at least $d$ powers of $(X - x)$. the goal is to show that each stratum containing fewer than $d$ powers of $(X - x)$ vanishes.

fix an element $w \in \{0, \ldots , d - 1\}$. the stratum corresponding to the power $(X - x)^w$ has the following representation. for each subset $I \subset \{0, \ldots , n - 1\}$ of cardinality exactly $n - w$, we can go through the full Leibniz expansion, and in each cell take the right summand $M_{i, j}(x)$ exactly when $i \in I$, and the left summand $(X - x) \cdot M'_{i, j}(X)$ otherwise. you can convince yourself that the resulting thing is a polynomial divisible by $(X - x)^w$ times the sum of all $n - w \times n - w$ minors of the specialized matrix $M(x)$ whose row-index set is $I$; we agreed that all of those vanish. now the $(X - x)^w$ stratum is simply the sum of all such things over choice of $I \subset \{0, \ldots , n - 1 \}$ of cardinality $n - w$. this proves the required statement.

there is probably a neater proof of this; will update accordingly

BD107
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