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Let $S = k[x_{i,j}\mid 1\leq i\leq n, 1\leq j\leq m]$ be a polynomial ring over an arbitrary field $k$. Let $M$ be a generic $n\times m$ matrix of indeterminates in the ring $S$ where $n\leq m$. For any $k\leq n$, let $M'$ denote the matrix $M$ with column $k$ removed and let $M_k$ denote the matrix $M$ restricted to columns $1, \ldots, k$. Consider the ideals $J_1 = I_n(M')$, the ideal of maximal minors of $M'$, and $J_2 = I_k(M_k)$, the ideal of maximal minors of $M_k$. I need to show that

$$ J_1\cap J_2 = J_1\cdot J_2 $$

or, equivalently,

$$ Tor_1(S/J_1,S/J_2) = 0. $$

I feel like this must follow somehow from the fact that NO determinants in $J_1$ contain variables from column $k$, but EVERY determinant in $J_2$ contains variables from column $k$, but I am not sure how to show it.

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1 Answer 1

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After some thought, I came up with the following answer to my own question.

First, observe that $$ \mathbf{x} = (x_{1,k}, \ldots, x_{n,k}) $$ is a regular sequence over $R/I_n(M')$. Take an element $f\in I_n(M')\cap I_k(M_k)$. Then $$ f = \sum_{I\in I_n(M')} c_I [i_1, \ldots, i_n] = \sum_{J\in I_k(M_k)} d_J [j_1, \ldots, j_k]. $$ Expanding the minor $[j_1,\ldots, j_k]$ along column $k$ using the Laplace expansion, one obtains the equality \begin{align*} \sum_{I\in I_n(M')} c_I [i_1, \ldots, i_n] &= \sum_{J\in I_k(M_k)} d_J \left(\sum_{\ell\in \{1,\ldots, n\}} (-1)^\ell \, x_{\ell, k} \, [1, \ldots, \hat \ell, \ldots, n\mid j_1, \ldots, j_{k-1}]\right)\\ &= \sum_{J\in I_k(M_k)} \sum_{\ell\in \{1,\ldots, n\}} (-1)^\ell d_J\cdot x_{\ell,k}\, [1, \ldots, \hat \ell, \ldots, n\mid j_1, \ldots, j_{k-1}]. \end{align*} Observe that since $\mathbf{x}$ is regular over $R/I_n(M')$, each $d_J\cdot [1, \ldots, \hat \ell, \ldots, n\mid j_1, \ldots, j_{k-1}]$ must be in $I_n(M')$. But $I_n(M')$ is a prime ideal, so either $d_J$ or $[1, \ldots, \hat \ell, \ldots, n\mid j_1, \ldots, j_{k-1}]$ must be in $M'$. Clearly $[1, \ldots, \hat \ell, \ldots, n\mid j_1, \ldots, j_{k-1}]$ is not in $M'$, so we have that $d_J$ must be in $I_n(M')$, so $I_n(M')\cap I_k(M_k) = I_n(M')\cdot I_k(M_k)$.

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