# When is the intersection of two determinantal ideals equal to the product?

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.

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 $$I_n(M')\cap I_k(M_k) = I_n(M')\cdot I_k(M_k)$$.