When two determinantal ideals together generate a power of the maximal ideal?

Question

(A somewhat technical question, but maybe it is well known.)

Consider matrices over the ring $k[[x_1,\dots,x_n]]$, whose entries vanish at the origin (i.e. belong to the maximal ideal $\mathfrak{m}$). Denote by $J(A_{k,l})$ the ideal of maximal minors of the matrix $A_{k,l}\in Mat(k,l,\mathfrak{m})$.

Given two generic enough (and mutually generic!) matrices, consider the ideal generated by all the maximal minors: $\Big( J(A_{k_1,l_1}),J(B_{k_2,l_2}) \Big)$. (Assume $k_1\le l_1$ and $k_2\le l_2$.)

'{\bf Q.}': When does this ideal contain $\mathfrak{m}^{k_1+k_2-1}$?

Of course, an obvious necessary condition is: $(l_1-k_1+1)+(l_2-k_2+1)\ge n$, but is this condition sufficient?

example 1. Suppose $k_1=l_1$, $k_2=l_2$, then for the generic case we have: det(A) is of order $k_1$ and $det(B)$ is of order $k_2$ and the lowest order parts of the two polynomials form a regular sequence. Thus, if $n\le 2$ we get: $(det(A),det(B))\supset\mathfrak{m}^{k_1+k_2-1}$.

example 2. Suppose $k_1=1$, take $A_{1,l}$ with generic entries of first order. Again, one can show that the needed property holds.

What about the general case? How to address such questions?

Answer 1

Let me discuss the graded case, that is the ring is the polynomial ring and the matrices have general homogeneous entries of degree $1$. The local version should follows by taking "lowest order part" as you do in example $1$.

Consider first of the case of the polynomial ring $S$ in variables $x_{ij}$ and $y_{ij}$ and matrices $X=(x_{ij})$ of size $a\times b$ and $Y=(y_{ij})$ of size $c\times d$ with $a\leq b$ and $c\leq d$. Then take the ideal $I$ of minors of size $a$ of $X$ and the ideal $J$ of minors of size $c$ of $Y$.

Set $A=S/I+J$.

The minimal free resolution of $A$ is obtained by taking the tensor product of the resolution of $S/I$ with that of $S/J$ because the ideals are generated by polynomials in different variables. We may use this fact, in combination with the fact that $I$ and $J$ are resolved by the Eagon-Northcot complex, to compute the Castelnuovo-Mumford regularity $\operatorname{reg}(A)$ of $A$, its projective dimension and its dimension. We have:

$\operatorname{reg}(A)=\operatorname{reg}(S/I)+\operatorname{reg}(S/J)=(a-1)+(b-1)$

$\dim(A) = ab+cd-(b-a+1)-(d-c+1)$

and $A$ is Cohen-Macaulay.

Now we specialize generically the $x_{ij}$'s and the $y_{ij}$'s generically to linear forms in variables $z_1,..,z_n$. The ring you want to understand gets identified with $A/L$ where $L$ is generated by $ab+cd-n$ general linear forms in the $x_{ij}$ and $y_{ij}$.

Now if $n\leq (b-a+1)+(d-c+1)$ then $ab+cd-n\geq ab+cd-(b-a+1)-(d-c+1)$ and $ab+cd-(b-a+1)-(d-c+1)$ of the general linear forms generating $L$ form a maximal regular sequence in $A$. Let $U$ be the ideal generated by $ab+cd-(b-a+1)-(d-c+1)$ of the general linear forms generating $L$.

Then $\operatorname{reg}(A/U)=\operatorname{reg}(A)=(a-1)+(b-1)$ and $\dim(A/U)=0$. The regularity for a $0$-dimensional module $M$ is the largest index $i$ such that $M_i\neq 0$. It follows that $(A/U)_i=0$ for $i>(a-1)+(b-1)$. And the same is true for $A/L$ (because it is a quotient of $A/U$). Hence we have $(A/L)_i=0$ for $i>(a-1)+(b-1)$. On the size of the $z$'s it says that the ideal $(z_1,..,z_n)$ to the power $(a-1)+(b-1)+1$ is contained in ideal of definition, which is exactly what you wanted to prove.