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

(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?

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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 fist of the case of the polynomial ring S in variables x_{ij} 's and y_{ij} 's and matrices X=(x_{ij}) of size axb and Y=(y_{ij}) of size cxd 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 are generated by polynmials in different varaibles. 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 reg(A) of A, its projective dimension and its dimension. We have:

reg(A)=reg(S/I)+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 udenrstand 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 reg(A/U)=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 the (A/U)_i=0 for i>(a-1)+(b-1). And the same is true for A/L (becasue it is a quotinet 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 power (a-1)+(b-1)+1 is contained in ideal of definiton, which is exactly what you wanted to prove.

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