(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?



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


