# Products of Determinantal Ideals

Let $$X=(x_{ij})_{1\leq i\leq m, 1\leq j\leq n}$$ where the $$x_{ij}$$ are variables. Now, for all $$d$$, we consider the ideal $$I_d$$ of the polynomial ring $$S=\mathbb{C}[x_{ij}:\, 1\leq i\leq m, 1\leq j\leq n]$$ that is generated by all $$d\times d$$ minors of $$X$$. After doing some experiments, I conjecture that we have $$I_{d-1}\cdot I_{d+1}\subset I_d^2$$ for all $$d$$. I suspect that this might simply follow from some well-known determinantal identities but I didn't make a find. Does anybody have a smart proof or a reference?

Edit: In order to resolve Darij's concern, let me give an example here. Let $$m=n=4$$, so we consider the matrix $$X=\begin{pmatrix} x_0& x_4& x_8& x_{12}\\x_1& x_5& x_9& x_{13}\\x_2& x_6& x_{10}& x_{14}\\x_3& x_7& x_{11}& x_{15} \end{pmatrix}.$$ The top left $$3\times 3$$ minor is $$h=-x_2 x_5 x_8+x_1 x_6 x_8+x_2 x_4 x_9-x_0 x_6 x_9-x_1 x_4 x_{10}+x_0 x_5 x_{10}.$$ Darij had some doubts that $$x_{15}\cdot h$$ is in the ideal $$I_2^2$$. Here is such a representation: $$x_{15}\cdot h=\frac{1}{2}((x _{11} x _{14}-x _{10} x _{15}) (x_1 x_4-x_0 x_5)-(x _{11} x _{13}-x_9 x _{15}) (x_2 x_4-x_0 x_6)-(x _{10} x _{13}-x_9 x _{14}) (x_3 x_4-x_0 x_7)-(x_7 x _{14}-x_6 x _{15}) (x_1 x_8-x_0 x_9)+(x_7 x _{13}-x_5 x _{15}) (x_2 x_8-x_0 x _{10})+(x_6 x _{13}-x_5 x _{14}) (x_3 x_8-x_0 x _{11})+(x_3 x _{14}-x_2 x _{15}) (x_5 x_8-x_4 x_9)-(x_3 x _{13}-x_1 x _{15}) (x_6 x_8-x_4 x _{10})-(x_2 x _{13}-x_1 x _{14}) (x_7 x_8-x_4 x _{11})).$$ I hope that no typo happened.

• Have you checked the case $m = 4$, $n = 4$ and $d = 1$? I'm a bit skeptical as to how the product of an entry of a $4\times 4$-matrix with the corresponding cofactor should be representable as a sum of products of $2\times 2$-minors. – darij grinberg Aug 20 '19 at 15:06
• Still weird. How do you write $x \det\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$ as a sum of products of $2\times 2$-minors of $\begin{pmatrix} u & a & b & c \\ v & d & e & f \\ w & g & h & i \\ x & y & z & w \end{pmatrix}$? – darij grinberg Aug 20 '19 at 15:43
• (Sorry, I meant $d=2$, not $d=1$ in my first comment.) – darij grinberg Aug 20 '19 at 15:44
• I have included your desired representation to the question. – Hans Aug 20 '19 at 16:14
• Oh! I didn't realize you can get mixed terms like $x_{11} x_{14}$ both from a single $2\times 2$-minor and from a product of two such. – darij grinberg Aug 20 '19 at 16:15

(Notation: Here $$m,n$$ are fixed positive integers and $$X = \{x_{i,j}\}$$ is an $$m \times n$$ matrix of indeterminates. Given $$k$$-tuples $$(a_{1},\dotsc,a_{k}) \in \{1,\dotsc,m\}^{k}$$ and $$(b_{1},\dotsc,b_{k}) \in \{1,\dotsc,n\}^{k}$$, they denote "$$[a_{1},\dotsc,a_{k}|b_{1},\dotsc,b_{k}]$$" the determinant of the $$k \times k$$ matrix whose $$(i,j)$$th entry is $$x_{a_{i},b_{j}}$$.)
(10.10) Lemma. Let $$\mathrm{F}(i,j)$$ be the $$\mathbb{Z}$$-submodule of $$\mathbb{Z}[\{x_{i,j}\}]$$ generated by the products $$\delta_{1}\delta_{2}$$ of the $$i$$-minors $$\delta_{1}$$ and the $$j$$-minors $$\delta_{2}$$. Then for $$u \le v-2$$ and $$\pi = [a_{1},\dotsc,a_{u}|b_{1},\dotsc,b_{u}]$$ and $$\rho = [c_{1},\dotsc,c_{v}|d_{1},\dotsc,d_{v}]$$, and $$\tilde{u} = \max(|\{a_{1},\dotsc,a_{u}\} \cap \{c_{1},\dotsc,c_{v}\}|,|\{b_{1},\dotsc,b_{u}\} \cap \{d_{1},\dotsc,d_{v}\}|)$$ one has $$(u+1-\tilde{u})!\pi\rho \in \mathrm{F}(u+1,v-1) \;.$$ (We include the case $$u=0$$, in which $$\pi=1$$ by convention.)