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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.

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  • $\begingroup$ 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. $\endgroup$ – darij grinberg Aug 20 '19 at 15:06
  • $\begingroup$ 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 $\begin{pmatrix} u & a & b & c \\ v & d & e & f \\ w & g & h & i \\ x & y & z & w \end{pmatrix}$? $\endgroup$ – darij grinberg Aug 20 '19 at 15:43
  • $\begingroup$ (Sorry, I meant $d=2$, not $d=1$ in my first comment.) $\endgroup$ – darij grinberg Aug 20 '19 at 15:44
  • $\begingroup$ I have included your desired representation to the question. $\endgroup$ – Hans Aug 20 '19 at 16:14
  • $\begingroup$ 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. $\endgroup$ – darij grinberg Aug 20 '19 at 16:15
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I think this follows from Lemma (10.10) of Bruns, Vetter, "Determinantal Rings" which I reproduce here, edited slightly.

(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.)

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