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.

  • $\begingroup$ 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

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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.