3
$\begingroup$

I have an $m \times 2m$ matrix of linear forms over $\mathbb{C}[x,y,z,w]$. It is of the form $$M = ( x I - A z -B w \mid y I - C z - D w).$$ Here $A,B,C$ and $D$ are $m \times m$ scalar matrices. Let $I$ be the ideal of maximal minors of $M$. I believe that the CM-regularity of this ideal is $m$. However I have no idea how to show it.

This is not a generic determinantal ideal, so many results for them do not apply in this case. However maybe there is a way to specialize and still keep the CM-regularity.

In general I want to show that for matrices of different size the closed subschemes of $\mathbb{P}^3$ defined by the ideals are non-isomorphic. Under certain assumptions the subscheme will be one dimensional.

I'm not very familiar with results about the CM-regularity (outside Eisenbud's "Commutative Algebra"), so a reference of something will be appreciated as well.

Thanks.

$\endgroup$

3 Answers 3

2
$\begingroup$

Let me first rephrase the question. You consider a morphism of vector bundles $$ O(-1)^{2m} \to O^{m} $$ on $P^3$ and ask about its degeneration scheme. The standard way to describe it as follows. Consider the product $P^3 \times P^{m-1}$ (with the second factor being the space of 1-dimensional quotients of $\mathbb{C}^m$. Then on the product consider the composition $$ O(-1,0)^{2m} \to O^{m} \to O(0,1), $$ where the second morphism is tautological. Let $Z$ be its zero locus. Then the degeneration scheme is the image of $Z$. In particular, if $m > 2$ then with generic choice of matrices $Z$ is empty (and hence so is the degeneration scheme). So, from now on I assume $m \le 2$ and the matrices are general.

Under these assumptions the scheme $Z$ has a Koszul resolution $$ \dots \to O(-2,-2)^{\binom{2m}{2}} \to O(-1,-1)^m \to O \to O_Z \to 0. $$ Pushing it forward to $P^3$ and using the cohomology of line bundles on $P^{m-1}$, one gets the following resolution for the degeneration scheme $D \subset P^3$: $$ \dots \to O(-1-m)^{\binom{2m}{m+1}} \to O(-m)^{\binom{2m}{m}} \to O \to O_D \to 0. $$ So, the CM-regularity of the ideal is $I$.

Of course, the case $m \le 2$ is not very interesting, but if you replace $P^3$ with a projective space of higher dimension, and then the same argument can be used for higher $m$ as well.

$\endgroup$
2
$\begingroup$

A minimal free resolution for an ideal generated by the maximal minors of a matrix whose entries are linear forms is given by the Eagon-Northcott complex, provided that the linear forms are generic enough. A reference for that fact is Exercise A2.19, b) in Eisenbud's book.

Note that the condition on the genericity of the linear forms is necessary, but it can written down explicitly (see e.g. the notion of 1-genericity in Eisenbud's book, Exercise A2.18).

Once you have proved that the Eagon-Northcott complex is a free resolution, then you can directly read the CM-regularity on the Eagon-Northcott complex, and this gives you indeed $m$.

$\endgroup$
0
$\begingroup$

One possible explanation for what you are claiming is that the ideal of m-minors is equal to the m-th power of the maximal ideal (x,y,z,w). Did you check if this is the case?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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