2
$\begingroup$

Consider a $n\times n$ matrix $M$ with entries in $R=\mathbb{C}[x_1,\dots,x_n]$. Let $I$ be the ideal of $(n-1)\times(n-1)$ minors of $M$. Is $\mathcal{O}_{\mathbb{C}^n}/I$ Cohen-Macaulay?If not, what additional assumptions we need for an affirmative answer?

$\endgroup$
4
  • $\begingroup$ I think in your case for $R/I$ to be CM it is necessary that proj. dim $R/I\leq4$. $\endgroup$ Mar 18 '12 at 16:37
  • $\begingroup$ the case that I'm more interested in is when I is of codimension 2. $\endgroup$ Mar 19 '12 at 15:23
  • $\begingroup$ @Mahdi: how do you get $4$? $\endgroup$ Mar 26 '12 at 5:37
  • $\begingroup$ A sufficient condition is that the quotient ring has the expected dimension. See mathoverflow.net/questions/44823/… $\endgroup$
    – jlk
    Mar 28 '12 at 17:45
2
$\begingroup$

In general, the answer is no: consider the ring $k[x_1, x_2] / (x_1^2, x_1 x_2)$ which is not Cohen-Macaulay. However, if you require that the matrix has homogeneous entries and a condition on the height of the ideal $I$, you will have that $k[x_1, \dots, x_n] / I$ is Cohen-Macaulay. You could read the first pages of Determinantal Ideals by R. Miro-Roig.

$\endgroup$
1
  • 1
    $\begingroup$ What is the condition on the height of $I$? $\endgroup$ Mar 18 '12 at 13:29
1
$\begingroup$

If you are dealing with codim two CM ideals then you can use Hilbert-Burch theorem. Those CM ideals of ht two are generated by the maximal minors of an n by n-1 matrix where n is the minimal number of generators for the ideals. Eisenbud's comm alg book should have a proof as well as the statement which is in a bit more general setting.

$\endgroup$
1
  • 1
    $\begingroup$ In Mike Artin's book, some part of the Hilbert-Burch theorem is also attributed to Schaps. $\endgroup$ Mar 26 '12 at 0:46

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.