Consider a $n\times n$ matrix $M$ with entries in $R=\mathbb{C}[x_1,\dots,x_n]$. Let $I$ be the ideal of $(n1)\times(n1)$ minors of $M$. Is $\mathcal{O}_{\mathbb{C}^n}/I$ CohenMacaulay?If not, what additional assumptions we need for an affirmative answer?

$\begingroup$ I think in your case for $R/I$ to be CM it is necessary that proj. dim $R/I\leq4$. $\endgroup$ – Mahdi MajidiZolbanin Mar 18 '12 at 16:37

$\begingroup$ the case that I'm more interested in is when I is of codimension 2. $\endgroup$ – Michele Torielli Mar 19 '12 at 15:23

$\begingroup$ @Mahdi: how do you get $4$? $\endgroup$ – Sándor Kovács 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
In general, the answer is no: consider the ring $k[x_1, x_2] / (x_1^2, x_1 x_2)$ which is not CohenMacaulay. 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 CohenMacaulay. You could read the first pages of Determinantal Ideals by R. MiroRoig.

1$\begingroup$ What is the condition on the height of $I$? $\endgroup$ – Michele Torielli Mar 18 '12 at 13:29
If you are dealing with codim two CM ideals then you can use HilbertBurch theorem. Those CM ideals of ht two are generated by the maximal minors of an n by n1 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.

1$\begingroup$ In Mike Artin's book, some part of the HilbertBurch theorem is also attributed to Schaps. $\endgroup$ – Jason Starr Mar 26 '12 at 0:46