Are the determinantal ideals prime?

I want to prove the determinantal ideals over a field are prime ideals. To be concrete:

For simplicity, let $I=(x_{11}x_{22}-x_{12}x_{21},x_{11}x_{23}-x_{13}x_{21},x_{12}x_{23}-x_{13}x_{22})$ be an ideal of the polynomial ring $k[x_{11},\ldots,x_{23}]$. I have no idea how to prove that $I$ is a radical ideal (i.e. $I=\sqrt{I}$). Could anyone give some hints?

Generally, let $K$ be an algebraically closed field, then $\{A\mid\mathrm{Rank}(A)\leq r\}\subseteq K^{m\times n}$ is an irreducible algebraic set (I first saw this result from this question). And I tried to prove this by myself, then I have proved it (when I see the "Segre embedding").

But I have no idea how to show that the "determinantal ideals" are radical ideals (I hope this is true). BTW, is the statement that the determinantal ideals over a field are prime ideals true ?

Thanks.

Solution 1:

There are several ways to prove that $I$ is radical. By the way, the statement that $I$ is prime is equivalent to $I$ being radical and the zero set of $I$ being an irreducible algebraic set.

An approach using Gröbner bases can be found in Chapter 16 of Miller-Sturmfels, Combinatorial Commutative Algebra

An approach using sheaf cohomology can be found in Sections 6.1-6.2 of Weyman, Cohomology of Vector Bundles and Syzygies. This requires a lot more background knowledge.

There is also the approach using induction on the size of the matrix and localization arguments in Chapter 2 of Bruns-Vetter, Determinantal Rings. Link to book: http://www.home.uni-osnabrueck.de/wbruns/brunsw/detrings.pdf

Solution 2:

This isn't really an answer but I can point you to a reference that might be of some help:

For a discussion of this example for the ideal generated by 2x2 minors of a 3x3 matrix, see this nice discussion in Eisenbud's Commutative Algebra, p 107.

It seems that the general case is much more difficult. Eisenbud also mentions that Bruns and Vetter, Determinantal Rings [1988] is a nice reference for the general case.

I hope someone else can come along to tell you something more useful!

Solution 3:

You can use the technique in section 18.3 of Eisenbud's commutative algebra to determine the reducedness of an ideal when it is Cohen-Macaulay. This approach even works when your ambient ring is only a quotient of a polynomial ring and your ideal is determinantal of maximal depth.

The philosophy is the following: Cohen-Macaulay implies no embedded components. No embedded components plus generically reduced imply reduced. To show generically reduced, you can show nonsingular in positive codimension, thus it boils down to showing that the Jacobian drops rank in positive codimension.

Here are more details:

Let $M$ denote the generic $(m\times n)$-matrix with $(x_{ij})$ in the polynomial ring $R = k[x_{ij}]$ for $m>n$. Let $I_r(M)$ denote the ideal of $(r\times r)$-minors of $M$. It is a well-known result of Eagon-Northcott that $I_n(M)$ is of maximal depth $m-n+1$. It follows from Buchsbaum-Rim's results that the $r$-th Buchsbaum-Rim complex $\mathcal{K}^r_\bullet$ resolves the factor ring $R/I_r(M)$, which implies that $R/I_r(M)$ is Cohen-Macaulay. Let $c = r(m-n+r)$ be the codimension of $I_r(M)$ and let $J$ be the $(c\times c)$-minor of the Jacobian, taken modulo $I_r(M)$. You need to check that $J$ is codimension $\ge 1$ in $R/I_r(M)$. Because determinants satisfy the minor expansion rule, you will see quickly that $J$ is supported on the $I_{r-1}(M)$ in the quotient $R/I_r(M)$, which corresponds to the geometric fact that $S_r$ is non-singular away from $S_{r-1}$.

The above gives reducedness of $R/I_r(M)$. Irreducibility follows from geometry.