TITLE: Are the determinantal ideals prime?
QUESTION [12 upvotes]: 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.

REPLY [2 votes]: 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.