# A problem related to parametrizing $\operatorname{rank}\le r$ matrices and Segre embedding

Given a field $k$. We denote $A_{mn}=k[\{X_{ij}\}_{1\le i\le m,1\le j\le n}]$ a polynomial ring of $mn$ variables. Given $m,n,r>0$, we have a natural homomorphism $\phi\colon A_{mn}\to A_{mr}\otimes_kA_{rn}$ induced by matrix multiplication: $Z_{ij}\mapsto\sum_{l=1}^rX_{il}Y_{lj}$. Let $I\subseteq A=A_{mn}$ be the ideal generated by all $(r+1)$-minors of the matrix $(X_{ij})_{1\le i\le m,1\le j\le n}$.

1. Is it true that $I$ is a prime ideal in $A$?
2. Is it true that $I=\ker\phi$?

When $r=1$, it's just something about Segre embedding and two propositions are all true. A proof could be found here. The essential point is that, $\phi$ induces a surjective homomorphism $\overline\phi\colon A/I\to B$ where $B$ is a subring of $A_{m1}\otimes_kA_{1n}$ which is generated by $a\otimes b$ where $\deg a=\deg b$, and $\overline\phi$ is in fact an isomorphism and we can easily construct an inverse of $\overline\phi$.

Any help is welcome. Thanks!

(If it's not considered a research level problem, moderators can tacitly migrate this post to MSE.)

• The answer to both questions is "yes". You might Google "generic determinantal variety" or "generic determinantal ideal". This is a very classic subject. One more recent article is the article by Israel Vainsencher where he writes down a log resolution of the generic determinantal ideal and uses his resolution to compute (what are now called) the discrepancies of this ideal. – Jason Starr Dec 10 '15 at 21:50