Is this quotient ring $\mathbb{C}[z_{ij}]/\ker\phi$ integrally closed? A few days ago, I asked a linear algebra question, but it seems that the notions are better stated in terms of algebraic geometry. I don't have much solid knowledge of algebraic geometry, so I'm wondering if there is a basic explanation for the following.
Suppose you have homomorphism given by
$$
\phi\colon\mathbb{C}[z_{11},\dots,z_{mn}]\to\mathbb{C}[x_1,\dots,x_m,y_1,\dots,y_n]: z_{ij}\mapsto x_iy_j.
$$
Then is $\mathbb{C}[z_{11},\dots,z_{mn}]/\ker\phi$ integrally closed or not?
By integrally closed, I mean that $\mathbb{C}[z_{11},\dots,z_{mn}]/\ker\phi$ is equal to its integral closure (the set of elements of $k$ integral over $\mathbb{C}[z_{11},\dots,z_{mn}]/\ker\phi$) in its quotient field $k$.
 A: Well, since nobody's taken a stab at it, I do have a sketch of what may be a proof.
First, the image consists of all polynomials whose terms each have equal numbers of $x$'s and $y$'s. The kernel, I'm sure, is generated by the elements of the form $z_{ij}z_{k\ell} - z_{i\ell} z_{kj}$. The verification would be to show that, modulo this ideal, a product of $z$'s is completely determined simply by the unordered multi-set of first indices and second indices, which corresponds to the appropriate product of $x$'s and $y$'s.
Now switch to the geometric view. If the system of equations $z_{ij}z_{k\ell} - z_{i\ell} z_{kj} = 0$ doesn't have a singularity, then the image of $\phi$ is, I believe, a regular ring. In particular, this implies it's integrally closed.
We can check if it has a singularity by adding in more equations that say the first-order partial derivatives of the defining equations are zero. i.e. every equation
$$ \frac{\partial}{\partial z_{uv}} (z_{ij}z_{k\ell} - z_{i\ell} z_{kj}) = 0 $$
The resulting system of equations does have a solution, if $m,n \geq 2$: it is $z_{ij} = 0$ for all $i,j$.
Now, I believe the following statements are true, if $m,n \geq 2$.


*

*The system of equations $z_{ij} z_{k\ell} - z_{i\ell} z_{kj}$ define an $(m+n)$-dimensional variety.

*The singular set of this variety is the single point defined by $z_{ij}=0$, and thus is zero dimensional.

*Because the singular set has codimension $> 1$, this implies the image of $\varphi$ is integrally closed.


If $m=1$ or $n=1$, then $\phi$ is injective, and so its image is integrally closed because its domain is.
A: Let's reformulate this problem in terms of commutative algebra (its second tag): for an arbitrary field $K$ the ring $K[z_{11},\dots,z_{mn}]/\ker\phi$ is isomorphic to $\text{Im}\ \phi$ which is obviously the subring of $K[x_1,\dots,x_m,y_1,\dots,y_n]$ generated by all the monomials $x_iy_j$. 
Now think in terms of affine semigroup rings:
$$K[x_1,\dots,x_m,y_1,\dots,y_n]=K[\mathbb{N}^{m+n}]$$ 
and 
$$K[x_iy_j: 1\le i\le m, 1\le j\le n]=K[S],$$ 
where $S\subset\mathbb{N}^{m+n}$ is the subsemigroup generated by the elements $(e_i,f_j)$. (Here we consider $e_i=(0,\dots,1,\dots,0)\in\mathbb{N}^m$ with $1$ on the place $i$, and $f_j=(0,\dots,1,\dots,0)\in\mathbb{N}^n$ with $1$ on the place $j$.) At this moment I leave you the pleasure to prove that $$S=\{(a_1,\dots,a_m,b_1,\dots,b_n)\in\mathbb{N}^{m+n}:a_1+\cdots+a_m=b_1+\cdots+b_n\}.$$
Theorem 6.1.4 from Bruns and Herzog, Cohen-Macaulay Rings, provides a criterion for the normality of affine semigroup rings. It says that $K[S]$ is normal if and only if $S$ is a normal semigroup, that is, if $n\in\mathbb{N}$, $n>0$, and $x\in\mathbb{Z}S$ (the subgroup of $\mathbb{Z}^{m+n}$ generated by $S$), then $nx\in S$ implies $x\in S$. In our case it is pretty clear that $S$ is normal.
Remark. Even simpler, we can think of $K[x_iy_j: 1\le i\le m, 1\le j\le n]$ as being the Segre product of two polynomial rings and use the fact that the Segre product of two normal rings is normal (why?).
