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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 $k$.

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It would probably be better if you just asked the algebraic parts without all the motivation. Are you asking: given the homomorphism $\phi:\mathbb C[z_{11},...,z_{mn}]\rightarrow C[x_1,...,x_m,y_1,...,y_n]$ defined by $z_{ij}\rightarrow x_iy_j$, then is $\mathbb C[z_{ij}]/ker \phi$ integrally closed? (Presumably, you mean as a sub-ring of $\mathbb C[x_1,..,x_m,y_1,...,y_n]$?) –  Thomas Andrews May 1 '12 at 20:00
    
@ThomasAndrews Thanks for the suggestion. I've tried to cut out the irrelevant details. –  hmIII May 2 '12 at 20:02
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You sure you didn't mean integrally closed in the absolute sense? i.e. integrally closed in its field of fractions? –  Hurkyl May 2 '12 at 20:22
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2 Answers 2

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.

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Here's a proof that the kernel is generated by the $z_{ij}z_{kl}-z_{il}z_{kj}$: We claim that for any products of z's, we can multiply them to get a product of $z_{i_aj_a}$ where $i_a$ and $j_a$ are both increasing. Note that swapping $z_{ij}z_{kl}$ with $z_{il}{kj}$ where $i \le k$ and $j \ge l$ increases the sum of $ij$ over all $z_{ij}$ in the product. This sum will always be an integer, and it's bounded above by (number of z's in the product) multiplied by mn, so we cannot keep on swapping indefinitely, and eventually we cannot make further swaps. This point must be something –  only May 4 '12 at 1:58
    
of the claimed form. Now note that all if the any two different products in that form give different polynomials with the substitution $z_{ij}$ = $x_iy_j,$ so if after the substitution we get 0, we must get 0 once we write all the products in this form. To see that two products of this form give different polynomials, note that the $x_i$ with i smallest must be paired up with the $y_j$ with j smallest here, so from the polynomial after substitution we can determine the values of $i_1$ and $j_1$, and then we can divide $z_{i_1j_1}$ out and repeat. –  only May 4 '12 at 2:00
    
@Hurkyl: having a small singular locus is not enough to be integrally closed (consider a smooth surface over $\mathbb C$ and identify two distinct points). This would suffice if the variety is, say, locally a complete interseciton (to insure the propery (S$_2$)). –  user18119 Nov 9 '12 at 11:25
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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?).

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