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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 $\mathbb C[x_1,..,x_m,y_1,...,y_n]$?)2012-05-01
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    @ThomasAndrews Thanks for the suggestion. I've tried to cut out the irrelevant details.2012-05-02
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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?2012-05-02

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