# Tensor in coordinate rings

Let $R$ be a commutative ring with $1$ , $I,J$ ideals of $R$ then:

$R/I \otimes_{R} R/J \cong R/(I+J)$.

Let $X$,$Y$ be affine varieties. I want to use this isomorphism to show that $A(X \times Y) \cong A(X) \otimes_{k} A(Y)$.

However using the result I end up with $A(X \times Y) \cong A(X) \otimes_{k[x,y]} A(Y)$? How to get $A(X) \otimes_{k} A(Y)$?

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(1) Do you know what $I(X \times Y)$ is? I don't think that it's just the ideal generated by $I(X)$ and $I(Y)$. (2) I know how to prove this in other ways; do you want to hear about those? – Dylan Moreland Feb 18 '12 at 4:05
Yea, this can be done using the universal property of the fiber product and of the tensor product. The commutative algebra result you cite won't give you what you want. – Parsa Feb 18 '12 at 4:12

In fact, a similar result can be used to prove $A(X\times Y)=A(X)\otimes_k A(Y)$. I assume here that the product $X\times Y$ of affine varieties is defined via the Segre embedding.
Let $X\subseteq\mathbb{A}^n$ and $Y\subseteq\mathbb{A}^m$ and let $A(\mathbb{A}^n)=k[\underline{x}]:=k[x_1,\ldots,x_n]$ as well as $A(\mathbb{A}^m)=k[\underline{z}]$ be the respective coordinate rings. Let $I=I(X)$ and $J=I(Y)$. We then define $I':=I\cdot k[\underline{x},\underline{z}]$ and $J':=J\cdot k[\underline{x},\underline{z}]$, the images of $I$ and $J$ in the coordinate ring of $\mathbb{A}^{n+m}$. This means $X\times\mathbb{A}^m=Z(I')$ and $\mathbb{A}^n\times Y=Z(J')$, so $X\times Y=(X\times\mathbb{A}^m)\cap(\mathbb{A}^n\times Y)$ is a closed subset of $\mathbb{A}^{n+m}$ with coordinate ring $A(X\times Y)=k[\underline{x},\underline{z}]/\sqrt{I'+J'}$. Now you have to show that this defines the tensor product $k[\underline{x}]/I \otimes_k k[\underline{z}]/J$, which is elementary and probably works very similar to the result you cited.
Why don't you have to take the radical of $I' + J'$? – Dylan Moreland Feb 18 '12 at 15:18
The whole point is that $I'+J'$ is already a radical ideal! This is the main part of the claim, and this is non-trivial. The same question has been asked a couple of times on SE, and answered. – Martin Brandenburg Sep 10 '13 at 20:28