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 $A(X) \otimes_{k} A(Y)$?
 A: 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.
