Let $X\subset\mathbb{A}^{n},Y\subset\mathbb{A}^{m}$ be affine varieties over $k$ algebraically closed. Then, $X\times Y\subset \mathbb{A}^{m+n}$ may be checked to be an affine variety as well; furthermore, it's just the product in the category of varieties. Then, because the functor $A(-)$ is an arrow-reversing equivalence of categories, it follows that $A(X\times Y)=A(X)\otimes_{k} A(Y)$, as tensor product is the coproduct in the category of finitely generated $k$-algebras which are domains.
Unless I have made a mistake, one can check directly from the universal property of tensor product that we also have \begin{equation*} k[x_{1},\ldots,x_{n}]/I(X)\otimes_{k}k[y_{1},\ldots,y_{m}]/I(Y)=k[x_{1},\ldots,x_{m},y_{1},\ldots,y_{m}]/(I(X)+I(Y)), \end{equation*}
and of course $I(X)+I(Y)\subset I(X\times Y)$, so it should be true that we in fact have equality $I(X)+I(Y)=I(X\times Y)$. Is there a direct way to see this? In particular, it seems like it should be true for arbitrary Zariski-closed $X,Y$.
(Source: Hartshorne, Exercise I.3.15)
