Generators for (radical) ideal of product of affine varieties

Question

If $X\subset \mathbb{A}^N$ and $Y\subset\mathbb{A}^M$ are affine varieties with $X=Z(f_1,\dots,f_n)$ and $Y=Z(g_1,\dots,g_m)$ then $X\times Y\subset\mathbb{A}^{N+M}$ is an affine variety with $X\times Y=Z(f_1,\dots,f_n,g_1,\dots,g_m)$. But if we take $f_1,\dots,f_n$ and $g_1,\dots,g_m$ to generate the ideal of $X$ and $Y$ respectively then must $f_1,\dots,f_n,g_1,\dots,g_n$ generate the ideal of $X\times Y$?

This is equivalent to asking if given $f_1,\dots,f_n\in\mathbb{k}[x_1,\dots,x_n]$ and $g_1,\dots,g_n\in\mathbb{k}[y_1,\dots,y_m]$ generating radical ideals whether $f_1,\dots,f_n,g_1,\dots,g_n\in\mathbb{k}[x_1,\dots,x_n,y_1,\dots,y_m]$ generate a radical ideal.

This is true, for example, if the ideals are monomial ideals. But in general I don't really have good intuition for this and the main reason I am asking is because it would make working with product varieties a lot easier.

Answer 1

If $k$ is algebraically closed, then for two $k$-algebras $A$ and $B$ which are also integral domains, we will have $A\otimes_kB$ is an integral domain. (Reference.)

Let $X=\{x_1,\ldots,x_n\},Y=\{y_1,\ldots,y_m\}$. Now suppose $I,J$ are radical ideals in $k[X]$ and $k[Y]$ respectively, and suppose $k$ is algebraically closed. We will prove that $(I,J)$ is a radical ideal in $k[X,Y]$.

There are finitely many prime ideals $\mathfrak{p}_i\subseteq k[X]$, $\mathfrak{q}_j\subseteq k[Y]$ respectively such that $I=\cap_i \mathfrak{p}_i$ and $J=\cap_j \mathfrak{q}_j$. Thus we have injections $k[X]/I\to \prod_ik[X]/\mathfrak{p}_{i}$ and $k[Y]/J\to \prod_jk[Y]/\mathfrak{q}_j$. Therefore, we have an injection $k[X]/I\otimes_k k[Y]/J\to (\prod_ik[X]/\mathfrak{p}_i)\otimes (\prod_jk[Y]/\mathfrak{q}_j)\cong \prod_{i,j}k[X,Y]/(\mathfrak{p}_i,\mathfrak{q}_j)$. Since $\prod_{i,j}k[X,Y]/(\mathfrak{p}_i,\mathfrak{q}_j)$ is reduced, $k[X,Y]/(I,J)=k[X]/I\otimes k[Y]/J$ is also reduced.

Edit For the injectivity of $k[X]/I\otimes_k k[Y]/J\to (\prod_ik[X]/\mathfrak{p}_i)\otimes (\prod_jk[Y]/\mathfrak{q}_j)$:

(a) $k[X]/I\to \prod_ik[X]/\mathfrak{p}_i$ is injective, tensoring $k[Y]/J$ gives that $k[X]/I\otimes_kk[Y]/J\to (\prod_ik[X]/\mathfrak{p}_i)\otimes_kk[Y]/J$ is injective.

(b)$k[Y]/J\to \prod_jk[Y]/\mathfrak{q}_j$ is injective, tensoring $\prod_ik[X]/\mathfrak{p}_i$ gives $(\prod_ik[X]/\mathfrak{p}_i)\otimes_kk[Y]/J\to (\prod_ik[X]/\mathfrak{p}_i)\otimes (\prod_jk[Y]/\mathfrak{q}_j)$ is injective.

Then the composition of two injections is injective.

Answer 2

A simpler answer:

Since $I=(f_1,\dots, f_n)$ and $J=(g_1,\dots,g_m)$ are radical ideals, then $k[X]/I$ and $k[Y]/J$ are reduced $k$-algebras. If we assume that $k$ is a perfect field (in particular, algebraically closed) we get that $k[X]/I\otimes_kk[Y]/J=k[X,Y]/(I,J)$ is also reduced (see the last part of this answer for perfect fields or Milne's book Algebraic Geometry, Proposition 4.15(a) for algebraically closed fields), and thus in this case the answer to the question is positive.