Let \[ X\subseteq\AA^n_\CC, \qquad Y\subseteq\AA^m_\CC \] be affine varieties. Show that \[ X\cross Y\subseteq\AA^{n+m}_\CC \] is an affine variety and that \[ \mco(X\cross Y) \cong \mco(X)\tensor_\CC\mco(Y). \]
Solution
Write coordinates on \(\AA^{n+m}_\CC\) as \[ (x_1,\ldots,x_n,y_1,\ldots,y_m). \]
1. The subset \(X\cross Y\) is Zariski closed in \(\AA^{n+m}_\CC\).
Let \[ I(X)\subseteq\CC[x_1,\ldots,x_n], \qquad I(Y)\subseteq\CC[y_1,\ldots,y_m] \] be the vanishing ideals. In \[ S=\CC[x_1,\ldots,x_n,y_1,\ldots,y_m] \] consider \[ J=I(X)S+I(Y)S. \] A point \((x,y)\) lies in \(V(J)\) exactly when every polynomial in \(I(X)\) vanishes at \(x\) and every polynomial in \(I(Y)\) vanishes at \(y\). Hence \[ V(J)=X\cross Y. \] Thus \(X\cross Y\) is closed.
2. If \(Z\subseteq X\cross Y\) is closed, then\[ X_Z = \{x\in X:\{x\}\cross Y\subseteq Z\} \]is closed in \(X\).
Choose finitely many polynomials \[ F_1,\ldots,F_r\in S \] whose common zero set on \(X\cross Y\) is \(Z\). Fix one such polynomial \(F\). Its image in \[ \CC[x_1,\ldots,x_n]\tensor_\CC\mco(Y) \] can be written \[ \sum_{j=1}^N a_j(x)e_j \] with \[ e_1,\ldots,e_N\in\mco(Y) \] \(\CC\)-linearly independent.
For a fixed \(x\in X\), the polynomial \(F(x,-)\) vanishes on all of \(Y\) exactly when its class in \(\mco(Y)\) is zero, that is, \[ \sum_{j=1}^N a_j(x)e_j=0. \] Linear independence gives \[ a_1(x)=\cdots=a_N(x)=0. \] Thus the condition that \(F\) vanish on the whole fibre \(\{x\}\cross Y\) is Zariski closed in \(X\). Intersecting these closed conditions for \(F_1,\ldots,F_r\) gives \(X_Z\).
3. The closed subset \(X\cross Y\) is irreducible.
Suppose \[ X\cross Y=Z_1\union Z_2 \] with \(Z_1\) and \(Z_2\) closed. For \(i=1,2\), put \[ X_i = \{x\in X:\{x\}\cross Y\subseteq Z_i\}. \] By step 2, each \(X_i\) is closed in \(X\).
Fix \(x\in X\). The fibre \[ \{x\}\cross Y \] is isomorphic to the irreducible variety \(Y\), and it is the union of its closed intersections with \(Z_1\) and \(Z_2\). Hence the whole fibre is contained in one \(Z_i\). Therefore \[ X=X_1\union X_2. \] Since \(X\) is irreducible, either \(X=X_1\) or \(X=X_2\). In the first case every fibre lies in \(Z_1\), so \[ X\cross Y=Z_1; \] similarly in the second case \[ X\cross Y=Z_2. \] Thus \(X\cross Y\) is irreducible.
4. The natural homomorphism\[ \mco(X)\tensor_\CC\mco(Y) \longrightarrow \mco(X\cross Y) \]is an isomorphism.
The quotient map \[ S \longrightarrow \mco(X)\tensor_\CC\mco(Y) \] has kernel \[ I(X)S+I(Y)S. \] By step 1 this ideal is contained in \[ I(X\cross Y). \]
For the reverse inclusion, let \[ F\in I(X\cross Y). \] Write its image in the tensor product as \[ \overline F = \sum_{j=1}^N a_j\tensor b_j, \] where \[ b_1,\ldots,b_N\in\mco(Y) \] are \(\CC\)-linearly independent. For every \(x\in X\), vanishing of \(F\) on the fibre \(\{x\}\cross Y\) gives \[ \sum_{j=1}^N a_j(x)b_j=0 \] in \(\mco(Y)\). Linear independence implies \[ a_j(x)=0 \] for every \(j\) and every \(x\in X\). Thus each \(a_j\) is zero in \(\mco(X)\), so \[ \overline F=0. \] Hence \[ I(X\cross Y)=I(X)S+I(Y)S. \] Taking quotients gives \[ \boxed{ \mco(X\cross Y) \cong \mco(X)\tensor_\CC\mco(Y). } \]
5. The product \(X\cross Y\) is an affine variety.
Step 1 shows that it is Zariski closed in affine space, and step 3 shows that it is irreducible. An irreducible Zariski-closed subset of affine space is an affine variety.
6. Q.E.D.
Step 5 proves that \(X\cross Y\) is an affine variety, and step 4 computes its coordinate ring.