Product of Algebraic Sets and its Coordinate Ring (15.1.28)

Prove that if $V$ and $W$ are algebraic sets, then so is $V×W$ and $k[V×W]≅k[V] ⊗_k k[W]$.

Proof: If $V=\mathcal{Z}(I')$ then also $V=\mathcal{Z}(I)$ where $I=\mathcal{I}(\mathcal{Z}(I'))$, with the additional helpful property that $\mathcal{I}(V)=I$. We shall assume the same for $W=\mathcal{Z}(J)$. Lemma: Let $V⊆\mathbb{A}^n$, $W⊆\mathbb{A}^m$ be algebraic sets. Then $\mathcal{I}(V×W)=\mathcal{I}(V)+\mathcal{I}(W)$. Proof: Clearly $⊇$ holds. Now assume $f∈\mathcal{I}(V×W)$ yet $f∉\mathcal{I}(V)+\mathcal{I}(W)$. (Following proof of lemma borrowed from MSE) We can write any polynomial in $k[x_1,...,x_n,y_1,...,y_m]/(\mathcal{I}(V)+\mathcal{I}(W))$ as $$\sum_{i=1}^r \overline{f_ig_i}$$ where $f_i∈k[x_1,...,x_n]$ and $g_i∈k[y_1,...,y_m]$ for all $i$. Let $f$ as above be minimal such that one of its reduction sums has the least summand terms $r$ of the form above of all the polynomials of its hypothesis. Now, assume $f_i(a)=0$ for all $a∈V$ and all $i$; then since $\mathcal{I}(V)=I$ we have $f_i∈I$, hence $f∈\mathcal{I}(V)+(\mathcal{I}(V)+\mathcal{I}(W))=\mathcal{I}(V)+\mathcal{I}(W)$, a contradiction. Therefore $f_k(a)≠0$ for some $k$ and $a∈V$, where we may reorder $f_1=f_k$. Now by the hypothesis for $f$, for all $b∈W$ we have $f(a)(b)=(\sum f_i(a)g_i(b))+t(a)(b)=\sum f_i(a)g_i(b)=0$ (for some negligible $t∈\mathcal{I}(V)+\mathcal{I}(W)$) hence $\sum f_i(a)g_i=p∈\mathcal{I}(W)$. Since $f_1(a)≠0$ write $g_1=\dfrac{p-f_2(a)g_2-...-f_r(a)g_r}{f_1(a)}$ to write $\overline{f}$ in strictly fewer summands, a contradiction.

Proceed with the exercise. To see that $V×W$ is an algebraic set, observe $V×W=\mathcal{Z}(\mathcal{I}(V)+\mathcal{I}(W))$, with $⊆$ being evident and $⊇$ resulting from $V=\mathcal{Z}(\mathcal{I}(V))$ and $W=\mathcal{Z}(\mathcal{I}(W))$. Now, define a $k$-bilinear map $k[V]×k[W]→k[V×W]$ by $(f,g)↦fg$ (well-defined since $\mathcal{I}(V)+\mathcal{I}(W)⊆\mathcal{I}(V×W)$) to induce a surjective group homomorphism $Φ~:~k[V] ⊗_k k[W]→k[V×W]$. This is also a ring homomorphism. To prove bijection, it suffices to check that the ring homomorphism $Φ'~:~k[x_1,...x_n,y_1,...,y_m]→k[V] ⊗_k k[W]$ given by $Φ'(x_i)=x_i ⊗ 1$ and $Φ'(y_i)=1 ⊗ y_i$ factors through $\mathcal{I}(V)+\mathcal{I}(W)$, as then it would be a two-sided inverse to $Φ$ when induced on $k[V×W]$. Indeed, when $f∈\mathcal{I}(V)$ or $g∈\mathcal{I}(W)$ is a generator of $\mathcal{I}(V×W)=\mathcal{I}(V)+\mathcal{I}(W)$, we see $Φ'(f)=f⊗1=0⊗1=0$ and $Φ'(g)=1⊗g=0$.$~\square$