TITLE: The vanishing ideal $I_{K[x,y]}(A\!\times\!B)$ is generated by $I_{K[x]}(A) \cup I_{K[y]}(B)$?
QUESTION [9 upvotes]: Let $K$ be a field, $x=(x_1,\ldots,x_m)$, $y=(y_1,\ldots,y_n)$, $A\!\subseteq\!\mathbb{A}^m_K$, $B\!\subseteq\!\mathbb{A}^n_K$. Does there hold $$I_{K[x,y]}(A\!\times\!B)=\langle\langle I_{K[x]}(A) \cup I_{K[y]}(B)\rangle\rangle?$$
Here $I_{K[x]}(A)=\{f\in K[x]; f(a)\!=\!0\text{ for all }a\!\in\!A\}$ is the vanishing ideal of set $A$, and $\langle\langle\ldots\rangle\rangle$ is the ideal, generated by $\ldots$. The inclusion $\supseteq$ is easy, but I don't see how to show $\subseteq$. 
If this equality does not hold, how else then can I prove (knowing $$K[x,y]/\langle\langle\mathfrak{a},\mathfrak{b}\rangle\rangle \,\cong\, K[x]/\mathfrak{a}\otimes_KK[y]/\mathfrak{b}$$ from this post), that there is an isomorphism of $K$-algebras (coordinate rings) $$K[A\!\times\!B] \,\cong\, K[A]\otimes_K K[B]?$$

REPLY [3 votes]: At the request of Martin Brandenburg, here is the result which may help you:


Let $k$ be a field and $V\subseteq \Bbb{A}_k^n$, $W\subseteq \Bbb{A}_k^m$  affine algebraic sets ( the proof I believe will also work if $V,W$ are  affine varieties). Then $$I(V \times W )= I(V) + I(W).$$ By $I(V)$ we mean now the extension of the ideal $I(V)$ in the polynomial ring $k[x_1,\ldots,x_{m+n}]$.


Proof: See the discussion in my question here.
Using this result we can now prove the following theorem. 


Theorem (Coordinate ring of a product): Let $V,W$ be as before. Then as $k$ - algebras we have $$k[V \times W] \cong k[V] \otimes_k k[W].$$


Proof: Let us write $I = \mathcal{I}(V)$ and $J = \mathcal{I}(W)$ and define $R = k[x_1,\ldots,x_n]$ and $S = k[x_{n+1},\ldots,x_{m+n}]$. Then by the lemma above, we have that $\mathcal{I}(V \times W) = I^e + J^e$ where the superscript denotes ideal extension in the ring $T = R \otimes_k S$, the polynomial algebra in $m+n$ variables. Now by the usual extension of scalars process we have $$I^e + J^e = I \otimes_k S +R \otimes_k J.$$
Thus to prove the theorem we need to prove that
$$\frac{T}{I \otimes_k S +R \otimes_k J} \cong \frac{R}{I} \otimes_k \frac{S}{J}.$$
The only tricky part in the proof is to construct a well-defined $k$ - algebra homomorphism 
$$f : \frac{R}{I} \times \frac{S}{J} \to \frac{ T}{I \otimes_k S + R \otimes_k J}$$
We can define one by declaring that $f$ sends $(\bar{a},\bar{b}) $ to $a\otimes b \mod{(I \otimes_k S + R \otimes_k J)}$. Is this well defined? Suppose $(\bar{a},\bar{b}) = (\bar{c},\bar{d})$. Then notice that
$$\begin{eqnarray*} a \otimes b - c\otimes d &=& a \otimes b + c\otimes b - c\otimes b - c\otimes d \\
&=& (a-c) \otimes b + c \otimes (b-d)\\
&=& 0 \mod{(I \otimes_k S+ R \otimes_k J)}\end{eqnarray*}$$
because $a -c \in I$ and $b-d \in J$. Thus we see that $f$ is well-defined and leave the rest of the details for your to fill in, including constructing an inverse map. If you have any questions, I can add more details.