TITLE: Question on an isomorphism in the proof that $k[V \times W] \cong k[V] \otimes_k k[W]$
QUESTION [7 upvotes]: First I should say that I am aware of the existence of this question here and this question here. My question is a little different from these two because I am asking about a certain detail in the proof and besides $V$ and $W$ are now just algebraic sets and not affine varieties.
Now $V \times W$ is naturally an algebraic set as follows. If $V$ is the zero locus of some $f_1,\ldots f_k\in k[\Bbb{A}^n]$ and $W$ of $g_1,\ldots,g_l$ in $k[\Bbb{A}^m]$ I believe that $V \times W$ is now the zero locus of $f_1,\ldots,f_k,g_1,\ldots,g_l$ now considered as polynomials in $k[\Bbb{A}^n \times \Bbb{A}^m]$. For notational purposes I will now define
$$\begin{eqnarray*} R &\stackrel{\text{def}}{\equiv}& k[x_1,\ldots,x_n]\\
S &\stackrel{\text{def}}{\equiv}& k[x_{n+1},\ldots,x_{m+n}]\\
T&\stackrel{\text{def}}{\equiv}& k[x_1,\ldots,x_{m+n}].\end{eqnarray*}$$
Let $\mathcal{I}(V)$ denote the ideal of functions that vanish on $V$ in $R$, $\mathcal{I}(W)$ an ideal of $S$ similarly defined. Then I have shown that as $k$ - algebras, we have $$\frac{T}{\mathcal{I}(V)T + \mathcal{I}(W)T} \cong \frac{R}{\mathcal{I}(V)} \otimes_k \frac{S}{\mathcal{I}(W)} $$
subject to the validity of: 


My question is: Is the isomorphism 
    $$\frac{T}{\mathcal{I}(V)T} \cong T \otimes_k \frac{R}{\mathcal{I}(V)}$$ of $k$ - algebras valid? I know of several related results concerning isomorphisms of polynomial algebras but they don't seem to apply to the result that I want above. 



Here's a proof that we have an isomorphism of $k$ - algebras as claimed if what I ask in my question is true. We have
$$\begin{eqnarray*} \frac{T}{\mathcal{I}(V)T + \mathcal{I}(W)T} &\cong& \frac{T}{\mathcal{I}(V)T} \otimes_T \frac{T}{\mathcal{I}(W)T} \\
&\cong& \left(T \otimes_k \frac{R}{\mathcal{I}(V)}\right) \otimes_T\frac{T}{\mathcal{I}(W)T}  \\
&\cong& \frac{R}{\mathcal{I}(V)} \otimes_k \frac{T}{\mathcal{I}(W)T}  \\
&\cong& \frac{R}{\mathcal{I}(V)} \otimes_k \left( T \otimes_k \frac{S}{\mathcal{I}(W)}  \right) \\
&\cong& \frac{R}{\mathcal{I}(V)} \otimes_k \frac{S}{\mathcal{I}(W)} \end{eqnarray*}$$
as $k$ - algebras.

REPLY [3 votes]: Take $R=k[X]$, $I=Xk[X]$ and $T=k[X,Y]$. Then
$$\frac{T}{I(V)T}=k[Y],$$
$$T \otimes_k \frac{R}{I} =k[X,Y],$$
hence your claim is not true.
For your problem, note that $T=R \otimes S$, and show that the maps
$$\frac{R \otimes S}{I \otimes S + R \otimes J} \rightarrow \frac{R}{I} \otimes \frac{S}{J}, x \otimes y \mod (I \otimes S + R \otimes J) \mapsto (x \mod I) \otimes (y \mod J), $$
$$\frac{R}{I} \otimes \frac{S}{J} \rightarrow \frac{R \otimes S}{I \otimes S + R \otimes J} , (x \mod I) \otimes (y \mod J) \mapsto x \otimes y \mod (I \otimes S + R \otimes J)$$
are well defined and inverse from each other.