decomposition of the polynomial ring in $n$ indeterminates I am wondering if the following isomorphism of rings is true.
Let $R =k[x_1, \ldots, x_n]$ be the polynomial ring in $n$-indeterminates over a field $k$. Consider $V$ the linear subspace $k \cdot x_1 + \cdots + k \cdot x_n$
and suppose that we can write $V = W \oplus W'$ for some subspaces $W, W'$. 
Let $I$ and $J$ be the ideals of $R$ generated by $W$ and $W'$ respectively. Then 
$R \cong R/I \otimes R/J$
as rings.
I have tried to establish a general isomorphism 
$R/I\cap J \cong R/I \otimes R/J$ for any two ideals $I,J$ and then use in my particular case that $I \cap J = \{0\}$ provided that $W \oplus W' = V$ by assumption. But I think we need more conditions over the ideals (as in the Chinese reminder theorem) .
 A: Yes, this is true (assuming that your tensor product is over $k$).  Here's one way to see it.  Note that any linear map $T:V\to V$ extends uniquely to a $k$-algebra homomorphism $T_*:R\to R$ (consider the unique $k$-algebra homomorphism such that $T_*(x_i)=T(x_i)$ for each $i$).  In particular, if $T$ is invertible, then $T_*$ is an automorphism (with inverse $(T^{-1})_*$).
Now choose an automorphism $T$ of $V$ such that $T(W)$ is spanned by $x_1,\dots,x_m$ and $T(W')$ is spanned by $x_{m+1},\dots,x_n$ (for $m=\dim W$).  We then see that $T_*(I)=(x_1,\dots,x_m)$ and $T_*(J)=(x_{m+1},\dots,x_n)$.  Thus $R/I\cong R/T_*(I)\cong k[x_{m+1},\dots,x_n]$ and $R/J\cong R/T_*(J)\cong k[x_1,\dots,x_m]$, and so $R\cong R/I\otimes R/J$.
In other words, by a linear change of variables on our polynomial ring, we can assume that $W$ and $W'$ actually just come from a partition of the set of variables, and so this statement reduces to the usual formula for a tensor product of polynomial rings.

As a side note, I don't think it's reasonable to hope for a result like $R\cong R/I\otimes R/J$ in any general setting, even with some strong hypotheses on $I$ and $J$.  The problem is that there are natural maps $R\to R/I$ and $R\to R/J$, but to get a map between $R$ and $R/I\otimes R/J$ you instead need maps $R/I\to R$ and $R/J\to R$ (which then combine to give a map $R/I\otimes R/J\to R$ by the universal property of tensor products).
In this case of polynomial rings, the natural statement would be to instead consider the subalgebras $k[W]$ and $k[W']$ of $R$ generated by $W$ and $W'$.  We then have a natural map $k[W]\otimes k[W']\to R$ induced by the inclusion maps which is an isomorphism.  The only way the quotients come up is that it happens to also be true in this case that the quotient map $R\to R/I$ restricts to an isomorphism $k[W']\to R/I$ and the quotient map $R\to R/J$ restricts to an isomorphism $k[W]\to R/J$, so these subrings can also be identified as quotients.
