Show that for any commutative ring $R$, $R[X] \otimes R[Y] \simeq R[X,Y]$. Show that for any commutative ring $R$, 
$$R[X] \otimes R[Y] \simeq R[X,Y].$$
I know that I have to first take a bilinear map from $R[X] \times R[Y]$ to $R[X,Y]$ and then it induces a linear map from $R[X] \otimes R[Y]$ to $R[X,Y]$. But what is the required bilinear map here that will work? I proceed by taking $(f(x),g(y)) \mapsto f(x)g(y)$.
Will that work? Please help me.
Thank you very much.
 A: I believe that your map should work. Indeed, the map is bilinear and thus induces a linear map (call it $\phi)$. We check that $\phi$ is an isomorphism. It is easy to see it is onto, so we only need to check whether it is 1-1. Notice that all tensors can be written as linear combinations of tensors of the form $x^i \otimes y^j$. So, suppose that $\phi(\displaystyle\sum_{i,j} a_{i,j}x^i\otimes y^j)=\phi(\displaystyle\sum_{i,j} b_{i,j}x^i\otimes y^j)$ (where the sums are finite and some of the $a_{i,j}$ and $b_{i,j}$ might be $0$). Then for each $i,j$, we get that $a_{i,j}=b_{i,j}$. So $\displaystyle\sum_{i,j} a_{i,j}x^i\otimes y^j=\displaystyle\sum_{i,j} b_{i,j}x^i\otimes y^j$. This means that $\phi$ is 1-1.
A: In my opinion showing something is isomorphic to the tensor product is easier if we just show that it has the universal property. Define the map $i:R[X]\times R[Y]\to R[X,Y]$ by $(p(x),q(y))\mapsto p(x)q(y).$ This is certainly a bilinear map. Now we want to show that if $\phi:R[X]\times R[Y]\to M$ is a bilinear map in $R$-Mod, there is a unique map $\bar{\phi}$ from $R[X,Y]\to M$ that has the property that $\bar\phi\circ i=\phi$, and $\bar\phi$ is linear. For any such map $\phi$, we can define $\bar\phi$ by $\bar\phi(x)=\phi(x)$, $\bar\phi(y)=\phi(y)$, $\bar\phi(1)=\phi(1)$, and extending via the universal property of the polynomial ring. So this map exists. Why is it unique? Any map which commuted would have to send $i(x)$ to $\phi(x)$, and similarly $i(y)$ to $\phi(y)$. So this object is the tensor product.
