Tensoring $k[x] \otimes_k k[y]$ Recalling result of tensor product of polynomial rings and Does the isomorphism $k[x]\otimes_k k[x]\cong k[x,y]$ hold? suggest that $k[x] \otimes_k k[y] = k[x,y]$.
I am sure that this is true if the tensor product is being done over $k[x,y]$. However, in $k[x] \otimes_k k[y]$ we have the non-equal tensors $x \otimes y$ and $ y \otimes x$. My confusion is that if I map $k[x] \times k[y] \to k[x,y]$, this factors through $k[x] \otimes_k k[y]$ with kernel containing $x \otimes y - y \otimes x$. Help?
 A: The tensor product over the field $k$ is the coproduct in the category of commutative $k$-algebras. This means that you are given maps $k[x] \to k[x] \otimes_k k[y]$ and $k[y] \to k[x] \otimes_k k[y]$, both given by $f(x) \mapsto f(x) \otimes 1$ and $g(y) \mapsto 1 \otimes g(y)$. Since you have inclusion maps $k[x] \to k[x,y]$ and $k[y] \to k[x,y]$, the universal property of the coproduct tells you that you get a map 
$$
k[x] \otimes_k k[y] \to k[x,y], \quad f(x) \otimes g(y) \mapsto f(x) g(y). 
$$
(The point of all this blabla is to show that this map is well-defined because you have to worry about this when defining maps where the domain is a tensor product. You could also do it "by hand", i.e. defining the corresponding $k$-balanced map and using the universal property of the tensor product.) 
Can you show that this map is an isomorphism? (Hint : find the inverse map using the fact that $\{ f(x) g(y) \, | \, f(x) \in k[x], g(y) \in k[y] \}$ spans $k[x,y]$ as a $k$-vector space.)
Hope that helps,
