Tensor product isomorphism $K[x] \otimes_K M \cong M[x]$ I am interested in proving the following isomorphism of tensor products. Let $M/K$ be an extension of fields. Then we have the $M$-algebra isomorphism
$$f : K[x] \otimes_K M \to M[x].$$
The most natural thing to do seems to be to define $f(g(x) \otimes m) = mg(x)$. If we linearly extend it then
$$f((g(x) \otimes m) + (h(x) \otimes m')) = mg(x) + m'h(x),$$
I can see that this map is surjective, and since it is bilinear it should respect the relations in $K[x] \otimes_K M$. Here's what I don't understand: Suppose we ask for such an isomorphism $f$ as $K$-vector spaces. My understanding of the tensor product is that there is a $K$-bilinear map $g : K[x] \times M \to K[x] \otimes_K M$ such that all $K$-linear maps $f : K[x] \otimes_K M \to M[x]$ are induced by $K$-bilinear maps from $K[x] \times M \to M[x]$. In this case the map from $K[x] \times M \to M[x]$ must be the multiplication map $(g(x),m) \to mg(x)$. But this map is not necessarliy surjective, which would mean our induced map $f$ is not surjective either. But $f$ is supposed to be an isomorphism.
 A: Think of $K[x]$ as
$$K[x]=\bigoplus_{n=0}^\infty Kx^n,$$
so a direct sum of 1-dimensional vector spaces. Direct sums commute
with tensor products so
$$K[x]\otimes_K M=\bigoplus_{n=0}^\infty (Kx^n\otimes_K M)
\cong\bigoplus_{n=0}^\infty Mx^n=M[x].$$
In general one has a bilinear map $A\times B\to A\otimes B$ which is universal
for bilinear maps. Usually this is not surjective, but its image
generates $A\otimes B$ as a module. Your conclusion that "$f$ is not surjective" is unwarranted.
A: In dealing with tensor products, it should always be your first instinct (and it is often your only recourse) to use the defining universal property of the tensor product.   Let us first deal with everything as $K$-- or $M$--vector spaces and ignore the algebra structure.  Thus $K[x]$ is a $K$--vector space, $K \subset M$ is an extension of fields, and thus $M$ is in particular a $K$--$M$ bimodule.   
Mini-course on tensor products: $T = K[x] \otimes_K M$ is characterized by the universal property:  (1)  there is a $K$--balanced $K$--$M$ bilinear map $j: K[x] \times M \to T$ such that (2)  whenever $V$ is an $M$--vector space and $h:  K[x] \times M \to V$ is a $K$--balanced $K$--$M$ bilinear map, then there is a unique $M$--linear map $\tilde h : T \to V$, such that $\tilde h \circ j = h$.  (Of course, it would be nicer if I would include commutative diagrams, but I don't want to take the time to create then and upload them.)  
It follows in the usual way from the universal property that the tensor product is unique up to isomorphism; even better, if you had two tensor products, say  $j: K[x] \times M \to T$ and  $j': K[x] \times M \to T'$, then the universal property would give maps $\varphi: T \to T'$ and $\varphi' : T' \to T$ such that $\varphi \circ j = j'$ and $\varphi' \circ j' = j$.  A little further argument using the universal property shows that $\varphi'\circ \varphi$ is the identity of $T$ and  $\varphi\circ \varphi'$ is the identity of $T'$. 
Now what you have to do is to show that your map $j' : K[x] \times M \to M[x]$ defined by $j'((g(x), m)) = m g(x)$ is $K$--balanced $K$--$M$ bilinear, and has the universal property.  Once you know that this is what has to be done, it is a matter of straightforward computation.
It then follows that there is a unique isomorphism of $M$ vector spaces $\varphi : T \to M[x]$ such that $\varphi \circ j = j'$.  Now you just have to check that $\varphi $ necessarily coincides with your map $f$.
You were puzzled that $f$ is supposed to be surjective while your bilinear map $j' : K[x] \times M \to M[x]$ is not surjective, and you suspected some contradiction because of this.  But there is no contradiction because all that is needed is that the linear span of the range of $j'$ is $M[x]$.
This answer is already long, so I will not discuss the algebra structures.  But see for example section 3.9 of Jacobsen, Basic algebra vol II.  
