Let $R,S$ be commutative rings with 1 satisfying $R \subseteq S$. I am looking for an alternative proof to the fact that $S \otimes_{R} R[x] \cong S[x]$ as $R$-modules.
This is a pretty easy corollary from the fact that if $M$ is any $R$-module and $F$ is a free $R$-module with basis $\{ e_{i} \}_{i \in I}$, then every element of $M \otimes_{R} F$ has a unique representation of the form $\sum_{i \in I} m_{i} \otimes e_{i}$, where $m_{i} = 0$ for all but finitely many $i \in I$. Using this coupled with the fact that $\{ x^i \}_{i \in \mathbb{N}}$ is a $R$-basis for $R[x]$, we know that every element of $S \otimes_{R} R[x]$ has a unique representation of the form $\sum_{i \in \mathbb{N}} s_{i} \otimes x^i$. We can then verify that the obvious map $\sum_{i \in \mathbb{N}} s_{i} \otimes x^i \mapsto \sum_{i \in \mathbb{N}} s_{i} x^i$ is an $R$-module isomorphism.
This proof is pretty slick, but is there a different way to prove this using the universal property of tensor products as the main machinery? To be specific, is there an "obvious" bilinear map $S \times R[x] \rightarrow S[x]$ that induces an isomorphism $S \otimes_{R} R[x] \rightarrow S[x]$.
EDIT: Upon Hurkyl's suggestion, I think I proved in the way I was trying. Also I think it establishes an isomorphism of $R$-algebras as Qiaochu suggested.
So let $B : S \times R[x] \rightarrow S[x]$ be defined by $(s, f) \mapsto sf$. It is easy to verify that $B$ is bilinear and so we get a (unique) $R$-linear map $T : S \otimes_{R} R[x] \rightarrow S[x]$ such that $T(s \otimes f) = sf$. Now define another map $U : S[x] \rightarrow S \otimes_{R} R[x]$ by $sx^i \mapsto s \otimes x^i$ and extending by linearity. It is easy to show that $T \circ U = \operatorname{id}_{S[x]}$ by showing that $T \circ U(sx^i) = sx^i$ for all $s \in S$ and $i \in \mathbb{N}$. Similarly, it is not too difficult to show that $U \circ T = \operatorname{id}_{S \otimes_{R} R[x]}$ on all elementary tensors $s \otimes f \in S \otimes_{R} R[x]$, which implies that $U \circ T = \operatorname{id}_{S \otimes R[x]}$ on all of $S \otimes_{R} R[x]$. This then implies that $S \otimes_{R} R[x] \cong S[x]$.
Is this correct?
