This is exercise $6$ in chapter $2$ on modules from Atiyah's and Macdonald's book.
Let $M$ be an $A$-module and let $M[x]$ be the set of all polynomials in $x$ with coefficients in $M$. Then $M[x]$ has structure of $A[x]$-module.
Show that $M[x] \cong A[x] \otimes_{A} M$.
Some questions:
1) We can view $M[x]$ as an $A$-module yes? via the map $A \times M[x] \rightarrow M[x]$ given by $(a,m_{0}+\cdots+m_{n}x^{n}) \mapsto am_{0}+\cdots+a_{n}m_{n}x^{n}$.
2) Can we proceed as follows? Note that $A[x] \cong \oplus A$ (infinite direct sum), note then:
$A[x] \otimes_{A} M \cong \oplus A \otimes_{A} M \cong \oplus (A \otimes_{A} M)$.
But $M$ is an $A$-module so $A \otimes_{A} M \cong M$.
Therefore $A[x] \otimes_{A} M \cong \oplus M$. But I think we also have that:
$M[x] \cong \oplus M$ as $A$-modules right?
So $A[x] \otimes_{A} M \cong M[x]$.
Is this OK?