Atiyah-Macdonald Exercise 2.6 I'm working on Exercise 2.6 from Atiyah-Macdonald's "Introduction to Commutative Algebra" and I had a question about the definition of bilinear maps.

Exercise 2.6: For any $A$-module $M$, let $M[x]$ denote the set of all polynomials in $x$ with coefficients in $M$, that is to say expressions of the form
$$ m_0 + m_1x + \dots + m_r x^r \qquad (m_i\in M)$$
Defining the product of an element of $A[x]$ and an element of $M[x]$ in the obvious way, show that $M[x]$ is an $A[x]$-module. Show that $M[x] \cong A[x]\otimes_A M$ (as $A[x]$-modules).

So I begin by trying to find explicit $A[x]$-linear maps from $M[x]\to A[x]\otimes_A M$ and then from $A[x]\otimes_A M \to M[x]$. I got a little confused about the definition of a bilinear map when trying to define the second map.
What is an $A[x]$-bilinear map on $A[x]\otimes_A M$? In particular, let $\phi: A[x]\times M \to M[x]$ be a map. Suppose I fix a first coordinate $f\in A[x]$, then what does it mean for $\phi(f,-) : M\to M[x]$ to be $A[x]$-linear? A priori, there is no defined multiplication of $A[x]$ onto $M$ right?
 A: First note that you want an $A[x]$-linear map from $A[x] \otimes_A M$ into $M[x]$, not an $A[x]$-bilinear map. This is because in the category of modules, the morphisms are the linear maps. Now in order to exploit the universal property of the tensor product in order to construct such a map, it is sufficient to start with a map on $A[x] \times M$ which is merely $A$-bilinear, because $A$ is the ring we take the tensor product over.
Such a map is given by
$$
(a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0, m) \mapsto a_n m x^n + a_{n-1} m x^{n-1} + \cdots + a_1 m x + a_0 m.
$$
Thus, the universal property of the tensor product implies that we get a map $A[x] \otimes_A M \to M[x]$, which a priori is only $A$-linear. But this map is in fact $A[x]$-linear because it's the inverse of a map which is $A[x]$-linear. Remember that the inverse of a module isomorphism is again a module isomorphism.
This $A[x]$-linear map is given by
$$
M[x] \to A[x] \otimes_A M, m_n x^n + \cdots + m_0 \mapsto x^n \otimes m_n + \cdots + 1 \otimes m_0.
$$
In order to show that these two maps are inverse to each other, it may be convenient to note that by linearity, it's sufficient to check this on convenient generating sets of each module.
