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?

  • $\begingroup$ There is an $A$-bilinear map $A[x]\times M\to M[x]$ which is just $(p(x),m)\mapsto p(x).m$. This will factor through $A[x]\otimes_A M$. Check that this is an isomorphism. $\endgroup$
    – shubhankar
    Jan 11, 2021 at 21:55
  • $\begingroup$ But I thought this isomorphism is an $A$-isomorphism, not an $A[x]$-isomorphism? $\endgroup$
    – klein4
    Jan 11, 2021 at 22:00
  • 1
    $\begingroup$ well both modules are isomorphic as $A$-modules. What you need to show is that this is compatible with the $A[x]$ action on both sides. What is the $A[x]$ action on $A[x]\otimes_A M$? What is it on $M[x]$? $\endgroup$
    – shubhankar
    Jan 11, 2021 at 22:09
  • $\begingroup$ Thanks, I got it now. $\endgroup$
    – klein4
    Jan 11, 2021 at 22:13

1 Answer 1


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.


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