# Tensor products and polynomials with coefficients in a module

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?

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Is the problem to show that these are isomorphic as $A[x]$-modules? Then there is more work to be done. – Dylan Moreland Jul 19 '11 at 1:48
Ah, it's ambiguous in Atiyah-Macdonald. I would assume that they want an $A[x]$-isomorphism. So you probably want your $A$-isomorphism to be more explicit. – Dylan Moreland Jul 19 '11 at 1:58
To your point 1 (undoubtedly Dylan wants you to address the same point): How does the element $x$ act on $M[x]$? How does a polynomial $a_0+a_1x+\cdots+a_nx^n$ act on $M[x]$? Until you have understood these questions, you will not know what is needed to prove the claimed isomorphism of $A[x]$-modules. – Jyrki Lahtonen Jul 19 '11 at 5:44
Any luck? Note that it's really enough to check that the action of $x$ on these two modules commutes with your map. – Dylan Moreland Jul 20 '11 at 1:56

## 1 Answer

Point 1): yes, you can do this. It's a "change of rings" arising from the ring map $A \to A[x]$ sending $a \in A$ to the constant polynomial $a$.

Point 2): as others pointed out, if you want an isomorphism of $A[x]$-modules what you wrote is not enough. If you would like to do this in a fancy way you can show that $M[x]$ satisfies the universal property of the induced module $M\uparrow _A ^{A[x]} = A[x] \otimes _A M$. If you prefer to do it by down-to-earth methods, let $\phi: A[x]\otimes_A M \to M[x]$ be defined by extending $ax^n \otimes m \mapsto amx^n$. You need to check this is a well-defined $A[x]$-module map and that it has an inverse described by $mx^n \mapsto x^n \otimes m$.

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