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I'm trying to solve the problems in the book of Atiyah and MacDonald. I want to verify my solution to the problem 2.6.

This is the exercise's statement:

2.6. For any $A$-module $M$, let $M[x]$ denote the set of all polynomials in $x$ with coefficients in $M$. Defining the product of an element $A[x]$ and an element of $M[x]$ in the obvious way, show that $M[x]$ is an $A[x]$-module.

Show that $M[x] \cong A[x] \otimes_{A} M.$

This is what I've tried: (It's straightforward that $M[x]$ is an $A[x]$-module)

Clearly, $M[x]=\oplus_{n=0}^{\infty} Mx^n$ and $A[x]=\oplus_{n=0}^{\infty} Ax^n.$ Furthermore, $Ax^n \cong A$ as $A$-modules, $Mx^n \cong M$ as $A$-modules and $A \otimes_{A} M \cong M$ as $A$-modules. Hence $$A[x] \otimes_{A} M \cong \oplus_{n=0}^{\infty} ((Ax^n)\otimes_{A}M) \cong \oplus_{n=0}^{\infty} (A\otimes_{A}M) \cong \oplus_{n=0}^{\infty} M \cong \oplus_{n=0}^{\infty} (Mx^n) = M[x].$$

End.

I don't know if the solution is correct, since I think that the isomorphisms I'm using to prove it could be needed to be as $A[x]$-modules and not as $A$-modules. What could be another solution to this problem? In MSE I only found the problem for $A$-algebras but not for $A$-modules.

Thanks in advance!

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    $\begingroup$ I would recommend you use a different approach, this one works out perfectly at the level of $A$-modules. However, your morphism is per se not canonical, and I suspect the exercise should ectually be that this iso is natural. so i would suggest you use the universal property of the tensor product,respectively polynomial algebra. Also, does M have a multiplicative structure (the above holds even in that case), because your iso does not necessarily preserve that. $\endgroup$ – Enkidu Feb 7 at 13:15
  • $\begingroup$ I've just found that my question is already answered in MSE. The link is "math.stackexchange.com/questions/52299/…" and is exactly my question and exactly your answer @Enkidu, thanks for your time! $\endgroup$ – DrinkingDonuts Feb 7 at 13:24
  • $\begingroup$ @Enkidu Why do you say the morphism is not canonical? $\endgroup$ – Arnaud D. Feb 7 at 13:52
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    $\begingroup$ because they are jotted down by hand and not coming from a universal property. I.e. might need some choices (like generating sets) $\endgroup$ – Enkidu Feb 7 at 14:00
  • $\begingroup$ This isn't the proper meaning of "canonical", @Enkidu. You mean to say that it is not visibly an $A\left[x\right]$-module map. But this is not a big issue: It still is an $A\left[x\right]$-module map; you just have to prove it separately. $\endgroup$ – darij grinberg Feb 7 at 20:43
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As an $A$-module, $M[x] \cong \oplus_{i \in \mathbb N}Mx^i$. Define action on $M[x]$ by $A[x]$ as $(\sum a_ix^i)(\sum m_j x^j)=\sum(\sum_{i+j=k} a_im_j).$ Then, make sure all the axioms of an $A[x]$-module is satisfied (distributivity, associativity.)

Next, define $\phi :M[x]\rightarrow A[x]\otimes_A M$ by $\phi(\sum m_j x^j)=\sum(x^j\otimes m_j).$ Check if this is $A[x]$-linear.

Then, define $\bar{\psi}:A[x]\times M\rightarrow M[x]$ by $\bar{\psi}(\sum a_ix^i,m)=\sum (a_im)x^i.$ This is obviously bilinear, and corresponds to a map $\psi:A[x]\otimes_A M\rightarrow M[x].$ Check $\psi$ and $\phi$ are inverse to each other.

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