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

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, $$M[x]=\bigoplus_{n=0}^{\infty} Mx^n$$ and $$A[x]=\bigoplus_{n=0}^{\infty} Ax^n.$$ 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 $$A[x] \otimes_{A} M \cong \bigoplus_{n=0}^{\infty} ((Ax^n)\otimes_{A}M) \cong \bigoplus_{n=0}^{\infty} (A\otimes_{A}M) \cong \bigoplus_{n=0}^{\infty} M \cong \bigoplus_{n=0}^{\infty} (Mx^n) = M[x].$$

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.

• 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. – Enkidu Feb 7 '19 at 13:15
• 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! – DrinkingDonuts Feb 7 '19 at 13:24
• @Enkidu Why do you say the morphism is not canonical? – Arnaud D. Feb 7 '19 at 13:52
• because they are jotted down by hand and not coming from a universal property. I.e. might need some choices (like generating sets) – Enkidu Feb 7 '19 at 14:00
• 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. – darij grinberg Feb 7 '19 at 20:43

As an $$A$$-module, $$M[x] \cong \oplus_{i \in \mathbb N}Mx^i$$. Define action on $$M[x]$$ by $$A[x]$$ as $$(\sum a_ix^i)(\sum m_j x^j)=\sum(\sum_{i+j=k} a_im_jx^k).$$ 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.