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


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!

  • 2
    $\begingroup$ 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 '19 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 '19 at 13:24
  • $\begingroup$ @Enkidu Why do you say the morphism is not canonical? $\endgroup$ – Arnaud D. Feb 7 '19 at 13:52
  • 1
    $\begingroup$ 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 '19 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 '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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.