Exercise 2.6

Question

For any $A$-module, let $M[x]$ denote the set of all polynomials in $x$ with coefficients in $M$, that is to say expressions of the form
  \begin{equation*}
    m_0 + m_1x + \cdots + m_rx^r \qquad (m_i \in M).
  \end{equation*}
  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.
  \par Show that $M[x] \cong A[x] \otimes_A M$.

Amit Awasthi · Accepted Answer

\begin{proof}
  $M[x]$ is clearly an abelian group, since addition is defined term-wise making $M \cong M \oplus xM \oplus x^2M \oplus \cdots \cong M \oplus M \oplus M \oplus \cdots$ as an abelian group.
  \par We check that it is an $A[x]$-module. Define
  \begin{equation*}
    \left( \sum_i a_ix^i ight) \left( \sum_j m_jx^j ight) \coloneqq \sum_{i + j = k} a_im_jx^k. 
  \end{equation*}
  We check this turns $M[x]$ into an $A[x]$-module. $1 \in A$ clearly acts as the identity, so we check the other axioms:
  \begin{align*}
    &\left( \sum_i a_ix^i ight) \left( \sum_j m_jx^j + \sum_j n_jx^j ight) = \left( \sum_i a_ix^i ight) \left( \sum_j (m_j+n_j)x^j ight)\
    &\qquad\qquad= \sum_k \sum_{i + j = k} a_i(m_j+n_j)x^k= \sum_k \sum_{i + j = k} a_im_jx^k + \sum_k \sum_{i + j = k} a_in_jx^k\
    &\qquad\qquad= \left( \sum_i a_ix^i ight) \left( \sum_j m_jx^j ight) + \left( \sum_i a_ix^i ight) \left( \sum_j n_jx^j ight)\
    &\left( \sum_i a_ix^i + \sum_i b_ix^i ight)\left( \sum_j m_jx^j ight) = \left( \sum_i (a_i+b_i)x^i ight)\left( \sum_j m_jx^j ight)\
    &\qquad\qquad= \sum_k \sum_{i+j=k} (a_i+b_i)m_jx^k= \sum_k \sum_{i + j = k} a_im_jx^k + \sum_k \sum_{i + j = k} b_im_jx^k\
    &\qquad\qquad= \left( \sum_i a_ix^i ight)\left( \sum_j m_jx^j ight) + \left( \sum_i b_ix^i ight)\left( \sum_j m_jx^j ight)\
    &\left(\left( \sum_i a_ix^iight)\left(\sum_j b_jx^j ight)ight)\left( \sum_k m_kx^k ight) = \left( \sum_\ell \sum_{i+j=\ell} a_ib_jx^\ell ight)\left( \sum_k m_kx^k ight)\
    &\qquad\qquad= \sum_p \sum_\ell \sum_{i+j=\ell} \sum_{\ell+k=p} a_ib_jm_kx^p = \sum_p \sum_{i+j+k=p} a_ib_jm_kx^p\
    &\qquad\qquad= \left( \sum_i a_ix^i ight) \left( \sum_\ell \sum_{j+k=\ell} b_jm_kx^\ell ight)\
    &\qquad\qquad= \left( \sum_i a_ix^i ight) \left( \left(\sum_j b_jx^j ight)\left( \sum_k m_kx^k ight) ight)
  \end{align*}
  \par We now show $M[x] \cong A[x] \otimes_A M$. By Exercise 2.5, we have $A[x] \cong A \oplus xA \oplus x^2A \oplus \cdots$, hence
  \begin{equation*}
    A[x] \otimes_A M \cong (A \oplus xA \oplus x^2A \oplus \cdots) \otimes_A M \cong M \oplus xM \oplus x^2M \oplus \cdots = M[x]
  \end{equation*}
  as $A$-modules, using Lemma in 2.4. Note that this isomorphism $\varphi$ is the map 
  \begin{equation*}
    \left(\sum a_ix^i \otimes might) \mapsto \sum (a_im)x^i
  \end{equation*}
  We need to check that this $A$-module isomorphism preserves the $A[x]$-module structure. Since $\varphi$ is $A$-linear, it suffices to check it respects multiplication by $A[x]$ on pure tensors:
  \begin{align*}
    \varphi\left(\left(\sum a_ix^iight)\left(\sum b_jx^j \otimes might)ight) &= \varphi\left(\left( \sum_{i+j=k} a_ib_jx^k ight) \otimes might)\
    &= \sum_{i+j=k} (a_ib_jm)x^k = \left(\sum_i a_ix^iight)\left(\sum b_jmx^jight)\
    &= \left(\sum_i a_ix^iight) \varphi\left( \sum b_jx^j \otimes m ight).
  \end{align*}
  Since $\varphi$ is an isomorphism of $A$-modules, it is bijective, and so it is a bijective $A[x]$-module homomorphism as well, hence an isomorphism of $A[x]$-modules.
  %Define the map
  %\begin{equation*}
  %  g\colon A[x] 	imes M 	o M[x] \qquad \left( \sum_i a_ix^i,m ight) \mapsto \sum_i a_imx^i
  %\end{equation*}
  %This is clearly $A$-bilinear by definition. It suffices by Prop.~2.12 to show that $M[x]$ satisfies the universal property for $A[x] \otimes_A M$, i.e., that for any $A$-bilinear map $f\colon A[x] 	imes M 	o P$, there exists a unique homomorphism $f'\colon M[x] 	o P$ making the diagram below commute:
  %\begin{equation*}
  %  \begin{tikzcd}
  %    A[x] 	imes M ar{g}\arrow{dr}[swap]{f} & M[x]\dar[dashed]{f'}\
  %    & P
  %  \end{tikzcd}
  %\end{equation*}
  %So, define the $A$-module homomorphism
  %\begin{equation*}
  %  f'\colon M[x] 	o P \qquad \sum_i m_ix^i \mapsto \sum_i f(x^i,m_i).
  %\end{equation*}
  %This makes the diagram commute since
  %\begin{align*}
  %  (f' \circ g)\left( \sum_i a_ix^i,m ight) &= f'\left( \sum_i a_imx^i ight) = \sum_i f(x^i,a_im)\
  %  &= \sum_i f(a_ix^i,m) = f\left( \sum_i a_ix^i,m ight).
  %\end{align*}
  %Moreover, $f'$ is unique since the $mx^i$ generate $M[x]$, and the action of $f'$ on these $mx^i$ is uniquely determined by $f$. Thus, $M[x] \cong A[x] \otimes_A M$.
\end{proof}

Exercise 2.6

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Exercise 2.6

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