Exercise 2.12

Question

Let $M$ be a finitely generated $A$-module and $\varphi\colon M 	o A^n$ a surjective homomorphism. Show that $\operatorname{Ker}(\varphi)$ is finitely generated.

Amit Awasthi · Accepted Answer

\begin{lemma}[Splitting lemma {\cite[2.10]{Rei95}}]\label{lem:splitting}
  Let $0 	o L \overset{\alpha}{	o} M \overset{\beta}{	o} N 	o 0$ be a short exact sequence of $A$-modules. Then, the following are equivalent:
  \begin{enumerate}[label=oman*)]
    \item there exists an isomorphism $M \cong L \oplus N$ under which $\alpha$ is given by $m \mapsto (m,0)$ and $\beta$ is by $(m,n) \mapsto n$;
    \item there exists a section of $\beta$, i.e., a map $s\colon N 	o M$ such that $\beta \circ s = \operatorname{id}_N$;
    \item there exists a retraction of $\alpha$, i.e., a map $r\colon M 	o L$ such that $r \circ \alpha = \operatorname{id}_L$.
  \end{enumerate}
  If this happens the sequence is a \emph{split exact sequence}, and we say the sequence \emph{splits}.
\end{lemma}
\begin{proof}[Proof of Lemma $ef{lem:splitting}$]
  $i) \Rightarrow ii)$, $i) \Rightarrow iii)$ are trivial by pulling back $m \mapsto (m,0)$ and $(m,n) \mapsto n$ through the isomorphism $M \cong L \oplus N$.
  \par $ii) \Rightarrow i)$. We claim $M \cong \alpha(L) \oplus s(N)$. Any $m \in M$ has the form $m = (m - s(\beta(m))) + s(\beta(m))$, where the second term is clearly in $s(N)$, and the first is in $\ker\beta$ since $\beta \circ s = \operatorname{id}_N$, and $\ker\beta = \alpha(L)$ by exactness. Moreover, $\alpha(L) \cap s(N) = 0$, since if $n \in N$ is such that $s(n) \in \alpha(L) = \ker\beta$ then $n = \beta(s(n)) = 0$. Finally, since $\alpha$ is injective by assumption and $s$ is injective since it has a left inverse, $\alpha(L) \cong L$ and $s(N) \cong N$, so $M \cong L \oplus N$.
  \par $iii) \Rightarrow i)$. We claim $M = \ker(r) \oplus \alpha(L)$. Any $m \in M$ has the form $M = (m - \alpha(r(m))) + \alpha(r(m))$, where the first term is clearly in $\alpha(L)$, and the second is in $\ker(r)$ since $r \circ \alpha = \operatorname{id}_L$. Moreover, $\ker(r) \cap \alpha(L) = 0$, since if $\ell \in L$ is such that $\alpha(\ell) \in \ker(r)$ then $\ell = r(\alpha(\ell)) = 0$. We now claim $\ker(r) \cong N$. Since $\beta$ is surjective, any $n \in N$ has preimage $m = m' + \alpha(\ell) \in M$ for some $\ell \in L$, where $m' \in \ker(r)$, so $n = \beta(m) = \beta(m')$ since $\beta \circ \alpha = 0$. So, $\beta\colon \ker(r) 	o N$ is surjective; it is injective since if $\beta(m) = 0$, then $m \in \alpha(L)$ by exactness, and $\ker(r) \cap \alpha(L) = 0$ from above. So, $\ker(r) \cong N$, and since $\alpha$ is injective by assumption, $\alpha(L) \cong L$ as well, so $M \cong L \oplus N$.
\end{proof}
\begin{proof}[Main Proof]
  We claim the sequence $0 	o \ker\varphi 	o M \overset{\varphi}{	o} A^n 	o 0$ splits. Let $e_1,\ldots,e_n$ be a basis of $A^n$ and choose $u_i \in M$ such that $\varphi(u_i) = e_i$ ($1 \le i \le n$). This defines a spliting $s\colon A^n 	o M$ by $e_i \mapsto u_i$, and so $M = \operatorname{Ker}(\varphi) \oplus \sum Au_i$ by the splitting lemma (Lemma ef{lem:splitting}). So $\operatorname{Ker}(\varphi) \cong M/A^n$. If $A^m 	woheadrightarrow M$, then $A^m 	woheadrightarrow M 	woheadrightarrow M/A^n = \operatorname{Ker}(\varphi)$ implies $\operatorname{Ker}(\varphi)$ is finitely generated by Prop.~2.3.
\end{proof}

Exercise 2.12

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Exercise 2.12

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