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\begin{document}
\centerline{Math 361 lecture for Monday, Week 5}
\bigskip

\centerline{Noetherian rings}
\bigskip

\begin{defn}
  An $R$-module~$M$ is \emph{Noetherian} if every submodule of~$M$ is finitely
  generated. 
\end{defn}

\begin{eg}
  As a special case, one may consider $R$ itself as an~$R$-module.  Its
  submodules are its ideals.  Thus, for example, if~$R$ is a PID, it is
  Noetherian since each of its submodules is generated by a single element.  For
  instance, $\Z$ is Noetherian, and the polynomial ring~$K[x]$ is Noetherian for
  any field~$K$.
\end{eg}

\begin{prop}
  The following are equivalent for an $R$-module $M$:
  \begin{enumerate}
    \item\label{m1:05mon} $M$ is Noetherian.
    \item\label{m2:05mon} $M$ satisfies the \emph{ascending chain condition} on submodules:
      every ascending chain of submodules of~$M$,
      \[
        N_{1}\subseteq N_{2}\subseteq\cdots,
      \]
      eventually \emph{stabilizes}.  In other words, there exists $k$ such that
      $N_{k}=N_{k+1}=\cdots$.
    \item\label{m3:05mon} Every  nonempty collection of submodules of~$M$ has a maximal element
      under inclusion.
  \end{enumerate}
\end{prop}
\begin{proof}
(\ref{m1:05mon} $\Rightarrow$ \ref{m2:05mon}) Suppose~$M$ is Noetherian and
let~$N_{1}\subseteq N_{2}\subseteq \cdots$ be a chain of submodules.
Let~$N:=\cup_{i\geq1}N_{i}$.  Then~$N$ is a submodule of $M$ (exercise).  Hence,
by assumption, $N$ is finitely generated by, say $n_{1},\ldots,n_{s}\in N$.
Since~$N=\cup_{i\geq1}N_{i}$, for each~$i=1,\ldots,s$, we have $n_{i}\in
N_{k_{i}}$ for some~$k_{i}$. Let~$k=\max\{k_{i}\}$.  Then~$n_{i}\in N_{k}$ for
all~$i$.  It follows that
\[
  N=\Span_{R}\{n_{1},\ldots,n_{s}\}=N_{k}=N_{k+1}=\cdots.
\]
\medskip

(\ref{m2:05mon} $\Rightarrow$ \ref{m3:05mon})  Suppose every ascending chain of
submodules of~$M$ stabilizes.  Let~$\mathcal{A}$ be a nonempty of submodules
of~$M$.  Pick~$N_{1}\in\mathcal{A}$. If~$N_{1}$ is not a maximal element
of~$\mathcal{A}$, there exists~$N_{2}\in\mathcal{A}$ with $N_{1}\subsetneq N_{2}$.
If $N_{2}$ is not a maximal element of~$\mathcal{A}$, then there
exists~$N_{3}\in\mathcal{A}$ such that $N_{1}\subsetneq N_{2}\subsetneq N_{3}$.
Repeat.  Since every ascending chain eventually stabilizes, we must eventually
reach a maximal element of~$\mathcal{A}$.
\medskip

(\ref{m3:05mon} $\Rightarrow$ \ref{m1:05mon}) Suppose that every nonempty
collection of submodules of~$M$ has a maximal element, and let $N$ be a
submodule of~$M$.  Let~$\mathcal{A}$ be the collection of all finitely generated
submodules of~$N$.  Then~$\mathcal{A}$ is nonempty since it contains the zero
module.  Take a maximal element~$N'\in\mathcal{A}$.  So $N'\subseteq N$.
For sake of contradiction, suppose that~$N'\neq N$, and take~$n\in N\setminus
N'$. Consider the module $N'':= N'+Rn$,
the smallest $R$-module containing both $N'$ and~$n$.  Since~$n\not\in N'$, we
have~$N'\subsetneq N''$.  However, $N''\in\mathcal{A}$, too, contradicting the
maximality of~$N'$.
\end{proof}
\bigskip

\textbf{Mappings of modules.} A sequence of $R$-module mappings
\[
  M'\xrightarrow{\phi}M\xrightarrow{\psi}M''
\]
is \emph{exact at~$M$} if $\im\phi=\ker\psi$.

\begin{eg}
  Consider the sequence of $R$-module mappings
  \[
    0\to M'\xrightarrow{\phi}M.
  \]
  There is only one choice for the mapping $0\to M'$.  It sends~$0$ to~$0\in
  M'$.  The sequence of mappings is exact at $M'$ if and only if $\phi$ is
  injective.  Similarly, a sequence of $R$-module mappings
  \[
    M\xrightarrow{\psi}M''\to0
  \]
  is exact at $M''$ if and only if~$\psi$ is surjective.
\end{eg}

\emph{Short exact sequences.} A \emph{short exact sequence} of $R$-modules is a
sequence of $R$-module mappings
\[
  0\to M'\xrightarrow{\phi} M\xrightarrow{\psi} M'' \to 0 
\]
if it is exact at $M'$, $M$, and $M''$.  In that case, $M'$ is isomorphic to its
image in~$M$, so we can identify~$M'$ with its image and write~$M'\subseteq M$.
We then have an isomorphism of $R$-modules
\[
  M/M'\xrightarrow{\sim} M''.
\]
where~$\phi(\overline{m}):=\phi(m)$.
\bigskip

\begin{prop}\label{prop:05mon}	
  Let
  \[
    0\to M'\xrightarrow{\phi} M\xrightarrow{\psi} M'' \to 0 
  \]
  be a short exact sequence of~$R$-modules.  Then $M$ is Noetherian if and only
  if $M'$ and $M''$ are Noetherian.
\end{prop}
\begin{proof}
  ($\Rightarrow$) Suppose that~$M$ is Noetherian.  We may assume $M'\subseteq
  M$.  Every submodule of~$M'$ is a submodule of~$M$, and hence is finitely
  generated.  Therefore, $M'$ is Noetherian.  Next, suppose that~$N$ is a
  submodule of~$M''$.  Then $\psi^{-1}(N)$ is a submodule of~$M$ (exercise).
  Since~$M$ is Noetherian, $\psi^{-1}(N)$ is finitely generated, say by
  $n_{1},\ldots,n_{k}$.  It is straightforward to check
  that~$\psi(n_{1}),\ldots,\psi(n_{k})$ generate $N$.  Hence, $M''$ is
  Noetherian.

  ($\Leftarrow$) Suppose that $M'$ and $M''$ are Noetherian, and let~$N$ be a
  submodule of~$M$.  Now $\psi(N)$ is a submodule of~$M''$, hence finitely
  generated.  Let $\overline{n_{1}},\ldots,\overline{n_{k}}$ be generators with
  corresponding $n_{1},\ldots.n_{k}\in M$ such
  that~$\psi(n_{i})=\overline{n_{i}}$ for all~$i$.  Next, identifying~$M'$ with
  its image $\phi(M')$, we have the submodule~$N\cap M'$ of $M'$. Since~$M'$ is
  Noetherian, $N\cap M'$ is finitely generated by, say, $v_{1},\ldots,v_{\ell}$.  

  We claim that~$\{v_{1},\ldots,v_{\ell},n_{1},\ldots,n_{k}\}$ generate~$N$.  To
  see this, let~$n\in N$.  We can then write~$\psi(n)=\sum_{i=1}^{k}
  r_{i}\overline{n_{i}}$ for some $r_{i}\in R$. We have~$\psi(n-\sum_{i=1}^{k}
  r_{i}n_{i})=0$.  So $n-\sum_{i=1}^{k} r_{i}n_{i}\in\ker\psi=\im\phi$ (and
  recall that we are identifying~$\im\phi$ with $M'$ since~$\phi$ is injective). 
  Thus, $n-\sum_{i=1}^{k} r_{i}n_{i}\in N\cap M'$.  So we can write
  \[
    n-\sum_{i=1}^{k} r_{i}n_{i} = \sum_{j=1}^{\ell} s_{j}v_{j}.
  \]
  So $n\in\Span_{R}\{v_{1},\ldots,v_{\ell},n_{1},\ldots,n_{k}\}$, as claimed.
\end{proof}

\begin{cor}\label{cor1:05mon}
  If $R$ is Noetherian, so is~$R^{n}$ for each $n\in N$.
\end{cor}
\begin{proof}
  If $n=0,1$, the result is trivial.  Let~$n>1$, and suppose the statement is true for~$R^{k}$
  with $0\geq k< n$.
  We have the short exact sequence
  \[
    0\to R\xrightarrow{\psi}R^{n}\xrightarrow{\phi}R^{n-1}\to0.
  \]
  where $\psi(r)=(r,0,\ldots,0)$ and
  $\phi(r_{1},\ldots,r_{n})=(r_{2},\ldots,r_{n})$.  The result now follows by
  induction and Proposition~\ref{prop:05mon}.
\end{proof}

\begin{cor}
  If $R$ is Noetherian and $M$ is an $R$-module.  Then $M$ is Noetherian if and
  only if $M$ is finitely generated.  In other words, a finitely generated
  module over a Noetherian ring is Noetherian.
\end{cor}

\begin{proof}
 ($\Rightarrow$) Suppose that~$M$ is Noetherian.  Then \emph{every} submodule
 of~$M$ is finitely generated, which includes~$M$, itself.

 ($\Leftarrow$) Now suppose that $M$ is finitely generated.
 Say~$M=\Span_{R}\{m_{1},\ldots,m_{n}\}$.  We then have a surjective
 homomorphism
 \begin{align*}
   \psi\colon R^{n}&\to M\\
   (r_{1},\ldots,r_{n})&\mapsto \sum_{i=1}^{n} r_{i}m_{i}.
 \end{align*}
 Let~$M':=\ker\psi$.  We have a short exact sequence
 \[
   0\to M'\to R^{n}\xrightarrow{\psi}M\to0
 \]
 where $M'\to M$ is the inclusion mapping. Since~$R$ is Noetherian, so
 is~$R^{n}$ by Corollary~\ref{cor1:05mon}.  Proposition~\ref{prop:05mon} then
 allows us to conclude that~$M$ is Noetherian.
\end{proof}

\begin{cor}
  Let $K$ be a number field.  Then its ring of integer~$\mathfrak{O}_{K}$ is
  Noetherian.
\end{cor}

\begin{proof}
 This follows since~$\Z$ is a PID, hence, Noetherian, and~$\mathfrak{O}_{K}$ is
 a finitely generated $\Z$-module.
\end{proof}

\begin{thm}
  Let $R$ be a Noetherian domain.  Then every nonzero non-unit element of~$R$ can be
  factored into irreducibles.
\end{thm}
\begin{proof}
  We start by considering a set of principal ideals in $R$.
 Let
 \[
   \mathcal{A} = \{(x): x\in R, x\neq0, \text{ $x$ is a non-unit and cannot be
   factored into irreducibles}\}.
 \]
 Our goal is to prove that~$\mathcal{A}=\emptyset$.  For sake of contradiction,
 suppose that it is not.  Since $R$ is Noetherian, $\mathcal{A}$ then has a
 maximal element~$(x)$ with respect to inclusion.  By definition
 of~$\mathcal{A}$, the element~$x$ is not irreducible.  Therefore, there exist
 non-units $y,z\in R$ such that~$x=yz$.  It follows that~$x\in (y):=\{ry: r\in
 R\}$, and hence, $(x)\subseteq (y)$.  Further, it is not possible for
 $(x)=(y)$, for otherwise, we would have $y\in(x)$.  This would mean there
 exists~$w\in R$ such that~$y=xw$.  However, then 
 \[
   y=xw=yzw\quad\Rightarrow\quad y(1-zw)=0.
  \]
  Since~$0\neq x=yz$, and~$R$ is a domain, we see $y\neq0$.  Then, since the
  cancellation law holds for domains, $y(1-zw)=0$
  implies $1-zw=0$, i.e.,~$zw=1$.  So~$z$ is a unit, which is a contradiction.
  
  We have shown that $(x)\subsetneq (y)$.  Similarly, $(x)\subsetneq (z)$.  My
  maximality of~$(x)$ in $\mathcal{A}$, we know $(y)$ and $(z)$ are not
  in~$\mathcal{A}$.  So $y$ and $z$ can be factored into irreducibles, say as
  $y=\prod_{i=1}^{m}y_{i}$ and $z=\prod_{i=1}^{n} z_{n}$.  However, it then
  follows that~$x$ factors into irreducibles:
  $x=yz=\prod_{i=1}^{m}y_{i}\prod_{i=1}^{n} z_{n}$.  That contradicts the fact
  that~$x$ cannot be factored into irreducibles.
\end{proof}

\begin{cor}\label{cor:05mon}
  Let~$K$ be a number field.  Then every element of its ring of
  integers,~$\mathfrak{O}_{K}$, can be factored into irreducibles in
  $\mathfrak{O}_{K}$.
\end{cor}

\begin{eg}
  Note that \emph{Corollary~\ref{cor:05mon} does not say that the ring of integers is
  a UFD.}  In homework, we considered the quadratic field~$\Q(\sqrt{-5})$, whose ring of
  integers is $\Z[\sqrt{-5}]$.  In this ring,
  \[
    6 = 2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5}),
  \]
  and we showed $2,3,1\pm\sqrt{-5}$ are non-units and irreducible.  So~$6$ has
  at least two factorizations into irreducibles in $\Z[\sqrt{-5}]$.  This result
  is consistent with Corollary~\ref{cor:05mon}.
\end{eg}
\end{document}
