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\begin{flashcard}[Hilbert-basis-theorem]
\begin{theorem*}[Hilbert's Basis Theorem]
    If $R$ is \cloze{a Noetherian ring, then
    $R[X]$ is also Noetherian.}
\end{theorem*}
\end{flashcard}

\begin{flashcard}[Hilbert-basis-theorem-proof]
\prompt{Proof of Hilbert's Basis Theorem? \\}
\begin{proof}
    \cloze{
    Assume $J \normalsub R[X]$ is not finitely
    generated. Choose $f_1 \in J$ of minimal
    degree. Then $(f_1) \subsetneq J$. Choose $f_2
    \in J \setminus (f_1)$ of minimal degree. Then
    $(f_1, f_2) \subsetneq J$. Choose $f_3 \in J
    \setminus (f_1, f_2)$ of minimal degree and so
    on. We obtain a sequence $f_1, f_2, \ldots$
    with $\deg f_i \le \deg f_{i + 1}$. Set $a_i
    \defeq \text{leading coefficient of $f_i$}$.
    We obtain $(a_1) \subseteq (a_1, a_2)
    \subseteq \cdots$, a chain of ideals in $R$.
    Since $R$ is Noetherian, there exists $m \in
    \NN$ such that $a_{m + 1} \in (a_1, \ldots,
    a_m)$. Let $a_{m + 1} = \sum_{i = 1}^m
    \lambda_i a_i$, $\lambda_i \in R$ and set
    \[ g = \sum_{i = 1}^m \lambda_i f_i X^{\deg
    f_{m - 1} - \deg f_i} \in (f_1, \ldots, f_m) \]
    Then $\deg f_{m + 1} = \deg g$ and they have
    the same leading coefficient $a_{m + 1}$. Then
    $f_{m + 1} - g \in J$ and $\deg (f_{m + 1} -
    g) < \deg f_{m + 1}$. Hence by minimality of
    degree of $f_{m + 1}$, we must have $f_{m + 1}
    - g \in (f_1, \ldots, f_m)$. But $g \in (f_1,
    \ldots, f_m)$, hence $f_{m + 1} \in (f_1,
    \ldots, f_m)$, contradiction. Thus $J$ is
    finitely generated, so $R[X]$ is Noetherian by
    Lemma 13.1.
}
\end{proof}
\end{flashcard}

\begin{corollary*}
    \begin{itemize}
        \item $\ZZ[X_1, \ldots, X_n]$ is
            Noetherian.
        \item $F[X_1, \ldots, X_n]$ Noetherian,
            $F$ a field.
    \end{itemize}
\end{corollary*}

\subsubsection*{Examples}

Let $R = \CC[X_1, \ldots, X_n]$. Let $V \subseteq
\CC^n$ be a subset of the form
\[ \{(a_1, \ldots, a_n) \mid f(a_1, \ldots, a_n) =
0, \forall f \in \mathcal{F}\} \]
where $\mathcal{F} \subset R$ is a possibly
infinite set of polynomials. Let
\[ I = \left\{ \sum_{i = 1}^m \lambda_i f_i \mid m
\in \NN, \lambda_i \in R, f_i \in \mathcal{F} \right\}  \]
Then $I \normalsub R$, so $I = (g_1, \ldots,
g_r)$, $g_i \in I$ (since $R$ Noetherian). Thus
\[ V = \{(a_1, \ldots, a_n) \mid g_i(a_1, \ldots,
a_n) = 0, i = 1, \ldots, n\} \]
i.e. $V$ is defined by finitely many polymonials.

\begin{flashcard}[Noetherian-quotient]
\begin{lemma}
    Let $R$ be a Noetherian ring and $I \normalsub
    R$. Then \cloze{$R / I$ is Noetherian.}
\end{lemma}
\end{flashcard}

\begin{flashcard}[Noetherian-quotient-proof]
\prompt{Proof that quotients of Noetherian rings
are Noetherian? \\}
\begin{proof}
    \cloze{
    Let $J_1' \subseteq J_2' \subseteq \cdots$ a
    chain of ideals in $R / I$. By the ideal
    correspondence we have $J_i' = J_i / I$ for
    some $J_1 \subseteq J_2 \subseteq \cdots$ a
    chain of ideals in $R$ (containing $I$). $R$
    Noetherian implies there exists $N \in \NN$
    such that $J_n = J_{n + 1}$ for all $n \ge N$,
    hence $J_n' = J_{n + 1}$ for all $n \ge N$.
    Thus $R / I$ is Noetherian.}
\end{proof}
\end{flashcard}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item $\ZZ[i] = \ZZ[X] / (X^2 + 1)$ is
        Noetherian.
    \item $R[X]$ Noetherian implies $R[X] / X$ is
        Noetherian.
\end{enumerate}

\subsubsection*{Examples of non-Noetherian Rings}

\begin{flashcard}[non-Noetherian-examples]
\prompt{Examples of non-Noetherian rings? \\}
\begin{enumerate}[(i)]
    \item \cloze{$R = \ZZ[X_1, X_2, \ldots] = \bigcup_{n
        \ge 1} \ZZ[X_1, \ldots, X_n]$. i.e.
        polynomials in countably many variables.
        But $(X_1) \subseteq (X_1, X_2) \subsetneq
        (X_1, X_2, X_3) \subsetneq \cdots$ is an
        infinite ascending chain, so $R$ is not
        Noetherian.}
    \item \cloze{$R = \{f \in \QQ[X] : f(0) \in \ZZ\} \le
        \QQ[X]$. But:
        \[ (X) \subsetneq \left( \half X \right)
        \subsetneq \left( \frac{1}{4} X \right)
        \subsetneq \left( \frac{1}{8} X \right)
        \subsetneq \cdots \]
        (each inclusion is strict because $2 \in
        R$ is not a unit).}
\end{enumerate}
\end{flashcard}

\newpage
\mychapter{Modules}

\newpage
\section{Modules}

\begin{flashcard}[module-defn]
\begin{definition*}[Module]
    Let $R$ be a ring. A module over $R$ is a
    triple $(M, +, \cdot)$ consisting of a set $M$
    and two operations
    \[ + : M \times M \to M, \qquad \cdot : R
    \times M \to M \]
    such that
    \begin{enumerate}[(i)]
        \item \cloze{$(M, +)$ is an abelian group,
            say with identity $0$ (=$0_M$).}
        \item \cloze{The operation $\cdot$
            satisfies:
            \begin{align*}
                (r_1 + r_2) \cdot m
                &= r_1 \cdot m + r_2 \cdot m
                &&\forall r_1, r_2 \in R, m \in M
                \\
                r \cdot (m_1 + m_2)
                &= r \cdot m_1 + r \cdot m_2
                &&\forall r \in R, m_1, m_2 \in M
                \\
                r_1 \cdot (r_2 \cdot m)
                &= (r_1 \cdot r_2) \cdot m
                &&\forall r_1, r_2 \in R, m \in M
                \\
                1_R \cdot m
                &= m
                &&\forall m \in M
            \end{align*}
            }
    \end{enumerate}
\end{definition*}
\end{flashcard}

\begin{remark*}
    Don't forget closure when checking $+$,
    $\cdot$ well-defined.
\end{remark*}

\begin{example*}
    \begin{enumerate}[(i)]
        \item Let $R = F$ be a field. Then an
            $F$-module is \emph{precisely the
            same} as a vector space over $F$.
        \item $R = \ZZ$, a $\ZZ$-module is
            \emph{precisely the same} as an
            abelian group, where $\cdot : \ZZ
            \times A \to A$ maps
            \[ (n, a) \mapsto \begin{cases}
                \overbrace{a + a + \cdots +
                a}^{n \text{ copies}} & n > 0 \\
                0 & n = 0 \\
                -(\ub{a + a + \cdots + a}_{n
                \text{ copies}}) & n < 0
            \end{cases} \]
        \item $F$ a field, $V$ a vector space over
            $F$ and $\alpha : U \to V$ a linear
            map. We can make $V$ an $F[X]$-module
            via
            \[ \cdot : F[X] \times V \to V \qquad
            (f v) \mapsto (f(\alpha)(v)) \]
            for example $(X^2 + !) \cdot v =
            (\alpha^2 + 1_V)(v)$.
            \begin{note*}
                Different choices of $\alpha$ make
                $V$ into different $F[X]$-modules.
                Sometimes we'll write $V =
                V_\alpha$ to make this clear.
            \end{note*}
    \end{enumerate}
\end{example*}

\subsubsection*{Examples}

General construction.

\begin{enumerate}[(i)]
    \item For any ring $R$, $R^n$ is an $R$-module
        via $r \cdot (r_1, \ldots, r_n) = (r_1,
        \ldots, rr_n)$. In particular, taking $n =
        1$, $R$ is an $R$-module.
    \item If $I \normalsub R$, then $I$ is an
        $R$-module (restrict the usual
        multiplication on $R$) and $R / I$ is an
        $R$-module via
        \[ r \cdot (s + I) = rs + I \]
    \item $\phi : R \to S$ a ring homomorphism,
        then any $S$-module $M$ may be regarded as
        an $R$-module:
        \[ R\times M \to M \qquad (r, m) \mapsto
        \phi(r) \cdot m \]
        In particular, if $R \le S$ then any
        $S$-module may be viewed as an $R$-module.
\end{enumerate}