% vim: tw=50
% 11/02/2023 12AM

\begin{definition*}
    A polynomial
    \[ f(X) = a_n X^n + a_{n - 1} X^{n - 1} +
    \cdots + a_0 \in R[X] \]
    if \emph{monic} if $a_n = 1_R$.
\end{definition*}

\begin{lemma}
    Let $R$ be an integral domain and $0 \neq f
    \in R[X]$. Let
    \[ \Roots(f) = \{a \in R \mid f(a) = 0\} \]
    Then $|\Roots(f)| \le \deg(f)$.
\end{lemma}

\begin{proof}
    Example Sheet 2.
\end{proof}

\begin{flashcard}[finite-subgroup-of-field-is-cyclic]
\begin{theorem}
    Let $F$ be a field. Then any finite subgroup
    $G \le (F^\times, \bullet)$ is \cloze{cyclic.}
\end{theorem}

\begin{proof}
    \cloze{
    $G$ is a finite abelian group. If $G$ not
    cyclic, then by Theorem 6.4 (structure theorem
    for finite abelian groups) there exists $H \le
    G$ such that $H \cong C_{d_1} \times C_{d_1}$
    for some $d_1 \ge 2$. But then the polynomial
    $f(X) = X^{d_1} - 1 \in F[X]$ has degree $d_1$
    and $\ge d_1^2$ roots, which contradicts Lemma
    9.2.
}
\end{proof}
\end{flashcard}

\begin{example*}
    $(\ZZ / p\ZZ)^\times$ is cyclic.
\end{example*}

\begin{flashcard}[finite-integral-domain-is-field]
\begin{proposition}
    Any finite integral domain is a \cloze{field.}
\end{proposition}

\begin{proof}
    \cloze{
    Let $R$ be a finite integral domain. Let $0
    \neq a \in R$. Consider map $\phi : R \to R$,
    $x \mapsto ax$. If $\phi(x) = \phi(y)$, then
    $a(x - y) = 0$ therefore $x - y = 0$ (since
    $R$ is an integral domain and $a \neq 0$),
    hence $x = y$.

    \myskip
    Thus $\phi$ is injective, and hence surjective
    since $R$ is finite. Hence there exists $b \in
    R$ such that $ab = 1$, i.e. $a$ is a unit.
    Thus $R$ is a field.
}
\end{proof}
\end{flashcard}

\begin{flashcard}[field-of-fractions-existence]
\begin{theorem}[Field of Fractions Existence]
    \cloze{
    Let $R$ be an integral domain. There exists a
    field $F$ such that
    \begin{enumerate}[(i)]
        \item $R \le F$.
        \item Every element of $F$ can be written
            in the form $ab^{-1}$ where $a, b
            \in R$ with $b \neq 0$.
    \end{enumerate}
    $F$ is called the \emph{field of fractions} of
    $R$.
}
\end{theorem}

\begin{proof}
    \cloze{
    Consider the set $S = \{(a, b) \mid a, b \in
    R, b \neq 0\}$ and the equivalence relation on
    $S$ given by
    \[ (a, b) \sim (c, d) \iff ad - bc = 0 \]
    Clearly reflexive and symmetric. For
    transitivity, if $(a, b) \sim (c, d) \sim (e,
    f)$, then
    \[ (ad)f = (bc)f = b(cf) = b(de) \implies d(af
    - be) = 0 \]
    Since $R$ an integral domain and $d \neq 0$,
    this gives $af - be = 0$, i.e. $(a, b) \sim
    (e, f)$. Let $F = S / \sim$ and write
    $\frac{a}{b}$ for $[(a, b)]$. Define
    \[ \frac{a}{b} + \frac{c}{d} = \frac{ad +
    bc}{bd} \]
    and
    \[ \frac{a}{b} \cdot \frac{c}{d} =
    \frac{ac}{bd} \]
    Can be checked that these operations are well
    defined and maps $F$ into a ring with $0_F =
    \frac{0_R}{1_R}$ and $1_F = \frac{1_R}{1_R}$.

    \myskip
    If $\frac{a}{b} \neq 0_F$, then $a \neq 0_R$
    and $\frac{a}{b} \cdot \frac{b}{a} =
    \frac{ab}{ba} = \frac{1_R}{1_R} = 1_F$. So $F$
    is a field and
    \begin{enumerate}[(i)]
        \item Identify $R$ with subring $\left\{
            \frac{r}{1_R} : r \in R \right\} \le
            F$.
        \item $\frac{a}{b} = a \cdot b^{-1}$.
    \end{enumerate}
}
\end{proof}
\end{flashcard}

\begin{example*}
    \begin{enumerate}[(i)]
        \item $\ZZ$ is an integral domain with
            field of fractions $\QQ$.
        \item $\CC[X]$ has field of fractions
            $\CC(X) = \text{field of rational
            functions in $X$}$.
    \end{enumerate}
\end{example*}

\begin{flashcard}[ideal-defn]
\begin{definition*}
    An ideal $I \normalsub R$ is maximal if
    \cloze{$I
    \neq R$ and if $I \subseteq J \normalsub R$
    then $J = I$ or $R$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[field-iff-ideals]
\begin{lemma}
    A (non-zero) ring $R$ is a field if and only
    if \cloze{its only ideals are $\{0\}$ and
    $R$.}
\end{lemma}

\begin{proof}
    \cloze{
    \begin{itemize}
        \item[($\Rightarrow$)] If $0 \neq I
            \normalsub R$, then $I$ contains a
            unit and hence $I = R$.
        \item[($\Leftarrow$)] If $0 \neq x \in R$,
            then the $(x)$ is non-zero hence $(x)
            = R$ and there exists $y \in R$ such
            that $xy = 1$, i.e. $x$ is a unit.
            \qedhere
    \end{itemize}
}
\end{proof}
\end{flashcard}

\begin{flashcard}[maximal-iff-quotient-is-field]
\begin{proposition}
    Let $I \normalsub R$ be an ideal. $I$ is
    maximised if and only if \cloze{$R / I$ is a
    field.}
\end{proposition}

\begin{proof}
    \cloze{
    \begin{align*}
        \text{$R / I$ is a field}
        &\iff \text{$I / I$ and $R / I$ are the
        only ideals in $R / I$} \\
        &\iff \text{$I$ and $R$ are the only
        ideals in $R$ containing $I$} \\
        &\iff \text{$I \normalsub R$ is maximal}
        \qedhere
    \end{align*}
}
\end{proof}
\end{flashcard}

\begin{flashcard}[prime-ideal-defn]
\begin{definition*}
    An ideal $I \normalsub R$ is prime if \cloze{$I \neq
    R$ and whenever $a, b \in R$ with $a, b \in
    I$, we have $a \in I$ or $b \in I$.}
\end{definition*}
\end{flashcard}

\begin{example*}
    The ideal $n\ZZ \normalsub \ZZ$ is a prime
    ideal if and only if $n = 0$ or $n = p$ is a
    prime number. If $ab \in p\ZZ$, then $p \mid
    ab$ so $p \mid a$ or $p \mid b$, so $a \in
    p\ZZ$ or $b \in p\ZZ$. Conversely, if $n = uv$
    with $u, v > 1$, then $uv \in n\ZZ$, but $u
    \not\in n\ZZ$, $v \not\in n\ZZ$.
\end{example*}

\begin{flashcard}[prime-iff-quotient-is-integral-domain]
\begin{proposition}
    Let $I \normalsub R$ be an ideal. Then $I$ is
    prime if and only if \cloze{$R / I$ is an integral
    domain.}
\end{proposition}

\begin{proof}
    \cloze{
    \begin{align*}
        \text{$I$ is prime}
        &\iff \text{whenever $a, b \in R$ with $ab
        \in I$, we have $a \in I$ or $b \in I$} \\
        &\iff \text{whenever $a + I, b + I \in R /
        I$ with $(a + I)(b + I) = 0 + I$} \\
        &\phantom{\iff{}}~\text{we have
        $a + I = 0 + I$ or $b = 0 + I$} \\
        &\iff \text{$R / I$ is an integral
        domain.}
    \end{align*}
}
\end{proof}
\end{flashcard}

\begin{remark*}
    Proposition 9.7 and 9.8 show that $I$ maximal
    implies $I$ is prime.
    \begin{hiddenflashcard}[maximal-implies-prime]
    Maximal \cloze{$\implies$} prime. \\
    Proof: \cloze{If $I$ is maximal, then $R / I$
    is a field, hence an integral domain, so $I$
    is a field.}
    \end{hiddenflashcard}
\end{remark*}
