% vim: tw=50
% 09/02/2023 12AM

\begin{flashcard}[rings-first-iso-thm]
\begin{theorem*}[First Isomorphism Theorem for Rings]
    \cloze{
    Let $\phi : R \to S$ be a ring homomorphism.
    Then $\Ker(\phi) \normalsub R$, $\Im(\phi) \le
    S$ and there exists isomorphism
    \[ R / \Ker(\phi) \cong \Im(\phi) \]
}
\end{theorem*}

\begin{proof}
    \cloze{
    Already saw that $\Ker(\phi) \normalsub R$
    (Lemma 8.2), and $\Im(\phi)$ is a subgroup of
    $(S, +)$. Now
    \begin{align*}
        \phi(r_1) \phi(r_2)
        &= \phi(r_1 r_2) \in \Im(\phi) \\
        1_S
        &= \phi(1_R) \in \Im(\phi)
    \end{align*}
    Thus $\Im(\phi)$ is a subring of $S$. Let $K =
    \Ker(\phi)$. Define
    \begin{align*}
        \Phi : R / K
        &\to \Im(\phi) \\
        r + K
        &\mapsto \phi(r)
    \end{align*}
    By the first isomorphism theorem for groups,
    this is well-defined, a bijection and a group
    homomorphism under $+$. Also $\Phi(1_R + K) =
    \phi(1_R) = 1_S$ and
    \begin{align*}
        \Phi((r_1 + K)(r_2 + K))
        &= \Phi(r_1 r_2 + K) \\
        &= \phi(r_1 r_2) \\
        &= \phi(r_1)\Phi(r_2) \\
        &= \Phi(r_1 + K) \Phi(r_2 + K)
    \end{align*}
    Thus $\Phi$ is a ring isomorphism.
}
\end{proof}
\end{flashcard}

\begin{flashcard}[second-iso-thm-for-rings]
\begin{theorem*}[Second Isomorphism Theorem for Rings]
    \cloze{
    Let $R \le S$ and $J \normalsub S$. Then $R
    \cap J \normalsub R$, $R + J = \{r + a \mid r
    \in R, a \in J\} \le S$, and
    \[ \frac{R}{R \cap J} \cong \frac{R + J}{J}
    \le \frac{S}{J} \]
}
\end{theorem*}

\begin{proof}
    \cloze{
    By second isomorphism theorem for groups, $R +
    S$ is a subgroup of $(S, +)$, and we have
    \[ 1_S = \ub{1_S}_{ \in R} + \ub{0_S}_{\in J}
    \in R + J \]
    If $r_1, r_2 \in R$ and $a_1, a_2 \in J$ then
    \[ (r_1 + a_1)(r_2 + a_2) = \ub{r_1 r_2}_{\in
    J} + \ub{r_1 a_2}_{\in J} + \ub{r_2 a_1}_{\in
    J} + \ub{a_1 a_2}_{\in J} \in R + J \]
    Thus $R + J \le J$. Let $\phi : R \to S / J$,
    $r \mapsto r + J$. This is the composite of
    inclusion $R \subset S$ and the quotient map
    $S \to S / J$ hence $\phi$ is a ring
    homomorphism.
    \begin{align*}
        \Ker(\phi)
        &= \{r \in R \mid r + J = J\}
        = R \cap J \normalsub R \\
        \Im(\phi)
        &= \{r + J \mid r \in R\}
        = \frac{R + J}{J} \le \frac{S}{J}
    \end{align*}
    Apply first isomorphism theorem.
}
\end{proof}
\end{flashcard}

\begin{note*}
    Let $I \normalsub R$. There exists bijection
    \begin{align*}
        \{\text{ideals in $R / I$}\}
        &\leftrightarrow
        \{\text{ideals in $R$ containing $I$}\} \\
        K
        &\mapsto \{r \in R \mid r + I \in K\} \\
        J / I
        &\mapsfrom J
    \end{align*}
\end{note*}

\begin{flashcard}[third-iso-thm-for-rings]
\begin{theorem*}[Third Isomorphism Theorem for Rings]
    \cloze{
    Let $I \normalsub R$, $J \normalsub R$ with $I
    \le J$. Then $J / I \normalsub R / I$ and
    \[ \frac{R / I}{J / I} \cong \frac{R}{J} \]
}
\end{theorem*}

\begin{proof}
    \cloze{
    Consider
    \begin{align*}
        \phi : R / I
        &\to R / J \\
        r + I
        &\mapsto r + J
    \end{align*}
    This is a surjective ring homomorphism
    (well-defined since $I \le S$).
    \[ \Ker(\phi) = \{r + I : r \in J\} = J / I
    \normalsub R / I \]
    Apply first isomorphism theorem.
}
\end{proof}
\end{flashcard}

\begin{example*}
    There is a surjective ring homomorphism $\phi
    : \RR[X] \to \CC$
    \[ f(X) = \sum_{n = 1}^m a_n X^n \mapsto f(i)
    = \sum_{n = 1}^m a_n i^m \]
    Proposition 7.1 implies $\Ker(\phi) = (X^2 +
    1)$. First isomorphism theorem implies $\RR[X]
    / (X^2 + 1) \cong \CC$.
\end{example*}

\begin{example*}
    $R$ a ring. Then there exists a unique ring
    homomorphism $i : \ZZ \to R$ given by
    \begin{align*}
        0
        &\mapsto 0_R \\
        1
        &\mapsto 1_R \\
        n
        &\mapsto (\ub{1_R + \cdots +
        1_R}_{\text{$n$ times}}) \\
        -n
        &\mapsto -(1_r + \cdots + 1_R)
    \end{align*}
    Since $\Ker(i) \normalsub \ZZ$, have $\Ker(i)
    = n\ZZ$ for $n \in \{0, 1, 2, \ldots\}$. By
    first isomorphism theorem, $\ZZ / n\ZZ \cong
    \Im(i) \le R$.

    \begin{definition*}
        We call $n$ the characteristic of $R$. For
        example $\ZZ, \QQ, \RR$ and $\CC$ have
        characteristic $0$, and $\ZZ / p\ZZ$ or
        $\ZZ / p\ZZ [X]$ have characteristic $p$.
    \end{definition*}
    \begin{hiddenflashcard}[characteristic-of-ring]
        \begin{definition*}[Characteristic of a Ring]
            \cloze{
            The \emph{characteristic} of a ring $R$
            is the smallest $n \in \ZZ_{\ge 0}$ such
            that
            \[ \ub{1_R + 1_R + \cdots + 1_R}_{\text{$n$
            times}} = 0_R \]
            }
        \end{definition*}
    \end{hiddenflashcard}
\end{example*}

\newpage
\section{Integral domains, maximal ideals and prime ideals}

\begin{flashcard}[integral-domain-defn]
\begin{definition*}[Integral Domain and Zero-Divisor]
    \cloze{
    An integral domain is a ring with $0 \neq 1$
    such that for $a, b \in R$, $ab = 0 \implies a
    = 0$ or $b = 0$. A \emph{zero-divisor} in a
    ring $R$ is a non-zero element $a$ such that
    $ab = 0$ for some $0 \neq b \in R$. So an
    integral domain is a ring with no
    zero-divisors.
}
\end{definition*}
\end{flashcard}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item All fields are integral domains (if $ab =
        0$ with $b \neq 0$, multiply by $b^{-1}$
        to get $a = 0$)
    \item Any subring of an integral domain is an
        integral domain, for example $\ZZ \le
        \QQ, \ZZ[i] \le \CC$.
    \item $\ZZ \times \ZZ$ is not an integral
        domain since $(1, 0)(0, 1) = (0, 0)$.
\end{enumerate}

\begin{lemma}
    $R$ an integral domain $\implies$ $R[X]$ an
    integral domain.
\end{lemma}

\begin{hiddenflashcard}[when-do-degrees-add]
When can we say that $\deg(fg) = \deg(f) +
\deg(g)$? \\
\cloze{When $R$ is an integral domain.}
\end{hiddenflashcard}

\begin{proof}
    Write $f(X) = a_m x^m + \cdots + a_1 X + a_0$,
    $a_m \neq 0$, $g(X) = b_n X^n + \cdots + b_1 X
    + b_0$, $b_n \neq 0$. Then
    \[ f(X) g(X) = a_m b_n X^n + \cdots \]
    where $a_m b_n \neq 0$ since $R$ is an
    integral domain. Thus $\deg(fg) = m + n =
    \deg(f) + \deg(g)$ and $fg \neq 0$.
\end{proof}
