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\newpage
\section{Homomorphisms, Ideals and Quotients}

\begin{flashcard}[ring-homomorphism]
\begin{definition*}
    Let $R$ and $S$ be rings. A function $\phi : R
    \to S$ is a ring \emph{homomorphism} if
    \cloze{
    \begin{enumerate}[(i)]
        \item $\phi(r_1 + r_2) = \phi(r_1) +
            \phi(r_2)$ for all $r_1, r_2 \in R$.
        \item $\phi(r_1 r_2) = \phi(r_1) \cdot
            \phi(r_2)$ for all $r_1, r_2 \in R$.
        \item $\phi(1_R) = 1_S$
    \end{enumerate}
    }
    A ring homomorphism that is also a bijection
    is called an \emph{isomorphism}.
\end{definition*}
\end{flashcard}

The kernel of $\phi$ is
\[ \ker(\phi) = \{r \in R \mid \phi(r) = 0_S\} \]

\begin{flashcard}[ring-homomorphism-injective-iff-ker-zero]
\begin{lemma}
    A ring homomorphism $\phi : R \to S$ is
    injective if and only if \cloze{$\ker(\phi) =
    0_R$.}
\end{lemma}

\begin{proof}
    \cloze{
    $\phi : (R, +) \to (S, +)$ is a group
    homomorphism.
}
\end{proof}
\end{flashcard}

\begin{flashcard}[ideal-defn]
\begin{definition*}
    A subset $I \in R$ is an ideal (written $I
    \normalsub R$) if
    \cloze{
    \begin{enumerate}[(i)]
        \item $I$ is a subgroup of $(R, +)$
        \item If $r \in R$ and $x \in I$, then $rx
            \in I$.
    \end{enumerate}
}
    We say $I$ is \emph{proper} if \cloze{$I \neq
    R$.}
\end{definition*}
\end{flashcard}

\begin{hiddenflashcard}[how-to-check-ideal]
How to check if $I$ is an ideal of $R$? \\
\cloze{
Check:
\begin{enumerate}[(i)]
    \item $0 \in I$
    \item $I$ closed under addition
    \item $\forall r \in R, x \in I$, $rx \in I$
\end{enumerate}
(we deduce that additive inverses exist by
property (iii))
}
\end{hiddenflashcard}

\begin{flashcard}[ker-is-ideal]
\begin{lemma}
    If $\phi : R \to S$ is a ring homomorphism,
    then $\ker(\phi)$ is \cloze{an ideal of $R$.}
\end{lemma}

\begin{proof}
    \cloze{
    $\phi : (r, +) \to (S, +)$ is a group
    homomorphism, $\ker(\phi)$ is a subgroup of
    $(R, +)$. If $r \in R$ and $x \in \ker(\phi$),
    then
    \[ \phi(rx) = \phi(r)\phi(x) = \phi(r) \cdot
    0_S = 0_S \]
    hence $rx \in \ker(\phi)$.
}
\end{proof}
\end{flashcard}

\begin{remark*}
    If $I$ contains a unit, then $1_R \in I$ and
    hence $I = R$. Thus if $I$ is a proper ideal,
    $1_R \not\in I$, so $I$ is not a subring.
\end{remark*}

\begin{hiddenflashcard}[ideal-proper-iff]
$I \normalsub R$ is a proper ideal if and only if
\cloze{$1_R \not\in I$.}
\end{hiddenflashcard}

\begin{lemma}
    The ideals in $\ZZ$ are
    \[ n\ZZ = \{\ldots, -2n, -n, 0, n, 2n,
    \ldots\} \]
    for $n = 0, 1, 2\ldots$.
\end{lemma}

\begin{proof}
    Certainly $n\ZZ \normalsub \ZZ$. Let $I
    \normalsub \ZZ$ be a non-zero ideal, and $n$
    the smallest positive integer in $I$. Then
    $n\ZZ \subset I$. If $m \in I$, then write $m
    = qn + r$ with $q, r \in \ZZ$. Then $r = m -
    qn \in I$. Contradicts choice of $n$ unless $r
    = 0$. But then $m \in n\ZZ$, i.e. $I \subset
    n\ZZ$.
\end{proof}

\begin{flashcard}[ideal-generated-by]
\begin{definition*}
    For $a \in R$, write $(a) = \cloze{\{ra : r \in R\}
    \normalsub R}$. This is \cloze{the \emph{ideal
    generated by $a$}}. More generally, if $a_1,
    a_2, \ldots, a_n \in R$, we write
    \[ (a_1, \ldots, a_n) = \cloze{\{r_1 a_1 + \cdots r_n
    a_n \mid r_i \in R\} \normalsub R} .\]
\end{definition*}
\end{flashcard}

\begin{flashcard}[principal-ideal]
\begin{definition*}
    Let $I \normalsub R$. We say $I$ is
    \emph{principal} if \cloze{$I = (a)$ for some $a \in
    R$.}
\end{definition*}
\end{flashcard}

\begin{theorem}
    If $I \normalsub R$ then the set $R / I$ of
    cosets of $I$ in $(R, +)$ forms a ring (called
    the quotient ring) with operations
    \begin{align*}
        (r_1 + I) + (r_2 + I)
        &= r_1 + r_2 + I \\
        (r_1 + I)(r_2 + I)
        &= r_1 r_2 + I
    \end{align*}
    and $0_{R / I} = 0_R + I$, $1_{R / I} = 1_R +
    I$. Moreover, the map $R \to R / I$, $r
    \mapsto r + I$ is a ring homomorphism with
    kernel $I$.
\end{theorem}

\begin{proof}
    Already know $(R / I, +)$ is a group. If $r_1
    + I = r_1' + I$ and $r_2 + I = r_2' + I$, then
    \[ r_1' = r_1 + a_1, \qquad r_2' = r_2 + a_2 \]
    for some $a_1, a_2 \in I$. Then
    \begin{align*}
        r_1' r_2'
        &= (r_1 + a_1)(r_2 + a_2) \\
        &= r_1 r_2 + \ub{r_1 a_2}_{\in I} +
        \ub{r_2 a_1}_{\in I} + a_1 a_2
    \end{align*}
    thus $r_1' r_2' + I = r_1 r_2 + I$. Remaining
    properties for $R / I$ follow from those for
    $R$.
\end{proof}

\begin{example*}
    \begin{enumerate}[(i)]
        \item $n\ZZ \normalsub \ZZ$. Quotient ring
            $\ZZ / n\ZZ$. $\ZZ / n\ZZ$ has
            elements $0 + n\ZZ, 1 + n\ZZ, \ldots,
            (n - 1) + n\ZZ$. Addition and
            multiplication carried out $\mod{n}$.
        \item Consider $(X) \subset \CC[X]$
            (polynomials with 0 constant term). If
            \[ f(X) = a_n X^n + r \cdots a_1 X +
            a_0, \qquad a_1 \in \CC \]
            then $f(X) + (X) = a_0 + (X)$. There
            is a bijection $\CC[X] / (X) \to \CC$,
            $f(X) + (X) \mapsto f(0)$, $a + (X)
            \mapsfrom a$. These maps are ring
            homomorphisms. Thus $\CC[X] / (X)
            \cong \CC$.
        \item Consider $(X^2 + 1) \normalsub
            \RR[X]$
            \[ \RR[X] / (X^2 + 1) = \{f(X) + (X^2
            + 1) : f(X) \in \RR[X]\} \]
            By proposition 7.1, $f(X) = q(X)(X^2 +
            1) + r(X)$ with $\deg r < 2$, i.e.
            $r(X) = a + bX$, $a, b \in \RR$. Thus
            \[ \RR[X] / (X^2 + 1) = \{a + bX +
            (X^2 + 1) : a, b \in \RR\} \]
            If $a + bX + (X^2 + 1) = a' + b'X +
            (X^2 + 1)$. Then $a = a' + (b - b')X =
            g(X)(X^2 + 1)$ for some $g(X) \in
            \RR[X]$. Comparing degrees, we see
            $g(X) = 0$ and $a = a'$, $b = b'$.
            Consider the bijection
            \[ \RR[X] / (X^2 + 1) \to \CC, \qquad
            a + bX + (X^2 + 1) \mapsto a + bi \]
            We show $\phi$ is a ring homomorphism
            It preserves additions and maps $1 +
            (X^2 + 1)$ to $1$. Now we check that
            it respects multiplication:
            \begin{align*}
                &~~~~\phi((a + bX + (X^2 + 1))(c + dX +
                (X^2 + 1))) \\
                &= \phi((a + bX)(c + dX) + (X^2 +
                1)) \\
                &= \phi(ac + (ad + bc)X +
                \cancel{bd(X^2 + 1)} - bd + (X^2 +
                1)) \\
                &= ac - bd + (ad + bc)i \\
                &= \phi(a + bX + (X^2 + 1)) \phi(c
                + dX + (X^2 + 1))
            \end{align*}
            Thus $\RR[X] / (X^2 + 1) \cong \CC$.
    \end{enumerate}
\end{example*}