\documentclass[12pt]{article}
\usepackage{fancyhdr,fancybox,amssymb,epsfig, amsmath, latexsym}
\usepackage[all]{xy}
\usepackage{tikz}
\usetikzlibrary{arrows}
\usetikzlibrary{decorations.markings}
\usetikzlibrary{decorations.pathreplacing}
\usetikzlibrary{patterns}
\usetikzlibrary{shapes.geometric}
\usepackage{tkz-graph}
\usepackage{tikz-cd}
\usepackage{enumerate}
\usepackage{multicol}
\usepackage{bm}

\topmargin=-.5in
\headsep=0.2in
\oddsidemargin=0in
\textwidth=6.7in
\textheight=9.2in
\footskip=.5in


% Useful math macros
\def\ds{\displaystyle}
\def\bold{\bm}     % Change ``bm'' to ''mathbf'' for upright boldface.

\newcommand{\vv}[2]{\begin{bmatrix} #1 \\ #2 \end{bmatrix}}
\newcommand{\vvv}[3]{\begin{bmatrix} #1 \\ #2 \\ #3 \end{bmatrix}}
\newcommand{\vvvv}[4]{\begin{bmatrix} #1 \\ #2 \\ #3 \\ #4 \end{bmatrix}}
\newcommand{\vvvvv}[5]{\begin{bmatrix} #1 \\ #2 \\ #3 \\ #4 \\ #5 \end{bmatrix}}
\def\<{\langle}
\def\>{\rangle}
\newcommand{\ceil}[1] {\left\lceil #1 \right\rceil}
\newcommand{\floor}[1] {\left\lfloor #1 \right\rfloor}
\def\normal{\lhd}
\def\normaleq{\unlhd}
\def\nnormal{\ntriangleleft}
\def\nnormaleq{\ntrianglelefteq}
\def\Dot{\bm{.}}

%\def\subgroup{\leqslant}
%\def\nsubgroup{\nleqslant}
%\def\normal{\leqslant}
%\def\nnormal{\nleqslant}

\newcommand{\rddots}{\rotatebox{90}{$\ddots$}}
\def\setmin{-}

\def\Sum{\sum\limits}
\def\Prod{\prod\limits}
\def\Cap{\mathop\cap\limits}
\def\Cup{\mathop\cup\limits}
\def\Times{\mathop\times\limits}
\def\Oplus{\mathop\oplus\limits}
\def\Lim{\mathop\lim\limits}

\def\And{\wedge}
\def\Or{\vee}
\def\Not{\neg}

\def\longto{\longrightarrow}
\def\into{\hookrightarrow}
\def\longinto{\longhookrightarrow}

\def\A{\mathbb{A}}
\def\C{\mathbb{C}}
\def\F{\mathbb{F}}
\def\HH{\mathbb{H}}
\def\N{\mathbb{N}}
\def\Q{\mathbb{Q}}
\def\R{\mathbb{R}}
\def\Z{\mathbb{Z}}

\DeclareMathOperator{\Id}{Id}
\DeclareMathOperator{\Lcm}{lcm}
\DeclareMathOperator{\del}{\nabla}
\DeclareMathOperator{\Trace}{tr}
\DeclareMathOperator{\Span}{Span}
\DeclareMathOperator{\Image}{Im}
\DeclareMathOperator\Perm{Perm}
\DeclareMathOperator\Char{char}
\DeclareMathOperator\Aut{Aut}
\DeclareMathOperator\Inn{Inn}
\DeclareMathOperator\Out{Out}
\DeclareMathOperator\orb{orb}
\DeclareMathOperator\Orb{Orb}
\DeclareMathOperator\stab{stab}
\DeclareMathOperator\Stab{Stab}
\DeclareMathOperator\fix{fix}
\DeclareMathOperator\Fix{Fix}
\DeclareMathOperator\Syl{Syl}
\DeclareMathOperator\Pre{Pre}
\DeclareMathOperator\Ker{Ker}
\DeclareMathOperator\Sym{Sym}
\DeclareMathOperator\Mat{Mat}
\DeclareMathOperator{\cl}{cl}
\DeclareMathOperator{\GL}{GL}
\DeclareMathOperator{\SL}{SL}
\DeclareMathOperator\AGL{AGL}
\DeclareMathOperator\PGL{PGL}
\DeclareMathOperator\PSL{PSL}
\DeclareMathOperator\PSU{PSU}
\DeclareMathOperator\PSP{PsP}
\DeclareMathOperator\Heis{Heis}
\DeclareMathOperator\Dih{D}
\DeclareMathOperator\Dic{Dic}
\DeclareMathOperator\SA{SA}
\DeclareMathOperator\SD{SD}
\DeclareMathOperator\Mod{Mod}
\DeclareMathOperator\Gal{Gal}
\DeclareMathOperator\Deg{Deg}

\DeclareMathOperator\Rect{\mathbf{Rect}}
\DeclareMathOperator\Tri{\mathbf{Tri}}
\DeclareMathOperator\Coin{\mathbf{Coin}}
\DeclareMathOperator\BOX{\mathbf{Box}}
\DeclareMathOperator\Light{\mathbf{Light}}
\DeclareMathOperator\Sq{\mathbf{Sq}}

% Colors 
\usepackage{xcolor}
\usepackage{color}
\definecolor{darkblue}{rgb}{0, 0, .6}
\definecolor{grey}{rgb}{.7, .7, .7}
\newcommand{\balert}[1]{\textcolor{blue}{#1}}
\newcommand{\galert}[1]{\textcolor{darkgreen}{#1}}
\newcommand{\palert}[1]{\textcolor{amethyst}{#1}}
\newcommand{\Alert}[1]{\textcolor{Red}{#1}}
\newcommand{\Balert}[1]{\textcolor{Blue}{#1}}
\newcommand{\Galert}[1]{\textcolor{darkgreen}{#1}}

%\hypersetup{colorlinks=true,linkcolor=darkblue,anchorcolor=darkblue,
%  citecolor=darkblue,pagecolor=darkblue,urlcolor=darkblue,
%  pdftitle={},pdfauthor={}
%}                               %% THIS CAUSES COMPILER ERROR
\colorlet{lightred}{red!30!white}
\colorlet{lightblue}{blue!30!white}
\colorlet{lightyellow}{yellow!30!white}
\colorlet{keylime}{green!10!white}
\colorlet{darkgreen}{green!50!black}
\colorlet{midgreen}{green!70!black}
\colorlet{lightgreen}{darkgreen!30!white}
\colorlet{lightpurple}{violet!60!white}
\colorlet{darkgrey}{black!70}
\colorlet{midgrey}{black!50}
\definecolor{Red}{rgb}{.9,0,0}
\definecolor{Blue}{rgb}{0,0,.9}
\definecolor{Green}{rgb}{0,.9,0}
\definecolor{Gold}{rgb}{.83,.67,.08}

\definecolor{conj-purple}{HTML}{E4B7FF}
\definecolor{conj-orange}{HTML}{FFD4B7}


\colorlet{lred}{red!50!white}
\definecolor{lblue}{rgb}{0.60, 0.75, 1}
\colorlet{lgreen}{green!40!white}
\colorlet{lyellow}{yellow!30!white}
%\definecolor{lpurple}{rgb}{0.6 0.35 0.8}
\definecolor{lpurple}{rgb}{0.7 0.45 0.9}
\colorlet{lorange}{orange!50!white}
\definecolor{lightgrey}{rgb}{.85, .85, .85}
\definecolor{faded}{rgb}{.75,.75,.75}
\definecolor{midgrey}{rgb}{.5,.5,.5}
%\definecolor{midgrey}{rgb}{.35, .35, .35}
%\definecolor{amethyst}{rgb}{0.6 0.4 0.8}
\definecolor{amethyst}{rgb}{0.54 0.17 0.89}
\definecolor{amber}{rgb}{1.0 0.49 0.0}
\definecolor{turq}{rgb}{0.12 0.69 0.71}


\colorlet{lightestblue}{blue!15!white}
\colorlet{midblue}{blue!45!white}

\colorlet{lightestred}{red!15!white}
\colorlet{midred}{red!45!white}


% COLORS FOR MULTIPLICATION TABLES  
\colorlet{tred}{red!70!white}
\colorlet{tblue}{blue!60!white}     %(higher number is more faded)
\colorlet{tgreen}{green!75!black}   %(higher number is brighter)
\colorlet{tyellow}{yellow!60!white}
\colorlet{tlime}{green!80!white}
\colorlet{tpink}{pink}
\colorlet{tpurple}{lpurple}
\colorlet{torange}{orange}


\newcommand*{\n}{5}   % counter for Cayley tables

%% CAYLEY DIAGRAM GLOBAL DEFAULTS
\tikzstyle{v} = [circle, draw, fill=lightgrey,inner sep=0pt, 
  minimum size=6mm]
\tikzstyle{r} = [draw, very thick, Red, -stealth]        % Red --->
\tikzstyle{b} = [draw, very thick, Blue, -stealth]       % Blue --->
\tikzstyle{g} = [draw, very thick, darkgreen, -stealth]  % Green --->
\tikzstyle{p} = [draw, very thick, amethyst, -stealth]     % Purple --->
\tikzstyle{rr} = [draw, very thick, Red]                 % Red ----
\tikzstyle{bb} = [draw, very thick, Blue]                % Blue ----
\tikzstyle{gg} = [draw, very thick, darkgreen]           % Green ----
\tikzstyle{oo} = [draw, very thick, orange]           % Orange ----
\tikzstyle{pp} = [draw, very thick, amethyst]           % Purple ----
\tikzstyle{tt} = [draw, very thick, turq]           % Turquoise ----
\tikzstyle{yy} = [draw, very thick, Gold]           % Gold ----


\tikzstyle{R} = [draw, very thick, Red, -stealth']        % Red --->
\tikzstyle{B} = [draw, very thick, Blue, -stealth']       % Blue --->
\tikzstyle{G} = [draw, very thick, darkgreen, -stealth']  % Green --->

\tikzset{r/.style={draw, very thick, Red, -stealth}}        % Red --->
\tikzset{R/.style={draw, very thick, Red, -stealth'}}        % Red --->

\tikzset{v/.style={circle, draw, fill=lightgrey,inner sep=0pt, 
  minimum size=6mm}}

\renewcommand\labelitemi{\rule[0.12em]{0.4em}{0.4em}}
\renewcommand\labelitemii{\normalfont\bfseries \rule[0.15em]{0.3em}{0.3em}}
\renewcommand\labelitemiii{\textasteriskcentered}
\renewcommand\labelitemiv{\textperiodcentered}

\renewcommand{\footrulewidth}{1pt}
\newenvironment{proof}[1][Proof.]{\smallskip\begin{trivlist}
    \item[\hskip \labelsep {\sffamily #1}]}{\qed\end{trivlist}\bigskip}
\newenvironment{sol}[1][Solution.]{\smallskip\begin{trivlist}
    \item[\hskip \labelsep {\sffamily #1}]}{\qed\end{trivlist}\bigskip}
\newenvironment{fminipage}%
{\setlength{\fboxsep}{15pt}\begin{Sbox}\begin{minipage}}%
{\end{minipage}\end{Sbox}\fbox{\TheSbox}}
\newcommand{\qed}{\hfill \ensuremath{\Box}}
\newcommand{\disp}{\displaystyle}


\pagestyle{fancy} \lhead{{\sf Homework 13 $|$ Due December 2 (Friday)
}} \rhead{\thepage} \cfoot{{\sf Math 4120/6120 $|$ Visual Algebra $|$
    Fall 2022 $|$ M.~Macauley}}

\begin{document}

$\;$ \vspace{-6mm}

\begin{enumerate}

  %%--------------------------------------------------------------------

%
\item The subgroup lattices of $\Z_4\times \Z_4$ and $\Z_6\times \Z_3$ are shown below. Re-draw these lattices with the subgroups written with generator(s), and then construct their \emph{subring lattices} by coloring each subgroup based on whether it is an \Balert{ideal}, \Alert{subring but not an ideal}, or {\color{faded}subgroup that is not a subring}.
  \[
  \hspace*{-8mm}
  \begin{tikzpicture}[shorten >= -2pt, shorten <= -2pt,scale=.93]
    \tikzstyle{every node}=[font=\normalsize]
    \begin{scope}[shift={(0,0)},scale=1.25,yscale=1.25]
      \node (G) at (0,4) {$\Z_4\!\times\!\Z_4$};
      \node (C4xC2-1) at (-1.5,3) {$\Z_4\!\times\!\Z_2$};
      \node (C4xC2-2) at (0,3) {$\Z_4\!\times\!\Z_2$};
      \node (C4xC2-3) at (1.5,3) {$\Z_4\!\times\!\Z_2$};
      \node (C4-1) at (-3,2) {$\Z_4$};
      \node (C4-2) at (-2.25,2) {$\Z_4$};
      \node (C4-3) at (-1.5,2) {$\Z_4$};
      \node (C4-4) at (1.5,2) {$\Z_4$};
      \node (C4-5) at (2.25,2) {$\Z_4$};
      \node (C4-6) at (3,2) {$\Z_4$};
      \node (V4) at (0,2) {$\Z_2^2$};
      \node (C2-1) at (-1.5,1) {$\Z_2$};
      \node (C2-2) at (0,1) {$\Z_2$};
      \node (C2-3) at (1.5,1) {$\Z_2$};
      \node (1) at (0,0) {$\Z_1$};
      \draw[faded] (G) to (C4xC2-1);
      \draw[faded] (G) to (C4xC2-2);
      \draw[faded] (G) to (C4xC2-3);
      \draw[faded] (C4xC2-1) to (C4-1);
      \draw[faded] (C4xC2-1) to (C4-2);
      \draw[faded] (C4xC2-1) to (V4); 
      \draw[faded] (C4xC2-2) to (C4-3);
      \draw[faded] (C4xC2-2) to (C4-4);
      \draw[faded] (C4xC2-2) to (V4); 
      \draw[faded] (C4xC2-3) to (C4-5);
      \draw[faded] (C4xC2-3) to (C4-6);
      \draw[faded] (C4xC2-3) to (V4);
      \draw[faded] (C4-1) to (C2-1); \draw[faded] (C4-2) to (C2-1);
      \draw[faded] (C4-3) to (C2-2); \draw[faded] (C4-4) to (C2-2);
      \draw[faded] (V4) to (C2-1); 
      \draw[faded] (V4) to (C2-2); \draw[faded] (V4) to (C2-3);
      \draw[faded] (C4-5) to (C2-3); \draw[faded] (C4-6) to (C2-3);
      \draw[faded] (C2-1) to (1); 
      \draw[faded] (C2-2) to (1); 
      \draw[faded] (C2-3) to (1);
    \end{scope}
    %%
    \begin{scope}[shift={(9,0)},scale=1.44,yscale=1.5]
      \newcommand\He{3} % order 18
      \newcommand\Hd{2.2} % order 9
     \newcommand\Hc{1.9} % order 6
     \newcommand\Hb{1.1} % order 3
      \newcommand\Ha{.8} % order 2
      %%
      \node (G) at (0,\He) {$\Z_6\!\times\!\Z_3$};
      \node (C3^2) at (-1.5,\Hd) {$\Z_3^2$};
      \node (C6-1) at (-.25,\Hc) {$\Z_6$};
      \node (C6-2) at (.75,\Hc) {$\Z_6$};
      \node (C6-3) at (1.75,\Hc) {$\Z_6$};
      \node (C6-4) at (2.75,\Hc) {$\Z_6$};
      %%
      \node (C3-1) at (-2.75,\Hb) {$\Z_3$};
      \node (C3-2) at (-1.75,\Hb) {$\Z_3$};
      \node (C3-3) at (-.75,\Hb) {$\Z_3$};
      \node (C3-4) at (.25,\Hb) {$\Z_3$};
      \node (C2) at (1.5,\Ha) {$\Z_2$};
      %%
      \node (1) at (0,0) {$\Z_1$};
      %%

      %%
      \draw[faded] (G) to (C6-1); \draw[faded] (G) to (C6-2); 
      \draw[faded] (G) to (C6-3); \draw[faded] (G) to (C6-4);
      \draw[faded] (G) to (C3^2);
      %%
      \draw[faded] (C2) to (C6-1); \draw[faded] (C2) to (C6-2);
      \draw[faded] (C2) to (C6-3); \draw[faded] (C2) to (C6-4);
      %%
      \draw[faded] (C6-1) to (C3-1); \draw[faded] (C6-2) to (C3-2);
      \draw[faded] (C6-3) to (C3-3); \draw[faded] (C6-4) to (C3-4);
      %%
      \draw[faded] (C3^2) to (C3-1); \draw[faded] (C3^2) to (C3-2);
      \draw[faded] (C3^2) to (C3-3); \draw[faded] (C3^2) to (C3-4);
      %%
      \draw[faded] (C3-1) to (1); 
      \draw[faded] (C3-2) to (1); 
      \draw[faded] (C3-3) to (1);
      \draw[faded] (C3-4) to (1);
      \draw[faded] (C2) to (1); 
    \end{scope}
    \end{tikzpicture}
    \]
    Finally, for each ring, write down the units, and the zero
    divisors. \medskip
  
    %%------------------------------------------------------------------

  \item All of the isomorphism theorems for groups have analogues for
    rings. The proofs just amount to showing that the group
    homomorphisms are additionally ring homomorphisms, and we will
    carry out these details in this problem.
    \begin{enumerate}
      \item The \emph{fundamental homomorphism theorem} (FHT) says that
        $R/\Ker(\phi)\cong\Image(\phi)$. To prove this, we construct a
        map
        \[
        \iota\colon R/I\longto\Image(\phi),\qquad\quad\iota(r+I)=\phi(r),
        \]
        which we already know is a well-defined group
        isomorphism. Show that it is also a ring homomorphism.
      \item By the \emph{correspondence theorem}, every subgroup of
        $R/I$ has the form $J/I$ for some $I\leq J\leq R$. Show that
        $J/I$ is an ideal of $R/I$ if and only if $J$ is an ideal of
        $R$.

      \item The \emph{fraction theorem} says that $(R/I)/(J/I)\cong R/J$. This is proven by constructing a map
        \[
        \phi\colon R/I\longto R/J,\qquad\quad \phi(r+I)=r+J
        \]
        which we already know is a group homomorphism
        with $\Ker(\phi)=J/I$. Show that it is also a ring
        homomorphism, and then apply the FHT. 
      \item The \emph{diamond theorem} says that $(S+I)/I\cong
        S/(S\cap I)$ for a subring $S$ and ideal $I$.
        \begin{enumerate}[(i)]
        \item Prove that $S\cap I$ is an ideal of $S$.
        \item We already know that
          \[
          \phi\colon S\longto (S+I)/I,\qquad\quad \phi(s)=s+I
          \]
          is a group homomorphism with $\Ker(\phi)=S\cap I$. Show that
          it is also a ring homomorphism, and then apply the FHT. 
        \end{enumerate}
          
    \end{enumerate}


    %%---------------------------------------------------------------------  
  
\item Let  $I$ and $J$ be ideals of a commutative ring $R$.
  \begin{enumerate}
  \item Show that $I+J$, $I\cap J$, and $IJ$ are ideals of $R$. Which
    of these remain ideals if the commutativity hypothesis is dropped?
  \item The set $(I:J):=\{r\in R\mid rJ\subseteq I\}$ is called the
    \emph{ideal quotient} or \emph{colon ideal} of $I$ and $J$. Show
    that $(I:J)$ is an ideal of $R$. Does this require commutativity?
  \item Determine $I+J$, $I\cap J$, $IJ$, and $(I:J)$ for the ideals
    $I=n\Z$ and $J=m\Z$ of $R=\Z$.
  \item Repeat Part~(c) for several pairs of ideals of $R=\Z_6\times
    \Z_4$, whose subring lattice is shown below.
  \item Describe how to find $IJ$ and $(I:J)$ by inspection,
    using only the subring lattice, if possible. \vspace{-8mm}
  \end{enumerate}

  \[
  \hspace*{-8mm}
   \newcommand\Hh{6-.1} % height of G
   \newcommand\Hg{4.75-.1} % height of \Z_{12},  \Z_6 x \Z_2 
   \newcommand\Hf{3.75} % height of \Z_4 x C2  
   \newcommand\He{3.5-.1} % height of \Z_6  
   \newcommand\Hd{2.5} % height of \Z_4, V_4
   \newcommand\Hc{2.2-.1} % height of \Z_3  
   \newcommand\Hb{1.25} % height of \Z_2 
   \newcommand\Ha{0} % height of \Z_1  
  \colorlet{Ideal}{Blue}
  \colorlet{Subring}{Red}
  \colorlet{Neither}{faded}
  \begin{tikzpicture}[shorten >= -2pt, shorten <= -2pt,scale=1.4]
    \tikzstyle{every node}=[font=\small]
    \begin{scope}[shift={(7,1.3)},xscale=1.5]
      \node[anchor=east] at (-3,\Hh) {\normalsize \textbf{Order} $=24$};
      \node[anchor=east]  at (-3,\Hg) {\normalsize $12$};
      \node[anchor=east]  at (-3,\Hf) {\normalsize $8$};
      \node[anchor=east]  at (-3,\He) {\normalsize $6$};
      \node[anchor=east]  at (-3,\Hd) {\normalsize $4$};
      \node[anchor=east]  at (-3,\Hc) {\normalsize $3$};
      \node[anchor=east]  at (-3,\Hb) {\normalsize $2$};
      \node[anchor=east]  at (-3,\Ha) {\normalsize $1$};
      %%
      \node[anchor=east] at (3.5,\Hh) {\normalsize \textbf{Index} $=24$};
      \node[anchor=east]  at (3.5,\Hg) {\normalsize $12$};
      \node[anchor=east]  at (3.5,\Hf) {\normalsize $8$};
      \node[anchor=east]  at (3.5,\He) {\normalsize $6$};
      \node[anchor=east]  at (3.5,\Hd) {\normalsize $4$};
      \node[anchor=east]  at (3.5,\Hc) {\normalsize $3$};
      \node[anchor=east]  at (3.5,\Hb) {\normalsize $2$};
      \node[anchor=east]  at (3.5,\Ha) {\normalsize $1$};
      %%
      \node[Ideal] (G) at (0,\Hh) {$\Z_6\times\Z_4$};
      \node[Ideal] (10-02) at (-1.25,\Hg) {$\<(10,02)\>$}; % Z_6 x Z_2
      \node[Subring] (11) at (0,\Hg) {$\<(11)\>$}; % Z_{12}
      \node[Ideal] (21) at (1.25,\Hg) {$\<(21)\>$}; % Z_{12}
      %%
      \node[Ideal] (30-01) at (1.25,\Hf) {$\<30,01\>$}; % Z_2 x Z_4
      %%
      \node[Ideal] (10) at (-2.5,\He) {$\<10\>$};  % Z_6
      \node[Subring] (12) at (-1.25,\He) {$\<12\>$};  % Z_6
      \node[Ideal] (22) at (0,\He) {$\<22\>$};  % Z_6
      \node[Ideal] (30-02) at (0,\Hd) {$\<30,02\>$}; % V_4
      \node[Subring] (31) at (1.25,\Hd) {$\<31\>$}; % \Z_4
      \node[Ideal] (01) at (2.5,\Hd) {$\<01\>$}; % \Z_4
      %%      
      \node[Ideal] (20) at (-1.25,\Hc) {$\<20\>$};  % \Z_3
      %%
      \node[Ideal] (30) at (-1.25,\Hb) {$\<30\>$};  % \Z_2 
      \node[Subring] (32) at (0,\Hb) {$\<32\>$};  % \Z_2 
      \node[Ideal] (02) at (1.25,\Hb) {$\<02\>$};  % \Z_2 
      %%
      \node[Ideal] (00) at (0,\Ha) {$\<00\>$}; % \Z_1
      %%
      \draw[faded] (G) to (10-02); \draw[faded] (G) to (21);  
      \draw[faded] (G) to (11);  \draw[faded] (G) to (30-01); 
      %%
      \draw[faded] (10-02) to (30-02); \draw[faded] (10-02) to (12);
      \draw[faded] (10-02) to (22); \draw[faded] (10-02) to (10);
      \draw[faded] (30-01) to (01); \draw[faded] (30-01) to (31); 
      \draw[faded] (30-01) to (30-02);
      \draw[faded] (10) to (20); \draw[faded] (10) to (30);
      \draw[faded] (22) to (20); \draw[faded] (22) to (02);
      \draw[faded] (12) to (20); \draw[faded] (12) to (32);
      \draw[faded] (01) to (02); \draw[faded] (31) to (02);
      \draw[faded] (21) to (22); \draw[faded] (21) to (01);
      \draw[faded] (11) to (22); \draw[faded] (11) to (31);
      %%
      \draw[faded] (30-02) to (30); \draw[faded] (30-02) to (02); 
      \draw[faded] (30-02) to (32);
      \draw[faded] (00) to (02); \draw[faded] (00) to (32); \draw[faded] (00) to (30); \draw[faded] (00) to (20); 
    \end{scope}
  \end{tikzpicture}
\]

%%------------------------------------------------------------------
  
\item Let $f\colon R\rightarrow S$ be a ring homomorphism between
  commutative rings.
  \begin{enumerate} 
  \item If $f$ is surjective and $I$ is an ideal of $R$, show that
    $f(I)$ is an ideal of $S$.
  \item Show that Part~(a) is not true in general when $f$ is not
    surjective.
  \item Show that if $f$ is surjective and $R$ is a field, then $S$ is
    a field as well. \medskip
  \end{enumerate}

%%---------------------------------------------------------------------
  
\item Let $R$ be a commutative ring with $1$.
  \begin{enumerate}
    \item Show that if $x$ is contained in every maximal ideal, then
      $1+x$ is a unit.
    \item A ring is \emph{local} if it has a unique maximal
      ideal. Show that $R$ is local if and only if the non-units form
      an ideal.
    \item Let $p$ be a fixed prime. Show that the ring 
      \[
      R=\big\{\tfrac{a}{b}\mid a,b\in\Z,\;(a,b)=1,\; p\nmid b\big\}\subseteq\Q
      \]
      is local, and characterize units and maximal unique ideal.
  \end{enumerate}
  
%%------------------------------------------------------------------
 
  
\end{enumerate}

\end{document}



