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\title{\vspace{-50pt}Elementary Number Theory\\
\normalsize Math 175, Section 30, Autumn 2010\\
\large Shmuel Weinberger ({\tt shmuel@math.uchicago.edu})\\
Tom Church ({\tt tchurch@math.uchicago.edu})\\
\url{www.math.uchicago.edu/~tchurch/teaching/175/}}
\author{}
\date{}


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\begin{document}
\maketitle
\vspace{-35pt}
\begin{center}
{\huge \bf Homework 5}\\\vspace{15pt}
{\Large Due Tuesday, November 9 in class.}
\end{center}

Recall that given $x,y\in \Z[i]$, we say that $y|x$ if there exists $z\in \Z[i]$ such that $x=yz$.
\begin{definitionq}[HW5.1]
Fix a nonzero element $x\in \Z[i]$. We say that two elements $y$ and $z$ are {\em congruent modulo $x$}
and write $y\equiv z\pmod{x}$ if and only if $x\,|\,(y-z)$.
\end{definitionq}
Congruence modulo $x$ is an equivalence relation.\footnote[1]{You may assume this without proving it.}

\begin{definitionq}[HW5.2]
For $y\in \Z[i]$, the \emph{residue class of $y$ modulo $x$} is the set of all elements $z\in \Z[i]$ which are congruent to $y$ modulo $x$:
\[[y]=\big\{z\in \Z[i]\big|y\equiv z\pmod{x}\big\}\]
\end{definitionq}\vskip15pt

\begin{question}
\label{finite}
Prove that for any nonzero $x\in \Z[i]$, there are only finitely many different residue classes modulo $x$.
\end{question}

\begin{question}
\label{size}
If $x=a+bi$, how many residue classes are there modulo $x$?
\end{question}\vskip25pt

\begin{definitionq}[HW5.3]
Fix a nonzero element $x\in\Z[i]$.
We define the number system $\Z[i]/(x)$ to be the set of residue classes modulo $x$. We define
addition and multiplication in $\Z[i]/(x)$ as follows:
\begin{align*}
[y]+[z]&=[y+z]\qquad\qquad\text{ for $y,z\in \Z$}\\
[y]\cdot[z]\ \,&=[y\cdot z]\qquad\qquad\ \,\text{ for $y,z\in \Z$}
\end{align*}
These operations are well-defined and make $\Z[i]/(x)$ into a commutative ring with identity; the additive identity is $[0]$, and the multiplicative identity is $[1]$.${}^{1}$
\end{definitionq}

\noindent(Thus Question~\ref{finite} asked you to prove that the ring $\Z[i]/(x)$ is finite, and Question~\ref{size} asked you to find its  cardinality $\big|\Z[i]/(x)\big|$.)

\begin{question} The following questions should be answered by concrete computations. For example, if one of these rings is not a field, you should give an explicit example of an nonzero element and a justification of why it does not have a multiplicative inverse.
\begin{enumerate}[a)]
\item For $x=2$, is $\Z[i]/(x)$ a field? 
\item For $y=3$, is $\Z[i]/(y)$ a field?
\item For $z=5$, is $\Z[i]/(z)$ a field?
\end{enumerate}
\end{question}\vskip40pt

In any commutative ring with identity, there are two related kinds of elements: \emph{irreducibles} and \emph{primes}. For $\Z$ and $\Z[i]$ these notions coincide and the terms can be (and are) used interchangably, but it is good to be familiar with the right general terminology.
\begin{definitionq}[HW5.4] Let $R$ be a commutative ring with identity.\newline
An element $x\in R$ is called \emph{prime} if \[x|ab\quad\text{ implies that either }\quad x|a\text{\ \ or\ \ }x|b.\]
An element $x\in R$ is called \emph{irreducible} if \[d|x\quad\text{ implies that either }\ \quad d|1\text{\ \ or\ \ }x|d.\] 
\end{definitionq}\vskip5pt
\begin{question}
We proved in Question HW3.1(c) that irreducible elements of $\Z[i]$ are prime. Prove the converse: if $x\in \Z[i]$ is prime, then $x$ is irreducible.
\end{question}\vskip20pt

\begin{question} Given $x\in \Z[i]$, prove that $\Z[i]/(x)$ is a field if and only if $x$ is a prime in $\Z[i]$.
\end{question}\vskip20pt

\begin{question} Prove that $2$ is irreducible in $\Z[\sqrt{-5}]$, but $2$ is not a prime in $\Z[\sqrt{-5}]$.\newline (So in general, the notions of ``irreducible'' and ``prime'' can be different.)
\end{question}


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