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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 1}\\\vspace{15pt}
{\Large Due Tuesday, October 12 in class.}
\end{center}

\vspace{30pt}
\begin{question}
The set $\Z[i]$ is the collection of formal expressions of the form $a+bi$, where $a\in \Z$ and $b\in \Z$. In this context, the plus sign and the symbol $i$ are inert symbols with no meaning.
For example, $2+0i$, $3+2i$, and $-2+7i$ are all elements of $\Z[i]$.\\

If $x\in \Z[i]$ and $y\in\Z[i]$ are two elements of $\Z[i]$, we define the operation of addition as follows:
\[\text{if }x=a+bi\quad\text{ and }\quad y=c+di\text{, \ \ then }\qquad x+y=(a+c)+(b+d)i\]
For example:
\begin{align*}(2+0i)+(3+2i)\ \ \,&=5+2i\\
(3+2i)+(-2+7i)&=1+9i\\
(2+0i)+(-2+7i)&=0+7i
\end{align*}
\begin{enumerate}[a)]
\item Prove that the addition operation on $\Z[i]$ is associative (that is, that $\Z[i]$ satisfies Axiom A2).
\item What is the additive identity in $\Z[i]$? Prove that the element you found satisfies Axiom A4: for any $x\in \Z[i]$, we have $x+0=x$ and $0+x=x$.
\item What is the additive inverse of $3+4i$? In general, if $x=a+bi$, what is the additive inverse of $x$? Prove that your answer satisfies Axiom A5: $x+(-x)=0$.
\end{enumerate}\pagebreak

For $x\in \Z[i]$ and $y\in\Z[i]$, we define the operation of multiplication as follows:
\[\text{if }x=a+bi\text{ and }y=c+di\text{, then }x\cdot y=(ac-bd)+(bc+ad)i\]
For example:
\begin{align*}
(3+2i)\cdot(1+4i)&\ \ =(3-8)\quad\ \,+(2+12)i\quad=-5+14i\\
(2+0i)\cdot(3+2i)&\ \ =(6-0)\quad\ \,+(0+4)i\quad\ \,=6+4i\\
(1-2i)\cdot(-2+5i)&\ \ =(-2+10)\ +(4+5)i\quad\ \,=8+9i
\end{align*}\vspace{2pt}

\begin{enumerate}
\item[d)] Prove that the multiplication operation on $\Z[i]$ is associative (that is, $\Z[i]$ satisfies Axiom M2).
\item[e)] What is the multiplicative identity in $\Z[i]$? Prove that the element you found satisfies Axiom M4: for any $x\in \Z[i]$, we have $x\cdot 1=x$ and $1\cdot x=x$.
\item[g)] Prove that multiplication distributes over addition: if $x=a+bi$, $y=c+di$, and $z=e+fi$ are elements of $\Z[i]$, prove that
\[x\cdot (y+z)=x\cdot y\ +\ x\cdot z\]
\end{enumerate}
\end{question}\vspace{15pt}

\begin{question}
The set $\Q[i]$ is the collection of formal expressions of the form $a+bi$, where $a\in \Q$ and $b\in \Q$. (Recall that $\Q$ is the set of rational numbers.)
For example, $\frac{1}{2}+0i$, $3+2i$, and $-2+\frac{7}{3}i$ are all elements of $\Z[i]$.
We define  addition and multiplication on $\Q[i]$ by the same formulas as before: if $x=a+bi$ and $y=c+di$, then
\[x+y=(a+c)+(b+d)i\qquad\qquad\qquad x\cdot y=(ac-bd)+(bc+ad)i\]\vspace{1pt}

\begin{enumerate}[a)]
\item First, you should check that your proofs for Question 1(a--g) apply to $\Q[i]$ as well. You do not need to write anything for this part.
\item What is the multiplicative inverse of $3+4i$?
\item Which elements of $\Q[i]$ have a multiplicative inverse? Prove your answer is correct. (You may want to return to this after Question 3.)
\end{enumerate}
\end{question}\pagebreak

\begin{question}If $a+bi$ is an element of $\Z[i]$, its \emph{norm} $N(a+bi)\in \Z$ is defined to be:
\[N(a+bi)=a^2+b^2\] If $a+bi$ is an element of $\Q[i]$, we define its norm $N(a+bi)\in\Q$ by the same formula: $N(a+bi)=a^2+b^2$.\vspace{10pt}

\begin{enumerate}[a)]
\item Find all elements $x\in \Z[i]$ with $N(x)=1$.
\item Find all elements $x\in \Z[i]$ with $N(x)=2$.
\item Prove that for any $x\in \Z[i]$ and $y\in \Z[i]$ we have \[N(x\cdot y)=N(x)\cdot N(y).\] (You should check that your proof also works for $\Q[i]$; you do not need to write the proof again.)
\end{enumerate}
\end{question}

\end{document}