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\begin{document}
\title{Final Exam solutions}
\author{Math 116: Finashin, Pamuk, Pierce, Solak}
\date{June 7, 2011, 13:30--15:30 (120 minutes)}
\maketitle

\begin{instructions}
Work carefully.
Your methods should be clear to the reader.
\end{instructions}

\begin{problem}[12 points]
Is $x^3+x+1$ reducible over:
\begin{enumerate}
\item
$\Q$ ?
%\vspace{6mm}
\item $\Z_3$ ?
%\vspace{6mm}
\item $\Z_5$ ?
%\vspace{6mm}
\end{enumerate}
\end{problem}

\begin{solution}
Since the polynomial (call it $f$) has degree $3$, it is reducible if and 
only if it has a root.
\begin{enumerate}
\item 
If $p/q$ is a root, then $p$ and $q$ divide $1$, so $p/q=\pm1$.
But $f(1)=3$ and $f(-1)=-1$, so $f$ has no roots and
is irreducible over $\Q$.
\item
$f(1)=0$, so $f$ is
reducible over $\Z_3$ (divisible by $(x-1)$).
\item
$f(0)=1$, $f(1)=3$, $f(2)=11$, $f(3)=31$,
$f(4)=69$; so $f$ is irreducible over $\Z_5$.
\end{enumerate}
\end{solution}

\begin{problem}[8 points]
Letting $f(x)=3x^4+5x^3+x^2+5x-2$, write $f(x)$ as a product of irreducible polynomials over $\Q$.
\end{problem}
\begin{solution}
If $p/q$ is a root of $f$, then $p\divides2$ and $q\divides 3$, so
$p\in\{\pm1,\pm2\}$ and $q\in\{\pm1,\pm3\}$, hence
$p/q\in\{\pm1,\pm2,\pm\frac13,\pm\frac23\}$.
\begin{gather*}
f(1)=3+5+1+5-2\ne0,\\
f(-1)=3-5+1-5-2\ne0,\\
f(2)=48+40+4+10-2\ne0,\\
f(-2)=48-40+4-10-2=0.
\end{gather*}
Since $-2$ is a root, $f(x)$ is divisible by
$(x+2)$. Using the division algorithm,
\begin{equation*}
f(x)=(x+2)(3x^3-x^2+3x-1).
\end{equation*}
Also $1/3$ is a root of $3x^3-x^2+3x-1$, and
\begin{equation*}
 3x^3-x^2+3x-1=(3x-1)(x^2+1).
\end{equation*}
Therefore \fbox{$f(x)=(x+2)(3x-1)(x^2+1)$}.
Since $(x+2)$ and
$(3x-1)$ have degree 1, they are irreducible.  Since
 $x^2+1>0$ for all $x$ in $\Q$, it has no rational roots and is therefore
 irreducible.
\end{solution}

\begin{problem}[5 points]
Is every integral domain a field?  Explain.
\end{problem}
\begin{solution}
No, consider the ring of integers $(\Z,+,\cdot)$ which is an integral
domain, since it has no zero divisors. But it is not a field, since
the non-zero elements except $\pm 1$ do not have multiplicative
inverses.  
\end{solution}

\begin{problem}[15 points]
Find a polynomial $f(x)$ of least positive degree with the given properties.
%(Write out $f(x)$ in the form $a_dx^d+a_{d-1}x^{d-1}+\dotsb+a_0$.)
(Your answer should show the coefficients of $f(x)$.)
\begin{enumerate}
\item
$f(x)$ is over $\C$, and $f(2\mi)=0=f(1+\mi)$.
\item
$f(x)$ is over $\R$, and $2\mi$ and $1+\mi$ are zeros of it.
\item
$f(x)$ is over $\Z_2$, and $1$ (that is, $[1]$) is a zero of multiplicity $4$.
\end{enumerate}
\end{problem}

%\fbox{
\begin{answers}
\mbox{}
\begin{enumerate}
\item
$f(x)=(x-2\mi)(x-1-\mi)=x^2-(1+3\mi)x+2\mi-2$ over $\C$.
\item
$f(x)=(x-2\mi)(x+2\mi)(x-1-\mi)(x-1+\mi)=(x^2+4)(x^2-2x+2)=x^4-2x^3+6x^2-8x+8$
  over $\R$.
\item
$f(x)=(x-1)^4=x^4-4x^3+6x^2-4x+1=x^4+1$ over $\Z_2$. 
\end{enumerate}
\end{answers}

\begin{problem}[5 points]
Over a field $K$, suppose $f(x)$ is a polynomial with no zeros in $K$.
Must $f(x)$ be irreducible over $K$?  Explain. 
\end{problem}
\begin{solution}
No. For example, the polynomial $x^4+2x^2+1$ over $\R$ has no zeros
in $\R$ but it is reducible over $\R$.
\end{solution}
\vspace{1cm}
\begin{problem}[15 points]
Working over $\Z_3$, letting
\begin{align*}
f(x)&=x^5+x+1,& g(x)&=x^2+1,
\end{align*}
find $s(x)$ and $t(x)$ such that $f(x)\cdot s(x)+g(x)\cdot t(x)=1$.
\end{problem}

\begin{solution}
  \begin{align*}
f(x)  =  x^5+x+1& =g(x)\cdot(x^2+2x)+2x+1\\
g(x) = x^2+1&=(2x+1)\cdot(2x+2)+2.
  \end{align*}
Then
\begin{align*}
 2&=g(x)-(2x+1)(2x+2)\\
  &= g(x)-[f(x)-g(x)(x^3+2x)](2x+2)\\
  &= f(x)(x+1)+g(x)[1+(x^3+2x)(2x+2)]\\
  &= f(x)(x+1)+g(x)(2x^4+2x^3+x^2+x+1)
\end{align*}
Hence,
\begin{equation*}
1=f(x)(2x+2)+g(x)(x^4+x^3+2x^2+2x+2).
\end{equation*}
So, 
\begin{align*}
s(x)&=2x+2,&t(x)&=x^4+x^3+2x^2+2x+2.
\end{align*}
\end{solution}

\begin{problem}[20 points]
Let $\rc$ be the subring $\{x+y\mi\colon x,y\in\Z\}$ of $\C$, and let
$\I$ be the ideal $\{x+y\mi\colon x,y\in2\Z\}$ of $\rc$. 
\begin{enumerate}
\item
How many additive cosets has $\I$ in $\rc$?  List them clearly.
\item
Is the quotient $\rc/\I$ cyclic as an additive group?  Explain.
\item
Show that the function $\phi$ from $\rc$ to $\Z_2$ given by
\begin{equation}\label{eqn}
\phi(x+y\mi)=[x+y]
\end{equation}
is a ring homomorphism.
\item
Does the same formula~\eqref{eqn} define a ring homomorphism from
$\rc$ to $\Z_3$?  Explain. 
\end{enumerate}
\end{problem}

\begin{solution}\mbox{}
  \begin{enumerate}
  \item 
Four cosets: $I$, $1+I$, $\mi+I$, and $1+\mi+I$.
\item
No: $R/I$ has order $4$, but each element has order $1$ or $2$.
\item
\hfill
$\begin{aligned}[t]
    \phi((a+b\mi)+(x+y\mi))
&=\phi(a+x+(b+y)\mi)\\
&=[a+x+b+y]\\
&=[a+b]+[x+y]\\
&=\phi(a+b\mi)+\phi(x+y\mi),
  \end{aligned}$\hfill\mbox{}

$\begin{aligned}[t]
    \phi((a+b\mi)\cdot(x+y\mi))
&=\phi(ax-by+(ay+bx)\mi)\\
&=[ax-by+ay+bx]\\
&=[ax+by+ay+bx]\\
&=[a+b]\cdot[x+y]\\
&=\phi(a+b\mi)\cdot\phi(x+y\mi).
  \end{aligned}$
\item
No, since $\phi(\mi^2)=\phi(-1)=[-1]=[2]$, while $\phi(\mi)^2=[1]^2=[1]$.
  \end{enumerate}
\end{solution}

\begin{remark}
  For part (d), many people observed that the computations of (c) for
  multiplication were not justified \emph{modulo} $3$.  This is
  correct, but one must show that the system of equations \emph{fails}
  in at least one case.  (In fact the system is correct when
  $3\divides by$, but fails in all other cases.)
\end{remark}
\end{document}