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\runningheader{Mathematics 403}{}{Palmieri}


\hspace{2.65in}Name: \underline{\hspace{0.5in}Solutions\hspace{2in}}

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\begin{center}
\textbf{{\large Mathematics 403 Exam}} \\
February 2, 2005
\end{center}

\vspace{0.1in}

\noindent 
\textbf{Instructions}: This is a closed book exam, no notes or
calculators allowed.  Justify all of your answers.  You may refer to
and use any result from the book, but for full credit, do not use
results from homework or practice problems -- if you need one of those
results, you should prove it.

$\Z$ denotes the ring of integers.  $\F{p}$ denotes the field with $p$
elements: $\F{p} = \Z /(p)$.  All rings mentioned in this exam are
assumed to be commutative.

This is a timed exam, so you may use abbreviations and symbols (such
as ``$\forall$''): as long as I can make sense of what you write
without struggling too much, it's okay.

\vspace{0.1in}

\addpoints

\begin{questions}

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\question[10]
Let $F$ be a field.  What are all of the ideals in $F[x]/(x^{2}-1)$?
(Your answer may depend on the characteristic of $F$.)

\answer. According to the ``correspondence theorem,'' the ideals of
$F[x]/(x^{2}-1)$ are in bijection with the ideals of $F[x]$ which
contain $(x^{2}-1)$.  The ideals of $F[x]$ are all principal and so
are of the form $(f(x))$ for some polynomial $f(x)$.  Furthermore,
$(f(x))$ contains $(x^{2}-1)$ if and only if $f(x)$ divides $x^{2}-1$.
The polynomial $x^{2}-1$ factors as $x^{2}-1=(x-1)(x+1)$, so its
divisors are (up to unit multiple)
\[
1, \, x-1, \, x+1, \, x^{2}-1.
\]
When the characteristic of $F$ is 2, the middle two terms are equal;
otherwise, they are all distinct.  Therefore when the characteristic
of $F$ is 2, there are three ideals in $F[x]/(x^{2}-1)$:
\[
(1), \ (\overline{x}+1), \ (\overline{x}^{2}+1)=(0).
\]
When the characteristic of $F$ is not 2, there are four ideals in
$F[x]/(x^{2}-1)$:
\[
(1), \ (\overline{x}-1), \ (\overline{x}+1), \ (\overline{x}^{2}-1)=(0).
\]
(I'm writing $\overline{x}$ for the residue of $x$ in the quotient
ring $F[x]/(x^{2}-1)$.)

\newpage

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\question[10]
Let $R$ be an integral domain.  Given nonzero polynomials $f(x), g(x)
\in R[x]$, prove that $\deg (f(x) g(x)) = \deg (f(x)) + \deg (g(x))$.

\answer. Let $f(x)=\sum_{i=0}^{m}a_{i}x^{i}$ with $a_{m} \neq 0$, so
that $\deg f(x) = m$.  Let $g(x)=\sum_{j=0}^{n} b_{j}x^{j}$ with
$b_{n}\neq 0$, so that $\deg g(x) = n$.  In the product $f(x)
g(x)$, the coefficient of $x^{k}$ is $\sum_{i+j=k} a_{i} b_{j}$.  This
means that when $k>m+n$, this coefficient is zero, and when $k=m+n$,
this coefficient equals $a_{m}b_{n}$.  Since $R$ is an integral domain
and since $a_{m}$ and $b_{n}$ are nonzero, their product is also
nonzero.  Thus the leading term of $f(x) g(x)$ is $a_{m} b_{n}
x^{m+n}$, so $\deg (f(x) g(x)) = m+n=\deg (f(x)) + \deg (g(x))$.

\vfill

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\question[5]
For which prime numbers $p$ is $\F{p}[x]/(x^{2}+1)$ a field?  

This is a hard question in general, so I would like you to make an
educated guess, and to provide justification and evidence for that
guess.

\answer. The answer is: all primes $p$ which are congruent to 3 mod
4.

To come up with this guess, $\F{p}[x]/(x^{2}+1)$ is a field if and
only if the ideal $(x^{2}+1)$ is maximal, which is true if and only if
the polynomial $x^{2}+1$ is irreducible.  Because this is a degree 2
polynomial, this is true if and only if it has no roots: if and only
if $x^{2} \neq -1$ for all $x \in \F{p}$.  In summary:
$\F{p}[x]/(x^{2}+1)$ is a field if and only if $-1=p-1$ is not a
square in $\F{p}$.

Now try some primes to see what happens.

\begin{itemize}
\item When $p=2$, $1=-1$ is a square, so we don't get a field.
\item When $p=3$, $0^{2}=0$, $1^{2}=1$, and $2^{2}=1$, so $-1=2$ is
not a square, so we get a field.
\item When $p=5$, $2^{2}=4=-1$, so we don't get a field.
\item When $p=7$, the squares are $0$, $1$, $2$, and $4$, so $-1=6$ is
not a square and we get a field.
\item When $p=11$, the squares are $0$, $1$, $4$, $5$, and $9$, so
$-1=10$ is not a square and we get a field.
\item When $p=13$, $5^{2}=25=-1$, so we don't get a field.
\item When $p=17$, $4^{2}=16=-1$, so we don't get a field.
\end{itemize}

So the good primes (so far) are $3$, $7$ and $11$.  These are all
congruent to 3 mod 4.  The bad primes (so far) are $2$, $5$, $13$,
$17$.  These are all congruent to 1 or 2 mod 4.

\vfill

\newpage

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\question[5]
The ring $\Z [x]/(9x^{2}+1)$ is isomorphic to a subring of the complex
numbers.  Which subring?

\answer. View this ring as being $\Z$ with an element $x$ adjoined
which satisfies the relation $9x^{2}+1=0$, or $9x^{2}=-1$, or
$x^{2}=-1/9$, so $x=\pm i/3$.  So I would guess that the answer is $\Z
[i/3]$; this is easy to verify with the first isomorphism theorem.

(Note, by the way, that $\Z [i/3]$ is \emph{not} equal to $\{
a+\frac{bi}{3} : a, b, \in \Z \}$.  Since $i/3$ and 3 are in our ring,
so is their product $i$, and hence so is $(i)(i/3) = -1/3$.  Note that
$-1/3$ is not in $\{ a+\frac{bi}{3} : a, b, \in \Z \}$.)

\vfill

\question[5]
Let $a$, $b$, and $k$ be positive integers, and let $n=a^{k}b$.  Show
that the residue of $ab$ is nilpotent in $\Z /(n)$.  (Recall that an
element of a ring is \emph{nilpotent} if some power of it is zero.)

\answer.  $(ab)^{k}= a^{k}b^{k}=(a^{k}b)b^{k-1} = n b^{k-1}$, so when
I reduce this mod $n$, I get zero.

\vfill

\question[5]
Let $R$ and $S$ be rings with identity elements $1_{R}$ and $1_{S}$,
respectively.  Suppose that $\varphi : R \rightarrow S$ is a function
with the properties $\varphi (a+b) = \varphi (a) + \varphi (b)$ and
$\varphi (ab) = \varphi (a) \varphi (b)$ for all $a, b \in R$.
Show that if $\varphi (1_{R})$ is not equal to $1_{S}$, then there is
a nonzero element $b \in S$ so that $\varphi (1_{R}) b = 0$.

\answer. Let $a = \varphi (1_{R})$.  Since $1_{R} 1_{R} = 1_{R}$, then
when I apply $\varphi$ to this, I get $a^{2}=a$, so $a^{2}-a=0$.  I
can rewrite this as $(a-1)a=0$.  If $\varphi (1_{R}) \neq 1$, then
$a-1 \neq 0$, so this is my element $b$.

\vfill

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\end{questions}

\end{document}
