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\lhead{MATH 522--01}
\chead{Problem Set \#3\answer{ solutions}}
\rhead{}% Your name here!
\lfoot{}
\cfoot{\answer{Page \thepage\ of \pageref{LastPage}}}
\rfoot{due Thursday, February 26, 2015}
\begin{document}
\begin{enumerate}
\item \question{For a field $F$, let $F(x)$ denote the field of
fractions of the ring $F[x]$; in other words, the field of
rational functions with coefficients in $F$. Prove that there is
no element $f\in F(x)$ such that $f^2=x$.}
\answer{Suppose there is an $f$ such that $f^2=x$. Let $f$ represent
the equivalence class given by some $(g,h)$ for $g,h\in F[x]$;
informally we would say $f=\frac{g}{h}$. Then
$(g,h)(g,h)=(g^2,h^2)$, which we know is in the same equivalence
class as $(x,1)$, so $g^2=xh^2$ by the equivalence criterion; note
that this is an equality among elements of the integral domain
$F[x]$ instead of its field of fractions. Note that if $\deg g=m$,
then $\deg (g^2)=\deg (g\cdot g)=(\deg g)+(\deg g)=2m$, which is
even, while if $\deg h=n$, then $\deg (xh^2)=\deg (x\cdot h\cdot
h)=\deg x+\deg h+\deg h=2n+1$ which is odd; thus $g^2$ cannot
equal $xh^2$.}
\item \question{Prove that, if $I$ is a prime ideal of a ring $R$,
then $I[x]$ is a prime ideal of $R[x]$.}
\answer{Since the sums and products of two elements of $I[x]$ are
determined by finite sums and products of individual coefficients
in $I$, and since $I$ is closed under addition and multiplication,
it is easy to show that the individual coefficients of sums and
products of elements of $I[x]$ will have coefficients in $I$ and
thus $I[x]$ is closed under addition and multiplication. In
addition, an additive inverse of an element of $I[x]$ is achieved
by performing termwise additive inverses on the coefficients; the
result of each individual procedure will lie in $I$, so the
polynomial so formed lies in $I[x]$. Thus $I[x]$ is a subring of
$R[x]$. It remains to show that $I[x]$ is absorbing and prime.
Consider some polynomial $i(x)\in I[x]$ and $r(x)\in R[x]$. Their
product has coefficients which are finite sums of the form
$\sum_{j=0}^ki_jr_j$ for some integer $k$ and each $i_j\in I$ and
$r_j\in R$. Note that since $I$ is absorbing, each $i_jr_j\in I$,
and since $I$ is closed under addition, the sum
$\sum_{j=0}^ki_jr_j\in I$. Thus each coefficient of $i(x)\cdot
r(x)$ is in $I$, so $i(x)\cdot r(x)\in I[x]$, demonstrating that
$I[x]$ is absorbing in $R[x]$.
Finally, let us prove primality. Suppose $a(x),b(x)\in R[x]-I[x]$;
we shall show that $a(x)b(x)\in R[x]-I[x]$. Let
$a(x)=a_mx^n+\cdots+a_0$ and $b(x)=b_nx^n+\cdots+b_0$. Since
$a(x)$ and $b(x)$ are not in $I[x]$, at least one of the
coefficients of each must be outside of $I$. Let $i$ be the
smallest integer such that $a_i\notin I$, and let $j$ be the
smallest integer such that $b_j\notin J$; now let us compute the
$i+j$th coefficient in $a(x)b(x)$:
\[a_0b_{i+j}+a_1b_{i+j-1}+\cdots+a_{i-1}b_{j+1}+a_ib_j+a_{i+1}b_{j-1}+\cdots+a_{i+j-1}b_1+a_{i+j}b_0\]
using the convention that the coefficients above the degree of a
polynomial are zero for simplicity. Note that
$a_0b_{i+j}+\cdots+a_{i-1}b_{j+1}\in I$ because $a_0,\ldots
a_{i-1}$ are in $I$ by the definition of $i$ above, and making use
of the absorbtion and additive closure properties. Likewise,
$a_{i+1}b_{j-1}+\cdots+a_{i+j-1}b_1+a_{i+j}b_0\in I$ because
$b_0,\ldots, b_{j-1}$ must be in $I$. However, $a_ib_j$ is
\emph{not} in $I$ because $a_i$ and $b_j$ are not in $I$ and $I$
is prime. Thus the complete sum, as the sum of an element of $I$
and a non-element of $I$, must not be an element of $I$ (by
additive closure with additive inverses). Since at least one
coefficient of $a(x)b(x)$ is not in $I$, $a(x)b(x)$ is not in
$I[x]$.}
\item \question{Describe a field of order 25 and a field of order 27.}
\answer{We know that for any finite field $F$ and irreducible
polynomial $f$ in $F$ of degree $n$, the quotient ring $F/\langle
f\rangle$ is a field of order $|F|^n$. Thus, since $25=5^2$ and
$27=3^3$, the question here can be satisfied by finding an
irreducible quadratic in $\mathbb Z_5[x]$ and an irreducible cubic
in $\mathbb Z_3[x]$. We know that a cubic or quadratic in $F[x]$
is irreducible iff it has no zero. Noting that $0$, $1$, and $4$
are the only perfect squares (a.k.a. quadratic residues) in
$\mathbb Z_5$, the quadratic $x^2+2$ will fit the bill; likewise
noting that $x^3=x$ for each $x\in\mathbb Z_3$, the polynomial
$x^3-x+1$ is nonzero everywhere in $\mathbb Z_3$. Thus, $\mathbb
Z_5[x]/\langle x^2+2\rangle$ will be a field of order 25, and
$\mathbb Z_3[x]/\langle x^3-x+1\rangle$ will be a field of order
27. Note that these are not the only possible answers; there are
other irreducible quadratics in $Z_5$ and cubics in $Z_3$ which
work equally well.}
\item \question{Let $\alpha\in\mathbb R$ be a \emph{transcendental
number}, i.e., there is no nonzero polynomial with integer
coefficients which has $\alpha$ as a zero. Prove that $\alpha^2$
cannot be written as a linear combination of $\alpha$ and a
rational number, i.e. $\alpha^2=\alpha x+y$ for $x,y\in\mathbb
Q$.}
\answer{Suppose there are rational numbers $\frac pq$ and $\frac rs$
such that $\alpha^2=\alpha\frac pq+\frac rs$. Then $\alpha^2-\frac
pq\alpha-\frac rs=0$; multiplying by $qs$, we fing that
$\alpha^2-(ps)\alpha-(qr)=0$. Considering the polynomial
$f(x)=x^2-psx-qr$, which is in $\mathbb Z[x]$, the above result
shows that $f(\alpha)=0$, contradicting the premise that $\alpha$
is transcendental.}
\item \question{Prove that for a principal ideal domain $D$ with $p$
an irreducible element of $D$, $D/\langle p\rangle$ is a
field. Show that if $D$ is merely an integral domain, $D/\langle
p\rangle$ may not be a field.}
\answer{Note that for any integral domain $D$, the quotient ring
$D/\langle p\rangle$ is a field if and only if $\langle p\rangle$
is maximal; we shall thus prove maximality of $\langle p\rangle$
for $p$ an irreducible element of $D$. Since $p$ is irreducible,
$p$ is not a unit, so $1\notin\langle p\rangle$ and thus $\langle
p\rangle\neq D$. Now let us suppose, contrary to the intended
maximality property, that there is an ideal $I$ such that $\langle
p\rangle\subsetneq I\subsetneq D$. Since $D$ is a principal ideal
domain, $I=\langle q\rangle$ for some $q\in D$. Since $p\in I$,
$p=kq$ for some $k\in D$. However, since $p$ is irreducible,
either $k$ or $q$ is a unit. Since $I\neq D$, $q$ cannot be a
unit, so $k$ must be a unit, but then $q=k^{-1}p$ so that $q\in
\langle p\rangle$ and so $I\subseteq \langle p\rangle$, leading to
the contradictory conclusion that $I=\langle p\rangle$.
A simple counterexample for an integral domain is $\mathbb Z[x]$,
which is an integral domain in which $x$ is irreducible, and yet
$\mathbb Z[x]/\langle x\rangle\cong \mathbb Z$ is not a field.}
\end{enumerate}
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