\documentclass{amsart}
\title{Algebra 2. Problem Set 9}
\author{}

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\begin{document}
\maketitle

\problem{1} Let $\fExt{E}{F}$ be a finite field extension, that is, $E$ is
finite--dimensional as an $F$--vector space. Prove that every element
of $E$ is algebraic over $F$.\\

\problem{2} Again assume that we have a field extension $\fExt{E}{F}$. Consider
the set $E^{\scriptstyle{\operatorname{alg}}}$ of all elements of $E$ which
are algebraic over $F$. Prove that $E^{\scriptstyle{\operatorname{alg}}}$
is a field containing $F$.\\
({\em When $F=\mathbb{Q}$ and $E=\mathbb{C}$, we find an extension, usually
denoted by $\ol{\mathbb{Q}}$, which is algebraic and not of finite degree.
This is a standard counterexample to the converse of Problem 1.})\\

\problem{3} Let $F$ be a field and $p(x)\in F[x]$. Let $E$ be the splitting
field of $p(x)$ over $F$. Prove that $(E:F)$ divides $n!$.\\

\problem{4} Let $K/k$ be a field extension and assume we have two finite subextensions
$E/k$ and $F/k$. That is, $k\subset E\subset K$ and $k\subset F\subset K$, and
both $E$ and $F$ are finite--dimensional as $k$--vector spaces. Let
$EF$ be the smallest subfield of $K$ containing both $E$ and $F$. Prove that
$(EF: k)\leq (E:k)(F:k)$.\\
\indent Further assume that
$(E:k)$ and $(F:k)$ are coprime. Prove that, in this case, the inequality
becomes an equality.\\

\problem{5} Let $E/F$ be a field extension such that $E = F(\alpha)$ for an
algebraic element $\alpha\in E$ whose minimal polynomial has odd degree.
Prove that $E = F(\alpha^2)$.\\

\problem{6} Again let $E/F$ be a field extension. Assume we have two
algebraic (over $F$) elements $\alpha,\beta\in E$. Let $f(x),g(x)\in F[x]$
be the minimal polynomials of $\alpha$ and $\beta$ respectively.
Prove that, if $\deg(f)$ and $\deg(g)$ are coprime, then $f(x)$
is irreducible in $F(\beta)[x]$.\\

\problem{7} Let $E/F$ be an algebraic extension. That is, every element
of $E$ is algebraic over $F$. Let $R\subset E$ be a subring containing $F$.
Prove that $R$ is a field.\\

\problem{8} Let $f(x) = x^6 + x^3 + 1 \in \Q[x]$. Prove that $f(x)$ is
irreducible. How many morphisms of fields are there from $\Q[x]/(f(x))$ to
$\C$?\\

\problem{9} Let $E$ be the splitting extension of $p(x) = x^5-7$ over $\Q$.
Compute $(E:\Q)$.\\

\problem{10} Determine the minimal polynomials, over $\Q$, of the following elements of $\C$.
Here $\iota^2 = -1$.
\[
\sqrt{2}+\sqrt{3}, \qquad
\frac{1+\iota \sqrt{3}}{2}, \qquad
\iota + 5^{\frac{1}{3}}.
\]


\hrule\hrule

\vspace*{0.2in}

\begin{center}
{\em The following problems are to be discussed during recitation.}\\
\end{center}

\vspace*{0.2in}

\noindent {\em Eisenstein's criterion.} Let $f(x)\in \Z[x]$ be the following
polynomial: $f(x) = a_nx^n + \ldots + a_0$. Assume that there exists a prime
$p$ such that 
\[
a_n\not\equiv 0 \text{ (mod $p$)}, \qquad
a_j \equiv 0 \text{ (mod $p$) for every $0\leq j\leq n-1$}, \qquad
a_0 \not\equiv 0 \text{ mod $p^2$}.
\]
Prove that $f(x)$ is irreducible in $\Q[x]$.\\

\newpage

\noindent {\bf Remarks.}

\begin{itemize}
\item The statement above can be generalized to any {\em unique factorization
domain} (see definition below). Namely, replace $\Z$ by a UFD $A$, and 
$\Q$ by the field of fractions of $A$. Instead of primes, you will need
prime ideals.\\

\item If $p\in \Z$ is prime, we can use the Eisenstein criterion to
prove that $f(x) = x^{p-1} + x^{p-2}+ \cdots +1$ is irreducible. The criterion
is to be applid to $f(x+1)$.\\
\end{itemize}

\noindent {\bf Definition.} Let $A$ be an integral domain. An element $a\in A$,
$a\not= 0$, is called irreducible if it is not a unit, and $a=bc$ implies
$b$ or $c$ is a unit. We say that $A$ is a {\em unique factorization domain}
if 
\begin{enumerate}
\item Every non--zero element $a\in A$ can be written as a product of 
irreducible elements and a unit: $a = u\prod_{i=1}^r p_1\cdots p_r$,
where $p_1,\ldots, p_r$ are irreducible elements and $u\in A$ is a unit.\\

\item  This expression is unique up to relabelling the
factors, and rescaling the factors by units of $A$. That is,
$a = u\prod_{i=1}^r p_i = u'\prod_{j=1}^s q_j$ implies that
$r=s$ and after permuting the indices, we have, for each $i$, $p_i = u_iq_i$ 
for some unit $u_i\in A$.\\
\end{enumerate}

\noindent {\em Unique factorization of polynomials.} Let $A = K[x]$
where $K$ is a field. Prove that any monic polynomial of degree at least $1$
can be written uniquely (upto the obvious shuffle of indices) as
a product of monic irreducible polynomials. Use the existence and
uniqueness of splitting extensions for this (in particular, make sure that
we didn't use the {\em unique factorization of polynomials} to prove the existence and
uniqueness of splitting extensions).
\end{document}
