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\begin{document}

\title{Math 81/111: Rings and Fields \\ Homework \#7}
\date{18 February 2015; due Wednesday, 25 February 2015}

\maketitle

\begin{problab}{7.1}
Let $n \in \Z_{\geq 2}$.  Let $F$ be a field with $\opchar F \nmid n$ in which $X^n-1$ splits.  Let $a \in F^{\times}$ and let $f(X)=X^n-a \in F[X]$.  Let $K$ be a splitting field for $f$ and let $\alpha \in K$ be a root of $F$.  Show that the following are equivalent:
\begin{enumalph}
\item $f \in F[X]$ is irreducible;
\item $a \not\in F^{\times d}$ for all $d \mid n$ with $d>1$; and
\item $n$ is the smallest positive integer such that $\alpha^n \in F$.
\end{enumalph}
\emph{[Hint: Inspiration for a direct proof is in Exercise 2.6.  A Galois-theoretic proof is given on page 71, but you will need to unpack this argument.]}
\end{problab}

\begin{problab}{7.2}
Let $K/F$ be a finite cyclic extension with $\Gal(K/F)=\langle \sigma \rangle \cong \Z/n\Z$.  Define the map
\begin{align*}
\Tr:K &\to F \\
\alpha &\mapsto \sum_{i=0}^{n} \sigma^i(\alpha) = \alpha+\sigma(\alpha)+\dots+\sigma^{n-1}(\alpha).
\end{align*}
Let $\alpha \in K$ (and check that indeed $\Tr(\alpha) \in F$).  Give two proofs of the additive version of Hilbert's Theorem 90:
\begin{center}
$\Tr(\alpha)=0$  if and only if there exists $\beta \in K$ such that $\alpha=\beta-\sigma(\beta)$.
\end{center}
First, argue directly with Dedekind's linear independence of characters.  Then rephrase this by proving (in a similar way) that $H^1(G,K)=\{0\}$ and deducing the result from this.
\end{problab}

\begin{problab}{7.3 (M5-2)}
Apply Hilbert's Theorem 90 to the extension $\Q(i)/\Q$ to prove that the rational solutions $x,y \in \Q$ of the Pythagorean equation $x^2+y^2=1$ are of the form
\[ x=\frac{s^2-t^2}{s^2+t^2}, \quad y=\frac{2st}{s^2+t^2}, \quad s,t \in \Q. \]
Deduce that a right triangle each with sides $a,b,c \in \Z$ with $\gcd(a,b,c)=1$ and $a,b < c$ has  
\[ a,b=m^2-n^2,2mn \quad \text{and} \quad c=m^2+n^2 \]
with $m,n \in \Z$.  What is the smallest such triple $(a,b,c)$ you have not seen before?
\end{problab}

\begin{problab}{7.4}
Show that the polynomial $f(X)=X^5+11X+11$ is not solvable by radicals over $\Q$, i.e., the roots of $f$ cannot be expressed in terms of radicals, with as nice an argument as possible.
\end{problab}

\begin{problab}{7.5}
Let $F$ an infinite field and let $K$ be an algebraic extension of $F$ (possibly infinite degree over $F$).  Show that $\#F=\#K$, i.e., $F$ and $K$ have the same cardinalities as sets.
\end{problab}

\end{document}