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\title{Math 210B: Homework 4}
\date{}

\maketitle

All rings are commutative, with 1. Some of the problems
make use of the norm map, so review Theorem~VI.5.1 of Lang's
\textit{Algebra}.

We will say a ring is of \textit{dimension 1} if it is not a field
and every nonzero
prime ideal is maximal. Recall that a Dedekind domain is an integral domain
that is integrally closed (meaning integrally closed in its field of
fractions), Noetherian, and of dimension~1.

\bigbreak\noindent
\textbf{Problem 1.} Let $E/F$ be a finite separable field extension and
let $N:E^\times\to F^\times$ be the norm map. Let $A$ be a
subring of $F$ that is integrally closed (in $F$) and let
$B$ be the integral closure of $A$ in $E$. Let $x\in B$. Show that
$N(x)\in A$. Moreover show that $x\in B^\times$ if and only if $N(x)\in A^\times$.

\medbreak\noindent

\bigbreak\noindent
\textbf{Problem 2.} Let $E/F$ be a finite separable field extension and
let $A$ be a Dedekind domain whose field of fractions is $F$. Let $B$ be the integral closure
of $A$ in $E$. Prove that $B$ is a Dedekind domain.

\bigbreak\noindent
\textbf{Problem 3.} Show that a principal ideal domain is a Dedekind
domain. Show that $\mathbb{Z}[\sqrt{-5}]$ is Dedekind domain that
is not a PID.

\bigbreak

Let $\mathfrak{o}$ be a Dedekind domain and let $K$ be its field
of fractions. Let $\mathfrak{a}$ be a nonzero $\mathfrak{o}$-submodule
of $K$. If there exists a nonzero $c\in\mathfrak{o}$ such that
$c\mathfrak{a}\subset\mathfrak{o}$,
then $\mathfrak{a}$ is called a \textit{fractional ideal}. Thus
a nonzero ideal of $\mathfrak{o}$ is a fractional ideal.

The next exercise continues (from Homework 3) Exercise VII.7 in Lang's {\textit{Algebra.}}

\vfill\eject
\noindent
\textbf{Problem 4.} Let $\mathfrak{o}$ be a Dedekind domain and let $K$
be its field of fractions. Let $\mathfrak{p}$ be a nonzero maximal ideal of
$\mathfrak{o}$. Define
\[\mathfrak{p}^{-1}=\{x\in K|x\mathfrak{p}\subseteq\mathfrak{o}\}.\]
\smallbreak\noindent
(a) Show that $\mathfrak{p}^{- 1}$
is a fractional ideal containing $\mathfrak{o}$ and that $\mathfrak{p}^{- 1}
\neq \mathfrak{o}$.
\smallbreak\noindent
(b) Continuing from the preceding problem, show
that $\mathfrak{p}^{-1}\mathfrak{p}=\mathfrak{o}$.
\smallbreak\noindent
(c) Show that every non-zero ideal is invertible by a fractional ideal.
Deduce that the fractional ideals of $\mathfrak{o}$ form
a group under multiplication.

\medbreak\noindent
\textbf{Hints}: For (a) try to use Problem~6 from
Homework~3. If $0\neq x\in\mathfrak{p}$ show that you can
find primes $\mathfrak{p}_i$ such that
$\mathfrak{p}_1\cdots\mathfrak{p}_r\subseteq x\mathfrak{o}\subseteq\mathfrak{p}$.
Show that one of the $\mathfrak{p}_i=\mathfrak{p}$. Then what?
For~(b), use the assumption that $\mathfrak{o}$ is
integrally closed to rule out the possibility that
$\mathfrak{p}^{-1}\mathfrak{p}=\mathfrak{p}$.

\bigbreak\noindent
\textbf{Problem 5.} Let $R$ be a principal ideal domain with a
unique nonzero prime ideal. Prove that $R$ is a valuation ring
of its field of fractions~$F$.

\medbreak
We will call a principal ideal domain with a unique nonzero
prime ideal a \textit{discrete valuation ring} (DVR).

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