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\title{Math 210B: Homework 3}
\date{}

\maketitle

Recall that if $\mathfrak{a}$ is an ideal of $A$ the {\textit{radical}} of
$\mathfrak{a}$ is
\[r(\mathfrak{a})=\left\{ f \in A| \text{$f^n \in \mathfrak{a}$ for some $n
> 0$} \right\}.\]
This is an ideal containing $\mathfrak{a}$, and it is easy to
see that $r (r (\mathfrak{a})) = r (\mathfrak{a})$. The ideal $\mathfrak{a}$
is called {\textit{radical}} if $\mathfrak{a}= r (\mathfrak{a})$.

Let $k$ be a field, which we assume to be algebraically closed. Let
$\mathbb{A}^n (k)$ be {\textit{affine $n$-space}} $k^n$. If $S$ is a subset of
the polynomial ring $k [X] = k [X_1, \cdots, X_n]$ then
\[ V (S) = \left\{ a = (a_1, \cdots, a_n) \in \mathbb{A}^n (k) | \text{$f (a)
   = 0$ for all $f \in S$} \right\} . \]
If $\mathfrak{a}$ is the ideal generated by $S$ and $r (\mathfrak{a})$ is the
radical of $\mathfrak{a}$ then $V (S) = V (\mathfrak{a}) = V (r
(\mathfrak{a}))$. Also if $X$ is a subset of $\mathbb{A}^n (k)$ let
\[ I (X) = \left\{ f \in k [X] | \text{$f (a) = 0$ for all $a \in
   \mathfrak{a}$} \right\} . \]
It is an ideal. By the Nullstellensatz $I (V (\mathfrak{a})) = r
(\mathfrak{a})$.

The sets $V (S)$ are called {\textit{algebraic sets}} and last week you proved
they form the closed sets in a topology, called the {\textit{Zariski topology}}
on $\mathbb{A}^n (k)$. The essential content of the Nullstellensatz is that
$\mathfrak{a} \mapsto V (\mathfrak{a})$ is a bijection between radical ideals
and Zariski closed sets.

A closed (i.e. algebraic) subset $X$ of $\mathbb{A}^n (k)$ is called
{\textit{reducible}} if there exist proper closed subsets $Y, Z$ of $X$ such
that $X = Y \cup Z$. If it is not reducible it is called {\textit{irreducible}}.
I will call an irreducible closed set a {\textit{variety}} though this
terminology is not universally adopted: often a variety is not required to be
irreducible.

\bigbreak\noindent
{\textbf{Problem 1}}. Prove that $V (\mathfrak{a})$ is irreducible if and
only if $r (\mathfrak{a})$ is prime.

\bigbreak\noindent
{\textbf{Problem 2.}} (a) Let $A \subset F$ where $F$ is a field, and let
$B$ be the integral closure of $A$ in $F$. Let $S \subseteq A$ be a
multiplicative set. Show that $S^{- 1} B$ is the integral closure of $S^{- 1}
A$ in~$F$.

(b) Let $A$ be an integral domain. We recall that we say $A$ is
{\textit{integrally closed}} if it is integrally closed in its field of
fractions~$F$. Show that $A$ is integrally closed if and only if
$A_{\mathfrak{m}}$ is integrally closed for every maximal ideal $\mathfrak{m}$
of $A$.

\medbreak\noindent
\textbf{Hint} for (b): Let $C$ be the integral closure of $A$ in $F$. Let $x \in
C$. Suppose that $x \notin A$. Let $\mathfrak{a}= \{ f \in A|f x \in A \}$.
Let $\mathfrak{m}$ be a maximal ideal of $A$ containing $\mathfrak{a}$. Then ...

\bigbreak

We recall the {\textit{Extension Theorem for valuations}} that was proved in
Week 2. If $F$ is a field, a {\textit{valuation ring of $F$}} is a subring $R$
such that if $x \in F$ then either $x \in R$ or $x^{- 1} \in R$. A valuation
ring is $R$ a local ring. Its maximal ideal $\mathfrak{p}$ may be
characterized as $\{ x \in R|x^{- 1} \notin R \}$ since in a local ring, the
nonunits comprise the unique maximal ideal.

\begin{theorem}
  Let $F$ be a field and $A$ a subring of $F$. Let $\phi : A \longrightarrow
  \Omega$ be a homomorphism of $A$ into an algebraically closed field
  $\Omega$. Then $\phi$ may be extended to a homomorphism $\Phi : R
  \longrightarrow \Omega$ where $R$ is a valuation ring of $F$.
\end{theorem}

This is Corollary~3.3 in Section VII.3 of Lang's {\textit{Algebra}}. It was also
proved in class and used to prove the Nullstellensatz. Use it in the next
Exercise.

\bigbreak\noindent
{\textbf{Problem 3}}. Let $F$ be a field and $A$ a subring of $F$. Show
that the integral closure of $A$ in $F$ is the intersection of all valuation
rings of $F$ containing~$A$.

\medbreak\noindent
{\textbf{Hint:}} To prove that if $x \in F$ is not integral over $A$ then
there is a valuation ring $R$ of $F$ such that $x \notin R$, show that $x^{-
1} A [x^{- 1}]$ is contained in a maximal ideal of $A [x^{- 1}]$, then find a
way to use the Extension Theorem.

\bigbreak\noindent
{\textbf{Problem 4.}} Let $F$ be a field and let $R$ be the polynomial
ring $F[X,Y]$ in two variables. Give examples of prime ideals $\mathfrak{p}$
and $\mathfrak{q}$ such that the local ring $R_{\mathfrak{p}}$ is a
valuation ring but the local ring $R_{\mathfrak{q}}$ is not.

\bigbreak\noindent
{\textbf{Problem 5.}} Let $A$ be an integrally closed integral domain. This
means that it is integrally closed in its field of fractions $K$. Let $L / K$
be a finite separable extension and let $B$ be the integral closure of $A$ in
$L$. Suppose that $A$ is Noetherian. Prove that $B$ is a finitely-generated
$A$-module and deduce that $B$ is also Noetherian.

\medbreak\noindent
{\textbf{Hint:}} This is Problem 3 in the exercises to Chapter VII of Lang
(page 353). See Lang for the hint.

\bigbreak

A commutative ring $R$ is called a {\textit{Dedekind domain}} if it is
integrally closed in its field of fractions, Noetherian and every nonzero
prime ideal is maximal. For example a principal ideal domain is a Dedekind
domain. The class of Dedekind domains is important because many important
rings are Dedekind domains. For example the integral closure of $\mathbb{Z}$
in a finite extension $E$ of $\mathbb{Q}$ is called the {\textit{ring of
algebraic integers}} in $E$ and it is a Dedekind domain. The affine algebra
(coordinate ring) of a nonsingular affine curve is a Dedekind domain.

The next exercise is part of Exercise VII.7 in Lang's {\textit{Algebra.}}
We'll return to this Exercise in Homework~4.

\bigbreak\noindent
{\textbf{Problem 6}}. Let $\mathfrak{o}$ be a Dedekind domain. Let $K$ be
its field of fractions. Given a nonzero ideal $\mathfrak{a}$ of $\mathfrak{o}$
prove that there exists a product of maximal ideals $\mathfrak{p}_1 \cdots
\mathfrak{p}_r \subseteq \mathfrak{a}$.

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