\documentclass[12pt]{article}
\usepackage{amsmath,amssymb,amsthm}

\newtheorem{theorem}{\textbf{Theorem}}
\newtheorem{proposition}[theorem]{\textbf{Proposition}}
\newtheorem{lemma}[theorem]{\textbf{Lemma}}

\theoremstyle{definition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}

\date{}

\begin{document}

\title{Homework 5}

\maketitle

All rings are commutative with unit.

Recall that an ideal $\mathfrak{q}$ in a ring $A$ is {\textit{primary}} if $x y
\in \mathfrak{q}$ implies that either $x \in \mathfrak{q}$ or $y^n \in
\mathfrak{q}$ for some $n$. Equivalently, every zero divisor in $A
/\mathfrak{q}$ is nilpotent.

\begin{lemma}
  If $\mathfrak{q}$ is primary then $r (\mathfrak{q})$ is prime.
\end{lemma}

\begin{proof}
  Suppose that $x y \in r (\mathfrak{q})$. Then $x^N y^N \in \mathfrak{q}$ for
  some $N$. Thus either $x^N \in \mathfrak{q}$ or $(y^N)^n \in \mathfrak{q}$
  for some $n$. Thus either $x \in r (\mathfrak{q})$ or $y \in r
  (\mathfrak{q})$.
\end{proof}

If $\mathfrak{q}$ is primary and $r (\mathfrak{q}) =\mathfrak{p}$ then we say
that $\mathfrak{q}$ is {\textit{$\mathfrak{p}$-primary}}.

\begin{lemma}
  Suppose that $\mathfrak{q}_1, \mathfrak{q}_2$ are $\mathfrak{p}$-primary
  ideals. Then $\mathfrak{q}_1 \cap \mathfrak{q}_2$ is also
  $\mathfrak{p}$-primary.
\end{lemma}

\begin{proof}
  We have $r (\mathfrak{q}_1 \cap \mathfrak{q}_2) = r (\mathfrak{q}_1) \cap r
  (\mathfrak{q}_2) =\mathfrak{p} \cap \mathfrak{p}=\mathfrak{p}$. To show
  $\mathfrak{q}_1 \cap \mathfrak{q}_2$ is $\mathfrak{p}$-primary, suppose that
  $x y \in \mathfrak{q}_1 \cap \mathfrak{q}_2$ but $x \notin \mathfrak{q}_1
  \cap \mathfrak{q}_2$. Then either $x \notin \mathfrak{q}_1$ or $x \notin
  \mathfrak{q}_2$. By symmetry, we may assume that $x \notin \mathfrak{q}_1$.
  Then since $x y \in \mathfrak{q}_1$ which is $\mathfrak{p}$-primary, we have
  $y \in \mathfrak{p}= r (\mathfrak{q}_1 \cap \mathfrak{q}_2)$.
\end{proof}

\begin{theorem}
  If $A$ is Noetherian, then every ideal may be expressed as a finite
  intersection of primary ideals.
\end{theorem}

\begin{proof}
  We will prove this in class. See Lang's {\textit{Algebra}}, Theorem~X.3.3 on
  page 423.
\end{proof}

Write the ideal $\mathfrak{a}$ as an intersection of primary ideals
$\mathfrak{q}_i$:
\[ \mathfrak{a}= \bigcap_{i = 1}^d \mathfrak{q}_i . \]
By Lemma~2 we may assume that the radicals $r(\mathfrak{q}_i)$ are all distinct. Moreover,
we may obviously discard any $\mathfrak{q}_i$ if that does not change the
intersection. Assuming this, the decomposition is called a (reduced)
{\textit{primary decomposition}}. The decomposition is not unique but has some
uniqueness properties. The primary decomposition is discussed in Lang's
{\textit{Algebra}} Chapter 10, but see the book of Atiyah and Macdonald, Chapter
4, for more information. Atiyah and Macdonald is available on-line through the
Stanford Libraries.

The notion of a primary ideal is needed in the dimension theory.

The following well-known fact may be useful in doing Problem~1. The set
$\mathfrak{N}$ of nilpotent elements of a ring $A$ is called the
{\textit{nilradical}}. It is an ideal, the radical of $(0)$.

\begin{proposition}
  The nilradical is the intersection of all prime ideals of $A$.
\end{proposition}

\begin{proof}
  See Lang's {\textit{Algebra}}, Corollary X.2.2 on page 417.
\end{proof}

\bigbreak\noindent
{\textbf{Problem 1}}. Show that if $r (\mathfrak{a})$ is maximal, then $A
/\mathfrak{a}$ is a local ring with a single prime ideal. The ideal
$\mathfrak{a}$ is primary.

\medbreak\noindent
One might hope from Problem~1 that if $r (\mathfrak{a})$ is only assumed to
be prime that $\mathfrak{a}$ might be primary. However the next problem shows
that this is not true.

\bigbreak\noindent
{\textbf{Problem 2}}. Let $A$ be the quotient of the polynomial ring
$\mathbb{C} [X, Y, Z]$ by the ideal $(X^2 - Y Z)$ and let $x, y, z$ be the
images of $X, Y, Z$ in $A$. Let $\mathfrak{p}$ be the ideal $(x, z)$. Show
that $\mathfrak{p}$ is prime and that the radical of $\mathfrak{p}^2$ is
$\mathfrak{p}$, but that $\mathfrak{p}^2$ is not primary.

\bigbreak\noindent
{\textbf{Problem 3}}. Let $A$ be the polynomial ring $k [x, y, z]$, where
$k$ is a field. Let $\mathfrak{p}_1 = (x, y)$, $\mathfrak{p}_2 = (x, z)$ and
$\mathfrak{m}= (x, y, z)$. Let $\mathfrak{a}=\mathfrak{p}_1 \mathfrak{p}_2$.
Prove that
\[ \mathfrak{a}=\mathfrak{p}_1 \cap \mathfrak{p}_2 \cap \mathfrak{m}^2 \]
is a primary decomposition.

\medbreak\noindent
The next problem is a strengthening from a fact from last week.

\bigbreak\noindent
{\textbf{Problem 4}}. Let $\mathfrak{o}$ be a Dedekind domain with a unique
maximal ideal $\mathfrak{p}$. Prove that $R$ is a discrete valuation ring.

\bigbreak\noindent
{\textbf{Problem 5.}} Let $A$ be the quotient of the polynomial ring $k [X,
Y]$ by the polynomial $Y^2 - X^3$ and let $x, y$ be the images of $X$ and $Y$
in $A$. Prove that $A$ is not integrally closed. If $\mathfrak{m}$ is the
maximal ideal generated by $x$ and $y$, determine the dimension of
$\mathfrak{m}^n /\mathfrak{m}^{n + 1}$ for all $n \geqslant 1$.

\bigbreak\noindent
{\textbf{Problem 6.}} Let $\lambda$ be a function defined on modules over a
ring taking values in $\mathbb{Z} \cup \{ \infty \}$ such that if $0
\longrightarrow M' \longrightarrow M \longrightarrow M'' \longrightarrow 0$ is
a short exact sequence then
\[ \lambda (M) = \lambda (M') + \lambda (M'') . \]
Prove that if $0 \longrightarrow M_1 \longrightarrow M_2 \longrightarrow M_3
\longrightarrow M_4 \longrightarrow 0$ is exact then
\[ \lambda (M_1) - \lambda (M_2) + \lambda (M_3) - \lambda (M_4) = 0. \]

\end{document}
