\documentclass{amsart}
\newenvironment{emquote}{\begin{quote}\em}{\end{quote}}
\newcommand{\superset}{\supset}
\newcommand{\bbZ}{{\mathbb Z}}
\newcommand{\bbA}{{\mathbb A}}
\DeclareMathOperator{\Spec}{Spec}
\DeclareMathOperator{\Ann}{Ann}
\DeclareMathOperator{\Fin}{{\bf Fin}}
\usepackage{hyperref}
\begin{document}
\title[Part III]{Primary Decomposition}
\author{Kapil Hari Paranjape}
\maketitle
%\appendix
\section{Decomposition into Irreducibles}
When an affine algebraic scheme $\Spec(R)$ can be written as
the ``categorical union'' of two closed proper subschemes $\Spec(R/I)$
and $\Spec(R/J)$, we say that
it is reducible; the term ``categorical union'' indicates the
smallest closed subscheme $\Spec(R/K)$ of $\Spec(R)$ such that
$\Spec(R/I)(A)\cup\Spec(R/J)(A)$ is contained in $\Spec(R/K)(A)$ for
every ring $A$ in $\Fin$. Assuming the result that the solutions in
finite rings determine the ideal, we see that reducibility is equivalent
to the condition that $(0)=I\ cap J$ with neither $I$ nor $J$ the zero
ideal. Thus we can define a ring $R$ to be irreducible if $(0)$ cannot
be written as the intersection of two non-zero ideals; an ideal $I$ in
$R$ is irreducible if $R/I$ is irreducible. From the Noetherian-ness of
the ring $R$ it follows that the zero ideal is a finite intersection of
irreducible ideals.

There are some problems with this decomposition. First of all there is
the lack of unique-ness. Secondly, the condition that an ideal is
irreducible seems somewhat difficult to characterise in terms of
elements. Finally, (and very importantly) the notion of categorical
irreducibility seems too strong a condition to impose. For example, even
the scheme $\Spec(k[x,y]/(x^2,xy,y^2)$ is reducible with this
definition; this seems rather counter-intuitive.

Another way to decide if a scheme is irreducible is as follows. Let
$f$ and $g$ be non-zero functions with $fg=0$, then $f=0$ and $g=0$
would define a decomposition of the space unless $f$ (or $g$ is
nilpotent). This corresponds to the condition that $(0)$ is a {\em
primary ideal}; an ideal $Q$ is primary if whenever $fg$ lies in $Q$ and
$g$ does not lie in $Q$, some power of $f$ lies in $Q$.

\subsection{Irreducibles are primary}
Let $R$ be an irreducible ring and $a$ be in $R$. If $a$ is not
nilpotent then $R_a=R[T]/(aT-1)$ is a non-zero ring. The kernel $J$ of
the natural homomorphism $R\to R_a$ is a proper ideal. Let
$J=(x_1,\dots,x_r)$ (since $R$ is Noetherian) and let $n$ be such that
$a^nx_i=0$ for all $i$. If $y$ is an element of $R$ such that $a^my=0$
for some $m$, then $y$ lies in $J$ and so $x^ny=0$. Now, suppose that
$c$ lies in $Ra^n \cap \Ann(a)$; we have $c=da^n$ and $ca=da^{n+1}=0$.
But then $d$ lies in $J$ and so $c=da^n=0$. Since we have written $(0)$
as an intersection of two ideals and $Ra^n$ is non-zero we see that
$\Ann(a)=0$. So we have shown that every non-nilpotent element of $R$ is
not a zero-divisor. Let $P$ be the ideal consisting of nilpotent
elements. If $xy$ lies in $P$ then $x^ny^n$ is zero for some $n$, so
either $x^n=0$ or $y$ is a zero divisor, hence nilpotent. Thus $P$ is a
prime ideal. Moreover, if $xy=0$ and $x$ does not lie in $P$ then $x$ is
not nilpotent and so we must have $y=0$. In other words, $(0)$ is a
$P$-primary ideal.

\subsection{Irredundant primary decomposition}
If $P$ is a prime ideal in $R$, then the intersection of finitely many
$P$-primary ideals in $R$ is again a $P$-primary ideal. Thus, given an
expression $\cap_i I_i=(0)$ of $(0)$ as an intersection of irreducible
ideals we can bunch together all the $I_i$'s which are primary for the
same prime ideal. This way we obtain an expression $\cap_j Q_j=(0)$,
where $Q_j$ are $P_j$-primary for some finite collection of distinct
prime ideals $P_j$ of $R$. Now, we can further assume that
$(0)\neq \cap_{j\neq k} Q_k$ for any $j$ else we can shorten the above
expression. Such an expression of $(0)$ as an intersection of primary
ideals that cannot be shortened further is called an irredundant primary
decomposition of $(0)$. 

Consider the ring $R=k[x,y]=k[X,Y]/(X^2,XY)$. We see that $(0)$ is the
intersection of the ideals $Rx\cap Ry^n$ for any $n$. This shows that we
cannot home for uniqueness of the primary decomposition. However, we
will see that some form of uniqueness can be proved.

\subsection{Annihilators which are primary}
Let $\cap Q_i=(0)$ be an irredundant primary decomposition of $(0)$ and
let $P_i$ be the prime ideal which is the radical of $Q_i$. By
irredundancy, the intersection $\cap_{j\neq k}Q_j$ is non-zero; let $c$
be an element of this intersection. Now suppose $xy$ lies in $\Ann(c)$
so that $xyc=0$. Thus $xyc$ lies in $Q_k$, if $x$ does not lie in $P_k$
then $yc$ lies in $Q_k$ {\em and} as a multiple of $c$ it lies in
$\cap_{j\neq k} Q_j$ as well. But then $yc$ lies in $(0)=\cap_j Q_j$ and
so is 0; in other words $y$ lies in $\Ann(c)$. We have shown that
$\Ann(c)$ is $P_j$-primary. It is also clear that $\Ann(c)$ contains
$Q_j$.

For any $d$ not in $\Ann(c)$ consider the proper ideal $\Ann(dc)$. If
$xy$ lies in this ideal then $xyd$ lies in $\Ann(c)$; so if $x$ is not
in $P_j$, then $yd$ lies in $\Ann(c)$ (by the $P_j$-primality of this
ideal) and so $y$ lies in $\Ann(dc)$. In other words $\Ann(dc)$ is also
$P_j$-primary. On the other hand, if $b$ lies in $P_j$ there is a
positive integer $n$ such that $b^{n+1}$ lies in $\Ann(c)$ but $b^n$
does not. Then $b$ lies in $\Ann(b^nc)$. Thus the maximal ideal of the
form $\Ann(dc)$ must contain and thus equal $P_j$.  We have therefore
shown that prime ideals that occur in an irredundant primary
decomposition of $(0)$ in the ring $R$ are all annihilators of some
element of $R$. 

\subsection{Associated primes}
The associated primes of a ring $R$ are the prime ideals $P$ which are
of the form $\Ann(c)$ for some $c$ in $R$. Suppose $P$ is such a prime.
Now consider an irredundant primary decomposition $\cap_j Q_j=(0)$.
Among the $Q_j$ let $A$ be the collection of those $j$ for which $c$
does not lie in $Q_j$ and let $I$ be their intersection. If $d$ lies in
$I$ then $dc$ lies in all the $Q_j$ and so must be 0; hence $d$ lies in
$\Ann(c)=P$; so $I\subset P$.  Consider some $Q_j$ which does not
contain $c$; let $P_j$ be the radical of $Q_j$. If $d$ lies in
$P=\Ann(c)$ then $dc=0$ lies in $Q_j$ and $c$ does not lie in $Q_j$ so
$d$ must lie in $P_j$. In other words $P$ is contained in the
intersection $\cap_{j\in A} P_j$. Since $P$ is prime it contains the
radical of $I$ which is the intersection $\cap_{j\in A} P_j$. Thus $P$
must be equal to one of the primes $P_j$.

We have shown that the associated primes are precisely the primes that
occur as the radicals of the primary ideals in an irredundant primary
decomposition.

\subsection{Minimal primes}
A prime ideal in $R$ that is not contained in any other prime ideal is
called a minimal prime of $R$. Let $\cap_j Q_j$ be an irredundant
primary decomposition of $(0)$ in $R$. The radical of $(0)$ consists of
all nilpotents and is the thus the intersection of the radicals $P_j$'s of
the $Q_j$'s. Since the nil radical of a ring $R$ is contained in all the
prime ideals of $R$ it follows that the minimal primes in $R$ are the
minimal elements in the collection of the $P_j$'s. The minimal prime
ideals of a ring $R$ are the minimal elements in the finite collection
of associated primes of the ring $R$. The non-minimal associated primes
are called {\em embedded} primes.

Let $P_k$ be an associated prime in $R$ and let $Q$ be the collection of
all $c$ such that there is some $d$ not in $P$ such that $dc=0$. For any
irredundant primary decomposition $\cap_j Q_j$, we see that $Q$ is
contained in $Q_k$. The ideal $I_k=\cap_{j\neq k} Q_j$ is non-zero and
its radical is $\cap_{j\neq k} P_j$. If $P_k$ is a minimal prime then it
does not contain the latter ideal and hence it does not contain $I_k$
either. It follows that there is a $d$ that lies in $I_k$ but does not
lie in $P_k$. For any $c$ in $Q_k$, the element $dc$ lies in $I_k\cap
Q_k=(0)$ so is 0. Thus we see that $Q=Q_k$. The primary ideal in an
irredundant primary decompostion that corresponds to a minimal prime in
$R$ is uniquely determined.

\subsection{Zero-divisors in a ring}
From the description given above we see that the associated prime ideals
consist of zero-divisors. Conversely, if $xy=0$ and $x$ does not lie in
an associated prime $P_j$ then $y$ must lie in the corresponding $Q_j$.
So if $x$ does not lie in any associated prime $P_j$ then $y$ lies in
all the $Q_j$'s and so must be 0. The collection of zero-divisors in $R$
is thus precisely the union of all its associated primes.

\subsection{Some geometry}
We have seen that the primary decomposition is related to the
decomposition of $\Spec(R)$ into irreducibles components. If we ignore
for the moment, the distinction between irreducible components which
differ only in their ``nilpotent thickenings'', then the primary
decomposition precisely corresponds to the decomposition into irreducible
components.

A zero-divisor $c$ should be seen as a function that vanishes on some
components but not others; the components on which it does not vanish
are defined by the vanishing of those functions $d$ such that $dc$ is
identically the zero function; that is, $\Ann(c)$. We have seen above
that for any component, there is a function $c$ such that $\Ann(c)$ is a
primary ideal associated to that component; this is a function that
vanishes on all components except the chosen one. This interprepretation
makes sense unless the corresponding prime ideal is embedded (i.~e.\
non-minimal). How can a function vanish on a component but apparently
not on a closed subset? What {\em can} and does happen is that the
function vanishes on the larger component but is nilpotent and non-zero
on the embedded component which is contained in this larger component.

The primes associated with a component consist of functions that are
nilpotent on that component. The above description also leads to
characterisation of functions $c$ that vanish on a non-embedded
component as those for which there is function $d$ which is not
nilpotent on this component such that $dc$ identically vanishes.

Finally, while dealing with primary decomposition, it is easy (but {\em
erroneous}) to loose track of the elements of $R$ which are not
zero-divisors. We will see these elements acquire importance in a
later section. 

\end{document}