\documentclass{amsart}
\newenvironment{emquote}{\begin{quote}\em}{\end{quote}}
\newcommand{\superset}{\supset}
\DeclareMathOperator{\Ann}{Ann}
\begin{document}
\title[Part I]{Finite Rings}
\author{Kapil Hari Paranjape}
\maketitle
\section{Finite (Dimensional) Rings}
Let us first examine a ring $A$ which is either finite or a finite
dimensional algebra over a field. It is clear that such a ring satisfies
the descending chain condition for ideals. That is to say any descending
chain of ideals $I_0\superset I_1 \superset\cdots$ is constant after
some $i$. A ring satisfying this condition is also called an Artinian
ring in honour of Emil Artin. The ring also satisfies the ascending
chain condition; any ascending chain of ideals is constant after a
finite stage. A ring satisfying this condition is called a Noetherian
ring in honour of Emmy Noether.
However, in this section we will only use the Artinian property to prove
a structure theorem. The Noetherian property will be a consequence! Thus
the results will apply to the larger class of all Artinian rings not
just finite rings. The proofs in the latter case could be marginally
simplified.
\subsection{Elements of an Artinian ring}
Let $x$ be an element of an Artinian ring $A$ and consider the sequence of
ideals $Ax \superset Ax^2 \superset \cdots$. By the descending chain
condition this sequence of ideals must become constant after some stage.
In other words, there must be an element $y$ in $A$ such that
$yx^{k+1}=x^k$ for some (sufficiently large) integer $k$.
Given such a $y$, let $k$ be the smallest non-negative integer for
which this equation holds (with the convention that $x^0=1$!). It is
clear that if $yx-1=0$ then $x$ is a unit; in other words $x$ is an
element of $A^{\times}$, the group of units of $A$. On the other hand if
$yx-1\neq 0$ then we see that $k>0$ and $(yx-1)x^{k-1}\neq 0$; since
$(yx-1)x^{k-1}\cdot x =0$, we see that $x$ is a zero divisor. Thus every
element of $A$ is either a unit or a zero-divisor.
\subsection{The Jacobson Radical of an Artinian ring}
If $x$ lies in an ideal $I$ and $(yx-1)$ lies in an ideal $J$, then
$I+J=A$. On the other hand, the equation $(yx-1)\cdot x^{k}=0$ shows
that if $(yx-1)$ does not lie in a prime ideal $P$ then $x$ must lie in
$P$. An element of a ring that does not lie in any maximal ideal is a
unit (by applying Zorn's lemma; however, this is not required for finite
rings or finite dimensional algebras). Thus $(yx-1)$ is a unit if and
only if $x$ lies in all maximal ideals; the above equation shows that
$x$ is nilpotent in this case.
Let $J$ be the intersection all maximal ideals of a ring $A$. This is
called the Jacobson radical of the ring in honour of Nathan Jacobson.
What we have shown above is that the Jacobson radical of an Artinian
ring consists of nilpotent elements.
\subsection{Prime ideals in an Artinian ring}
Now, if $P$ is a prime ideal of $A$ to which $x$ does not belong, then
the equation $yx^{k+1}=x^k$ says that $(yx-1)$ belongs to $P$ so that $x$
becomes a unit modulo $P$. We can argue similarly for every element that
does not belong to $P$. Thus every prime ideal of an Artinian ring is
maximal. We can also see this by noting that the ideal consisting of all
nilpotent elements of a ring is the intersection of all its prime
ideals.
\subsection{Some basic facts about co-prime ideals}
Let $I$ and $J$ be ideals in a ring $A$. We say that they are {\em
co-prime} if $I+J=A$. Let us state some basic facts about co-primality:
\begin{enumerate}
\item The product of co-prime ideals equals their intersection,
i.~e.\ $I\cap J=I\cdot J$.
\item Powers of co-prime ideals are co-prime, i.~e.\ $I^k+J^l=A$
for any positive integers $k$ and $l$.
\item If $I$ is co-prime to each of the ideals $J_i$,
$i=1,\dots,n$, then $I$ is also co-prime to the product
$\prod_i J_i$.
\end{enumerate}
All the above facts can be proved easily by observing that co-primality
of $I$ and $J$ is equivalent to the existence of an element $x\in I$
(respectively $y\in J$) such that $x$ acts as identity modulo $J$
(resp. $y$ acts as identity module $I$).
For our purpose we note that this implies that for any collection $M_1$,
\dots, $M_k$ of distinct maximal ideals we have
\[
M_1^{n_1} \cap \cdots \cap M_k^{n_k} =
M_1^{n_1} \cdot \cdots \cdot M_k^{n_k}
\]
Moreover, if $M$ is a maximal ideal distinct from these maximal ideals
then $M^n\cap M_1^{n_1} \cdot \cdots \cdot M_k^{n_k}$ is a proper
sub-ideal of $M_1^{n_1} \cdot \cdots \cdot M_k^{n_k}$. By the descending
chain condition applied in the case where the $n_i=1$ we see that there
are only finitely many maximal ideals in an Artinian ring.
\subsection{Annihilators and the ratio ideal}
For an ideal $I$ in a ring $A$ we use $\Ann_A(I)$ or, where there is no
ambiguity about the ring, $\Ann(I)$ to denote the annihilator of $I$;
this consists of the elements $x$ in $A$ such that $x\cdot I=0$.
This is a particular case of the ``ratio'' of ideals; if
$I\subset J$ are ideals of $A$ we define $(I:J)$ as the ideal consisting of
elements $x$ in $A$ such that $xJ\subset I$; note that $\Ann(I)=(0:I)$.
The inclusion $I\subset J$ is strict if and only if $(I:J)$ is strictly
contained in $A$. Moreover, if $J_1\subset J_2$ then $(I:J_2)\subset
(I:J_1)$.
\subsection{Minimal elements}
Let $x_0$ be a non-zero element of an Artinian ring $A$. If $I=\Ann(Ax_0)$ is
not a maximal ideal, then there is a proper sub-ideal of $A$ such that
$J$ properly contains $I$; thus $Jx_0$ is a non-zero proper sub-ideal of
$Ax_0$. Let $x_1$ be a non-zero element of $Jx_0$. Then $Ax_1$ is a
non-zero proper sub-ideal of $Ax_0$. Continuing this process, we see
that, the descending chain condition for $A$ implies that there must
exist some multiple $y$ of $x_0$ with the property that $\Ann(Ay)$ is a
maximal ideal. We call such an element a minimal element.
More generally, if $I$ is any proper sub-ideal of $A$, then $A/I$ is
also a non-zero Artinian ring. If $x$ is an element of $A$ whose image
is a minimal element of $A$, then $(I:Ax+I)$ is a maximal ideal.
\subsection{Nakayama-type lemma for Artinian rings}
Let $J$ again denote the Jacobson radical of the ring $A$. As seen
above $\Ann(J\cdot I)\superset\Ann(I)$; for an Artinian ring we will
show that that this is a strict inclusion whenever $I$ is not the
zero-ideal. It will follow that $J\cdot I$ is strictly contained in $I$
whenever $I$ is not the zero ideal. A result of this type for Noetherian
(semi-)local rings is called Nakayama's lemma.
Since $I$ is not the zero ideal, $\Ann(I)$ is a proper sub-ideal of the
ring $A$. As seen above there is an element $x$ of $A$ such that
$(\Ann(I):Ax+\Ann(I))$ is a maximal ideal $M$. It follows that $x\cdot
M\cdot I=0$ so that $x$ lies in $\Ann(M\cdot I)$ but not in $\Ann(I)$.
Since $J\cdot I=K\cdot M\cdot I$ where $K$ is the product of the
remaining maximal ideals of $A$, we see that the result follows.
\subsection{The nilpotency of the Jacobson radical}
As above let $J$ denote the Jacobson radical of an Artinian ring $A$. By
the result just proved $J\cdot J^r$ is a proper sub-ideal of $J^r$ if
$J^r$ is a non-zero ideal. The descending chain condition for ideals
thus implies that $J^r$ must be the zero ideal for some $r$.
By the co-primality of the maximal ideals of $A$ we see that
\[ A \to (A/M_1^r) \times \cdots \times (A/M_k^r) \]
is an isomorphism when the product on the right runs over all maximal
ideals of $A$.
\subsection{Artinian rings are Noetherian rings}
If $A$ is a ring and $M$ is a maximal ideal, then $W=M^k/M^{k+1}$ is a
vector space over the field $A/M$. Any strictly decreasing sequence of
subspaces $V_i \superset V_{i+1}$ in $W$ is of the form
$V_i=I_i/M^{k+1}$, where $I_i \superset I_{i+1}$ is a decreasing
sequence of ideals. Thus, if $A$ is an Artinian ring, this sequence must
be constant after a finite stage. It follows that $W$ is a finite
dimensional vector space; we then conclude easily that any ideal of
$A/M^k$ is finitely generated. Since $A$ is itself a product of such
rings, we see that any ideal in $A$ is finitely generated.
A ring where every ideal is finitely generated satisfies the ascending
chain condition for ideals. That is to say given an ascending chain of
ideals $I_0\subset I_1 \subset\cdots$, this sequence is constant after
some $i$. A ring satisfying this condition is also called an Noetherian
ring in honour of Emmy Noether. As we shall see later such rings arise
naturally in algebraic geometry.
\subsection{The Nil radical of a Ring}
If $x$ is a non-nilpotent element of a ring $R$, then $R_x$ is a
non-zero ring. By an application of Zorn's lemma (which is not required
if the ring $R$ (and hence $R_x$) is Noetherian), there is a maximal
proper ideal $M$ of $R_x$. The kernel of the natural homomorphism $R\to
R_x/M$ is a prime ideal since the latter ring is a field. We have thus
shown that any non-nilpotent element is in the complement of some prime
ideal; the converse, that a nilpotent element belongs to all prime
ideals, is also clear. Hence the intersection of all prime ideals is
also the ideal consisting of all nilpotent elements of a ring; this
ideal is called the nil radical of the ring.
\subsection{The Nilpotency of the Nil radical of a Noetherian Ring}
When the ring under consideration is Noetherian, the nil radical is
finitely generated so there is an upper bound $k$ on the index of
nilpotency of the generators. Let $x_1$, \dots, $x_l$ be the generators
of the nil radical and consider an expression of the form
\[
(a_{11} x_1 + \dots + a_{1l} x_l)
(a_{21} x_1 + \dots + a_{2l} x_l) \cdots
(a_{r1} x_1 + \dots + a_{rl} x_l)
\]
When $r>l(k-1)$ we see that each term in the expanded expression {\em
must} contain the $k$-th power of $x_i$ for at least one $i$. It follows
that this expression must be zero whatever the choice of the $a_{ij}$.
This shows that the $l(k-1)+1$-th power of the nil radical
is the zero ideal. In other words the nil radical of a Noetherian ring
is also nilpotent.
\subsection{When are Noetherian rings Artinian?}
Let $R$ be a Noetherian ring in which there is just one prime ideal $M$;
then that prime ideal is also a maximal ideal. As we have seen above this prime
ideal is the nil radical and is nilpotent; hence $R=R/M^s$ for some
natural number $s$. If $x_1$, \dots, $x_r$ generate $M$ then
monomials of degree $k$ in the $x_i$ generate its $k$-th power $M^k$. This
shows that $M^k/M^{k+1}$ is a finite dimensional vector space over $R/M$
for every $k$. It now follows that $R$ is an Artinian ring.
More generally, suppose $R$ is a Noetherian ring in which every prime
ideal is a maximal ideal; let $M$ be such an ideal. By what has been
proved above we see that the localisation $R_M$ of $R$ with respect to
the multiplicative set $R\setminus M$ is an Artinian ring; moreover we
have $R_M=R_M/M^kR_M=R/M^k$ for some integer $k$. Let $y_M$ be an
element of $R$ whose image in this Artinian ring is a minimal element;
in other words $\Ann_{R_M}(y_M)=MR_M$. It follows that $\Ann_R(y_M)=M$.
Consider the collection of elements $y_M$ of $R$ as $M$ varies over
maximal ideals of $R$. By the Noetherian-ness of $R$ there is a finite
collection $y_i=y_{M_i}; 1\leq i\leq p$, such that every $y_M$ is
a linear combination of these $y_i$; but then $M=\Ann(y_M)$ contains the
intersection of $\Ann(y_i)=M_i$. It follows that $M=M_i$ for some $i$.
This proves that there are only finitely many maximal (prime) ideals in
$R$.
As seen above the nil radical of a Noetherian ring is nilpotent. Since
every prime ideal in our ring is maximal and there are only finitely
many maximal ideals the nil radical is of the form $M_1\cdots M_p$. Thus
there is an integer $k$ such that $M_1^k\cdots M_p^k=0$. But then we see
that $R=R/M_1^k \times \dots \times R/M_p^k$. Since each of the latter
rings is Artinian we see that $R$ is Artinian.
\subsection{Summary of Results}
To summarize the results of this section:
\begin{emquote}
Finite rings and finite dimensional rings are examples of
Artinian rings.
\end{emquote}
\begin{emquote}
The Jacobson radical of an Artinian ring is the product of
its (finite collection of) maximal ideals and is a nilpotent
ideal; each prime ideal is maximal and consists of
zero-divisors; the complement of the union of all maximal ideals
consists of units.
\end{emquote}
Essentially equivalently:
\begin{emquote}
An Artinian ring $A$ can be written in the form
\[ (A/M_1^r) \times \cdots \times (A/M_k^r) \]
where the $M_i$ are the (finite collection of) maximal ideals of
$A$. The groups $M_i^p/M_i^{p+1}$ are finite dimensional vector
spaces over the field $A/M_i$ for each $i$ and for each $p$.
\end{emquote}
Consequently:
\begin{emquote}
An Artinian ring is also Noetherian.
\end{emquote}
Finally:
\begin{emquote}
A Noetherian ring in which every prime is maximal is Artinian.
\end{emquote}
\appendix
\section{Some common applications of the structure of Artinian rings}
The structure result for Artinian rings says that an Artinian ring has
the form
\[ A \cong (A/M_1^{q_1}) \times \cdots \times (A/M_p^{q_p}) \]
This result says that the idempotents of
the ring $A/M_1 \times \cdots \times A/M_p$ can be lifted to idempotents
in the ring $A$. Most results on Artinian rings say that such liftings
are possible for various algebraic structures.
\subsection{Hensel's Lemma}
Let $f(X)$ be a monic polynomial with coefficients in an Artinian
local ring $A$. Suppose that there are polynomials $g(X)$ and $h(X)$ in
$A[X]$ such that $f(X)-g(X)h(X)$ lies in $M[X]$ where $M$ is the
Jacobson radical of $M$. Further assume that $g(X)$ and $h(X)$ are
co-prime modulo $M$; in other words we can find $a(X)$ and $b(X)$ such
that
\[ 1 \equiv a(X)g(X) + b(X)h(X) \text{~in~} (A/M)[X] \]
We also get a decomposition
\[
(A/M)[X]/(\overline{f}(X)) =
(A/M)[X]/(g(X)) \times (A/M)[X]/(h(X))
\]
Since $f(X)$ is a monic polynomial $A[X]/(f(X))$ is isomorphic to
$A^{\deg(f)}$ as an $A$-module; it follows that it is Artinian as an
$A$-module and thus also as a ring. We will now see how the idempotents
corresponding to the above decomposition can be explicitly lifted by
factoring $f(X)$.
Suppose we have found $g_k(X)$ and $h_k(X)$ in $A[X]$ such that
$f(X)-g_k(X)h_k(X)$ lies in $M^{k}[X]$. Moreover, assume that the images
of $g_k(X)-g(X)$ and $h_k(X)-h(X)$ lie in $M[X]$.
Consider $f(X)-g_{k}(X)h_{k}(X)$
as an element of $(M^k/M^{k+1})[X]$; the latter is a module over
$(A/M)[X]$. Thus
\begin{multline*}
f(X)-g_{k}(X)h_{k}(X) = c_{k}(X)\\
\equiv c_{k}(X)a(X)g(X) + c_{k}(X)b(X)h(X) \pmod{M^{k+1}}\\
= c_{k}(X)a(X)g_{k}(X) + c_{k}(X)b(X)h_{k}(X)\pmod{M^{k+1}}
\end{multline*}
Now put
\begin{eqnarray*}
g_{k+1}(X) & = & g_{k}(X) + c_{k}(X)b(X) \\
h_{k+1}(X) & = & h_{k}(X) + c_{k}(X)a(X)
\end{eqnarray*}
We see that $f(X)-g_{k+1}(X)h_{k+1}(X)$ lies in $M^{k+1}[X]$. Starting
with $k=1$ we see that we can factor the image of $f(X)$ in
$(A/M^k)[X]$ for all $k$.
Since $A$ is Artinian local we see that $A=A/M^{k}$ for some $k$.
It follows that we have a factoring of $f(X)$ in $A[X]$.
\subsection{Weierstrass' preparation theorem}
Let $f(X)$ be a formal power series over a ring $A$. If the constant
term of $f$ is a unit in $A$, then by multiplying by its inverse we can
assume that the constant term of $f$ is 1; thus $f(X)=1-Xf_0(X)$. If we
write the formal inverse
\[
(1-Xf_0(X))^{-1} = 1 + Xf_0(X) + X^2f_0^2(X) + \dots
\]
we can see that, for each positive integer $k$,
the coefficient of $X^k$ involves only finitely many coefficients of
$f_0(X)$. It can thus be computed as a formal power series $g(X)$ in
$A$. To summarise, a formal power series with coefficients in $A$
whose leading coefficient is a unit in $A$, is also a unit in the ring of
formal power series over $A$.
Now, let $K$ be a field and $f(X)$ be any formal power series over $K$.
Consider the first integer $n$ such that the coeffient of $X^n$ is
non-zero. Then $f(X)=X^nf_0(X)$ and by what has been seen above $f_0(X)$
is a unit (since any non-zero element of a field is a unit).
Weierstrass' preparation theorem generalises this to an appropriate
statement for all power series over an Artinian local ring.
Let $f(X)$ be a formal power series with coeffients in an Artinian local
ring $A$. Assume that $n$ is the first integer such that the coefficient
of $X^n$ in $f(X)$ is a unit. For any power series $g$ let $L(g)$ be the
polynomial of degree at most $n-1$ consisting of terms of degree at most
$n-1$ of of $g$ and $H(g)=(g-L(g))/X^n$ be the high degree terms of $g$.
The given condition on $f$ can be restated by saying that $H(f)$ is a
unit in the ring of formal power series and $L(f)$ is in $M[X]$ where
$M$ is the maximal ideal of the ring $A$. Let $m=-L(f)H(f)^{-1}$; this
is a formal power series all of whose coefficients are in $M$; we have
$X^n = H(f)^{-1}(X)f(X) + m(X)$. Given any
formal power series $g(X)$ let us write $M(g)=H(mg)$. We can write
\begin{multline*}
g(X) = X^nH(g)(X) + L(g)(X) = \\
H(g)(X)H(f)^{-1}(X)f(X) + H(g)(X)m(X) + L(g)(X)
\end{multline*}
We have thus written $g(X)=h_1(X)f(X) + g_1(X) + l_1(X)$, where
$h_1(X)$ is a power series,$g_1(X)$ is a power series with coeffients in
$M$ and $l_1(X)$ is a polynomial of degree at most $(n-1)$. Suppose that
we have written $g(X)=h_k(X)f(X) + g_k(X) + l_k(X)$ where $h_k(X)$ is a
power series, $g_k(X)$ is a power series with coefficients in $M^k$ and
$l_k(X)$ is a polynomial of degree at most $(n-1)$. We use the above
formalism to write
\begin{multline*}
g_k(X) = X^nH(g_k)(X) + L(g_k)(X) =
H(g_k)(X)H(f)^{-1}(X)f(X) + H(g_k)(X)m(X) + L(g_k)(X)
\end{multline*}
So that putting $h_{k+1} = h_k + H(g_k)H(f)^{-1}$,
$g_{k+1}(X)=H(g_k)(X)m(X)$ and $l_{k+1}=l_k + L(g_k)$, we see that we
have the same formulae for $g$ with $k$ replaced by $k+1$; further note
that $g_{k+1}(X)$ has coefficients in $M^{k+1}$ since the coefficients
of $H(g_k)$ are the same as the coefficients of $g_k$ and lie in $M^k$
while the coefficients of $m(X)$ are in $M$. Since $A$ is an Artinian
ring $A=A/M^k$ for some $k$; so $g_k(X)=0$. Thus we have obtained an
expression
\[ g(X) = h(X) f(X) + l(X) \]
where $h(X)$ is a power series and $l(X)$ is a polynomial of degree at
most $n-1$.
Applying this to $g(X)=X^n$ we see that $X^n = u(X) f(X) + r(X)$.
Comparing coeffiecients of degree at most $n-1$ on both sides we see
that the coefficients of $r(X)$ are linear combinations of coefficients
of degree at most $n-1$ of $f(X)$; thus $r(X)$ lies in $M[X]$. Further
comparing the coeffiecients of $X^n$ on both sides we see that the
constant coefficient of $u(X)$ multiplied by the coefficient of $X^n$ in
$f(X)$ differs from 1 by an element of $M$. Thus the leading coefficient
of $u(X)$ is a unit and so $u(X)$ is a unit. We can also re-write the
above equation as $u(X)f(X)=X^n-r(X)$; in words, any power series, with
coefficients in an Artinian ring $A$, which is not zero modulo $M$, is
the multiple of a monic polynomial by an invertible power series.
\end{document}