\documentclass{article}
\newtheorem{problem}{Problem}
\newcommand{\ann}{\mbox{Ann}}
\title{CHM: A.7}
\author{Po-Shen Loh}
\date{24 March 2001}
\begin{document}
\maketitle

\begin{problem}[A.6]
  Let $R$ be a finite commutative ring.  Prove that $R$ has a
  multiplicative identity element (1) if and only if the annihilator
  of $R$ is 0 (that is, $aR = 0$, $a \in R$ imply $a = 0$).
\end{problem}

First, observe that there exist elements $e, x \in R$ such that $xe =
x$: we can do the following ``chain multiplication.''  Start with an
element $a_1 \in R$.  By the given information, there exists another
element $a_2 \in R$ such that $a_1a_2 \neq 0$.  Now by a similar
argument there exists $a_3 \in R$ such that $a_1a_2a_3 \neq 0$.
Repeating this process, we can get arbitrarily long chains.  For a
given chain, let us define the partial products $\pi_n = \Pi_1^n a_i$.
We have a finite ring $R$, so at some point our partial products will
repeat; therefore, we have the form $\pi_n \alpha = \pi_n$ where
$\alpha$ is a substring of the chain.  Therefore, the opening claim is
true.

Now we use an extremal argument.  Let us select the $e \in R$ for
which the ideal $I = \{x | xe = x\}$ is maximal.  Proceed by indirect
proof; suppose, for the sake of contradiction, that $I \neq R$.  We
show that we can choose $r, s \in R$ such that $rs \neq 0$, $rI = 0$,
and $se = 0$.  Observe that there exists some element $y \in R$ for
which $ye \neq y$.  Let $r = ye-y \neq 0$.  But for any $x \in I$,
$yex = yx \rightarrow rx = (ye-y)x = 0$, so $r$ annihilates $I$.  Use
indirect proof to show that we can find our $r, s$ as claimed: suppose
that for all $r$ found above, $rs = 0$ for all $s$ that satisfy $se =
0$.  This is equivalent to the statement $R - I \subset
\ann(\ann(e))$.

We have finite rings, so by Lagrange's theorem applied to the additive
groups, $|R-I| \geq |R|/2$ and $|\ann(\ann(e))| \leq |R|/2$;
therefore, the inclusion above must be exact, so $R-I =
\ann(\ann(e))$.  This is not true, however, because $0 \in I
\rightarrow 0 \not \in R-I$ but $0 \in \ann(\ann(e))$.  Our
supposition is false, so our claim is indeed true.

We return to chain multiplication.  Start with $r$ and $s$, and build
the chain as above, but this time start with $a_0 = r$, $a_1 = s$ and
maintain the definition of $\pi_n = \Pi_1^n a_i$.  By the same
argument as above, we will be able to find a form $r\pi_n \alpha =
\pi_n$.  This translates into a relation $ye' = y$, where $y = \pi_n$
and $e' = r\alpha$.  Since $r|e'$, we find that $e'$ annihilates all
of $I$, and similarly $s|y \rightarrow ey = 0$.  Also, since $rI = 0$,
by construction $\pi_n$ is not contained in $I$.  Observe that we have
discovered that for all $x \in I$, $(e+e')x = ex + e'x = x + 0 = x$
and for all $z$ in the ideal generated by $y$, $(e+e')z = ez + e'z = 0
+ z$.

Therefore, the element $(e+e')$ is a better identity than $e$; it acts
trivially on both $I$ and the ideal generated by $y$.  This violates
our maximality condition, so our assumption that $I \neq R$ is false,
and we are done.  The identity is $e$.

\end{document}
