\documentclass{article}
\newtheorem{problem}{Problem}
\title{CHM: A.3}
\author{Po-Shen Loh}
\date{24 March 2001}
\begin{document}
\maketitle

\begin{problem}[A.3]
  Let $R = R_1 \oplus R_2$ be the direct sum of the rings $R_1$ and
  $R_2$, and let $N_2$ be the annihilator ideal of $R_2$ (in $R_2$).
  Prove that $R_1$ will be an ideal in every ring $\tilde{R}$ containing
  $R$ as an ideal if and only if the only homomorphism from $R_1$ to
  $N_2$ is the zero homomorphism.
\end{problem}

\noindent \textbf{Solution:}

Clearly, $R_1$ is closed under addition, so it suffices to show that
it is closed under scalar multiplication.  Since $R$ is an ideal in
$\tilde{R}$, multiplication of any element of $R_1$ by some $x \in
\tilde{R}$ sends it to some element in $R_1 \oplus R_2$.  Since we
have a direct sum, there exists a well-defined projection homomorphism
$\pi : R \rightarrow R_2$.  Therefore, we can define the family of
functions $\phi_x(r_1) = \pi(xr_1)$.  We now show that the $\phi_x$
are homomorphisms as well.

In order for multiplication by $x$ to be defined, we need the ring
axioms to hold.  In particular, the associative law must be true (we
now use ordered-pair notation for elements of our direct sum:

\begin{eqnarray*}
  x \cdot ((r_1, 0) (r_1', r_2')) &=& (x \cdot (r_1, 0)) (r_1',
  r_2') \\
  x \cdot (r_1r_1', 0) &=& (\alpha, \phi_x(r_1)r_2') \quad \mbox{for
  some $\alpha \in R_1$}\\
  \phi_x(r_1r_1') &=& \phi_x(r_1)r_2'.
\end{eqnarray*}

The above result must hold for all choices of $r_1$, $r_1'$, and
$r_2'$.  Suppose we choose $r_2' = 0$; then we discover that for all
$r_1$ and $r_1'$, $\phi_x(r_1r_1') = 0$.  Hence for all $r_1$, $r_2'$,
$\phi_x(r_1)r_2' = 0 \Rightarrow \phi_x(r_1) \in N_2$.  Now we can
show that $\phi_x$ is a homomorphism.

It is clearly true that $\phi_x(a + b) = \phi_x(a) + \phi_x(b)$, so it
suffices to show that $\phi_x(ab) = \phi_x(a)\phi_x(b)$.  But from
above, we see that $\phi_x(ab) = 0$, and since $\phi_x(a) \in N_2$ and
$\phi_x(b) \in R_2$, the RHS is also zero.  Therefore we do have a
homomorphism, as claimed.

This immediately solves our problem; any $x-$multiplication
automatically induces a homomorphism from $R_1$ to $N_2$.  $R_1$ is an
ideal in the larger ring if and only if all $\phi_x$ are zero;
therefore, we are done.

\end{document}
