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Subrings, ideals and quotients


\begin{de}
A subring of a ring $R$\ is any subset $S\subseteq R$\ which forms a ring
under the same operations.
\end{de}


\begin{ex}
$\mathbb{Z}$\ is a subring of $\mathbb{Q}$\ is a subring of $\mathbb{R}$\ is a subring of
$\mathbb{C}$\ is a subring of $\mathbb{H}$.
\end{ex}


\begin{ex}
Every non-trivial ring $R$\ has at least two subrings -- $\{0\}$\ and $R$;
all the other subrings are called {\em proper}.
\end{ex}

As in groups, we can reduce the number of axioms one has to check when proving that something is a subring.


\begin{thm}
A non-empty subset $S$\ of a ring $R$\ is a subring if and only if
$...
...ightarrow -x\in S$) and multiplication
$x,y\in S\Rightarrow xy\in S$).
\end{thm}


\begin{proof}
% latex2html id marker 4720($\Rightarrow$) This is obvious.
\par...
... and distributivity laws are
then automatically inherited from $R$.
\end{proof}


\begin{exc}
Prove that the set of matrices
\begin{displaymath}
S=\left\{ \left(\...
...ht\}
\end{displaymath}forms a subring of the ring $M_2({\mathbb{R}})$.
\end{exc}


\begin{ex}
The set of matrices
\begin{displaymath}
\left\{ \left(\begin{array}{c...
...$A,-A\in GL(2,{\mathbb{R}})$, but $A+(-A)=0\not\in GL(2,{\mathbb{R}})$.
\end{ex}


\begin{de}
A subring $I$\ of a ring $R$\ is called an {\em ideal} if
it satisfie...
...laymath}we say that $I$\ is {\em stable} under multiplication from $R$.
\end{de}


\begin{re}
Since stability implies closure under multiplication,
to check that a...
...at $I$\ is a subgroup of the additive group $R$,
and that it is stable.
\end{re}


\begin{ex}
The set $m{\mathbb{Z}}$\ is an ideal of $\mathbb{Z}$.
Indeed, if $ma\...
...}}$\ and $n\in {\mathbb{Z}}$\ the
$n(ma)=(ma)n=m(na)\in m{\mathbb{Z}}$.
\end{ex}


\begin{ex}
The set $x{\mathbb{R}}[x]$\ is an ideal of ${\mathbb{R}}[x]$.
\end{ex}


\begin{ex}
% latex2html id marker 4758The subring $S$\ of $M_2({\mathbb{R}})$\...
...\begin{array}{cc}2&2\\ 1&1\end{array}\right)\not\in S.
\end{displaymath}\end{ex}


\begin{ex}
$\{0\}$\ and $R$\ are ideals of $R$.
\end{ex}


\begin{thm}
A field $F$\ has no proper ideals.
\end{thm}


\begin{proof}
Let $I$\ be an ideal of $F$, and assume that $I\neq\{0\}$.
Choose ...
...1=baa^{-1}=a(ba^{-1})\in I.
\end{displaymath}This proves that $I=F$.
\end{proof}

Ideals play for rings the role that normal subgroups play for groups.


\begin{thm}
Let $R$\ be a ring, and let $I$\ be an ideal of $R$.
The set
\begin{...
...displaymath}
(a+I)+(b+I)=(a+b)+I,\ (a+I)(b+I)=(ab)+I.
\end{displaymath}\end{thm}


\begin{proof}
% latex2html id marker 4781By Theorem \ref{thm76} it follows tha...
...ociativity and distributivity is now easy,
and is left for exercise.
\end{proof}



Edmund F Robertson

11 September 2006