<?xml version="1.0" encoding="UTF-8"?>

<record version="7" id="3167">
 <title>annihilator is an ideal</title>
 <name>AnnihilatorIsAnIdeal</name>
 <created>2002-07-15 09:03:29</created>
 <modified>2007-01-18 05:23:25</modified>
 <type>Theorem</type>
<parent id="996">annihilator</parent>
 <creator id="2760" name="yark"/>
 <author id="2760" name="yark"/>
 <author id="225" name="saforres"/>
 <classification>
	<category scheme="msc" code="16D10"/>
	<category scheme="msc" code="16D25"/>
 </classification>
 <preamble>\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
</preamble>
 <content>\PMlinkescapeword{addition}
\PMlinkescapeword{equivalent}
\PMlinkescapeword{multiplication}
\PMlinkescapeword{right}

The right annihilator of a right $R$-module $M_R$ in $R$ is an ideal.

{\bf Proof:}\\
By the distributive law for modules, it is easy to see that $\operatorname{r.ann}(M_R)$ is closed under addition and right multiplication.
Now take $x \in \operatorname{r.ann}(M_R)$ and $r \in R$.

Take any $m \in M_R$.  Then $mr \in M_R$, but then $(mr)x = 0$ since $x \in \operatorname{r.ann}(M_R)$.  So $m(rx)=0$ and $rx \in \operatorname{r.ann}(M_R)$.

An equivalent result holds for left annihilators.</content>
</record>
