How to show that the annihilator of an element of a left module is a left ideal but not necessarily a two-sided ideal. When does this become an ideal?
$\begingroup$
$\endgroup$
Add a comment
|
$\begingroup$
$\endgroup$
2
Hints:
Let $m$ be an element of your module.
Suppose $x,y\in ann(m)$. Compute $(x+y)m$. What does this say about $x+y$?
Let $r\in R$ be arbitrary. Compute $(rx)m$. What does this say about $rx$?
Now, supposing we tried to compute $(xr)m$, we would find that we have trouble concluding that it is zero. (We only know that $xm=0$, but that multiplication isn't happening here!)
This suggests that it might not be true after all. In fact, there is a counterexample. Let $M=R=M_2(F)$ be the square matrix ring over a field $F$, and look at the (left) annihilator of $\begin{bmatrix}1&0\\0&0\end{bmatrix}$. You'll find out it is the set $\{\begin{bmatrix}0&a\\0&b\end{bmatrix}\mid a,b\in F\}$. Show that this isn't a right ideal of $R$.
When does the annihilator become a two-sided ideal?
Obviously, if $R$ is commutative, it is true.
More interestingly, you should try to prove that the annihilator of any left $R$ module is a two-sided ideal of $R$. The commutative result above can be relaxed a little bit by noticing that the annihilator is an ideal if $ann(x)=ann(Rx)\lhd R$. One case this occurs, for example, is if $xR=Rx$.
