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?


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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 $xR=Rx$.

  • $\begingroup$ I'm having a trouble with matrix multiplication or you meant $\left\{\begin{bmatrix}0&a\\0&b\end{bmatrix}\mid a,b\in F\right\}$. $\endgroup$
    – rgm
    May 20, 2017 at 13:23
  • $\begingroup$ @rgm You're right about the typo: corrected $\endgroup$
    – rschwieb
    May 20, 2017 at 15:35

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