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In R. Ash, Abstract Algebra in the chapter on non-commutative rings (9.2.2), the following exericse occurs:

Let $M$ be a nonzero cyclic module. Show that $M$ is simple if and only if $\operatorname{ann} M$, the annihilator of $M$, is a maximal left ideal.

As the chapter says, the rings are not assumed to be commutative, the modules are assumed to be left modules, and the annihilator is the set of all $r \in R$ such that $rm = 0$ for all $m \in M$.

Now I tried messing around with the map $f_x(r) = rx$ for $x \ne 0$, which is surjective by simplicity. Then I tried looking at its kernel, and surely we have $\operatorname{ann} M \subseteq \operatorname{ker}(f)$. But I do not see the converse inclusion (which should hold according to the solution given, which I looked up). If $R$ is not commutative, we cannot conclude from $rx = 0$ that $r(sx) = 0$ for some $s \in R$? All I see is that $$ \operatorname{ann}(M) = \bigcap_{x\in M} \operatorname{ker}(f_x) $$ and that if $y = sx$ we have $r \in \mbox{ker}(f_y) \Leftrightarrow ry = 0 \Leftrightarrow (rs)x = 0 \Leftrightarrow rs \in \mbox{ker}(f_x)$ hence $\mbox{ker}(f_{sx}) = \{ r \in R \mid rs \in \mbox{ker}(f_x) \}$. But I do not see that the annihilator coincides with any kernel?

StefanH
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    The exercise claims a false statement. See this question: https://math.stackexchange.com/questions/716659/is-every-left-maximal-ideal-the-annihilator-of-a-simple-left-module It is clear that annihilators of modules are two sided ideals and one can cook up examples of two sided maximal ideals strictly contained in maximal left ideals. From these you can build a simple module with annihilator the given two sided maximal ideal, which is therefore not a maximal left ideal. See in particular rschwieb 's comment. – guest Oct 23 '17 at 17:01

1 Answers1

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Let $R=M_2(\mathbb R)$ and $L$ be any nonzero left submodule. Then the left annihilator of $L$ is $\{0\}$ since $R$ is a simple ring, and in particular the annihilator is not a maximal left ideal.

This is true even for the $L$'s which are simple, such as $\left\{\begin{bmatrix}x&0\\y&0\end{bmatrix}\middle|\, x,y\in\mathbb R\right\}$.

So the statement is false.

But if $S$ is any simple left $R$ module, $S$ is cyclic, so you can write $S=Rx$ for some $x$ (and in fact, any nonzero element of $S$ will do). Then you have a natural homomorphism $R\to Rx$ given by $r\mapsto rx$, whose kernel is a maximal left ideal of $R$ by the first homomorphism theorem.

So, the left annihilator of any nonzero element of a simple left module is a maximal left ideal.

rschwieb
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  • I am not sure about your example, as $R = M_2(\mathbb R)$ is simple in the sense that it has no two-sided ideals, but not in the sense of the definiton given by Ash that for a (left) module simplicity means that it has no non-trivial submodules, which correspont to left ideals in the ring (viewed as a module). So simplicity for a ring (by forbidding two-sided ideals) is a much stronger claim. Does your remaks then still apply? – StefanH Oct 24 '17 at 09:35
  • @StefanH Everything I said applies... I don't understand your concern. I gave an example of a ring $R$ and a simple left module $L$ (in fact, a minimal and maximal left ideal of $R$) such that $ann(L)$ is not a maximal left ideal (the annihilator is properly contained in $L$, for example.) – rschwieb Oct 24 '17 at 10:43
  • In your example R is simple, meaning it has no two-sided ideal, but the exercise talks about simple modules, which have no left-sided ideals, hence if we view $R$ as a module saying it is simple according to the exercise also means it is not allowed to non-trivial have left-sided ideals, so your $L$ is not allowed! Is it now clearer? It is about the difference what it means for a ring to be simple vs what it means for a module to be simple. – StefanH Oct 24 '17 at 12:10
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    Okay, sorry you apply everything to $L$, I somehow thought you want to apply your argument to $R$ considered (as a module). But $L$ is your counterexample of a simple left module! But your $L$ is not cyclic, or? – StefanH Oct 24 '17 at 12:20
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    Ok, sorry it is cyclic by simplicity! Sorry, was somehow confused your example is totally fine! – StefanH Oct 24 '17 at 12:23
  • @StefanH Glad to be of help! – rschwieb Oct 24 '17 at 13:05