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I'm trying to that prove $Rm$ is a simple ring $\iff Ann(m)$ is a left ideal of $R$.

I did a couple of questions earlier that I believe could be useful to show this:

$(Q1)$ Quotient of cyclic $R$-module is cyclic.

$(Q2)$ Isomorphism and cyclic modules

And I have the following theorem from the book:

$$\text{Let M be a bilateral ideal of a ring R. Then the following is equivalent:}\\ (1) \text{ M is maximal}\\(2) \text{The ring }\frac{R}{M}\neq 0\text{ and is simple} $$

I believe the former theorem could be applied changing bilateral idea by left ideal, right?.

Then if $Rm$ simple by $(Q2)$ is $\frac{R}{Ann(m)}\simeq Rm$ (I would like to prove this using left $R-$modules, but is not explicit in $Q2$). By the last isomorphism is $\frac{R}{Ann(m)}$ simple and by the theorem follows $Ann(m)$ left maximal (instead of bilateral using left).

Conversely, if $Ann(m)$ is a left maximal, the ring $\frac{R}{Ann(m)}$ is simple, and by the ismorphism follows $Rm$ simple.

Now, what's left and I'm couldn't do is prove the former isomorphism: $\displaystyle\frac{R}{Ann(m)}\simeq Rm$. How could I do this?, and was the proof above right?.

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Cure
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  • Isn't ann(M) always a left ideal??? – Marc Nov 04 '14 at 19:15
  • If it were correctly stated it would be a duplicate of http://math.stackexchange.com/q/969493/29335 , but there appear to be deeper issues with the poster's understanding that probably make it a nonduplicate worth keeping. – rschwieb Nov 04 '14 at 20:21

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I'm trying to that prove $Rm$ is a simple ring $\iff$ $Ann(m)$ is a left ideal of $R$.

Thanks for including your work. It tells us where a couple problems are. First, I think you might be obfuscating simple rings with simple modules. Let's also be clear that (from context) you mean the annihilator of something called $m$ apparently in a left $R$ module.

You mention "if $Ann(m)$ is a left maximal, the ring $R/Ann(m)$ is simple," but $R/Ann(m)$ won't be a ring unless $Ann(m)$ is a two-sided ideal, and it usually won't be. But $R/Ann(m)$ certainly is a left module. Another thing is that the left annihilator of $m$ will always be a left ideal of $R$, regardless of what $Rm$ looks like. So there is something more wrong with the statement.

You can say this:

$Rm$ is a simple left $R$ module $\iff$ $Ann(m)$ is a maximal left ideal of $R$

The proof is to notice that $Rm\cong R/Ann(m)$ by a natural map: right multiplication by $m$. If $Ann(m)$ is a maximal left ideal, then by submodule correspondence, the module $Rm$ is simple.

It's also true that $R/I$ is a simple ring iff $I$ is a maximal two-sided ideal. You can't tie this to the annihilator of an element in a module, though, since the annihilator is not generally a two-sided ideal.

For commutative rings, of course, $Ann(m)$ will be an ideal, and the two cases merge.

rschwieb
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  • Thanks. I missed the "maximal" when I copied the statement!, thanks for noticing that!. Regarding the isomorphism, if I understand you correctly the iso could be defined by $\phi:Rm\to R/Ann(m)$ with $\phi(m)=r\cdot m$?. For this I'd say that is injective since $\phi(m)=0\iff r\cdot m = 0 \iff m=0$, but how could I prove it's surjective?. – Cure Nov 04 '14 at 21:56
  • @cure Actually, if you interchange the domain and range on your map $\phi$ you will have better luck. – rschwieb Nov 05 '14 at 03:51
  • @rschwieb does correspondence theorem theorem also work on one sided ideal? – 123 Apr 20 '15 at 04:41
  • @123 of course it does: we are applying the correspondence theorem for modules, not rings. If K is a left ideal or two sided ideal, the left module R/K does not care a bit if K is also a right module or not. – rschwieb Apr 20 '15 at 10:02