Proving that $annM$ is a prime ideal for ireducible $M_R$ If we let $R$ be a ring with $1$ and $M_R$ to be an irreducible right $R$-module then I want to show that $annM]\{r\in R|Mr=0\}$ is a prime ideal.
So if we take $a,b\in R$ with $aRb\subset annM$ then we need to show that either $a\in annM$ or $b\in annM$.
Now as $1\in R$ we have that $ab\in annM$ so that we have $M(ab)=0$. Now suppose that neither $Ma=0$ or $Mb=0$ then as $M$ is irreducible we must have $Ma=M$ and $Mb=M$ but this then gives $M(ab)=(Ma)b=Mb=m$ a contradiction and so one of $a$ or $b$ is in $annM$ and so $ann M$ is prime.
Is this proof correct?
Thanks for any help.
 A: It's 90% of the way there, but there is a mistake in the reasoning at this part: "since $M$ is irreducible, $Ma$ is either $\{0\}$ or $M$."
The problem is that $Ma$ need not be a submodule of $M$.
However, $MaR$ is a submodule of $M$, and the logic should be reworked this way: if $MaR=\{0\}$ then we are done because $a\in aR$ annihilates $M$. On the other hand if $MaR=M$, we have immediately that $(MaR)b=M(aRb)=\{0\}$, so that $b$ annihilates $M$.

I sometimes see students taking this path of taking a hypothesis and then immediately turning it into a "more familiar" hypothesis. In this case, $aRb=\{0\}$ was used to draw the (much weaker) conclusion $ab=0$, but there did not seem to be any indication this should be done. In fact the full power of $aRb=\{0\}$ was needed for the proof, so we lost a lot by changing it to merely $ab=0$!
So this brings us to a pretty good piece of advice: if you find that you didn't use the full strength of the hypotheses, you should double check the work. On one hand it might have been necessary, and a mistake was made. On the other hand, it might truly be unnecessary, and the statement of the problem could be improved.
A: This is to write the proof another way(but very much the same reasoning). For simplicity of notation, let me assume that the ring acts on the module on the left instead.
The ring $R/(Ann\ M)$ acts on $M$ with zero kernel, i.e., no nonzero element in $R/(Ann\ M)$ acts on $M$ by the zero endomorphism; i.e., if $a \in R/(Ann\ M)$ and $aM= 0$, we get $a = 0$. What we have to prove is that this quotient ring is an integral domain.
Now suppose $a, b$ to be two nonzero elements in this quotient ring $R/(Ann\ M)$. Since $a \neq 0$, $aM \neq 0$. But, irreducibility of $M$ means that, $aM$ , being a nonzero submodule of $M$, has to be $M$ itself. Now since $b \neq 0$, we again have $b(aM) \neq 0$. So, $ba \neq0$. So this ring is a domain, and therefore the ideal $Ann\ M$ is prime.
