Let $R$ be a commutative ring, $M$ be an $R$ module and $m\in M$.
I am trying to prove or disprove that $M/ann(m)$ is an injective $R$ module. Any suggestions are welcome.
I am expecting the answer to be yes. I was trying to prove that given any $R$ module $M$ there exists an $R$ module $I$ with an injective morphism $M\rightarrow I$ i.e., to prove that there are enough injectives in the category of $R$ modules. Idea is to prove that the above quotient is an injective module and so is the product of quotients over all elements of $M$ with natrural (which I expect to be injective)map $M\rightarrow \prod_{m\in M} M/ann(m)$.
This is usually not true (indeed, it's harder to come up with a nontrivial example where it is true than one where it isn't true). For instance, take $R=M=\mathbb{Z}$ and $m=1$. Then $ann(m)=0$, so $M/ann(m)=\mathbb{Z}$ which is not an injective $\mathbb{Z}$-module.
For an example where $ann(m)M$ is nonzero, take $R=\mathbb{Z}$, $M=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}$, and $m=(1,0)$. Then $ann(m)=2\mathbb{Z}$ and $M/ann(m)=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ is not injective.
Really, I cannot see any reason why you would expect something like this to be true. If you do not have any motivating examples or any inkling of how to prove a similar statement, there is no reason at all to imagine you are on the right track.