Let $S$ be a surface and $C$ be an effective divisor in $S$. That is $C$ is a curve in $S$ and $i:C\longrightarrow S$ is the inclusion morphism. Let $E$ be line bundle over $C$, so $i_*E$ is a coherent sheaf over $S$. Since $E$ is a line bundle, $E$ is simple. Does that mean $i_*E$ is simple too?
In commutative algebra terms, it translates to the following. Let $f: A\longrightarrow A/I$ be a ring homomorphism, and $M$ is a free module of rank 1 over $A/I$. $M$ is simple as an $A/I$ module, that is it has no nontrivial endomorphisms. Is $M$ simple as an $A$ module?
Any help will be appreciated!
If $R$ is a commutative ring with unity when can $R$ be a simple $R$-module? If it is so, then $R\simeq R/\mathfrak m$ with $\mathfrak m$ a maximal ideal, hence $\operatorname{Ann}_R(R)=\operatorname{Ann}_R(R/\mathfrak m)$ and thus we have $\mathfrak m=(0)$. It follows that $R$ is a field.
If $R=A/I$ we get $I$ is maximal, and since $M\simeq A/I\Rightarrow$ $M$ is a simple $A$-module.