Is φ a bijective homomorphism between simple $R$-modules?

[mathmari]

Hey!

Let $R$ be a commutative ring with unit and $M$ be a $R$-module.

Let $\phi : M\rightarrow M'$ be a non-zero homomorphism of simple $R$-module.
I want to show that $\phi$ is an isomorphism.

To show that we have to show that $\phi$ is bijective, right? (Wondering)

What exactly is the definition of $M'$ ? (Wondering)

 

[Euge]

To show that we have to show that $\phi$ is bijective, right? (Wondering)

Yes, that's right.

What exactly is the definition of $M'$ ? (Wondering)

$M'$ is a simple $R$-module.

 

[mathmari]

We have that $\text{Im}\phi $ is a $R$-submodule of $M'$. Since $M'$ is simple, it follows that $\text{Im}\phi=O$ or $\text{Im}\phi=M'$.
Since $\phi$ is a non-zero homomorphism, it must be $\text{Im}\phi=M'$, right? (Wondering)
If this is correct, it follows that $\phi$ is onto, or not? (Wondering)

We also have that $\ker\phi $ is a $R$-submodule of $M$. Since $M$ is simple, it follows that $\ker\phi=O$ or $\ker\phi=M$.
The kernel is equal to the set of elements mapped to $0$. Since $\phi$ is non-zero, it must be that $\ker\phi=O$, right? (Wondering)
If this is correct, it follows that $\phi$ is 1-1, or not? (Wondering) Therefore, $\phi$ is bijective.

 

[Euge]

You are correct on all points. (Nod)