Are all injective endomorphisms of a module automorphisms?

Question

As far as I understand, an automorphism is an isomorphism from a set to itself. If we have a homomorphism $f:M\rightarrow M$, then, from the first isomorphism theoreom, $im(f)$ is a submodule of $M$. As it is injective, the kernel is zero, and hence the image is isomorphic to $M$.

Does this not show that $f$ is an isomorphism from $M$ to itself? What am I missing?

Answer 1

Not in general. For example, if your base ring is $\mathbb{Z}$, and you take the map from $\mathbb{Z}$ to itself multiplying every element by $2$, this map is injective, but is not an automorphism.

However, the statement is true if $M$ is an Artinian module, as can be seen in this question.

Answer 2

The isomorphism theorem states that $im(f) \cong M/ ker( f )$, but this isomorphism does not need to be the injection of the submodule.

In the example above, $2\mathbb{Z} \cong \mathbb{Z}$, but not by the injection.