1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Consider $\phi : \mathbb{Z} \to \mathbb{Z} : x \mapsto 2x$. $\endgroup$ – anakhro Jul 9 '15 at 16:31
  • $\begingroup$ But wouldn't that have an inverse? When you apply the function, you multiply by 2. In the inverse, you divide by 2, but you only define it for even numbers. $\endgroup$ – man_in_green_shirt Jul 9 '15 at 16:33
  • $\begingroup$ Or is that why you can't consider it an isomorphism? Does every element in target set need to have an inverse? $\endgroup$ – man_in_green_shirt Jul 9 '15 at 16:34
  • 1
    $\begingroup$ If you define a function by "dividing by $2$", then what is the image of $1$? An isomorphism is a morphism which has an inverse; this inverse must be defined on the whole module. $\endgroup$ – Pierre-Guy Plamondon Jul 9 '15 at 16:35
  • 1
    $\begingroup$ I guess the counter-intuitive fact here (but fact nonetheless) is that a module $M$ can have a strictly contained submodule $N$ which is isomorphic to $M$. The example with $\mathbb{Z}$ illustrates this. $\endgroup$ – Pierre-Guy Plamondon Jul 9 '15 at 16:44
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.