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?

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. — man_in_green_shirt, Jul 09 '15 at 16:33
Or is that why you can't consider it an isomorphism? Does every element in target set need to have an inverse? — man_in_green_shirt, Jul 09 '15 at 16:34
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. — Pierre-Guy Plamondon, Jul 09 '15 at 16:35
It does not have an inverse. If you believe it does, what is $\phi ^{-1}(1)$? — anak, Jul 09 '15 at 16:36
1 isn't even - that's why I said 'you only define it for even numbers'. But I guess you can't do that an it needs to be defined on the whole target set — man_in_green_shirt, Jul 09 '15 at 16:36
Makes sense now, I realise that my notion of 'isomorphism' wasn't very clear — man_in_green_shirt, Jul 09 '15 at 16:37
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. — Pierre-Guy Plamondon, Jul 09 '15 at 16:44

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

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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.

edited Apr 13 '17 at 12:21

Community

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answered Jul 09 '15 at 16:31

Pierre-Guy Plamondon

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score 0 · Answer 2 · answered Jul 09 '15 at 16:34

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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.

answered Jul 09 '15 at 16:34

sebigu

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Are all injective endomorphisms of a module automorphisms?

2 Answers2