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What is $\mathrm{Hom}(\mathbb Q;\mathbb Q)$? Is it isomorphic to $\mathbb Q$? If yes, what is the isomorphism?

Is there a reference where the hom functor is explicitly studied with concrete examples like the above one?

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    i can see the homomorphisms multiplication by a rational $r$ i.e the homomorphism $q\mapsto rq$ so we have a homomorphism for each $r$ and then $\mathbb Q\subset Hom(\mathbb Q;\mathbb Q)$ . The problem in this question is that $\mathbb Q$ is not finitely generated as an additive group so we can not enumarate homomorphisms by counting them by their values on generators like we do for $\mathbb Z$ or $\mathbb Z /n\mathbb Z$2011-05-23

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So, you already know that you have at least one homomorphism for every rational number, given by "multiplication by $r$". You might also want to notice that composition of these homomorphisms coincides with multiplication of rationals; that is, you seem to have the multiplicative semigroup $\mathbb{Q}$ at the very least.

The question then is whether these are all the homomorphisms...

Suppose that $f\colon\mathbb{Q}\to\mathbb{Q}$ is an additive homomorphism, and $f(1)=a$. This clearly tells you what $f$ restricts to in the integer (namely, $f(m) = am$). Does it tell you anything else?

Well, what is $f(\frac{1}{2})$? Whatever it is, we have: $f(1) = f\left(\frac{1}{2}+\frac{1}{2}\right) = f\left(\frac{1}{2}\right)+f\left(\frac{1}{2}\right) = 2f\left(\frac{1}{2}\right).$ So in fact, we also know what the value of $f$ is at $\frac{1}{2}$.

Can we figure out how much $f$ is at other rationals, given how much it is at $1$? What does this tell us?

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    @palio: Of course, you need to show that $f(1/q) = (1/q)f(1)$ for that argument (which, granted, is not hard).2011-05-25