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Regarding $\mathbb{Q}$ as a field, there is only the trivial automorphism. What happens if we only consider $(\mathbb{Q},+,0)$ as an abelian group? Is there a description or way to compute the ring of endomorphisms $\mathrm{End}(\mathbb{Q},+,0)$?

madison flom
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1 Answers1

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Yes. $\def\Q{\mathbb{Q}}$ $\def\End{\text{End}}$Suppose that $f\in\End(\Q)$. Then, you may note that since $f(n)=nf(1)$ for all $n\in\mathbb{Z}$ we clearly have

$$qf\left(\frac{p}{q}\right)=f(p)=pf(1)$$

so that

$$f\left(\frac{p}{q}\right)=\frac{p}{q}f(1)$$

Thus, we have a bijection $\End(\Q)\to\Q:f\mapsto f(1)$. Since, if $f(x)=ax$ and $g(x)=bx$ for $a,b\in\Q$ we have that

$$f(g(x))=f(bx)=bf(x)=baf(x)$$

and

$$(f+g)(x)=ax+bx=(a+b)(x).$$

Thus, we see that $\End(\Q)\to\Q$ is actually a ring isomorphism.

This more generally shows that if $R$ is an integral domain then, as rings, $\End_R(\text{Frac}(R))\cong \text{Frac}(R)$.

Alex Youcis
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    That's even more general - if $R$ is a domain and $(R)$ is its field of fractions, than for any $(R)$-modules $V, W$ we have that $Hom _{(R)}(V, W) \simeq Hom _{R}(V, W)$. :) – Piotr Pstrągowski Mar 10 '13 at 02:05
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    @PiotrPstragowski The $R$-module isomorphism, yes. But not the ring structure :) – Alex Youcis Mar 10 '13 at 02:05