9367 Q51

[yoyobarn]

Hi, what are the two distinct automorphisms?

I think one of them is the identity map: phi(n)=n

Thank you very much for help.

[cincodemayo5590]

The only field automorphism $$\phi : \mathbb{Q} \rightarrow \mathbb{Q}$$ is the identity. This can be seen since $$\phi(1) = \phi(1) \phi(1) \implies \phi(1) = 1$$ or $$\phi(1) = 0$$, but the latter does not give an injective map, since $$\phi(0) = \phi(0) + \phi(0) \implies \phi(0) = 0$$; then, $$\phi(n) = n \phi(1)$$ for integers $$n$$; and thus, for integers $$p$$, $$q$$, since $$\phi(q) \phi \left( \frac{p}{q} \right) = \phi(p)$$, we have $$\phi \left( \frac{p}{q} \right) = \frac{ \phi(p)}{\phi(q)} = \frac{p}{q}$$.

Is there a solution manual that says there are two field automorphisms on the rationals? That's wrong.

[aaroncraig]

The answer key at the end has "B" (i.e. there is exactly one automorphism) as the answer

[yoyobarn]

Thanks for the detailed explanation!

Sorry my bad, I mentally associated "B" with 2..

[DanielMcLaury]

Expanding on this, nontrivial field automorphisms have to fix some proper subfield, but the rationals don't contain subfields -- $$\mathbb{Q}$$ its own prime subfield (which is essentially what cincodemayo5590 shows above).