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.
9367 Q51

 Posts: 11
 Joined: Tue Jan 24, 2012 6:43 pm
Re: 9367 Q51
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.
Is there a solution manual that says there are two field automorphisms on the rationals? That's wrong.

 Posts: 7
 Joined: Thu Feb 16, 2012 2:07 pm
Re: 9367 Q51
The answer key at the end has "B" (i.e. there is exactly one automorphism) as the answer
Re: 9367 Q51
Thanks for the detailed explanation!
Sorry my bad, I mentally associated "B" with 2..
Sorry my bad, I mentally associated "B" with 2..

 Posts: 12
 Joined: Mon Nov 21, 2011 10:42 pm
Re: 9367 Q51
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).