9367 Q51

Forum for the GRE subject test in mathematics.
yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

9367 Q51

Postby yoyobarn » Sun Feb 19, 2012 11:33 am

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
Posts: 11
Joined: Tue Jan 24, 2012 6:43 pm

Re: 9367 Q51

Postby cincodemayo5590 » Sun Feb 19, 2012 1:33 pm

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
Posts: 7
Joined: Thu Feb 16, 2012 2:07 pm

Re: 9367 Q51

Postby aaroncraig » Sun Feb 19, 2012 4:42 pm

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

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

Re: 9367 Q51

Postby yoyobarn » Mon Feb 20, 2012 2:02 am

Thanks for the detailed explanation!

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

DanielMcLaury
Posts: 12
Joined: Mon Nov 21, 2011 10:42 pm

Re: 9367 Q51

Postby DanielMcLaury » Mon Feb 20, 2012 9:59 pm

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).




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