Automorphisms of Algebraic Numbers

Can you find an automorphism [itex]\mathbb{Q}[\sqrt{2}]\rightarrow \mathbb{Q}[\sqrt{3}][/itex]?? Can you extend this to [itex]\mathbb{A}[/itex]??

 

Are you sure? Q[sqrt(2)] has an element such that x^2=2. Q[sqrt(3)] doesn't. I only know two automorphisms of A. Am I confused? I know lots of automorphisms of subfields of A. But not of A.

 

Ah yes

Maybe you can do something like [itex]\sqrt{2}\rightarrow -\sqrt{2}[/itex]??

 

Q[sqrt(2)]=Q[-sqrt(2)]. Need to do better than that.

 

Yes, but it will be a nontrivial automorphism [itex]\mathbb{Q}[\sqrt{2}]\rightarrow \mathbb{Q}[\sqrt{2}][/itex]. Or am I just too tired??

 

True, I think you are right. But how to extend that to A? Maybe I'm too tired as well. Like I said before, I know lots of automorphisms of subfields of A. But the full field A? zzzzzzz!

 

Haha. I am totally in your class. Math 257 with PSally?

 

Well you got the following theorem:

If [itex]\phi:F_1\rightarrow F_2[/itex] is an isomorphism between fields and if [itex]F_i\subseteq K_i[/itex] is the algebraic closure, then [itex]\phi[/itex] can be extended to an isomorphism [itex]K_1\rightarrow K_2[/itex].​

This theorem is true, but its proof is nonconstructive: it uses the lemma of Zorn. So you can show that there exists nontrivial automorphisms of A, but you can't construct one explicitly.

I think this provides the OP with enough hints. So he should try to figure out the details now.