# Automorphisms of Algebraic Numbers

Gold Member

## Homework Statement

1) Find more than two automorphisms of A.
2) Do automorphisms of C fix AR?

N/A

## The Attempt at a Solution

I managed to figure out the second question since a map which preserves the additive structure of C will fix Q. And since the maps preserves multiplicative structure as well, it is not difficult to show that the automorphism will fix AR. So I think I have this part covered.

So right now, I could use some help managing the first part of this problem. My professor indicated that there are a lot of automorphisms of A, so I'm thinking that there might be a family of automorphisms which aren't too difficult to construct. Any pointers on how to get started with this part are appreciated.

Thanks!

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

Dick
Homework Helper
Can you find an automorphism $\mathbb{Q}[\sqrt{2}]\rightarrow \mathbb{Q}[\sqrt{3}]$?? Can you extend this to $\mathbb{A}$??
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.

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?
Ah yes

Maybe you can do something like $\sqrt{2}\rightarrow -\sqrt{2}$??

Dick
Homework Helper
Ah yes

Maybe you can do something like $\sqrt{2}\rightarrow -\sqrt{2}$??
Q[sqrt(2)]=Q[-sqrt(2)]. Need to do better than that.

Q[sqrt(2)]=Q[-sqrt(2)]. Need to do better than that.
Yes, but it will be a nontrivial automorphism $\mathbb{Q}[\sqrt{2}]\rightarrow \mathbb{Q}[\sqrt{2}]$. Or am I just too tired??

Dick
Homework Helper
Yes, but it will be a nontrivial automorphism $\mathbb{Q}[\sqrt{2}]\rightarrow \mathbb{Q}[\sqrt{2}]$. 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?

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!
Well you got the following theorem:

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

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.

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

micromass said:
I think this provides the OP with enough hints. So he should try to figure out the details now.
Indeed! Thanks!