Automorphisms of Algebraic Numbers

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  • #1
jgens
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Homework Statement



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

Homework Equations



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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Answers and Replies

  • #2
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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]??
 
  • #3
Dick
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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.
 
  • #4
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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 :blushing:

Maybe you can do something like [itex]\sqrt{2}\rightarrow -\sqrt{2}[/itex]??
 
  • #5
Dick
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Ah yes :blushing:

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.
 
  • #6
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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??
 
  • #7
Dick
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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!
 
  • #8
Haha. I am totally in your class. Math 257 with PSally?
 
  • #9
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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 [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. :frown:

I think this provides the OP with enough hints. So he should try to figure out the details now.
 
  • #10
jgens
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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!
 

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