Prove/Disprove: F1 and F2 Isomorphic as Fields

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Discussion Overview

The discussion revolves around the statement regarding the isomorphism of two finite field extensions F1 and F2 of a field K, specifically whether the condition of having the same degree over K implies that F1 and F2 are isomorphic as fields. The scope includes theoretical exploration and counterexamples related to field theory.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant seeks to prove or disprove that if [F1:K] = [F2:K], then F1 and F2 are isomorphic as fields.
  • Another participant asserts that the statement is obviously false and suggests looking at examples.
  • A participant expresses confusion about extending the proof from field extensions to fields, noting the difficulty with general field isomorphisms.
  • A counterexample is provided involving the field extensions \(\mathbb{Q}(i)\) and \(\mathbb{Q}(\sqrt{3}i)\), both of degree 2, to illustrate that they are not isomorphic as fields.
  • One participant acknowledges the counterexample and appreciates the insight into exploiting isomorphism properties.
  • A hint is given to consider a base field like \(\mathbb{Q}\) where every isomorphism is the identity, suggesting that similar reasoning applies to finite fields with the same number of elements.

Areas of Agreement / Disagreement

Participants generally disagree on the original statement, with some asserting it is false and providing counterexamples, while others are exploring the implications and conditions surrounding the statement.

Contextual Notes

The discussion highlights the dependence on definitions of isomorphism and the specific nature of field extensions versus fields, as well as the need for careful consideration of counterexamples.

T-O7
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Hey all,
I need to prove (or disprove) the following statement:

F1 and F2 are two finite field extensions of a field K. Assume [F1:K]=[F2:K]. Then F1 and F2 are isomorphic as fields.

Some help would be much appreciated.
I know the statement is false if i replace "isomorphic as fields" by "isomorphic as field extensions", but that's all i can think of so far.
 
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thats obviously false. look at any example at all.

just extend your proof from field extensions, to fields.
 
Sorry...i don't know what you mean.
I can come up with counter examples if it said "isomorphic as field extensions", because then i can just use the result that F(a) and F(b) are F-isomorphic iff a and b have the same irreducible poly.
But I'm stuck in this case, with general field isomorphisms...
 
Ok here's a counterexample. Take the two field extension of the rationals [tex]\mathbb{Q}(i)[/tex] and [tex]\mathbb{Q}(\sqrt{3}i)[/tex]. Both are of degree 2. But then there is an isomorphism [tex]f(0) = f(i^2 +1) = (f(i))^2 +f(1) = (f(i))^2 +1[/tex] But there are no elements in [tex]\mathbb{Q}(\sqrt{3}i)[/tex] that satisfies this.
 
Nice...that's wonderful. Thanks snoble! Exploit the isomorphism property, i get it. :smile:
 
my hint was meant to get you to consider a base field like Q where every isomorphism is the identity, then your same proof works as for extensions. see?

by the way try the same problem for finite fields. i.e. suppose two finite fields have the same number of elements, what then?
 
Last edited:

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