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Possible Duplicate:
Finiteness of the Algebraic Closure

For short, I wonder if there are other fields $F\subset \mathbb{C}$ rather than $\mathbb{R}$, with finite index $[\mathbb{C}:F]$.

Since $\mathbb{C}=\mathbb{R}+\mathbb{R}\sqrt{-1}$, we naively hope that if we are given a $p$th primitive unit root then we "should" find a subfield $F\subset \mathbb{C}$, such that $F(\omega)=\mathbb{C}$.

Applying Zorn's lemma, we do be able to find a maximal subfield $F$ with respect to $\omega\notin F$. And for any finite algebraic extension $E/F$ with $E\subset \mathbb{C}$, we can show that $E/F$ is Golois with cyclic Galois group. Because if $L/F$ is finite Galois then pick an $\sigma\notin Gal(L/F(\omega))$, we obtain $Gal(L/F)=(\sigma)$.

However, I do not know how to go forward to get a finite index subfield yet.

Just for curiousness.

Thanks.

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marked as duplicate by Qiaochu Yuan Oct 28 '12 at 17:01

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EDIT: The original answer given below is wrong, as was pointed out by Qiaochu Yuan. There are uncountably many such subfields (using the axiom of choice), and by the Artin-Schreier theorem they are real closed fields of index 2 in $\mathbb{C}$.


Original wrong answer

Apparently there are no other such subfields, by the Artin-Schreier theorem. See https://mathoverflow.net/questions/8756/examples-of-algebraic-closures-of-finite-index

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    $\begingroup$ @Lukas: this is not what the Artin-Schreier theorem tells you here, and in fact this statement is false. If $\sigma : \mathbb{C} \to \mathbb{C}$ is any automorphism whatsoever, then $\sigma(\mathbb{R})$ is another subfield of index $2$ in $\mathbb{C}$, and there are uncountably many such automorphisms (assuming the axiom of choice). If such an automorphism preserves $\mathbb{R}$ then it is either the identity or complex conjugation. $\endgroup$ – Qiaochu Yuan May 10 '15 at 4:08
  • $\begingroup$ @QiaochuYuan: Thanks for the correction, I updated the answer. $\endgroup$ – Lukas Geyer May 11 '15 at 16:58

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