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Let $\mathbb R$ be the field of real numbers. Its algebraic closure, the field $\mathbb C$, is a finite extension of $\mathbb R$, which has degree 2.

Are there other examples of fields (not algebraic closed) such that its algebraic closure is a finite extension?

user26857
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zacarias
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1 Answers1

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The Artin-Schreier theorem asserts that these are precisely the real closed fields, which roughly speaking are the fields which behave like $\mathbb{R}$, and that their algebraic closures have degree $2$ and are given by adjoining a square root of $-1$. The Wikipedia article gives several examples; the simplest one is probably the real algebraic numbers $\mathbb{R} \cap \overline{\mathbb{Q}}$.

Qiaochu Yuan
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    What is the artin-schreier? I think it is about characteristic p. What is the statement? – nicksohn Jan 02 '16 at 08:50
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    @nicksohn: "Artin-Schreier" is attached to two different things which are unrelated. The Artin-Schreier theorem asserts that $k$ is a field whose algebraic closure is a nontrivial finite extension of $k$ iff $k$ is a real closed field; in this case the algebraic closure is $k[i]$ where $i^2 = -1$. – Qiaochu Yuan Jan 02 '16 at 09:19
  • May I ask you the source? I want to study this theorem, statement, PROOF, and more examples... I can't find related articles.. Thank you! – nicksohn Jan 02 '16 at 09:37
  • @nicksohn: this is Theorem 171 in Pete Clark's notes on field theory: http://math.uga.edu/~pete/FieldTheory.pdf – Qiaochu Yuan Jan 02 '16 at 09:40
  • Thanks! Just to be sure, In our case, Because (iv) implies (iii), Right? – nicksohn Jan 02 '16 at 09:53
  • Wow, This is hard for me. But, yes I'll study. Just for now, Can we prove that this implication only? OR (Grand) Artin-Schreier is not overkill? – nicksohn Jan 02 '16 at 10:01