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I'm trying to determine if there is a field, F, such that $\mathbb{R}$ $\subsetneq$ F $\subsetneq$ $\mathbb{C}$ where F is not the same as $\mathbb{R}$ or $\mathbb{C}$.

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2 Answers 2

There is not. The extension $[\mathbb C : \mathbb R]$ has degree 2, so there cannot be a proper intermediate extension.

Assume such an extension $F$ exists. Then by the tower theorem, we have $[\mathbb C:F][F:\mathbb R]=2$. For it to be a proper extension, each factor must be greater than one. This is a contradiction.

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That's what my intuition was telling me. Thanks. –  user61657 Feb 10 '13 at 2:15
    
@user61657 I edited in a proof, and I am pinging you to make sure you see it. –  Potato Feb 10 '13 at 2:16
    
It's worth noting that while this isn't true under the canonical embedding. We do have weird things like $\mathbb R \subset \mathbb R(t) \subset \mathbb C(t) \subset \mathbb C$. –  JSchlather Feb 10 '13 at 2:28

Suppose $F$ is a field such that $\mathbb{R} \subsetneq F \subseteq \mathbb{C}.$ Then $F$ contains at least one element $a+bi$ where $b\neq 0.$ Since $F$ contains all the reals, it contains $b^{-1}$ and $-a,$ so it contains $b^{-1}\left((a+bi)-a\right)=i.$ Hence it contains all complex numbers and $F=\mathbb{C}.$

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