# $\mathbb{R}\subset\mathbb{F}\subset\mathbb{C}$ implies $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$?

Let $\mathbb{F}$ a field such that $\mathbb{R}\subset\mathbb{F}\subset\mathbb{C}$. Prove that $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$.

What I know is, that if $\mathbb{F}\neq\mathbb{R}$ it means that exists $a\in\mathbb{F}$ such that $a\in\mathbb{C}$ and $a\notin\mathbb{R}$. But I don't know how to use it. since I don't know the actions of $\mathbb{F}$ can't it just be defined with the special case of $a$ in mind and still hold all the axioms?

Also, this is from homework in Linear Aglebra (1), where the rest of the questions are about bases of vector spaces and dimensions and such, but there are no vector spaces here so I'm not really sure how it fits.

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$\mathbb{F}$ and $\mathbb{C}$ are vector spaces over $\mathbb{R}$. – Ted Dec 5 '12 at 9:38
Use dimensions. – Ittay Weiss Dec 5 '12 at 9:39

The following steps lead to a solution:

1. $\Bbb{F}$ as a field can be considered as a vector space over $\Bbb{R}$.

2. The fact that $\Bbb{F}$ is contained in $\Bbb{C}$ implies that its real dimension is either $1$ or $2$.

3. Use the fact that if a subspace $U$ of a vector space $V$ is such that $\dim U = \dim V$ then in fact $U = V$.

4. Conclude.

Alternatively $\Bbb{C}/\Bbb{R}$ is a Galois extension and so by the fundamental theorem of Galois theory we have any intermediate subfield of $\Bbb{C}/\Bbb{R}$ being either $\Bbb{C}$ or $\Bbb{R}$ since there are only subgroups of $\Bbb{Z}/2\Bbb{Z}$.

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Oh yeah... You reminded me that we did prove that fields are vector spaces above subfields of themselves. This really does make it a lot simpler. Thank you! – Nescio Dec 5 '12 at 9:46

Recall that $[\mathbb{C}:\mathbb{R}]=2$ and thus $[\mathbb{F}:\mathbb{R}]\leq2$. What happens if $[\mathbb{F}:\mathbb{R}]=1$ ? what can you say when $[\mathbb{F}:\mathbb{R}]=2$ ?

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You can also proceed the way you intended to. Suppose $\mathbb F\neq\mathbb R$. Then there must be a $a+bi\in\mathbb F$ such that $b\neq 0$. Using $\mathbb R\subseteq\mathbb F$ you can now easily show that $i\in\mathbb F$ and from this it easily follows that every complex number is contained in $\mathbb F$.

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hmm.. Interesting. But I'm not sure how to conclude $i\in\mathbb{F}$ from that. mind clarifying? – Nescio Dec 5 '12 at 10:20
@Nescio - its closed under addition so $a+bi-a\in F$ and closed under devision so since $b\neq 0$ you can divide by it (+1) – Belgi Dec 5 '12 at 10:23