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Does there exist a proper subfield of $\mathbb{R}$ whose algebraic closure is $\mathbb{C}$ ?

A weaker question: Does there exist a proper subfield $F$ of $\mathbb{R}$ such that $\mathbb{R}$ is algebraic over $F$?

Watson
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Shubhodip Mondal
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2 Answers2

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The two conditions you cited are equivalent, since ${\bf C}$ is the algebraic closure of ${\bf R}$

Yes, there is such a field. Just take a transcendental basis $B$ of $\bf R$ over ${\bf Q}$, and then ${\bf Q}(b)_{b\in B}\subseteq{\bf R}$ is algebraic and $\bf C$ is the algebraic closure of ${\bf Q}(b)_{b\in B}$.

I'm not sure if this is true without axiom of choice, so it might be hard to give a “concrete” example.

tomasz
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    I'm not too comfortable with it at the moment, but how are you sure that $Q(b)_{b \in B}$ is proper subfield of $\mathbb{R}$? – Shubhodip Mondal Feb 26 '14 at 17:35
  • @ShubhodipMondal: well, ${\bf Q}$ is a subfield of $\bf R$ and $B\subseteq {\bf R}$, so it's clearly a subfield. What's confusing you? – tomasz Feb 26 '14 at 17:37
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    @Shubhodip: $2$ is not a square in $\mathbf{Q}(B)$, for example. –  Feb 26 '14 at 17:38
  • @ShubhodipMondal: Ah, that's simple. No polynomial in ${\bf Q}[X]$ has a root in that field if it doesn't have one in ${\bf Q}$, as the extension is purely transcendental. Putting it differently, $\bf R$ is not a purely transcendental extension of rationals (in fact, it's not a purely transcendental extension of any field at all) – tomasz Feb 26 '14 at 17:39
  • I'm sorry. But what is wrong in the following ? Suppose $\sqrt{2} \not\in Q(B)$. Then $Q(B) (\sqrt{2})$ has a nontrivial automorphism. ($\sqrt{2} \to -\sqrt{2}$). But as $Q(B) (\sqrt{2})$ is dense in $R$, this is a contradiction ? – Shubhodip Mondal Feb 26 '14 at 17:55
  • There's no contradiction. Why would there be one? – tomasz Feb 26 '14 at 18:08
  • As $Q(B) (\sqrt{2})$ is dense in $R$, so all automorphism has to be trivial. But we just constructed a nontrivial automorphism. So $\sqrt{2} \in Q(B) $? – Shubhodip Mondal Feb 26 '14 at 18:11
  • No. That would be true for automorphisms extending to automorphisms of reals. Density is completely unrelated, as we're disregarding topology anyway. If your argument worked, it'd show that ${\bf Q}[\sqrt 2]$ (or any real extension of rationals, really), has no automorphisms, and that is clearly false. – tomasz Feb 26 '14 at 18:14
  • Yes. That is a stupid mistake I just made :( Sorry. – Shubhodip Mondal Feb 26 '14 at 18:16
  • and thanks for the help. – Shubhodip Mondal Feb 26 '14 at 18:27
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Yes, there is.

Pick you favorite algebraic number $a \notin \mathbb Q$. For example $a= \sqrt{2}$.

Consider now $$\mathcal{F} := \{ K | K \mbox{ is a subfield of } \mathbb R , \mbox{ and } a \notin K \}$$

Now, $(\mathcal{F}, \subset)$ is a partially ordered set, which is non-empty (as $\mathbb Q \in \mathcal F$). Moreover, if $(K_{j})_j$ is a chain in $\mathcal F$, it is straightforward that $\bigcup K_j \in \mathcal{F}$.

Therefore, by Zorn Lemma, $\mathcal{F}$ has a maximal element $\mathbb{K}$.

The maximality of $\mathbb K$ implies that $\mathbb R$ is algebraic over $\mathbb K$. Indeed, if $t \in \mathbb R$ is transcendental over $\mathbb K$ then we cannot have $a \in \mathbb K(t)$ which contradicts the maximality of $\mathbb K$.

Indeed, if $a \in \mathbb K(t)$ then writing $a=\frac{P(t)}{Q(t)}$ we get that $P(t)-aQ(t)=0$ which means that $t$ is the root of $P-at \in \mathbb K(a)$. Then $t$ is algebraic over $\mathbb K(a)$ and $\mathbb K(a)$ is finite over $K$ which implies that $t$ is algebraic over $\mathbb K$.

Finally, as $\mathbb R$ is algebraic over $\mathbb K$, $\mathbb C$ is also algebraic over $\mathbb K$.

N. S.
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