Is there a different copy of $\Bbb{R}$ in $\Bbb{C}$ such that the extension is algebraic? Since $\Bbb{C}$ contains a non-trivial copy of itself, I know that there are multiple subfields of $\Bbb{C}$ isomorphic to $\Bbb{R}$. But these inclusions make the extension non-algebraic. So are there other ways of including $\Bbb{R}$ in $\Bbb{C}$ (as a field) keeping the extension algebraic? Or is it that, given $\Bbb{C}$, there is an algebraic way to define the usual copy of $\Bbb{R}$?
I can see that this is equivalent to asking for an order 2 automorphism of $\Bbb{C}$ which is not the usual conjugation. But I could neither come up with one, nor prove it does not exist.
More generally, if $\overline F$ is finite over $F$, is the inclusion unique?
 A: The question about embeddings of the real numbers such that the extension of the complex numbers are algebraic over the image is different from the question of order 2 automorphisms.

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*All maps $\tau:\Bbb{R} \to \Bbb{C}$ turning $\Bbb{C}/\Bbb{R}$ into an algebraic extension agree up to the action of $Aut(\Bbb{C})$.

The first is a question about Galois theory. Let $\tau:\Bbb{R} \to \Bbb{C}$ be such a map. Then it extends to an automorphism of $\Bbb{C} = \tau(\Bbb{R})(i)$. Let us call this $\gamma$. Then the complex conjugation associated to $\tau$ is just the $\gamma$-conjugate of the usual one.
Now comes the confusing part.


*There are index two subfields of $\Bbb{C}$ not isomorphic to $\Bbb{R}$, but they are all real closed.

Theorem(Artin-Schreier): Let F be a field whose algebraic closure is a finite extension. Then F is either algebraically closed or real closed.
However, while real closed fields share many properties of the reals, they are not necessarily isomorphic even when they have the same cardinality. Note that the hyperreal numbers are real closed of continuum cardinality, but they are not isomorphic as abstract field to the real numbers, since such an isomorphism would respect the order(positive elements are precisely the squares). However, the algebraic closure of the hyperreals is a quadratic extension and by cardinality reasons isomorphic to $\Bbb{C}$. Taking the complex conjugation yields a nonconjugate involution of $\Bbb{C}$.


*Involutions need not extend to involutions on the algebraic closure.

By the way there is no guarantee that an involution extends to an involution on the algebraic closure. Just take $\Bbb{Q}(i)(t_1,t_2)$ and the involution exchanging the $t_i$. Then an extension to an involution of $\Bbb{C}$ would contradict the Artin-Schreier theorem.
