So, what are the subfields of C that are isomorphic to R?

You can make some of them this way: Choose a maximal algebraically independent subset of the reals, call it S. "Algebraically independent" means that no non-trivial polynomial over the algebraic numbers is 0.

Then, if we choose another algebraically independent subset T of the reals, of the same cardinality as S, then any bijection f: S → T induces an isomorphism Q(S) → Q(T).

The algebraic closure of Q(S), cl(Q(S)), must contain R, so it is C.

What's the size of S? Since C has cardinality 2^ℵ and the cardinality of an infinite field is the same as the cardinality of its algebraic closure, Q(S) must have cardinality 2^ℵ. So S has cardinality 2^ℵ.

So, the cardinality of isomorphisms of Q(S) to Q(T) is 2^2ℵ, since there are 2^2ℵ bijections between S and T.

Any isomorphism of Q(S) to Q(T) can be extended to an isomorphism of cl (Q(S)) = C → cl (Q(T)) by the isomorphism extension theorem.

Suppose f, g are distinct isomorphisms Q(S) → Q(T) that have been extended in this way to isomorphisms f, g: C → cl (Q(T)). Then the fields f(R) and g(R) must be distinct. If f(R) = g(R) then f ^-1 g would be a non-identity automorphism of R, which is not possible.

So, there are 2^2ℵ subfields of C isomorphic to R. And also 2^2ℵsubfields of C of index 2 in C, isomorphic to R! The cardinality of the set of such subfields can't be any greater than 2^2ℵ, since that is also the cardinality of subsets of C.

But might there be subfields of C isomorphic to R, that aren't obtainable this way? This method produces a field isomorphic to R with an uncountable intersection with R.

But you can also find fields isomorphic to R that intersect R only in the algebraic numbers. First, split S, the maximal algebraically independent subset of the reals, into two sets X and Y of the same cardinality. Let f be a bijection X → Y. Then if α is real and algebraic over Q({x + iy, with x ∈ X and y = f(x)}), α is algebraic over Q.

Proof: α is also algebraic over Q({x - iy, with x ∈ X and y = f(x)}).

The sets {x + iy, with x ∈ X and y = f(x)} and {x - iy, with x ∈ X and y = f(x)} are algebraically independent from each other.

α is algebraic over Q(x ∈ X, iy for y ∈ Y).

There's a lemma in Lang's Algebra which says: Let K be algebraically closed in extension L. Let α be some element of an extension of L, but algebraic over K. Then [K(x):K] = [L(x):L].

This implies that if you have a field K, and S₁ and S₂ are transcendental over K and algebraically independent of each other, then cl(K(S₁) ∩ cl(K(S₂) = cl(K). That's because if α satisfies a polynomial over K(S₁), then this polynomial is also the minimal polynomial for α over K(S₁, S₂). And α must also satisfy a polynomial of the same degree over K(S₂). This means that the minimal polynomial for α over K(S₁, S₂). can be taken to have coefficients in K.

Suppose that f(R) is of index 2 in C. Since R is of index 2 in C, is it possible to say anything about the index of f(R) ∩ R in C?

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