1998/2007

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 and the cardinality of an infinite field is the same as the cardinality of its algebraic closure, Q(S) must have cardinality 2. So S has cardinality 2.

So, the cardinality of isomorphisms of Q(S) to Q(T) is 22, since there are 22 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 22 subfields of C isomorphic to R. And also 22subfields of C of index 2 in C, isomorphic to R! The cardinality of the set of such subfields can't be any greater than 22, 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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