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
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 2ℵ and the cardinality of an infinite field is the same as the cardinality of its algebraic closure, Q(S) must have cardinality 2ℵ. So S has cardinality 2ℵ.
So, the cardinality of isomorphisms of Q(S) to Q(T) is 22ℵ, since there are 22ℵ bijections between S and T.
Any isomorphism of Q(S) to Q(T) can be extended to an isomorphism of
Suppose f, g are distinct isomorphisms
So, there are 22ℵ subfields of C isomorphic to R. And also 22ℵsubfields of C of index 2 in C, isomorphic to R! The cardinality of the set of such subfields can't be any greater than 22ℵ, 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
Proof: α is also algebraic over
α is algebraic over
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
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
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