Prove that any subfield of must contain .

By the previous exercise, contains a unique inclusion-smallest subfield which is isomorphic either to for a prime or to .

Suppose the unique smallest subfield of is isomorphic to , and let in this subfield be nonzero. Then in , and since is a unit, , a contradiction.

Thus the unique smallest subfield of is isomorphic to . In particular, any subfield of contains a subfield isomorphic to .

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## Comments

You could also go about the proof thusly:

Let F be any subfield of the real numbers. Since F is a Subfield of the Reals it contains at least the identity element, 1. Since F is an abelian group under addition it follows that it contains all the integers. Since F-{0} is an abelian group under multiplication it contains all the inverses of the integers. Because F is an abelian group under addition, it follows that F contains the Rationals.