Every subfield of the real numbers must contain the rational numbers

Prove that any subfield of \mathbb{R} must contain \mathbb{Q}.


By the previous exercise, \mathbb{R} contains a unique inclusion-smallest subfield which is isomorphic either to \mathbb{Z}/(p) for a prime p or to \mathbb{Q}.

Suppose the unique smallest subfield of \mathbb{R} is isomorphic to \mathbb{Z}/(p), and let a in this subfield be nonzero. Then pa = 0 in \mathbb{R}, and since p \in \mathbb{R} is a unit, a = 0, a contradiction.

Thus the unique smallest subfield of \mathbb{R} is isomorphic to \mathbb{Q}. In particular, any subfield of \mathbb{R} contains a subfield isomorphic to \mathbb{Q}.

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Comments

  • dpb492  On March 19, 2011 at 8:36 pm

    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.

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