TITLE: Are $\mathbb{R}$ and $\mathbb{Q}$ the only nontrivial subfields of $\mathbb{R}$?
QUESTION [17 upvotes]: I've been asked to prove that any subfield of $\mathbb{R}$ contains $\mathbb{Q}$, and I know how to do it, but it made me wonder if there were subfields of $\mathbb{R}$ that strictly contained $\mathbb{Q}$

REPLY [14 votes]: Not only are there nontrivial finite dimensional extensions of $\mathbb Q$ inside $\mathbb R$, there are $2^{\mathfrak c}$ infinite dimensional extensions indexed by the subsets of a transcendence basis of $\mathbb R$ over $\mathbb Q$. 
Transcendence bases play a significant role in several areas of algebra. A reference for them is in David Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry, Springer Verlag Graduate Texts in Mathematics vol 150, 1994, page 562. A finite set $\mathfrak f$ of elements of a field, say $\mathbb R$, over a subfield, say $\mathbb Q$, is called algebraically independent iff $\mathbb Q\left[\mathfrak f\right]$ is a free commutative algebra, that is a ring of polynomials, in $\mathfrak f$ with coefficients in $\mathbb Q$ so the field generated by $\mathbb Q$ is rational functions in $\mathfrak f$. As seen in the Eisenbud text, this definition can be extended to infinite sets and a maximal algebraically independent set is a transcendence basis. Distinct subsets of this transcendence basis will generate distinct subfields.