# Are $\mathbb{R}$ and $\mathbb{Q}$ the only nontrivial subfields of $\mathbb{R}$?

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 $\mathbb{Q}$

• Consider $\mathbb{F} = \{a + b \sqrt{2} | a, b \in \mathbb{Q} \}$. Jan 24, 2013 at 0:25

There are infinitely many such fields. A particular example is $\{x+y\sqrt 2\mid x,y\in \mathbb Q \}$. It is easy to verify that this is a proper subfield of $\mathbb R$ that properly contains $\mathbb Q$. More generally, let $\alpha \in \mathbb R$ be any irrational number. Then, since the intersection of any family of subfields of a given field is again a field, there exists the smallest subfield of $\mathbb R$ containing $\alpha$. This field is usually denoted by $\mathbb Q(\alpha)$. It certainly contains $\mathbb Q$ and it is not difficult to show (take it as a nice exercise) that it must be properly contained in $\mathbb R$.

The above already gives you a huge (infinite) repository of intermediate fields. But there are more. Recall that an algebraic real number is a real number that is the root of a polynomial with integer coefficients. It requires a bit of general field theory to prove that the collection of all real algebraic numbers forms a field. The existence of transcendental numbers show that this field is properly contained in $\mathbb R$. And there are even more intermediate fields.

Just to place things in the right context, recall that any field $F$ containing $\mathbb Q$ can be seen as a vector space of $\mathbb Q$. Every vector space has a dimension. The dimension of $\mathbb R$ over $\mathbb Q$ is infinite. So in that sense the field extension $\mathbb R:\mathbb Q$ is very large. For every natural number $n$ there is an intermediate field $\mathbb Q\subseteq F\subseteq \mathbb R$ whose dimension over $\mathbb Q$ is $n$.

• $\{x+y\sqrt 2\mid x,y\in \mathbb R \}$? or $\{... \in \mathbb Q \}$? Jan 24, 2013 at 5:33
• thank you so much @alancalvitti !! (corrected) Jan 24, 2013 at 5:39
• @alancalvitti, $\in \mathbb{Q}$, otherwise you just get back $\mathbb{R}$. Jan 25, 2013 at 16:23
• @vonbrand, I know. Jan 25, 2013 at 18:07

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.

• The strongest result I know of was given by Richard Schmutt in Solution to Monthly Problem #6548, American Mathematical Monthly 96 #6 (June-July 1989), 533-535. Schmutt proved the following two results: (1) There exist continuum many pairwise non-isomorphic countable subfields of $\mathbb R$, all of whose corresponding additive groups are isomorphic to each other and all of whose corresponding multiplicative groups are isomorphic to each other. (continues) Jan 24, 2013 at 16:23
• (continuation) (2) There exist $2^c$ many pairwise non-isomorphic subfields of $\mathbb R$ with cardinality continuum, all of whose corresponding additive groups are isomorphic to $\left({\mathbb R},+\right)$ and all of whose corresponding multiplicative groups are isomorphic to $\left({\mathbb R},*\right).$ Jan 24, 2013 at 16:24

There are definitely subfields of $\mathbb{R}$ that properly contain $\mathbb{Q}.$ The common example is $\mathbb{Q}[\sqrt{2}]$. This is one of the smallest field extension of $\mathbb{Q}$, where "smallest" is defined in terms of the degree of the field extension which is simply the dimension of the extension viewed as a vector space (trivial verification that this is possible). $\mathbb{Q}[\sqrt{2}]$ is definitely contained in $\mathbb{R}$. You can read more about field extensions here.