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 $\mathbb{Q}$
[Math] Are $\mathbb{R}$ and $\mathbb{Q}$ the only nontrivial subfields of $\mathbb{R}$
abstract-algebrafield-theory
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Best Answer
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 $n$.