TITLE: Dimension of $\Bbb Q(e)$ over $\Bbb Q$?
QUESTION [8 upvotes]: The dimension of $\Bbb Q(\sqrt{2})$ over $\Bbb Q$ is finite since $\sqrt2$ is algebraic over $\Bbb Q$.
But what about any transcendental number (say $e$)?
Which is the smallest field containing $\Bbb Q$ and $e$?
What is the dimension of that field over $\Bbb Q$?

REPLY [11 votes]: The dimension is countably infinite. Here is a nice exercise: let $k$ be any field. Then $k(x)$ has a basis consisting of the functions $1, x, x^2, ...$ together with the functions $\frac{1}{f(x)^k}$, where $f(x) \in k[x]$ is monic and irreducible. This is essentially partial fraction decomposition. (Note that the condition that $\mathbb{Q}(e) \cong \mathbb{Q}(x)$, with the isomorphism sending $e$ to $x$, is one equivalent way to state the condition that $e$ is transcendental.) For $k$ an infinite field, this implies that $\dim_k k(x)$ is the cardinality of $k$.