Dimension of $\Bbb Q(e)$ over $\Bbb Q$?

Question

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 $\Bbb Q$?

Answer 1

As pointed out in Adam's comment, the dimension will be infinite. The reason is as follows: suppose that the dimension was finite, say $n$. Then any $n+1$ elements of the field must be linearly dependent over $\Bbb Q$; in particular, $$1\,,\ e\,,\ e^2\,,\ e^3\,,\ldots,\ e^n$$ are linearly dependent; so we have $$a_0+a_1e+a_2e^2+a_3e^3+\cdots+a_ne^n=0$$ for some rational numbers $a_0,a_1,a_2,a_3,\ldots,a_n$; and this is by definition impossible for a transcendental number.

Answer 2

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 $k$.