field extension that is finitely generated but not finite dimensional

Question

Give an example of a field extension that is finitely generated but not finite dimensional.

I'am really getting stack to find such an example.

I would appreciate any help or hints with that.

Thank you in advance

Answer 1

Let $F=\mathbb{Q}(\pi)$, $K=\mathbb{Q}$. Clearly $F$ is finitely generated since the set $\mathcal{B}=\{1,\pi\}$ is a basis for $F$ with coefficient in $K$. If $[F:K]=n,$ then that would mean there exists $p(x) \in K[x]$ of degree $n$ such that $p(x)=0$, which is impossible. Hence, $[F:K]>n$ for all positive integer $n,$ i.e, $F$ has no finite dimensional.

Answer 2

Well, we don't really need fancy stuff like transcendental numbers or whatever.

Definition: Let $K\subseteq F$ be a subfield. The field $K(x_1,x_2,...,x_n)$, where $x_i\in X$ and $|X|=n<\infty$, for some subset $X$ of $F$, is said to be finitely generated.

Let $F_1$ be a field, e.g. $\mathbb Q$.

Let $F_1[t]$ be the polynomial ring of one indeterminate $t$ over $F_1$.

Let $F_1(t)$ be the fraction field of $F_1[t]$, i.e. the set derived from adding all inverses of non-zero polynomial in $F_1[t]$ to $F_1[t]$ itself. (One can check that the fraction field is exactly the generated field here.)

In the definition above, taking $F:=F_1(t,s)$ and $K:=F_1$ and $X:=\{t\}$, we see $F_1(t)$ is finitely generated, where we add one more indeterminate $s$ just because we don't want $F_1(t)=F$.

$F_1(t)$ (as a $K$-module or $K$-vector space) has infinite dimension over $K$ because we cannot find a finite subset of $F_1(t)$, say $E$, such that span$_K(E)=F_1(t)$.