field not finitely generated over a non-field subring

Let \(K\) be a field, \(A \hookrightarrow K\) a subring, which is not a field.
Then \(K\) is not a finitely generated module over \(A\)

w.l.o.g. we may assume that \(A\) is a local ring:
Choose a prime ideal \(\mathfrak{p} \subseteq A\) with \(\mathfrak{p} \neq 0\).

If \(K\) was finitely generated as \(A\)-module, then it is also finitely generated as \(A_{\mathfrak{p}}\) module.
So it suffices to show that \(K\) is not finitely generated as \(A_{\mathfrak{p}}\) module.

Note that

\begin{align*} \mathfrak{p} K = K \end{align*}

as we may pick an element \(p \in \mathfrak{p} \setminus \{0\}\) and it acts by an isomorphism on \(K\), since \(K\) is a field.

If \(K\) were finitely generated, then Nakayama's Lemma for local rings would show that \(K = 0\).
Of course there does not exist a field with one element.