Finitely generated as a field vs. as an algebra? Let $K$ be a field, $L$ a $K$-algebra. For $x_1, \ldots, x_n \in L$, denote $K[x_1, \ldots, x_n]$ for the smallest subalgebra of $L$ containing $K$ and $x_1, \ldots, x_n$. If $L$ is a field, denote $K(x_1, \ldots, x_n)$ for the smallest subfield of $L$ containing $K$ and $x_1, \ldots, x_n$. Clearly $K[x_1, \ldots, x_n] \subseteq K(x_1, \ldots, x_n)$.
If $L$ is a field and $L = K(x_1, \ldots, x_n)$ for some $x_1, \ldots, x_n \in L$, can we conclude that $L = K[y_1, \ldots, y_m]$ for some $y_1, \ldots, y_m \in L$? Both hints and full answers are appreciated.

No. For example, $\;L=\Bbb Q(\pi)\;$ is a field, but it can't  be that $\;L=\Bbb Q[\alpha]\;$ , as:
(1) If $\;\alpha\;$ is algebraic over $\;\Bbb Q\;$ then $\;\Bbb Q[\alpha]=\Bbb Q(\alpha)\neq\Bbb Q(\pi)\;$ , as the last one is infinite dimensional over $\;\Bbb Q\;$ whereas the first one is finite dimensional, and
(2) if $\;\alpha\;$ is transcendental over $\;\Bbb Q\;$ then $\;\Bbb Q[\alpha]\;$ is not even a field and in fact $\;\Bbb Q[\alpha]\cong\Bbb Q[x]=$ the "usual" polynomial ring of rational polynomials
