Let $k$ be a field and $L$ a finitely generated $k$-algebra, which is also a field. Then $L$ is finite over $K$
Finitely generated means $L = k[x_1, \dots, x_n]$ ($x_i \in L$), and finite over $k$ means $L$ is a finitely generated $k$-module.
I am struggling to understand how to prove this result. It seems to me that $\mathbb{Q}[\pi]$ is a counterexample.
The notes I'm reading prove this by induction on $n$, and I can't work out how to do the base case.
A hint would be the most helpful answer.
This is a key step in many proofs of Hilbert's Nullstellensatz.
$\Bbb Q[\pi]$ is not a field. It consists of the polynomials $a_0+a_1\pi+\cdots +a_n\pi^n$ with the $a_i$ rational. Were it a field, $1/\pi\in\Bbb Q[\pi]$ so $1/\pi=a_0+a_1\pi+\cdots +a_n\pi^n$ for rational $a_i$ contradicting the transcendence of $\pi$.
The base case of the induction $n=1$ essentially reduces to this example. If $k\subseteq L$ are fields and $y\in L$, then either $k[y]$ is finite over $k$, or $k[y]$ is a polynomial ring, and so $k[y]$ isn't a field.