Zariski's lemma

Remarks

By Noether normalization, $K$ is finite over a polynomial ring $k[y_1, \ldots, y_d]$. By integral extensions of fields, $k[y_1, \ldots, y_d]$ is then a field, which forces $d = 0$, since $y_1$ has no inverse in a polynomial ring when $d \ge 1$. So $K$ is finite over $k$. Another proof uses the Artin–Tate lemma.

Example: $\Q(\sqrt{2}) = \Q[\sqrt{2}]$ is a field generated by one element as a $\Q$-algebra, and it is finite over $\Q$ of degree $2$.

The lemma is about generation as an algebra. The field $k(t)$ of rational functions is generated by $t$ as a field, but it is not a finitely generated $k$-algebra, and it is not finite over $k$.

Going deeper

For a student

Zariski's lemma is the algebraic core of the weak Nullstellensatz. If $\mathfrak{m}$ is a maximal ideal of $k[x_1, \ldots, x_n]$, then $k[x_1, \ldots, x_n]/\mathfrak{m}$ is a field finitely generated as a $k$-algebra, so it is finite over $k$. When $k$ is algebraically closed, this residue field is $k$ itself, and $\mathfrak{m}$ is the ideal of a point.