---
tag: YBJ2VQHFDQ
kind: statement
role: lemma
name: Zariski's lemma
msc: ["13E15", 12F05]
uses: [04C5ZBG1WT, AC5TFFQT9M, 0C1Y04SZ1W]
about: [AC5TFFQT9M, 0C1Y04SZ1W]
sources:
  - url: https://en.wikipedia.org/wiki/Zariski's_lemma
same_as:
  wikidata: Q8066794
  wikipedia: Zariski's lemma
---

## Statement

Let $k$ be a [[field|04C5ZBG1WT]] and let $K$ be a field extension of $k$ that is [[finitely generated|AC5TFFQT9M]] as a $k$-algebra.
Then $K$ is a [[finite extension|0C1Y04SZ1W]] of $k$.

## Remarks

By [[Noether normalization|7Z28JY8QW4]], $K$ is finite over a polynomial ring $k[y_1, \ldots, y_d]$.
By [[integral extensions of fields|5GCANV7337]], $k[y_1, \ldots, y_d]$ 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|4MBR0W0S6N]].

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 $k$.

## Explanations

### L1 human

A field that can be built from a smaller field with finitely many elements, using only addition and multiplication, is a finite extension of it.
Inverses come for free in such a field, so no transcendental element can occur.

### L2 human

Zariski's lemma is the algebraic core of the [[weak Nullstellensatz|ZWS0YWGF17]].
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.