Proof of Zariski's Lemma

Zariski's lemma is a key ingredient in the proof of the Hilbert Nullstellensatz, showing that finitely generated field extensions of algebraically closed fields are trivial.

Statement

Theorem5.7Zariski's lemma

If $K$ is a field that is a finitely generated algebra over a field $k$ (i.e., $K = k[a_1, \ldots, a_n]$ for some $a_i \in K$ ), then $K$ is a finite (algebraic) extension of $k$ . If $k$ is algebraically closed, then $K = k$ .

Proof

We prove that each $a_i$ is algebraic over $k$ . Suppose not; after reordering, let $a_1, \ldots, a_m$ be algebraically independent over $k$ and $a_{m+1}, \ldots, a_n$ algebraic over $k(a_1, \ldots, a_m)$ , with $m \geq 1$ .

Then $K$ is a finite algebraic extension of $k(a_1, \ldots, a_m)$ , the field of rational functions. So $K = k(a_1, \ldots, a_m)[a_{m+1}, \ldots, a_n]$ and each $a_j$ ( $j > m$ ) satisfies a polynomial equation over $k(a_1, \ldots, a_m)$ .

Since $K = k[a_1, \ldots, a_n]$ , every element of $K$ is a polynomial in $a_1, \ldots, a_n$ over $k$ . In particular, the coefficients of the minimal polynomials of $a_{m+1}, \ldots, a_n$ over $k(a_1,\ldots,a_m)$ are rational functions in $a_1, \ldots, a_m$ .

Let $d$ be the product of all denominators appearing in these minimal polynomial coefficients. Then $d \in k[a_1, \ldots, a_m] \setminus \{0\}$ and $K = k[a_1, \ldots, a_n, 1/d_1, \ldots]$ for finitely many elements.

Now we reach a contradiction: $K$ is a finitely generated $k$ -algebra containing $k(a_1, \ldots, a_m)$ , but $k(a_1, \ldots, a_m)$ is not a finitely generated $k$ -algebra (since $k[a_1, \ldots, a_m]$ has infinitely many non-associate irreducible elements, and we cannot invert them all with finitely many generators).

More precisely: if $1/d \in k[a_1, \ldots, a_n]$ for some nonzero $d \in k[a_1, \ldots, a_m]$ , this means $d$ divides $1$ in some polynomial extension, which forces $d \in k^\times$ . But rational function coefficients may have non-constant denominators.

Formal argument (Noether normalization style): Since $K$ is a finitely generated $k$ -algebra that is a field, the Noether normalization lemma gives $k[y_1, \ldots, y_m] \hookrightarrow K$ with $K$ integral (finite) over $k[y_1, \ldots, y_m]$ and $y_i$ algebraically independent. If $m \geq 1$ , then $k[y_1, \ldots, y_m]$ is not a field, but $K$ is a field, so the map $k[y_1, \ldots, y_m] \to K$ is an integral extension of a non-field into a field. By the going-up theorem, this means $k[y_1, \ldots, y_m]$ is also a field -- contradiction. So $m = 0$ , and $K$ is integral (algebraic) over $k$ . $\blacksquare$

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Applications

ExampleDeriving the weak Nullstellensatz

If $\mathfrak{m}$ is a maximal ideal of $k[x_1,\ldots,x_n]$ with $k$ algebraically closed, then $K = k[x_1,\ldots,x_n]/\mathfrak{m}$ is a field that is a finitely generated $k$ -algebra (generated by $\bar{x}_i$ ). By Zariski's lemma, $K = k$ . So each $\bar{x}_i = a_i \in k$ , giving $\mathfrak{m} = (x_1 - a_1, \ldots, x_n - a_n)$ .

RemarkGeneralizations

Zariski's lemma generalizes to: if $A \subseteq B$ are integral domains with $B$ a finitely generated $A$ -algebra and $B$ a field, then $A$ is also a field and $B$ is a finite extension of $A$ . This is a special case of the going-up and going-down theorems in commutative algebra.