Proof of the Weak Hilbert Nullstellensatz

The weak Hilbert Nullstellensatz identifies the maximal ideals of a polynomial ring over an algebraically closed field. It says that these ideals are exactly the ideals of polynomials that vanish at individual points. The proof reduces the claim to Zariski’s lemma about fields that are finitely generated as algebras.

Throughout, let $k$ be an algebraically closed field and let $R=k[x_1,\ldots,x_n]$. Rings are assumed commutative and have an identity.

The statement

For every maximal ideal $\mathfrak m$ of $R$, there are scalars $a_1,\ldots,a_n\in k$ such that

$$\mathfrak m=(x_1-a_1,\ldots,x_n-a_n).$$

Conversely, every ideal of this form is maximal: evaluating a polynomial at $(a_1,\ldots,a_n)$ gives a surjective ring homomorphism $R\to k$ whose kernel is that ideal, so its quotient is the field $k$.

A related formulation is that every proper ideal $I\subsetneq R$ has a common zero in $k^n$. Indeed, a proper ideal is contained in a maximal ideal, and the maximal-ideal statement supplies a point where every element of that maximal ideal—and therefore every element of $I$—vanishes.

The key lemma: Zariski’s lemma

Lemma. If a field $L$ is finitely generated as a $k$-algebra, then $L$ is a finite algebraic extension of $k$.

Here “finitely generated as a $k$-algebra” means that $L=k[b_1,\ldots,b_m]$ for finitely many elements $b_i$. This is stronger than saying that $L$ is finitely generated as a field extension; the distinction matters.

To prove the lemma, choose a largest algebraically independent subset of the generators, say $b_1,\ldots,b_r$. The remaining generators are algebraic over $K=k(b_1,\ldots,b_r)$. Since there are only finitely many of them, they generate a finite extension of $K$ as a field.

Suppose, for a contradiction, that $r>0$. Each remaining generator satisfies a monic polynomial over $K$. The coefficients of these finitely many equations are rational functions in $b_1,\ldots,b_r$. Choose a nonzero polynomial $f\in k[b_1,\ldots,b_r]$ that clears all their denominators, and set

$$A=k[b_1,\ldots,b_r,1/f].$$

Each remaining generator is then integral over $A$. Also, because $L$ is a field and $f\ne 0$, the element $1/f$ belongs to $L$. Thus $L=A[b_{r+1},\ldots,b_m]$ is integral over $A$.

If a field is integral over a subring, that subring must itself be a field. In fact, for any nonzero $a\in A$, its inverse in $L$ is integral over $A$. A monic equation for $a^{-1}$, multiplied by a suitable power of $a$, expresses $a^{-1}$ as an element of $A$.

But $A$ is not a field when $r>0$. To see this, regard $f$ as a polynomial in $b_1$ over the field $k(b_2,\ldots,b_r)$ (interpreted as $k$ if $r=1$). Since $k$ is infinite, choose $c\in k$ that is not a root of this nonzero polynomial. Then $q(b_1)=b_1-c$ is irreducible in $k[b_1]$ and does not divide $f$. It remains a nonunit after localizing at $f$: no power of $f$ is divisible by $q(b_1)$. This contradicts that $A$ is a field. Therefore $r=0$, so all the generators are algebraic over $k$. Finitely many algebraic generators give a finite algebraic extension, proving the lemma.

Proof of the weak Nullstellensatz

Let $\mathfrak m$ be a maximal ideal of $R$, and form the quotient

$$L=R/\mathfrak m.$$

Because $\mathfrak m$ is maximal, $L$ is a field. It is generated as a $k$-algebra by the residue classes $\bar x_1,\ldots,\bar x_n$ of the variables. Zariski’s lemma therefore says that $L$ is a finite algebraic extension of $k$.

Since $k$ is algebraically closed, it has no proper algebraic field extensions. Hence $L=k$. In particular, each coordinate class $\bar x_i$ is equal to some scalar $a_i\in k$. This means $x_i-a_i\in\mathfrak m$ for each $i$, so

$$(x_1-a_1,\ldots,x_n-a_n)\subseteq\mathfrak m.$$

The ideal on the left is maximal, since its quotient is $k$. A maximal ideal cannot be properly contained in another proper ideal, so equality follows. This proves the theorem.

A concrete example

Take $R=\mathbb C[x,y]$ and $\mathfrak m=(x-2,y+1)$. The quotient sends $x$ to $2$ and $y$ to $-1$, giving

$$\mathbb C[x,y]/\mathfrak m\cong\mathbb C.$$

Thus $\mathfrak m$ is maximal and corresponds to the point $(2,-1)$. Every polynomial in $\mathfrak m$ vanishes at that point. For example, $x^2+xy-2y$ has value $4-2+2=4$ there, so it does not belong to this ideal. This check illustrates the criterion: membership in this maximal ideal requires evaluation to be zero. By contrast, $x+y-1$ does belong, because its value there is $2-1-1=0$.

Common mistakes

• Using algebraic closedness too early. The quotient by a maximal ideal is first shown to be a finitely generated field extension. Zariski’s lemma makes it algebraic; only then does algebraic closedness force it to equal $k$.
• Confusing algebra generation with field generation. The lemma applies to a field that is finitely generated as a ring over $k$, not merely as a field extension.
• Claiming every proper ideal is maximal. The theorem classifies maximal ideals. A general proper ideal need not be maximal; it is contained in at least one maximal ideal.
• Dropping the field hypothesis. For a field $k$ that is not algebraically closed, maximal ideals can have quotients that are proper finite algebraic extensions of $k$. The algebraically closed assumption is what ensures the quotient is exactly $k$.