Today I was sitting in my Algebraic Geometry class and we were discussing Hilbert’s Nullstellensatz. I have seen proofs of it before using Noether normalization, but today we saw a slick proof of it I thought was particularly satisfying. I wanted to share it here. I thank Leonardo Constantin Mihalcea for the proof.

The one piece of input we take as given is Zariski’s lemma: if a field \(K\) is finitely generated as an algebra over a field \(k\), then \(K\) is a finite (hence algebraic) extension of \(k\). This is where the real depth of the Nullstellensatz lives; the argument below shows how quickly everything else follows from it.

Weak Nullstellensatz

Let \(k\) be an algebraically closed field. Then

1) All maximal ideals of \(k[x_1, \ldots, x_n]\) are of the form \((x_1 - a_1, \ldots, x_n - a_n)\) for some \(a_i \in k\).

2) For all proper ideals \(\f{a} \subset k[x_1, \ldots, x_n]\), the variety \(V(\f{a})\) is non-empty.

Proof:

1) If \(\f{m}\) is a maximal ideal then \(k[x_1, \ldots, x_n]/\f{m}\) is a field that is finitely generated as a \(k\)-algebra, so by Zariski’s lemma it is an algebraic field extension of \(k\). As \(k\) is algebraically closed this field extension is trivial, so \(k[x_1, \ldots, x_n]/\f{m} \cong k\). Then \(\f{m}\) is the kernel of the surjective map \(k[x_1, \ldots, x_n] \to k\) sending \(x_i \mapsto a_i\) for some \(a_i \in k\). Thus \(\f{m} = (x_1 - a_1, \ldots, x_n - a_n)\).

2) If \(\f{a}\) is a proper ideal then \(\f{a}\) is contained in some maximal ideal \(\f{m}\). By part 1 we have \(\f{m} = (x_1 - a_1, \ldots, x_n - a_n)\) for some \(a_i \in k\). By the order reversing nature of the \(V(\cdot)\) map we have

\[\{ (a_1, \ldots, a_n)\} = V(\f{m}) \subset V(\f{a}),\]

which gives the desired conclusion. \(\square\)

Strong Nullstellensatz

Thm: (Hilbert’s Nullstellensatz) Let \(k\) be an algebraically closed field and \(\f{a} \subset k[x_1, \ldots, x_n]\) be an ideal. Then \(I(V(\f{a})) = \sqrt{\f{a}}\).

Proof: If \(\f{a} = k[x_1, \ldots, x_n]\) then \(V(\f{a}) = \emptyset\) and \(I(V(\f{a})) = k[x_1, \ldots, x_n] = \sqrt{\f{a}}\). So assume \(\f{a}\) is a proper ideal.

We know already that \(\sqrt{\f{a}} \subset I(V(\f{a}))\), so we need to show the reverse inclusion. Let \(f \in I(V(\f{a}))\). If \(f = 0\) then \(f \in \sqrt{\f{a}}\) trivially, so assume \(f \neq 0\).

By Hilbert’s Basis Theorem \(\f{a}\) is finitely generated, say \(\f{a} = (f_1, \ldots, f_r)\). The slick step, known as the Rabinowitsch trick, is to add a new variable \(y\) and consider the ideal \(\f{b} = (f_1, \ldots, f_r, 1 - yf) \subset k[x_1, \ldots, x_n, y]\). If \(\f{b}\) were a proper ideal then by the weak Nullstellensatz there would be some \((a_1, \ldots, a_n, b) \in k^{n+1}\) such that \(f_i(a_1, \ldots, a_n) = 0\) for all \(i\) and \(1 - b f(a_1, \ldots, a_n) = 0\). But \((a_1, \ldots, a_n) \in V(\f{a})\) and \(f \in I(V(\f{a}))\), so \(f(a_1, \ldots, a_n) = 0\) and hence \(1 - b \cdot 0 = 1 \neq 0\), a contradiction. So \(\f{b}\) must be the unit ideal. Thus there are polynomials \(g_1, \ldots, g_r, h \in k[x_1, \ldots, x_n, y]\) such that

\[1=f_1 g_1 + \cdots + f_r g_r + (1 - yf)h.\]

Applying the ring homomorphism \(k[x_1, \ldots, x_n, y] \to k(x_1, \ldots, x_n)\) that fixes each \(x_i\) and sends \(y \mapsto 1/f\) gives

\[1 = f_1 g_1(x_1, \ldots, x_n, 1/f) + \cdots + f_r g_r(x_1, \ldots, x_n, 1/f),\]

since the last term becomes \((1 - f/f)\,h(x_1, \ldots, x_n, 1/f) = 0\). Multiplying by a sufficiently high power \(f^m\) to clear denominators gives \(f^m = f_1 h_1 + \cdots + f_r h_r\) for some \(h_i \in k[x_1, \ldots, x_n]\). Thus \(f^m \in \f{a}\), so \(f \in \sqrt{\f{a}}\). \(\square\)