Nilpotents are invisible
Consider k[x]/(x^2). The class of x is nonzero yet squares to zero — a nilpotent. Geometrically it is invisible: as a function on the single point Spec, it vanishes. Commutative algebra makes this precise. The nilradical of a ring R is nil(R) = { a ∈ R : a^n = 0 for some n ≥ 1 }, the set of all nilpotents. It is an ideal, and R/nil(R) — the reduced ring — is the largest quotient with no nonzero nilpotents.
More generally, the radical of an ideal I is √I = { a : a^n ∈ I for some n }, so nil(R) = √(0). A clean structural fact pins it down: √I is the intersection of all [[prime-ideal|prime ideals]] containing I. In particular the nilradical is the intersection of all primes of R.
Why nil(R) = intersection of all primes P:
(c) If a is nilpotent, a^n = 0 in every P, and P prime + a^n=0 => a in P.
(>) If a is NOT nilpotent, the set {1, a, a^2, ...} avoids 0.
Localize / use Zorn: among ideals disjoint from {a^n} pick a
maximal one P. One checks P is prime and a not in P.
So a escapes some prime, hence a not in the intersection.
Example: R = Z/12Z. 12 = 2^2 * 3.
Nilpotents: classes a with a^n = 0 mod 12 => need 6 | a.
So nil = (6) = {0,6}. Indeed 6^2 = 36 = 0 mod 12.
Primes of Z/12Z: (2) and (3). Their intersection (2)n(3) = (6). Checks out.The Nullstellensatz: weak and strong
Over an algebraically closed field k, Hilbert's Nullstellensatz turns the radical into geometry. The weak form: a maximal ideal of k[x_1,…,x_n] is exactly (x_1−a_1,…,x_n−a_n) for a point a ∈ k^n. So maximal ideals ↔ points. Equivalently, a proper ideal always has a common zero — you cannot write 1 as a combination of polynomials with no shared root.
The strong form pins down which functions vanish on a variety. Write V(I) for the vanishing set of I and I(V) for the ideal of functions vanishing on V. Then I(V(J)) = √J. So the only obstruction to recovering an ideal from its zero set is — exactly — the radical. Radical ideals ↔ affine varieties is a perfect dictionary.
Why it is true: the Rabinowitsch trick
The strong form follows from the weak one by a single slick move. The weak form itself is usually deduced from Noether normalization (next-but-one guide) or Zariski's lemma: a field that is finitely generated as a k-algebra is finite over k. Granting the weak form, here is the bridge to the strong form.
Goal: if g vanishes wherever f_1,...,f_r all vanish, then g in sqrt(f_1,...,f_r). Rabinowitsch trick: add a fresh variable t and work in k[x_1,...,x_n, t]. Consider the ideal J = (f_1, ..., f_r, 1 - t*g). Any common zero of J would have all f_i = 0 (so g = 0 there by hypothesis) yet 1 - t*g = 1 - 0 = 1 != 0. Impossible => V(J) = empty. Weak Nullstellensatz => J = (1), the whole ring. So write 1 = sum h_i * f_i + h_0 * (1 - t*g) in k[x,t]. Now substitute t = 1/g and clear denominators by multiplying by g^N: g^N = sum (h_i with t=1/g) * f_i in k[x]. Hence g^N in (f_1,...,f_r), i.e. g in sqrt(f_1,...,f_r). QED