- State the Weak and Strong forms of Hilbert’s Nullstellensatz over an algebraically closed field \(k\).
- State the geometric bijection between radical ideals in \(k[x_1, \dots, x_n]\) and algebraic sets in \(\mathbb{A}^n(k)\).
- Sketch the proof of the Nullstellensatz (Zariski’s Lemma / Noether Normalization, and the Rabinowitsch trick).
Solution
Let \(k\) be algebraically closed and \(R=k[x_1,\dots,x_n]\).
Weak Nullstellensatz. Every maximal ideal of \(R\) has the form \[ (x_1-a_1,\dots,x_n-a_n),\qquad a_i\in k. \] Equivalently, every proper ideal of \(R\) has a common zero in \(k^n\).
Strong Nullstellensatz. For every ideal \(I\subseteq R\), \[ I(V(I))=\sqrt I. \] Consequently, \(I\mapsto V(I)\) and \(X\mapsto I(X)\) give inclusion-reversing bijections between radical ideals of \(R\) and algebraic subsets of \(\mathbb A^n(k)\). Under this correspondence, maximal ideals correspond to points and prime ideals to irreducible algebraic sets.
For the weak theorem, if \(\mathfrak m\) is maximal then \(R/\mathfrak m\) is a field finitely generated as a \(k\)-algebra. Zariski’s lemma makes it algebraic over \(k\), hence equal to \(k\) because \(k\) is algebraically closed. The images of the \(x_i\) therefore give the point \((a_1,\dots,a_n)\).
For the strong theorem, let \(g\) vanish on \(V(I)\) and consider \[ J=(I,1-yg)\subset k[x_1,\dots,x_n,y]. \] Then \(V(J)=\varnothing\), so the weak theorem gives \(J=(1)\). Writing \(1\) as a combination of generators of \(I\) and \(1-yg\), then substituting \(y=g^{-1}\) and clearing denominators, yields \(g^N\in I\) for some \(N\). Hence \(g\in\sqrt I\).