Let \(k\) be algebraically closed and let \(S\coloneqq k[x_0,\ldots,x_n]\). Prove the homogeneous Nullstellensatz: if \(\mfa \subseteq S\) is a homogeneous ideal and \(f \in S\) is a homogeneous polynomial with \(\deg f > 0\) such that \(f(P) = 0\) for all \(P \in Z(\mfa)\) in \(\PP^n\), then \(f^q \in \mfa\) for some \(q > 0\).
(The positive-degree hypothesis)
For \(\mfa=(x_0,\ldots,x_n)\), the projective zero set is empty, so the constant polynomial \(1\) vanishes there vacuously. No power of \(1\) belongs to this proper ideal. Thus the positive-degree hypothesis cannot be dropped; in the affine argument it supplies vanishing at the cone vertex.
Hint
Interpret the problem in the affine \((n+1)\)-space whose affine coordinate ring is \(S\), and apply the usual Nullstellensatz.
Solution
Write \(V_a(\mfa)\subseteq\AA_k^{n+1}\) for the affine common zero set and \(Z(\mfa)\subseteq\PP_k^n\) for the projective common zero set. The notation \(f(P)=0\) for a homogeneous polynomial means vanishing at any nonzero homogeneous-coordinate representative of \(P\). It is independent of that choice, since scaling by \(\lambda\in k^\times\) multiplies the value by \(\lambda^{\deg f}\).
1. Every nonzero point \(v\in V_a(\mfa)\) determines a point \([v]\in Z(\mfa)\).
The ideal \(\mfa\) is generated by homogeneous polynomials. For each such generator \(g\), the equation \(g(v)=0\) holds by the definition of the affine zero set. These are exactly the equations defining membership of \([v]\) in the projective zero set. Conversely, every nonzero representative of a point of \(Z(\mfa)\) satisfies the generators and therefore every element of \(\mfa\), so lies in \(V_a(\mfa)\). The zero vector is not used as a projective point.
2. The polynomial \(f\) vanishes on all of \(V_a(\mfa)\).
At a nonzero point, step 1 and the assumed projective vanishing give \(f(v)=0\). At the zero vector, if it belongs to \(V_a(\mfa)\), the same equality follows because \(f\) is homogeneous of positive degree. These two cases exhaust the affine zero set. They also cover the case where the projective zero set is empty. If \(\mfa=S\), the affine zero set is empty as well and this vanishing is vacuous.
3. Q.E.D.
By step 2 and the affine Nullstellensatz, \[ f\in I(V_a(\mfa))=\sqrt{\mfa} \] [@Har10a, Theorem I.1.3A]. By the definition of the radical, there is an integer \(q\ge1\) for which \(f^q\in\mfa\), as required.