Let \(k\) be algebraically closed, let \(S\coloneqq k[x_0,\ldots,x_n]\), and let \(\mfa\subseteq S\) be a homogeneous ideal. Show that the following conditions are equivalent.
\(Z(\mfa) = \emptyset\).
\(\sqrt{\mfa}\) is either \(S\) or the irrelevant ideal \(S_+ = \bigoplus_{d>0} S_d\).
\(\mfa \supseteq S_d\) for some \(d > 0\).
Solution
Write \(J=\mfa\) and distinguish its affine zero set \(V_a(J)\subseteq\AA_k^{n+1}\) from its projective zero set \(Z(J)\subseteq\PP_k^n\). The vector subspace \(S_d\) consists of the homogeneous polynomials of degree \(d\).
1. Conditions (i) and (ii) are equivalent.
If \(J=S\), then \(Z(J)=\varnothing\) and \(\sqrt J=S\), so both conditions hold. Now assume \(J\) is proper. It contains no nonzero constant, and homogeneity implies \(J\subseteq S_+\). Thus the origin belongs to \(V_a(J)\), and the relation between the affine and projective equations is \[ V_a(J)=\{0\}\cup\{v\in k^{n+1}\setminus\{0\}:[v]\in Z(J)\}. \] Indeed, vanishing of homogeneous generators is unchanged by multiplying a nonzero vector by a nonzero scalar. Therefore \(Z(J)=\varnothing\) exactly when \(V_a(J)=\{0\}\). The affine Nullstellensatz identifies the latter condition with \[ \sqrt J=I(V_a(J))=I(\{0\})=(x_0,\ldots,x_n)=S_+ \] [@Har10a, Theorem I.1.3A]. Conversely, if \(\sqrt J=S_+\), a point vanishes on \(J\) exactly when it vanishes on its radical, so \(V_a(J)=\{0\}\) and \(Z(J)=\varnothing\). Finally \(\sqrt J=S\) forces \(1\in J\), which is the unit-ideal case already treated.
2. Condition (ii) implies (iii).
For \(J=S\), any positive \(d\) works. If \(\sqrt J=S_+\), choose integers \(N_i\ge1\) with \(x_i^{N_i}\in J\) for \(0\le i\le n\). Set \[ D=1+\sum_{i=0}^n(N_i-1)>0. \] Every monomial of total degree \(D\) has an exponent at least \(N_i\) for some \(i\); otherwise its degree would be at most \(\sum_i(N_i-1)=D-1\). It is therefore divisible by one of the powers \(x_i^{N_i}\) and belongs to \(J\). Since these monomials span \(S_D\), the whole graded piece \(S_D\) is contained in \(J\).
3. Condition (iii) implies (i).
If \(S_d\subseteq J\) for a positive \(d\), then \(x_i^d\in J\) for every \(i\). At a point of \(\PP^n\), at least one coordinate is nonzero, and its \(d\)th power is nonzero as well. Thus these polynomials have no common projective zero, so neither does \(J\).
4. Q.E.D.
Step 1 proves (i) equivalent to (ii), and steps <1>2–<1>3 prove (ii) implies (iii) implies (i). These implications establish all three equivalences, including the unit ideal.