Let \(k\) be algebraically closed, let \(S\coloneqq k[x_0,\ldots,x_n]\), and let \(S^h\) denote its set of homogeneous elements.
If \(T_1 \subseteq T_2\) are subsets of \(S^h\), then \(Z(T_1) \supseteq Z(T_2)\).
If \(Y_1 \subseteq Y_2\) are subsets of \(\PP^n\), then \(I(Y_1) \supseteq I(Y_2)\).
For any two subsets \(Y_1, Y_2 \subseteq \PP^n\), \(I(Y_1 \union Y_2) = I(Y_1) \intersect I(Y_2)\).
If \(\mfa \subseteq S\) is a homogeneous ideal with \(Z(\mfa) \neq \emptyset\), then \(I(Z(\mfa)) = \sqrt{\mfa}\).
For any subset \(Y \subseteq \PP^n\), \(Z(I(Y)) = \bar{Y}\).
(The nonempty-locus hypothesis)
For \(\mfa=S_+=(x_0,\ldots,x_n)\), the projective zero set is empty. Thus \(I(Z(\mfa))=I(\varnothing)=S\), whereas \(\sqrt{\mfa}=S_+\), since \(S/S_+=k\) is a field. This shows why (d) excludes the empty projective zero set; its possible radicals are computed in P-AGH22IRRELEVANT.
Solution
For \(T\subseteq S^h\), let \(Z(T)\) be the set of projective points where every member of \(T\) vanishes. For a subset \(Y\subseteq\PP_k^n\), write \[ I(Y)_d=\{f\in S_d:f(v)=0\text{ for every nonzero representative }v\text{ of a point of }Y\}, \qquad I(Y)=\bigoplus_{d\ge0}I(Y)_d. \] Homogeneity makes vanishing independent of the representative. Each \(I(Y)_d\) is a vector subspace, and \(S_e I(Y)_d\subseteq I(Y)_{d+e}\), so this direct sum is a homogeneous ideal. It is exactly the ideal generated by all homogeneous polynomials vanishing on \(Y\), the convention in [@Har10a, Chapter I, §2].
1. The operators \(Z\) and \(I\) reverse inclusions, proving (a) and (b).
If every member of \(T_2\) vanishes at a point, then every member of its subset \(T_1\) vanishes there. Thus \(Z(T_2)\subseteq Z(T_1)\). Likewise, a homogeneous polynomial vanishing at every point of \(Y_2\) vanishes at every point of \(Y_1\subseteq Y_2\). This gives \(I(Y_2)_d\subseteq I(Y_1)_d\) in every degree, and hence \(I(Y_2)\subseteq I(Y_1)\).
2. One has \(I(Y_1\cup Y_2)=I(Y_1)\cap I(Y_2)\), proving (c).
For \(f\in S_d\), vanishing on \(Y_1\cup Y_2\) is equivalent to vanishing on each of \(Y_1\) and \(Y_2\). Therefore \(I(Y_1\cup Y_2)_d=I(Y_1)_d\cap I(Y_2)_d\). Membership in a homogeneous ideal is equivalent to membership of every homogeneous component, so these degreewise equalities imply equality of the ideals.
3. The radical of a homogeneous ideal is homogeneous.
Let \(J\) be homogeneous and let \(0\ne f\in\sqrt J\); the zero polynomial needs no argument. Choose \(m\ge1\) with \(f^m\in J\) and write \(f_D\) for the highest nonzero homogeneous component of \(f\). The degree-\(mD\) component of \(f^m\) is \(f_D^m\), and belongs to \(J\) because \(J\) is homogeneous. Hence \(f_D\in\sqrt J\). The radical is an ideal, so \(f-f_D\in\sqrt J\). Repeating on the strictly shorter sum of components proves that every homogeneous component of \(f\) lies in \(\sqrt J\). This proves the assertion.
4. For the ideal in (d), \(I(Z(\mfa))=\sqrt{\mfa}\).
Both ideals are homogeneous, the second by step 3, so it suffices to compare homogeneous elements. If a homogeneous \(f\) lies in \(\sqrt{\mfa}\), some power \(f^m\) belongs to \(\mfa\) and vanishes on \(Z(\mfa)\). At each nonzero coordinate representative its value is a power in a field; \(f^m(v)=0\) implies \(f(v)=0\). Thus \(f\in I(Z(\mfa))\).
Conversely, let \(f\) be homogeneous and vanish on \(Z(\mfa)\). For positive degree, the homogeneous Nullstellensatz gives \(f^m\in\mfa\) for some \(m\ge1\), hence \(f\in\sqrt{\mfa}\). For degree zero, the assumed nonemptiness of \(Z(\mfa)\) forces the constant \(f\) to be zero, which also belongs to \(\sqrt{\mfa}\). These cases prove both inclusions.
5. For any \(Y\subseteq\PP^n\), one has \(Z(I(Y))=\overline Y\), proving (e).
By definition, every homogeneous member of \(I(Y)\) vanishes on \(Y\). Consequently \(Y\subseteq Z(I(Y))\). The latter is projectively closed, so \(\overline Y\subseteq Z(I(Y))\).
For the reverse inclusion, let \(W\) be any closed subset containing \(Y\). Write \(W=Z(T)\) for a family \(T\) of homogeneous polynomials, using the projective Zariski topology. Every member of \(T\) vanishes on \(Y\) and therefore belongs to \(I(Y)\). Part (a), proved in step 1, gives \(Z(I(Y))\subseteq Z(T)=W\). Intersecting all closed sets \(W\) containing \(Y\) gives \(Z(I(Y))\subseteq\overline Y\). For \(Y=\varnothing\), the same argument applies; explicitly \(I(Y)=S\) and \(Z(S)=\varnothing\).
6. Q.E.D.
Step 1 proves (a) and (b), step 2 proves (c), step 4 proves (d), and step 5 proves (e).