[proofplan] We define the projective vanishing ideal $I_+(X)$ and projective zero set $V_+(I)$, then use the [projective Nullstellensatz](/theorems/2135) as the main input. The key formula is $I_+(V_+(I))=\sqrt{I:S_+^\infty}$ for every homogeneous ideal $I\trianglelefteq S$. This formula recovers exactly the radical saturated homogeneous ideals from their projective zero sets, while $V_+(I_+(X))=X$ follows because $X$ is already projectively algebraic. The empty set is handled separately, since many homogeneous ideals have empty projective zero set but the chosen saturated representative is the unit ideal. [/proofplan]

[step:Define the projective zero set and vanishing ideal] Let \begin{align*} S := k[x_0,\dots,x_n] \end{align*} and let \begin{align*} S_+ := (x_0,\dots,x_n) \end{align*} be the irrelevant ideal. For a homogeneous ideal $I\trianglelefteq S$, define \begin{align*} V_+(I) := \{[a_0:\cdots:a_n]\in \mathbb P_k^n : F(a_0,\dots,a_n)=0 \text{ for every homogeneous } F\in I\}. \end{align*} This definition is independent of the chosen representative of the projective point, because if $F$ is homogeneous of degree $d$ and $\lambda\in k^\times$, then \begin{align*} F(\lambda a_0,\dots,\lambda a_n)=\lambda^dF(a_0,\dots,a_n). \end{align*} For a subset $X\subseteq \mathbb P_k^n$, define $I_+(X)\trianglelefteq S$ to be the homogeneous ideal whose degree-$d$ part consists of the homogeneous forms of degree $d$ vanishing on $X$: \begin{align*} I_+(X)_d := \{F\in S_d : F(a)=0 \text{ for every } [a]\in X\}. \end{align*} Thus \begin{align*} I_+(X)=\bigoplus_{d\ge 0} I_+(X)_d. \end{align*} For the empty subset we use the convention \begin{align*} I_+(\varnothing)=S. \end{align*} [/step]

[step:Verify that projective vanishing ideals are proper radical saturated homogeneous ideals]Let $X\subseteq \mathbb P_k^n$ be nonempty and projectively algebraic. By construction, $I_+(X)$ is homogeneous. It is proper because $1\notin I_+(X)$: if $[a]\in X$, then $1(a)=1\ne 0$. To see radicality and saturation, use the cone over $X$. Define \begin{align*} C(X):=\{0\}\cup \{\lambda a\in k^{n+1}: \lambda\in k,\ [a]\in X\}\subseteq \mathbb A_k^{n+1}. \end{align*} The affine vanishing ideal $I(C(X))\trianglelefteq S$ is radical because it is the ideal of all polynomials vanishing on a subset of [affine space](/page/Affine%20Space). Since $C(X)$ is stable under scalar multiplication, $I(C(X))$ is homogeneous. A homogeneous form $F\in S$ vanishes on $C(X)$ if and only if it vanishes at every projective point of $X$, so \begin{align*} I(C(X))=I_+(X). \end{align*} Therefore $I_+(X)$ is radical. Now let $f\in I_+(X):S_+^\infty$. By definition of saturation, there exists $m\in\mathbb N$ such that \begin{align*} fS_+^m\subseteq I_+(X). \end{align*} It is enough to check homogeneous $f$, because both $I_+(X)$ and $I_+(X):S_+^\infty$ are homogeneous ideals. Fix a homogeneous $f\in I_+(X):S_+^\infty$ and a point $[a]\in X$. Choose an index $i\in\{0,\dots,n\}$ with $a_i\ne 0$. Since $x_i^m\in S_+^m$, we have \begin{align*} fx_i^m\in I_+(X). \end{align*} Evaluating at $a$ gives \begin{align*} f(a)a_i^m=0. \end{align*} Because $a_i\ne 0$, it follows that $f(a)=0$. Since $[a]\in X$ was arbitrary, $f\in I_+(X)$. Hence \begin{align*} I_+(X):S_+^\infty\subseteq I_+(X). \end{align*} The reverse inclusion is immediate from the definition of saturation, so \begin{align*} I_+(X)=I_+(X):S_+^\infty. \end{align*}[/step]

[guided]We must show that the ideal attached to a nonempty projective algebraic set has exactly the algebraic properties appearing in the statement: homogeneous, proper, radical, and saturated. The homogeneity is built into the definition. We defined $I_+(X)$ degree by degree: \begin{align*} I_+(X)_d := \{F\in S_d : F(a)=0 \text{ for every } [a]\in X\}. \end{align*} Thus $I_+(X)$ is a [direct sum](/page/Direct%20Sum) of homogeneous pieces. It is proper because $X$ has at least one point. If $[a]\in X$, then the constant polynomial $1$ satisfies $1(a)=1$, so $1$ does not vanish on $X$. Therefore \begin{align*} 1\notin I_+(X). \end{align*} For radicality, the clean way to avoid ambiguity about evaluating nonhomogeneous polynomials on projective points is to pass to the affine cone. Define \begin{align*} C(X):=\{0\}\cup \{\lambda a\in k^{n+1}: \lambda\in k,\ [a]\in X\}. \end{align*} The affine vanishing ideal $I(C(X))$ is radical because if $G^r$ vanishes at every point of $C(X)$, then for every $b\in C(X)$ one has \begin{align*} G(b)^r=0. \end{align*} Since $k$ is a field, this implies $G(b)=0$ for every $b\in C(X)$, so $G\in I(C(X))$. Now compare $I(C(X))$ with $I_+(X)$. If $F$ is homogeneous and vanishes on $X$, then for every $\lambda\in k$ and every representative $a$ of a point of $X$, \begin{align*} F(\lambda a)=\lambda^{\deg F}F(a)=0. \end{align*} So $F$ vanishes on the whole cone $C(X)$. Conversely, if a homogeneous form vanishes on $C(X)$, then it vanishes on every representative of every point of $X$. Hence the homogeneous elements of $I(C(X))$ are precisely the homogeneous elements of $I_+(X)$, and because $C(X)$ is closed under scalar multiplication the ideal $I(C(X))$ is homogeneous. Therefore \begin{align*} I(C(X))=I_+(X). \end{align*} This proves that $I_+(X)$ is radical. It remains to prove saturation. Let $f\in I_+(X):S_+^\infty$. By definition, there is an integer $m\in\mathbb N$ such that \begin{align*} fS_+^m\subseteq I_+(X). \end{align*} Because the ideals involved are homogeneous, it suffices to prove that every homogeneous component of $f$ lies in $I_+(X)$; equivalently, we may assume $f$ itself is homogeneous. Fix a point $[a]\in X$. Since $[a]$ is a projective point, at least one coordinate is nonzero. Choose $i\in\{0,\dots,n\}$ with $a_i\ne 0$. The monomial $x_i^m$ lies in $S_+^m$, so \begin{align*} fx_i^m\in I_+(X). \end{align*} Evaluating at the representative $a$ gives \begin{align*} f(a)a_i^m=0. \end{align*} Since $a_i\ne 0$, also $a_i^m\ne 0$, and therefore $f(a)=0$. This holds for every $[a]\in X$, so $f\in I_+(X)$. We have proved \begin{align*} I_+(X):S_+^\infty\subseteq I_+(X). \end{align*} The opposite inclusion always holds for an ideal and its saturation, hence \begin{align*} I_+(X)=I_+(X):S_+^\infty. \end{align*}[/guided]

[step:Recover a saturated radical homogeneous ideal from its projective zero set] We use the projective Nullstellensatz, in the following form: for every homogeneous ideal $I\trianglelefteq S$, \begin{align*} I_+(V_+(I))=\sqrt{I:S_+^\infty}. \end{align*} This is a cited result not yet in the wiki: Projective Nullstellensatz. Let $I\trianglelefteq S$ be a proper radical homogeneous ideal satisfying \begin{align*} I=I:S_+^\infty. \end{align*} Then the projective Nullstellensatz gives \begin{align*} I_+(V_+(I))=\sqrt{I:S_+^\infty}. \end{align*} Using the saturation hypothesis and radicality, \begin{align*} \sqrt{I:S_+^\infty}=\sqrt I=I. \end{align*} Thus \begin{align*} I_+(V_+(I))=I. \end{align*} In particular, if such an ideal $I$ is not the unit ideal, it is recovered exactly from its projective zero set. [/step]

[step:Recover a nonempty projective algebraic set from its vanishing ideal] Let $X\subseteq \mathbb P_k^n$ be a nonempty projective algebraic set. By definition of projective algebraic set, there exists a homogeneous ideal $J\trianglelefteq S$ such that \begin{align*} X=V_+(J). \end{align*} Since every [homogeneous polynomial](/page/Homogeneous%20Polynomial) vanishing on $X$ is, by definition, an element of $I_+(X)$, we have \begin{align*} X\subseteq V_+(I_+(X)). \end{align*} Conversely, because every homogeneous element of $J$ vanishes on $X$, one has \begin{align*} J\subseteq I_+(X). \end{align*} The projective zero-set construction is inclusion-reversing, so \begin{align*} V_+(I_+(X))\subseteq V_+(J). \end{align*} Since $V_+(J)=X$, the two inclusions give \begin{align*} V_+(I_+(X))=X. \end{align*} [/step]

[step:Handle the empty projective zero set by saturation to the unit ideal] By convention, \begin{align*} I_+(\varnothing)=S. \end{align*} Applying the projective Nullstellensatz to a homogeneous ideal $I\trianglelefteq S$ gives \begin{align*} I_+(V_+(I))=\sqrt{I:S_+^\infty}. \end{align*} If $V_+(I)=\varnothing$, then the left-hand side is $I_+(\varnothing)=S$, hence \begin{align*} \sqrt{I:S_+^\infty}=S. \end{align*} This is equivalent to \begin{align*} I:S_+^\infty=S. \end{align*} Conversely, if \begin{align*} I:S_+^\infty=S, \end{align*} then the same formula gives \begin{align*} I_+(V_+(I))=S. \end{align*} Since $1\in I_+(V_+(I))$, the constant polynomial $1$ vanishes on $V_+(I)$, which is possible only when \begin{align*} V_+(I)=\varnothing. \end{align*} Thus the empty projective algebraic set is represented by the unit ideal as the chosen saturated representative, while any homogeneous ideal whose saturation is $S$ defines the same empty zero set. [/step]

[step:Check that the two assignments reverse inclusions] Let $X,Y\subseteq \mathbb P_k^n$ be projective algebraic subsets with \begin{align*} X\subseteq Y. \end{align*} If a homogeneous form $F\in S$ vanishes on $Y$, then it vanishes on the smaller set $X$. Therefore \begin{align*} I_+(Y)\subseteq I_+(X). \end{align*} Let $I,J\trianglelefteq S$ be homogeneous ideals with \begin{align*} I\subseteq J. \end{align*} Every homogeneous equation belonging to $I$ also belongs to $J$ only in the sense that $J$ imposes at least the equations imposed by $I$. Hence any point satisfying all homogeneous equations in $J$ also satisfies all homogeneous equations in $I$, so \begin{align*} V_+(J)\subseteq V_+(I). \end{align*} Together with the recovery statements above, this proves the claimed mutually inverse inclusion-reversing correspondence. [/step]