projective irreducibility homogeneous prime

Every homogeneous polynomial vanishing on $X$ vanishes on $C$. Conversely, [given, algebra] write a polynomial vanishing on $C$ as $P={\sum }_d{P}_d$ with each $P_d$ homogeneous. For a representative $a$ of a point of $X$ and every $\lambda \in k$, one has $0=P(\lambda a)={\sum }_d{\lambda }^d{P}_d(a)$. Since the algebraically closed field $k$ is infinite, each $P_d(a)$ is zero. Thus all homogeneous components of $P$ lie in $I_{+}(X)$, and $I(C)=I_{+}(X)$.

Suppose $X$ is irreducible and homogeneous $F,G$ satisfy [given, algebra] $FG\in I_{+}(X)$. Then $X\subseteq V_{+}(F)\cup V_{+}(G)$, so irreducibility gives $F\in I_{+}(X)$ or $G\in I_{+}(X)$. A homogeneous ideal is prime exactly when this test holds for homogeneous elements, so $I_{+}(X)$ is prime.

Conversely, suppose $I_{+}(X)$ is prime and $X=Y\cup Z$ with $Y,Z$ [given, algebra] projective algebraic subsets of $X$. If both are proper, choose $p\in X\setminus Y$ and $q\in X\setminus Z$. Homogeneous defining equations give $F\in I_{+}(Y)$ with $F(p)\mathrm{\ne }0$ and $G\in I_{+}(Z)$ with $G(q)\mathrm{\ne }0$. Then $FG$ vanishes on $X$, contrary to primality of $I_{+}(X)$. Hence $X=Y$ or $X=Z$, so $X$ is irreducible.

Now assume the Axiom of Choice and let $J$ be homogeneous radical with [step 1.1, algebra] $V_{+}(J)\mathrm{\ne }\mathrm{\varnothing }$. Its affine zero locus is the cone over $V_{+}(J)$. The affine Nullstellensatz gives $I(V(J))=\sqrt{J}=J$, while step 1.1 gives $I(V(J))=I_{+}(V_{+}(J))$. Conversely, every $I_{+}(X)$ is homogeneous and radical, because $F^r$ vanishing on $X$ forces $F$ to vanish there. This is the stated radical-ideal correspondence.