projective variety cone irreducible

By the irreducibility clause of projective irreducibility homogeneous prime, $I_{+}(X)$ is prime. This clause and its cone-ideal argument require no AC. Homogeneous defining equations show that $C(X)$ consists exactly of zero and the nonzero representatives of points of $X$. Decomposing a polynomial vanishing on this cone into homogeneous components and evaluating on all scalar multiples shows, over the infinite field $k$, that each component vanishes on $X$. Hence $I(C(X))=I_{+}(X)$.

Suppose $C(X)=A\cup B$ for two proper relatively closed subsets. Choose a point of $C(X)\setminus A$ and a polynomial $f$ from defining equations of $A$ that is nonzero there. Likewise choose a polynomial $g$ vanishing on $B$ but not on all of $C(X)$. Then $fg$ vanishes on the cone, so $fg\in I(C(X))=I_{+}(X)$, although neither factor belongs to that prime ideal. This is impossible.

The cone is nonempty since it contains zero, and step 2.1 excludes a union of two proper closed subsets. Thus it is irreducible.