Let \(k\) be an algebraically closed field, and let \[ X=V_+(P)\subseteq\PP^n_k \] be a nonempty projective variety, where \[ P\subseteq k[x_0,\ldots,x_n] \] is a homogeneous prime ideal. Show that \(X\) is irreducible.
Solution
Let \[ \widehat X=V(P)\subseteq\AA^{n+1}_k \] be the affine cone over \(X\).
1. The affine cone \(\widehat X\) is irreducible.
Its coordinate ring is \[ k[\widehat X] \cong k[x_0,\ldots,x_n]/P. \] Since \(P\) is prime, this quotient is an integral domain. An affine algebraic set has integral-domain coordinate ring exactly when it is irreducible. Therefore \(\widehat X\) is irreducible.
2. The punctured cone\[ \widehat X^\times \coloneqq \widehat X\sm\{0\} \]is a nonempty irreducible topological space.
The origin is closed in affine space, so \(\widehat X^\times\) is open in \(\widehat X\). A nonempty open subset of an irreducible topological space is irreducible.
It is nonempty because \(X\) is nonempty: any projective point of \(X\) has a nonzero affine representative lying in \(\widehat X\).
3. The projectivization map\[ q:\widehat X^\times\longrightarrow X, \qquad (a_0,\ldots,a_n) \longmapsto [a_0:\ldots:a_n] \]is continuous and surjective.
Every point of \(X\) has a nonzero homogeneous representative in the affine cone, so \(q\) is surjective.
To check continuity, let \[ Y=V_+(J)\subseteq X \] be projectively closed, with \(J\) homogeneous. Then \[ q^{-1}(Y) = V(J)\intersect\widehat X^\times, \] which is closed in the subspace \(\widehat X^\times\). Thus inverse images of closed subsets are closed, so \(q\) is continuous.
4. The projective variety \(X\) is irreducible.
By step 2, \(\widehat X^\times\) is irreducible. By step 3, \[ X=q(\widehat X^\times) \] is its continuous image. A continuous image of an irreducible space is irreducible. Hence \[ \boxed{X\text{ is irreducible}.} \]
5. Q.E.D.
Steps <1>1–<1>4 show that \(V_+(P)\) is irreducible for every homogeneous prime \(P\) with \(V_+(P)\neq\varnothing\).