Let \[ X\subseteq\PP^n_k \] be a nonempty projective algebraic set over the algebraically closed ground field \(k\), and let \[ I(X)\subseteq k[x_0,\ldots,x_n] \] be its homogeneous vanishing ideal. Show that \[ X\text{ is irreducible} \quad\Longleftrightarrow\quad I(X)\text{ is prime}. \]
Solution
1. Suppose that \(X\) is irreducible. Then \(I(X)\) is prime.
Assume for contradiction that the homogeneous ideal \(I(X)\) is not prime. A homogeneous ideal \(I\) is prime if and only if, for all homogeneous \(f,g\), \(fg\in I\) implies \(f\in I\) or \(g\in I\). Hence there exist homogeneous polynomials \[ f,g\in k[x_0,\ldots,x_n] \] such that \[ fg\in I(X), \qquad f\notin I(X), \qquad g\notin I(X). \]
Because \(fg\) vanishes at every point of \(X\), for each \[ p\in X \] one has \[ f(p)g(p)=0. \] Since \(k\) is a field, either \(f(p)=0\) or \(g(p)=0\). Therefore \[ X = \bigl(X\intersect V_+(f)\bigr) \union \bigl(X\intersect V_+(g)\bigr). \] Both terms are Zariski closed in \(X\) because \(f\) and \(g\) are homogeneous.
The first term is proper: since \[ f\notin I(X), \] there is a point of \(X\) where \(f\) does not vanish. Likewise the second term is proper because \[ g\notin I(X). \] Thus \(X\) is the union of two proper closed subsets, contradicting irreducibility. Hence \(I(X)\) is prime.
2. Suppose that \(I(X)\) is prime. Then \(X\) is irreducible.
Since \(X\) is Zariski closed, \[ X=V_+\bigl(I(X)\bigr). \] The ideal \(I(X)\) is homogeneous and prime, and \(V_+(P)\) is irreducible for every homogeneous prime \(P\) with \(V_+(P)\) nonempty P-AGXVARPROJIRR. Therefore \(X\) is irreducible.
3. Therefore\[ \boxed{ X\text{ is irreducible} \quad\Longleftrightarrow\quad I(X)\text{ is a homogeneous prime ideal}. } \]
Step 1 proves the forward implication. Step 2 proves the converse. The ideal \(I(X)\) is homogeneous by its projective vanishing-ideal construction.
4. Q.E.D.
Step 3 is the required equivalence.