INTRODUCTION TO ALGEBRAIC GEOMETRY Note del corso di ...

24 MezzettiProof. Note first that if U ⊂ X is open and U ∩Y = ∅ then U ∩Y = ∅. Otherwise,if P ∈ U ∩Y , let A be an open neighbourhood of P: then A∩Y ≠ ∅. In particular,U is an open neighbourhood of P so U ∩ Y ≠ ∅.Let Y be irreducible. If U and V are open subsets of X such that U ∩Y ≠ ∅,V ∩ Y ≠ ∅, then U ∩ Y ≠ ∅ and V ∩ Y ≠ ∅ so Y ∩ U ∩ V ≠ ∅ by irreducibility ofY . Hence Y ∩ (U ∩ V ) ≠ ∅. So Y is irreducible. If Y is irreducible, we get theirreducibility of Y in a completely analogous way.□6.4. Corollary. Let X be an irreducible topological space and U be a non–emptyopen subset of X. Then U is irreducible.Proof. By Proposition 6.2 U = X which is irreducible. By Proposition 6.3 U isirreducible.□For algebraic sets (both affine and projective) irreducibility can be expressedin a purely algebraic way.6.5. Proposition. Let X ⊂ A n ( resp. P n ) be an algebraic set. X is irreducibleif and only if I(X) (resp. I h (X)) is prime.Proof. Assume first that X is irreducible, X ⊂ A n . Let F, G polynomials ofK[x 1 , . . ., x n ] such that FG ∈ I(X): thenV (F) ∪ V (G) = V (FG) ⊃ V (I(X)) = Xhence either X ⊂ V (F) or X ⊂ V (G). In the former case, if P ∈ X then F(P) = 0,so F ∈ I(X), in the second case G ∈ I(X); hence I(X) is prime.Assume now that I(X) is prime. Let X = X 1 ∪X 2 be the union of two closedsubsets. Then I(X) = I(X 1 ) ∩ I(X 2 ) (see §4). Assume that X 1 ≠ X, then I(X 1 )strictly contains I(X) (otherwise V (I(X 1 )) = V (I(X)). So there exists F ∈ I(X 1 )such that F ∉ I(X). But for every G ∈ I(X 2 ), FG ∈ I(X 1 )∩I(X 2 ) = I(X) prime:since F ∉ I(X), then G ∈ I(X). So I(X 2 ) ⊂ I(X) hence I(X 2 ) = I(X).If X ⊂ P n , the proof is similar, taking into account the following:6.6. Lemma Let P ⊂ K[x 0 , x 1 , . . ., x n ] be a homogeneous ideal. Then P is primeif and only if, for every pair of homogeneous polynomials F, G such that FG ∈ P,either F ∈ P or G ∈ P.Proof of the Lemma. Let H, K be any polynomials such that HK ∈ P. LetH = H 0 + H 1 + . . . + H d , K = K 0 + K 1 + . . . + K e (with H d ≠ 0 ≠ K e ) be theirexpressions as sums of homogeneous polynomials. Then HK = H 0 K 0 + (H 0 K 1 +H 1 K 0 ) + . . . + H d K e : the last product is the homogeneous component of degreed + e of HK. P being homogeneous, H d K e ∈ P; by assumption either H d ∈ P orK e ∈ P. In the former case, HK − H d K = (H − H d )K belongs to P while in thesecond one H(K − K e ) ∈ P. So in both cases we can proceed by induction. □