Lecture 5: More on Finite Morphisms and Irreducible Varieties

Lecture 5 18.725 Algebraic Geometry I Lecture 5: More on Finite Morphisms and Irreducible Varieties Lemma 1. Let f : X → Y be a finite map of varieties and Z1 ( Z2 irreducible subvarieties of X. Then f (Z1 ) ( f (Z2 ). Proof. We can assume WLOG that f : X = Spec(A) → Spec(B) = Y is surjective and Z2 = X. Pick a nonzero function g ∈ I(Z1 ). Since f is finite, the ring map B → A turns A into a finitely-generated B-module. In particular, the B-subalgebra of A generated by g is finitely-generated as a B-module. Hence, n−1 n−1 X X n i n g = hi g for some natural number n and h0 6= 0. Since h0 = g − hi g i vanishes on Z1 , h0 vanishes i=0 i=1 on f (Z1 ). Lemma 2. If f : X → Y is a finite surjection of varieties, then dim(X) = dim(Y ). Proof. Let X0 ( X1 ( ... ( Xn be any chain of non-empty irreducible closed subsets of X. Set Yi = f (Xi ). Since f is continuous, {Yi } are irreducible and since f is finite {Yi } are closed. By the previous lemma, the sequence Y0 ⊂ ... ⊂ Yn is strictly increasing. Hence, dim(Y ) ≥ dim(X). Conversely, let Y0 ( Y1 ( ... ( Ym be a chain of non-empty irreducible closed subsets of Y . We wish to show that there is a sequence (of non-empty irreducible closed subsets) X0 ( ... ⊂ Xm of X such that f (Xi ) = Yi . Write f −1 Ym as a union of irreducible components V1 ∪ ... ∪ Vt . Since f is surjective and finite, Ym = f (V1 ) ∪ ... ∪ f (Vt ), where f (Vt ) are closed and irreducible. Since Ym is irreducible, we must have Ym = f (Vj ) for some index j. By induction on m, we may find a chain of non-empty closed irreducibles X0 ( ... ( Xm−1 of Vj with f (Xi ) = Yi . Then X0 ( ... ( Xm−1 ( Vj is the desired sequence in X. Theorem 1.1. dim(An ) = n Proof. dim(An ) ≥ n is clear. Suppose Z0 ( ... ( Zm is a saturated chain of non-empty closed irreducible subsets of An . We need to show that m ≤ n. Then Zm = An and Zm−1 is a closed, proper subset of An . In particular, one can find a non-constant function g ∈ k[X1 , ..., Xn ] such that Zm−1 ⊆ Z(g). By (the proof of) Noether normalization, there is a finite surjective morphism Z(g) → An−1 . Then the previous lemma implies dim(Z(g)) = dim(An−1 ). Inducting on n, we can assume dim(An−1 ) = n − 1. Hence m − 1 ≤ dim(Z(g)) = dim(An−1 ) = n − 1, which completes the proof. Corollary 1. If X is a hypersurface in An defined by a non-constant polynomial then dim(X) = n − 1. Corollary 2. Every variety has finite dimension. We now return to curves. Proposition 1. All irreducible curves over a given field (or even various fields of equal cardinality!) are homeomorphic Proof. From the definition of dimension it is clear that a closed irreducible subset of an irreducible curve X is either zero dimensional or X. Any proper closed subset of X is therefore finite. Hence, any bijection between irreducible curves is a homeomorphism. But a curve over a field k has as many points as k. The proposition follows. Definition 1. Let X ⊂ An be a hypersurface defined by a polynomial g. Write g as a sum of homogenous components g = gm + gm+1 + ... with gm 6= 0. If 0 ∈ X, the multiplicity of X at 0 is defined to be the natural number m. The multiplicity at p ∈ X is the multiplicity at 0 after applying a linear change of coordinates mapping p to 0. Definition 2. Let X, Y be two curves in A2 with no common component and (a, b) be an intersection point. If IX and IY are the ideals in k[x, y] defining X and Y , respectively. Then V = k[x, y]/(IX + IY ) is a finite dimensional vector spaces and multiplication by x,y induce two commuting operators on V . The multiplicity of intersection of X and Y at (a, b) is defined as dimension of the common generalized eigenspace of the two operators, with eigenvalues a,b respectively. 1 Lecture 5 18.725 Algebraic Geometry I Theorem 1.2 (Bezout). Let X, Y ⊂ P2 be curves without a common component, of degree d and e, respectively. Then X ∩ Y contains de points, counted with multiplicities. Proof. Proof in lecture notes from 11/5. Theorem 1.3 (Pascal). Let Q be a circle in P2 and X a hexagon inscribed in C. Then the three pairs of opposite sides of X intersect at three points which lie on a straight line. Proof. Let A, B, C be linear equations of three pairwise nonintersecting sides of our hexagon inscribed in Q and A0 , B 0 , C 0 be the equations of the remaining three ones with A0 opposite to A etc. Pick a 7th point on Q and consider a degree 3 homogeneous polynomial P=ABC - t A’B’C’ where t is such that P vanishes at the chosen 7th point. By Bezout’s theorem, the intersection of Q with a deg 3 curve has at most 6 points, unless they have a common component. Since P has at least 7 zeroes, the latter must be true. Hence, the vanishing locus of P is the union of Q with some other component, which has to be a line L by a degree count. Now the intersection point of A and A0 has to lie on L, as well as that of B with B 0 and C with C 0 . Theorem 1.4. Let X be an irreducible variety of dimension n and let g be a non-constant function on X. Then any irreducible component of Z(g) has dimension n − 1. Lemma 3. dim(Z(g)) ≥ n − 1. Proof. The special case X = An is proved above. We will reduce to this special case by Noether’s lemma: choose B = k[x1 , ..., xn ] ⊂ k[X] = A such that A is a finitely-generated B-module. Then g is the root of some monic irreducible polynomial P ∈ B[t] = k[x1 , ..., xn , t]. Write P = a0 + a1 t + ... + tn with ai ∈ B. The inclusion B ⊂ A descends to a map B/(a0 ) → A/(g). It is enough to show that the map of spectra Spec(A/(g)) → Spec(B/(a0 )) is surjective. Let C = B[t]/(P ) and factor B ⊂ A as B ⊂ C ⊂ A. Spec(C) is irreducible of dimension n. Thus π : Spec(A) → Spec(C) is onto, so the preimage π −1 (Z(t)) = Z(g) maps onto Z(t). But B/(a0 ) ⊂ C/(t) = B/(free terms of polynomials in P ). Lemma 4. Let X be an irreducible variety and U ⊂ X a non-empty open subset. Then dim(U ) = dim(X). Proof. If we replace X by An the lemma is clear: dim(U ) ≤ dim(X) since U ⊆ X and the chain (point in U) ( line ( ... ( An of closed irreducibles in U shows that dim(U ) ≥ dim(X). For X affine, use Noether’s lemma to get a finite surjection π : X → An . Since π is closed, V = An − π(X − U ) is open. Let U 0 = π −1 V . Then π : U 0 → V is a finite surjection. Hence, dim(U 0 ) = dim(V ) = n. On the other hand, U 0 ⊆ U so dim(U 0 ) ≤ dim(U ) ≤ dim(X) = n. So dim(U ) = n as desired. For general X, reduce to the affine case by using dim(X) = max {dim(U ); U affine}. Proof of Theorem 1.4. Assume Z is a component of Z(g) and dim(Z) ≤ dim(X) − 2. We can find an open affine subvariety U of X such that U ∩ Z(g) = Z ∩ U is non-empty. Then by lemma 4 we have dim(U ∩ Z) = dim(Z) ≤ dim(X) − 2 = dim(U ) − 2. Then by lemma 3, g|U is constant. But U is an open subset in an irreducible variety and therefore dense, so continuity implies g is globally constant. 2