DISCLAIMER: These unpolished lecture notes were designed for in-class use only, not as reference material. They are only web-posted in response to student request. 1 Dimension Def. The dimension of an irreducible qp var X is the transcendence degree of k(X) over k. The dimension of a general qp var is the maximum dim of each of its components Note: if U ⊂ X is open, then k(U ) = k(X), so dim(U ) = dim(X) Recall that if k ⊂ K is a field extension, the transcendence degree tr deg(K/k) of K over k is the maximum number of elts of K which are alg indep over k If tr deg(K/k) = n and x1, . . . , xn ∈ K are alg indep over k, then K is an alg ext of k(x1, . . . , xn) Example. Since k(An) = k(x1, . . . , xn), we have tr deg(k(An/k)/k) = n and dim An = n. Since An ⊂ Pn is open, it follows that dim Pn = n Def. If Y ⊂ X are qp vars, the codimension of Y in X is dim X − dim Y 2 Main Results of this section: Thm 0. If f : X → Y is finite, then dim X = dim Y Thm 1. If X ⊂ Y closed subset, then dim X ≤ dim Y . If Y is irred and dim X = dim Y , then X = Y Thm 2. (purity) Every irreducible component of a hypersurface in An or Pn has codim 1 Thm 3. (Geometric version of Krull’s Hauptidealsatz) Suppose X ⊂ An is closed with all components of dimension n − 1. Then X is a hypersurface and IX is principal Similarly, there is a projective version of Theorem 3, which generalized to products of projective spaces Thm 4. Let X ⊂ Pn1 × · · · × Pnk be closed with all P components of dimension ni − 1. Then X is defined by a single equation, separately homogeneous in each of the k sets of homogeneous coordinates of the respective Pni 3 Examples. 1. Irred affine plane curves have dimension 1. Hence all plane curves have dim 1 2. If X is a point, then k(X) = k and so dim X = 0. Therefore the dim of any finite set of points is zero. Conversely, if dim X = 0, show X is a finite set WLOG, may assume X ⊂ An Coord fcns xi ∈ k[X] are algebraic over k Therefore only finitely many values are possible 3. dim(X × Y ) = dim X + dim Y . Idea: get a transcendence basis for k(X × Y ) by concatenating the pullbacks to X × Y of transcendence bases for k(X) and k(Y ) See Shafarevich Example 4 P. 68 for details 4 4. dim G(k, n) = k(n − k) Exhibit an open subset of G(k, n − k) isomorphic to Ak(n−k) Illustrate with G(2, 4) ,→ P5 Denote the Pl¨ucker coords by wij , i < j Let Uij ⊂ P5 be the affine piece wij 6= 0 Claim: G(2, 4) ∩ Uij ' A4 Just check for U12 Consider φ : A4 → G(2, 4) (a1, a2, a3, a4) 7→ span(e1 + a1e3 + a2e4, e2 + a3e3 + a4e4) To compute Pl¨ucker coordinates, write (e1 + a1e3 + a2e4) ∧ (e2 + a3e3 + a4e4) = e1∧e2+a3e1∧e3+a4e1∧e4−a1e2∧e3−a2e2∧e4+(a1a4−a2a3)e3∧e4 i.e. (1, a3, a4, −a1, −a2, a1a4 − a2a3) ∈ U12 ⊂ P5 Clearly a regular embedding Explicit computation in proof of closedness of Pl¨ucker image shows precisely that U12 ∩ G(2, 4) = φ(A4) 5 Thm 0. If f : X → Y is finite, then dim X = dim Y Pf. WLOG may assume X and Y are irred and affine Then k[X] is a finite k[Y ]-module k(X) is a finite extension of k(Y ) Tr deg(k(X)) = Tr deg(k(Y )) Thm 1. If X ⊂ Y closed, then dim X ≤ dim Y . If Y is irred and dim X = dim Y , then X = Y Pf. WLOG may assume X and Y are irred and affine X ⊂ Y ⊂ AN , dim Y = n Any t1, . . . , tn+1 ∈ k[Y ] must be alg dep over k Same conclusion for restrictions to t1, . . . , tn+1 ∈ k[X] dim X = Tr deg(k(X)) ≤ n = dim Y Now suppose dim X = dim Y , Y irred To show X = Y suffices to show IX ⊂ k[Y ] is the zero ideal Let 0 6= u ∈ k[Y ] with u = 0 on X Pick some coord fcns t1, . . . , tn ∈ k[X] alg indep (therefore still alg indep viewed in k[Y ]) u is alg over k(t1, . . . , tn) U = u solves an irred eqn F (U ) = 0, F ∈ k[Y ][U ] Restrict to X, get contradiction; see text for details 6 Thm 2. (purity) Every irreducible component of a hypersurface in An or Pn has codim 1 Pf. Only need do An. WLOG X = Z(F ), F irreducible Claim: X is irred If X = X1 ∪ X2, take Fi ∈ IXi − IX Then F1F2 vanishes on X By Nullstellensatz, (F1F2)k ∈ IX = (F ) F divides F1 or F2, contradiction WLOG may assume xn occurs in F Let ti ∈ k[X] restriction of xi to X Claim: t1, . . . , tn−1 alg indep in k[X] This implies dim X = n − 1 Suppose G(t1, . . . , tn−1) = 0 Then G(x1, . . . , xn−1) vanishes on X F divides G Impossible because F contains xn and G does not 7 Thm 3. (Geometric version of Krull’s Hauptidealsatz) Suppose X ⊂ An is closed with all components of dimension n − 1. Then X is a hypersurface and IX is principal Pf. WLOG may assume X irred X 6= An since dim X = n − 1 So IX 6= 0. Let 0 6= F ∈ IX Since X is irred, some factor of F vanishes on X So may assume F irred Then X ⊂ Z(F ), Z(F ) proven to be irred Since dim X = dim Z(F ), conclude X = Z(F ), a hypersurface To show IX is principal, recall r I(X) = I(Z(F )) = (F ) So if G ∈ IX we have F divides Gk for some k Therefore F divides G IX = (F ) 8 Hypersurface sections (intersections with a hypersurface) Let X ⊂ Pn be a proj var F ∈ k[x0, . . . , xn] homog, not identically zero on X XF := X ∩ Z(F ) ⊂ X hypersurface section of X We might think dim XF = dim X − 1, but this is false Ex. X = Z(x1x2), F = x1 dim XF = dim X = 1 What went “wrong” is that F vanishes identically on a component of X This is essentially the only thing that can go wrong Thm. Let X ⊂ Pn irred, proj, F a homog poly not identically zero on X. Then dim XF = dim X − 1 Furthermore, if X = ∪Xi is a union of irreducible components, either Xi ⊂ Z(F ) or not Will see that if Xi ⊂ Z(F ), then dim(Xi ∩ Z(F )) = dim Xi since they are equal Otherwise we are in the case of the theorem, and dim((Xi)F ) = dim Xi − 1 9 Example. In P3 consider F1 = x0x3 − x1x2, F2 = x0, F3 = x1 Then X1 := Z(F1) = (P3)F1 is a quadric hypersurface (isomorphic to P1 × P1 by Segre, hence irred) dim X1 = 2 = dim P3 − 1 Next, consider X2 := (X1)F2 = Z(F1, F2) Note (x0x3 − x1x2, x0) = (x0, x1x2) So X2 = L1 ∪ L2 is reducible, the union of the two lines Li = Z(x0, xi) for i = 1, 2 Consistent with the theorem: dim X2 = 1 = dim X1 − 1 Now, F3 = x1 is not identically zero on X2 so we can define X3 = (X2)F3 = Z(F1, F2, F3) But (x0x3 − x1x2, x0, x1) = (x0, x1), so X3 = L1 dim X3 = dim X2 No contradiction to theorem, since X2 is reducible Note (L2)F3 = {(0, 0, 0, 1)} and so dim(L2)F3 = 0 = dim L2 − 1 10 More interesting example: twisted cubic C = {(s3, s2t, st2, t3)} ⊂ P3 Can check that IC = (x0x3 − x1x2, x0x2 − x21, x1x3 − x22) First hypersurface section: X1 := Z(x0x3−x1x2), smooth irred quadric isomorphic to P1 × P1 Second hypersurface section: X2 = Z(x0x3 − x1x2, x0x2 − x21) not only contains C, but it also contains the line L = Z(x0, x1). In fact, not hard to check X2 = C ∪ L Third hypersurface section: X3 = C In fact, is clearly not possible to write IC = (G1, G2) for any G1, G2 Even though C has codim 2, we need at least 3 equations Projective varieties X ⊂ Pn of codimension r with IX = (F1, . . . , Fr ) are called complete intersections The twisted cubic is not a complete intersection Complete intersections are relatively easy to analyze, as they are iteratively understood as hypersurface sections 11 There is a weaker notion of a set-theoretic complete intersection: X = Z(F1, . . . , Fr ), codim(X) = r Example: Z(x0x2 − x21, x0) ⊂ P3 is the line L given by x0 = x1 = 0 However, (x0x2 − x21, x0) = (x0, x21) 6= (x0, x1) 12 The theorem has many easy useful corollaries Cor. 1. A projective variety X has closed subvarieties of any dimension s < dim X Pf. Iterate the process of taking hypersurface sections and irred comps thereof Cor. 2. dim X is the maximum number n of irred closed subsets Xi of X, X ⊃ X1 ⊃ X2 ⊃ · · · ⊃ Xn , all inclusions proper Cor. 3. Let X ⊂ PN be projective and let s be the maximal dimension of a linear subspace E ⊂ X which is disjoint from X. Then dim X = N − s − 1 Pf. Let n = dim X and suppose E is a linear subspace of dim s ≥ N − n E can be defined by ≤ n lin eqns so dim(X ∩ E) ≥ n−n=0 i.e. X ∩ E is nonempty To finish proof, need only exhibit an E of dim N − n − 1 disjoint from X Get this by applying theorem repeatedly to linear forms 13 Example. Twisted cubic curve C = {(s3, s2t, st2, t3) has dimension 1: Any hyperplane ax0 + bx1 + cx2 + dx3 = 0 intersects C when (s, t) ∈ P1 satisfies as3 + bs2t + cst2 + dt3 = 0 C is disjoint from line x0 = x3 = 0 Cor. 4. Given X ⊂ Pn and F1, . . . , Fr homog, then dim (X ∩ Z(F1) ∩ . . . ∩ Z(Fr )) ≥ dim X − r This corollary has an immediate corollary, so powerful that it is elevated to a Proposition Prop. Any r ≤ n homogeneous forms on Pn have a common solution Example. Any two plane curves have at least one intersection point In fact, B´ezout’s thm says that the number of intersection points, counted by multiplicities, is the product of the degrees of the curves 14 Now let’s prove the theorem: Thm. Let X ⊂ Pn irred, proj, F a homog poly not identically zero on X. Then dim XF = dim X − 1 Pf. Choose forms F0 = F, F1, . . . , Fn inductively as follows: Letting Xi = X ∩ Z(F0, . . . , Fi), choose Fi+1 so as to not vanish identically on any component of Xi This implies that each component of Xi+1 is a proper closed subset of a component of Xi In particular, dim Xi+1 < dim Xi Therefore Xn is empty Then have reg finite map f = (F0, . . . , Fn) : X → Pn Therefore surjective If thm is not true, then dim XF ≤ n−2 and consequently Xn−1 is empty Therefore f −1(0, 0, . . . , 0, 1) is empty Contradicts surjectivity of f 15 A stronger result is in fact true Thm. Every component of XF has dimension dim X −1 Pf. Shafarevich Pp. 74–75 Cor. If X ⊂ Pn is an irred qp var, F a form not vanishing identically on X, then every irred comp of XF has codimension 1 Pf. Apply the theorem to X ⊂ Pn Each component of X F has codim 1 Since Z(F ) is closed and nonempty open subsets of X are dense in X, we see that XF = X ∩ Z(F ) = X ∩ Z(F ) Let Z ⊂ XF be an irred comp Then Z is an irred comp of XF dim Z = dim Z = dim X − 1 Example. X = A20, F = x0, XF empty, no contradiction Cor. Let X ⊂ Pn be an irred qp var of dim n and let F1, . . . , Fr be forms. The every comp of X∩Z(F1, . . . , Fr ) has dim at least n − r Pf. Induction on r 16 Prop. Let X, Y ⊂ PN be irred qp vars of dims m, n. Then each comp of X ∩ Y has dim at least m + n − N . If X and Y are proj, and m + n ≥ N , then X ∩ Y is nonempty. Pf. Via the diagonal embedding X ,→ ∆ ⊂ X × X, identify X ∩ Y with (X × Y ) ∩ ∆. dim(X × Y ) = n + m Since dimension is local, may assume X, Y ⊂ AN ∆ ⊂ AN × AN is a (complete) intersection of N hypersurfaces xi = yi, i = 1, . . . , N If X, Y are proj, construct the affine cones C(X) and C(Y ) in AN +1 The affine cone C(X) is defined as Z(I(X)), where I(X) is the homogeneous ideal of X Z(I(X)) ⊂ AN +1 is understood as the zero set of the polynomials in (x0, . . . , xN ), now identified as the ordinary (non-homogeneous) coords of AN +1 C(X) and C(Y ) are qp vars in PN +1 of dims m+1, n+1 dim(C(X) ∩ C(Y )) ≥ (m + 1) + (n + 1) − (N + 1) = m+n−N +1 If m + n ≥ N , then dim C(X) ∩ C(Y ) ≥ 1 C(X) ∩ C(Y ) contains a nonzero point of AN +1 17 The coordinates of this point are then the homog coords of a point of X ∩ Y 18 We now turn to the dimension of the fibers of a reg map and apply it to study lines on surfaces Let X, Y be qp vars and f : X → Y regular Def. The fiber of f over y ∈ Y is f −1(y) Example. Consider X ⊂ P2 × P2 with equation x0y0 + x1y1 + x2y2 = 0. with projection π : X → P2, π(x, y) = y The fiber π 1(y), y = (y0, y1, y2), is identified with line Ly = Z(x0y0 + x1y1 + x2y2 = 0) ⊂ P2 (y0, y1, y2) in the above equation are fixed elts of k Thus, y ∈ P2 (the dual P2) parametrizes lines in P2 X describes the family of all lines in P2 Since X ⊂ P2 × P2 is a hyperplane section in the Segre embedding P2 ×P2 ,→ P8, we have dim X = 2+2−1 = 3. 19 Theorem. Let f : X → Y be a reg surjective map, dim X = n, dim Y = m. Then n ≥ m and • dim F ≥ n − m for any component F of any fiber of f • There is a nonempty open subset U ⊂ Y such that the fiber over any y ∈ U has dimension exactly n − m. In the previous example, the fibers of π : X → P2 are lines, of dimension 1 = 3 − 2 Continuing with the example above, we have U = Y = P2 and all fibers are irreducible Cor. The sets Yk = {y ∈ Y | dim f −1(y) ≥ k} are closed in Y Pf of Cor. The theorem says Yn−m = Y and there is a proper closed set Y 0 = Y − U ⊂ Y such that Yk ⊂ Y 0 if k >n−m Let Zi be the irred comps of Y 0, so closed in Y Consider the restriction fi : f −1(Zi) → Zi The result follows by induction on dim Y 20 Thm. Let f : X → Y reg surjective map, X, Y projective. Suppose Y is irred and the fibers of f are all of the same dimension and irreducible. Then X is irreducible. The proofs of the last two theorems will be deferred until after powerful applications are given Now we show how these results easily imply existence and non-existence of lines on surfaces! Thm. For any m > 3, there exist surfaces in P3 of degree m which do not contain any lines. Moreover, such surfaces correspond to an open subset of PN . Thm. Every cubic surface contains at least one line. There is a nonempty open subset U ⊂ P19 parametrizing cubic surfaces with only finitely many lines. Remark. In fact, U contains the open set Usm parametrizing smooth cubic surfaces, and any smooth cubic surface contains exactly 27 lines. 21 Let S ⊂ P3 be a surface of degree m. We have seen how to identify S with a point of PN , N = m+3 −1 3 Let L ⊂ P3 be a line Can identify L with a point of G = G(2, 4): P1s contained in P3 are in 1-1 correspondence with 2dimensional subspaces of k 4 22 Let I be the incidence correspondence N I = (L, S) ∈ G × P | L ⊂ S ⊂ G × PN Lemma. I is a closed subset of G × PN , hence is a projective variety Pf. Identify L = span(v, w) with the projectivization of P v ∧ w = i<j pij ei ∧ ej in terms of Pl¨ucker coords pij Put pij = −pji for i > j and put pii = 0 Equivalently, pij = xiyj − xj yi, where v = (x0, . . . , x3) and w = (y0, . . . , y3) Claim. L consists of all points (z0, . . . , z3) with zj = P i ai pij as the ai range over k For example, consider z = yk v − xk w ∈ L Then zj = yk xj − xk yj = −pkj which is of claimed form, ak = −1 and ai = 0 for i 6= k The general case is obtained from these by linear combinations 23 Now let F be a form of deg m defining S P Then L ⊂ S iff F ( i aipij ) = 0 as in identity in ai This is a homog poly eqn in the ai whose coeffs are homog polys in the pij and linear in the coeffs of F (the homog coords on PN ) I is obtained by setting all of these coeffs to zero Resulting equations are bihomogeneous in the Pl¨ucker coords and the homog coords of PN , so I closed Lemma. I is irreducible of dimension N + 3 − m Pf. Let π : I → G be the projection. π is surjective. The fiber over L ∈ G is the set of hypersurfaces of deg m which containing L Letting L = Z(`1, `2) with `1, `2 linear forms, we let G, H be any homog forms of deg m − 1 Then F = `1G + `2H defines a hyp of deg m containing L 24 Claim. The fibers of π are all projective spaces of dim N −m−1 The lemma follows immediately from this claim Since the F as above form a vector space, the fibers are projective space and we just have to compute the dimension G, H are chosen from a vector space of dim m+2 3 But we have overcounted, since forms `1`2K with deg K = m − 2 can be written in either the form `1G or `2H So the set of all F forms a vector space of dimension m + 2 m + 1 m + 3 −m−1 = N −m = − 2 3 3 3 and the claim follows Thm. For any m > 3, there exist surfaces in P3 of degree m which do not contain any lines. Moreover, such surfaces correspond to an open subset of PN . Pf. Consider the projection φ : I → PN Since I is projective, the image of φ is closed and has dim ≤ dim I = N + 3 − m < N φ is not surjective 25 Thm. Every cubic surface contains at least one line. There is a nonempty open subset U ⊂ P19 parametrizing cubic surfaces with only finitely many lines. Pf. Repeating the above argument, we first have to show that φ is surj For m = 3 have dim I = N If φ not surj, the image has dim < N , so all fibers have positive dim Get a contradication by exhibiting just one zero-dimensional fiber Ex: x1x2x3 = x30 contains exactly 3 lines No lines intersecting affine piece A30: x1x2x3 = 1 has no nontrivial linear parametric solns xi = ai t + bi Hyperplane at infinity x0 = 0 has 3 lines x0 = xi = 0, i = 1, 2, 3 So φ is surjective There is a U ⊂ PN = P19 over which all fibers are zero-dimensional, hence finite 26 We now return to prove the results that the analysis of lines on surfaces was based on Theorem. Let f : X → Y be a reg surjective map, dim X = n, dim Y = m. Then n ≥ m and 1. dim F ≥ n − m for any component F of any fiber of f 2. There is a nonempty open subset U ⊂ Y such that the fiber over any y ∈ U has dimension exactly n − m. Pf. (i) This statement is local WLOG may assume Y affine, Y ⊂ AN We showed earlier in affine space that we can take a succession of hyperplane sections, each decreasing the dimension by 1 Starting with Y , have g1 . . . , gm ∈ k[Y ] s.t. dim Z(g1, . . . , gm) = m−m=0 Can further choose each gi with gi(y) = 0 Z(g1, . . . , gm) is a finite set containing y By removing some points from Y , can assume Z(g1, . . . , gm) = {y} Then f −1(y) is defined by m eqns f ∗(g1), . . . , f ∗(gm) ∈ k[X] So each component of f −1(y) has dim ≥ n − m 27 (ii) (Sketch) First, reduce to the affine case: Take W ⊂ Y affine, then choose V ⊂ f −1(W ) ⊂ X affine Consider f : V → W By surjectivity, k[W ] → k[V ] is injective View k[W ] ⊂ k[V ], k(W ) ⊂ k(V ) Tr degk(W )(k(V )) = n − m Choose generators vi for k[V ] with v1, . . . , vn−m alg indep over k(W ) Remaining generators vi = vn−m+1, . . . are all alg dependent Satisfy polynomial eqns Fi(v1, . . . , vn−m, vi) = 0 with coefficients in k[W ]. In particular, have leading coefficients hi ∈ k[W ] Show U = W − Z((hn−m+1, . . .)) satisfies the conclusion 28 Thm. Let f : X → Y reg surjective map, X, Y projective. Suppose Y is irred and the fibers of f are all of the same dimension and irreducible. Then X is irreducible. Pf. Express X = ∪Xi as the union of its irred comps Xi is projective, so f (Xi) is closed and Y = ∪f (Xi) Y is irred, so Y = f (Xi) for some i For each i, let fi : Xi → Y be the restriction of f If f (Xi) = Y then fi still surj By theorem just proven have nonempty Ui ⊂ Y and integer ni s.t. dim fi−1(y) = ni all y ∈ Ui If f (Xi) 6= Y put Ui = Y − f (Xi) Pick y ∈ ∩Ui f −1(y) is irred by assumption So f −1(y) contained in some Xi Then f −1(y) = fi−1(y) for that value of i n = ni Now fi−1(y) ⊂ f −1(y) for any y ∈ Y dim fi−1(y) ≥ ni = n by previous thm dim f −1(y) = n Therefore fi−1(y) = f −1(y) So X = Xi, and X is irreducible 29

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