Math 145 Solution 5 Problem 1

Home , Affine space, Projective variety, Quadric

Math 145 Solution 5 Problem 1. (Products of affine varieties.) Let X ⊂ An and Y ⊂ Am be affine algebraic sets. (i) Show that X × Y ⊂ An+m is also an affine algebraic set. Answer: (i) Let X ⊂ An and Y ⊂ Am be affine algebraic sets such that

X = Z(f1, . . . , fr),Y = Z(g1, . . . , gs) where fi ∈ k[X1,X2,...,Xn] and gj ∈ k[Y1,Y2,...,Ym] .

We regard fi and gj as polynomials in the variables X1,...,Xn,Y1,...,Ym. We claim that n+m X × Y = Z(f1, . . . , fr, g1, . . . , gs) ⊂ A . This will show that X × Y is an afﬁne algebraic set. To prove the claim, observe that

(x1, . . . , xn, y1, . . . , ym) ∈ Z(f1, . . . , fr, g1, . . . , gs)

⇔ fi(x1, . . . , xn) = 0 and gj(y1, . . . , ym) = 0 for all i, j

⇔ (x1, . . . , xn) ∈ X and (y1, . . . , ym) ∈ Y ⇔ (x1, . . . , xn, y1, . . . , ym) ∈ X × Y. Problem 2. (Intersections with hypersurfaces.) (i) Let H ⊂ Pn be a linear hyperplane in Pn (e.g. H is cut out by one linear homogeneous equation in Pn). Show that the quasiprojective algebraic set Pn\H is isomorphic to An. (ii) Show that if X is a degree d hypersurface in Pn then Pn\X admits a nonconstant morphism to an afﬁne space. (In fact, its not much harder to show that Pn\X is isomorphic to an afﬁne variety). (ii) Conclude that if X is a hypersurface in Pn and Y ⊂ Pn is a projective variety which is not a point, then the intersection of X and Y is non-emtpy. (iv) If X is a degree d hypersurface in Pn and L is a line in Pn not contained in X, show that L and X intersect in at most d points. In fact, they intersect in exactly d points counted with multiplicity. 0 0 n Answer: (i) Write temporarily X0,...,Xn for the coordinates on P . Assume that the hyperplane H is cut out by the linear equation ( n ) X 0 H = aiXi = 0 . i=0

Without loss of generality, we may assume an 6= 0. Deﬁne n X 0 0 Xn = aiXi,Xi = Xi for 0 ≤ i ≤ n − 1. i=0 n Then it’s easy to see that Xi are new coordinates on P . Indeed, the change of coordinates matrix  1 0 ... 0 0   0 1 ... 0 0  C =    0 0 ... 1 0  a0 a1 . . . an−1 an

is invertible having determinant equal to an 6= 0. In the new coordinates, H is deﬁned by

Xn = 0 . 1 2

The map n n X0 Xn−1 π : P \H → A , [X0 : ... : Xn] → ,..., Xn Xn n is well-deﬁned because Xn 6= 0 on P \H. Let

n n φ : A → P \H, (Y1,...,Yn) → [Y1 : ... : Yn : 1] Clearly π and φ are inverses, and therefore π is an isomorphism. (ii) Suppose X = Z(f) where

X i0 in f = ai0i1...in X0 ··· Xn = 0 . i0+···+in=d Let v be the d-fold Veronese embedding n N v : P → P . It is easy to check that v is injective (this follows from the deﬁnitions, we need to consider quotients of the monomials od degree d.) N Let Y0,...,YN be the coordinates on P . For points in the image of the Veronese embedding, Y0,...,YN is a list of all degree d monomials

i0 in X0 ··· Xn ,

in some order. To fix notation, we write Yi0...in for the coordinate corresponding to the i0 in monomial X0 ··· Xn . 0 N Let H be the hyperplane in P defined by the linear equation with coefficients ai0...in e.g. X ai0...in Yi0...in = 0. Clearly, v(X) ⊂ H0. Then n N 0 N v : P \ X → P \ H =∼ A , where in the last equation we used (ii). Therefore, we obtained a non-constant (in fact injective) morphism from Pn \ X and AN . (iii) Suppose the intersection of X and Y is empty, then n Y ⊂ P \X. Let n N v : P \X → A

be the morphism described in (ii) and πi be the projection map from N 1 πi : A → A to the i-th coordinate. Then n 1 πi ◦ v : P \X → A . 1 Hence πi ◦ v restricts to a morphism from Y → A . Because Y is a projective variety, πi ◦ φ must be a constant. This is true for every i, therefore v is a constant morphism on Y . But since v is injective, it follows that Y is a point. 3

(iv) After a change of coordinates, we can assume that L is cut out by the equations

X2 = ... = Xn = 0 . Indeed, pick any two points p and q on L, change coordinates such that p and q become [1 : 0 : ... : 0] and [0 : 1 : ... : 0]. It is clear that the line L that p and q determine is given by X2 = ... = Xn = 0. Recall that X i0 in f = ai0i1...in X0 ··· Xn . i0+···+in=d Let X i0 i1 F (X0,X1) = f(X0,X1, 0,..., 0) = ai0i10...0X0 X1 . i0+i1=d Assuming x = [x0 : x1 : ··· : xn] ∈ L ∩ X

then x2 = ... = xn = 0 and f(x) = 0. Therefore

X i0 i1 F (x) = ai0i10...0x0 x1 = 0 . i0+i1=d Note that if F =∼ 0 it follows that L ⊂ X, which is impossible. Therefore F is a non-zero homogenous polynomial of degree at most d. If ad0...0 = 0, then F = x0G where G has degree d − 1. It follows by induction that G has at most d − 1 roots and thus F has at most d roots, namely the zeros of F and [0 : 1 : ... : 0]). n Assume now ad0...0 6= 0. If x1 = 0, this implies x0 = 0, which is not in P . Therefore we can assume x1 = 1 and x0 is a solution of

X i0 ai0i10...0X0 = 0 i0+i1=d This is a nonzero degree d polynomial, with at most d solutions. Therefore there are at most d points in the intersection L ∩ X. Problem 3. (Rational varieties.) Two projective varieties X and Y are birational if there are rational maps f : X / Y , g : Y / X. which are rational inverses to each other. We say that X is rational if X is birational to Pn for some n. (i) Show that a birational isomorphism f : X / Y induces an isomorphism of the ﬁelds of rational functions f ? : K(Y ) → K(X). The converse is also true but you don’t have to prove it.

(ii) Show that Pn × Pm is rational, by constructing an explicit birational isomorphism with Pn+m. Show that if X and Y are rational, then X × Y is rational.

Answer: (i) For h ∈ K(Y ), deﬁne f ?(h) = h ◦ f. This is a rational function on X. This fact is explained for instance in the text, section 4.10. In the terminology used there, f is a dominant map. In short, the idea is as follows. We have seen that h is deﬁned on an open set Dom(h) ⊂ Y . By assumption, f establishes 4

an isomorphism between nonempty open sets U and V in X and Y . Then f ?h is defined on the intersection f −1(V ∩ Dom(h)). To show this intersection is non-empty, we need to argue that the nonempty open sets V and Dom(h) intersect. But if they do not, the proper closed sets Y \ V and Y \ Dom(h) cover Y contradicting the fact that Y is irreducible. Therefore f ? : K(Y ) → K(X) is a well defined field homomorphism. Similarly, the birational inverse morphism g : Y / X induces a field homomor- ? ? ? phism g : K(X) → K(Y ). We claim that f and g are inverses. Indeed, for h1 ∈ K(Y ), h2 ∈ K(X), ∗ ∗ ∗ f g (h2) = f (h2 ◦ g) = h2 ◦ g ◦ f = h2 ◦ id = h2 , ∗ ∗ ∗ g f (h1) = g (h1 ◦ f) = h1 ◦ f ◦ g = h1 ◦ id = h1 . Therefore K(X) =∼ K(Y ). n m (ii) Let U ⊂ P be the open set where the coordinate x0 6= 0. Similarly, let V ⊂ P be the n+m open set where the coordinate y0 6= 0, and let W be the open set in P where the first coordinate is non-zero Define φ : U × V → Pn+m as x1 x2 xn y1 y2 yn [x0 : x1 : ... : xn] × [y0 : y1 : ... : ym] → 1 : : : ... : : : ,... : . x0 x0 x0 y0 y0 y0 It is easy to check that φ establishes an isomorphism between U ×V and W , with inverse

ψ : W → U × V, [1 : z1 : ... : zn+m] → [1 : z1 : ... : zn] × [1 : zn+1 : ... : zn+m]. Therefore φ and ψ deﬁne birational isomorphisms between Pn × Pm and Pn+m. Finally, if X and Y are birational to Pn andPm, then X × Y is birational to Pn × Pm which in turn is birational to Pn+m. Therefore, X × Y is rational. Problem 4. (Quadrics are rational.) (i) Show that a non-singular irreducible quadric Q in P3 can be written in the form xy = zw after a suitable change of homogeneous coordinates. Combining this result with the Segre embedding, conclude that any quadric in P3 is isomorphic to P1 × P1, hence it is rational. (ii) In general, show that any non-singular irreducible quadric Q ⊂ Pn+1 is birational to Pn.

Answer: (i) A homogeneous quadratic polynomial f(x, y, z, w) can be written as x y  f = [x y z w] A   z  w where A is a symmetric 4 × 4 matrix. Because A is symmetric, it can be diagonalized. Thus we can ﬁnd one set of coordinates x1, y1, z1, w1 such that     λ1 0 0 0 x1  0 λ2 0 0  y1  2 2 2 2 f = [x1 y1 z1 w1]     = λ1x1 + λ2y1 + λ3z1 + λ3w1.  0 0 λ3 0  z1  0 0 0 λ4 w1 5

We claim that λi 6= 0, for all 1 ≤ i ≤ 4. Indeed, if for instance λ1 = 0, then [1 : 0 : 0 : 0] is a singular point of the quadric which is impossible. We can change coordinates again, setting p p p p x2 = λ1x1 + i λ2y1, y2 = λ1x1 − i λ2y1, p p p p z2 = i( λ3z1 + i λ4w1), w2 = i( λ3z1 − i λ4w1). This is a change of coordinates since the matrix of coefﬁcients √ √  √λ1 i √λ2 0 0  λ1 −i λ2 0 0   √ √   0 0 i√λ3 −√ λ4 0 0 i λ3 λ4

is invertible. Indeed, the determinant equals 4λ1λ2λ3λ4 6= 0. Then

f = x2y2 − z2w2 . (ii) Similar to (i), we can assume that the quadric Q is deﬁned by

2 2 2 x0x1 − x2 − x3 − . . . xn = 0.

Pick the point p = [1 : 0 : ... : 0] ∈ Q, and let H be the hyperplane X0 = 0. The line passing through p = [1 : 0 : ... : 0] and q = [x0 : ... : xn] is

[r + sx0 : sx1 : ... : sxn].

This line intersets the hyperplane H when r + sx0 = 0. Therefore the line intersects H at [0 : sx1 : ... : sxn] = [0 : x1 : ... : xn]. This may be undeﬁned when x1 = x2 = ··· = xn = 0, i.e. when q = p. We obtain a morphism

f : Q \{p} → H, [x0 : x1 : ... : xn] → [0 : x1 : ... : xn]. The rational inverse of f is given by 2 2 x2 + ··· xn g : H / Q , [x1 : x2 : ... : xn] = : x1 : x2 : ... : xn . x1

This may be undeﬁned at the points where x1 = 0. Since f has a rational inverse, Q is birational to H.

Problem 5. (Quadrics and curves.) (i) Show that two quadrics in P3 intersect in an elliptic curve. More precisely, consider the following two quadrics in P3:

Q1 = Z(xw − yz),Q2 = (yw − (x − z)(x − λz)). Show that the intersection of the two quadrics is in bijective correspondence to the elliptic curve Eλ. (ii) Show that the twisted cubic in P3 is the intersections of the three quadrics 2 2 Q1 = Z(xz − y ),Q2 = Z(xt − yz),Q3 = Z(yt − z ). Show that any two of these quadrics will not intersect in the twisted cubic. 6

Answer: (i) Suppose p = [x : y : z : w] ∈ Q1 ∩ Q2, then xw = yz, yw = (x − z)(x − λz). Multiplying the ﬁrst equation by y and the second by x and substracting, we have y2z = x(x − z)(x − λz). We deﬁne f : Q1 ∩ Q2 → Eλ, [x : y : z : w] → [x : y : z] .

Note that f is not well-defined at [0 : 0 : 0 : 1] ∈ Q1 ∩ Q2. In this case, we set f([0 : 0 : 0 : 1]) = [0 : 0 : 1]. We claim that f is bijective and its inverse is given by yz g : E → Q ∩ Q , [x : y : z] → [x : y : z : ]. λ 1 2 x Note that g is undefined at [0 : 0 : 1] and [0 : 1 : 0]. Since [0 : 1 : 0] = f([0 : 1 : 0 : 0]), and [0 : 0 : 1] = f([0 : 0 : 0 : 1]), we set g([0 : 1 : 0]) = [0 : 1 : 0 : 0], g([0 : 0 : 1]) = [0 : 0 : 0 : 1]. It is straightforward to check that f and g are inverses. It is more delicate to check that f and g are actually morphisms, and we will not pursue this here in detail. If you wish to prove this fact, then you need to define ( [x : y : z] if (x, y, z) 6= (0, 0, 0) f([x : y : z : w]) = . [w(x − z)(x − λz): z(y − w)(y − λw): w3] if w 6= 0

You will need to make sure that the function is defined everywhere on Q1 ∩ Q2, and well-defined on overlaps. This last statement is equivalent to the equalities of rational functions w(x − z)(x − λz) z(y − w)(y − λw) w3 = = x y z which can be derived from manipulating the equations of the quadric. The argument for the inverse is similar. We define ( [x2 : xy : xz : yz] when [x : y : z] 6= [0 : 0 : 1] or [0 : 1 : 0], g([x : y : z]) = . [xy : y2 : yz :(x − z)(x − λz)] when [x : y : z] 6= [1 : 0 : 1] or [1 : 0 : λ]

Since on Eλ: y2z = x(x − z)(x − λz), it follows that [x2 : xy : xz : yz] = [xy : y2 : yz :(x − z)(x − λz)] on overlaps (indeed, the coordinates are proportional to each other with the proportion- ality factor y/x). Therefore, g is a well-deﬁned morphism. (ii) The twisted cubic is given by [x : y : z : t] = [a3 : a2b : ab2 : b3] for some [a : b] ∈ P1. Clearly, points of these type satisfy the equations xz = y2, xt = yz, yt = z2,

so the twisted cubic is contained in the intersection Q1 ∩ Q2 ∩ Q3. 7

Conversely, given a point p = [x : y : z : t] in the intersection of the three quadrics Q1,Q2,Q3, set [a : b] = [x : y] if x 6= 0. Otherwise, if x = 0, then y = z = 0 and set [a : b] = [0 : 1]. We claim [x : y : z : t] = [a3 : a2b : ab2 : b3] e.g. p lies on the twisted cubic. This is clear for the case x = 0. If x 6= 0, then a 6= 0, and from the ﬁrst and third equations we have b y = x · . a Using the ﬁrst and second equations, we solve b2 b3 z = x · , t = x · . a2 a3 Then, b b2 b3 [x : y : z : t] = x : x : x : x = [a3 : a2b : ab2 : b3]. a a2 a3

Finally, note that [0 : 0 : 1 : 0] ∈ Q1 ∩ Q2 but not in Q3; [0 : 1 : 0 : 0] ∈ Q2 ∩ Q3 but not in Q1; [1 : 0 : 0 : 1] ∈ Q3 ∩ Q2 but not in Q1. So any of the two quadrics can’t generate the twisted cubic. 3 Problem 6. (Introduction to moduli theory.) Show that for any 3 lines L1,L2,L3 in P , there is a quadric Q ⊂ P3 containing all three of them. (i) First, observe that any homogeneous degree 2 polynomial in 4 variables has 10 coefficients. These coefficients can be regarded as a point in the projective space P9. Show that this point only depends on the quadric Q and not on the polynomial defining it. Let us denote this point by pQ. Show that any point p ∈ P9 determines a quadric in P3. (ii) Consider a line L ⊂ P3. Show that there is a codimension 3 projective linear subpace 9 HL ⊂ P such that L ⊂ Q iff and only if pQ ∈ HL. (iii) Show that any three codimension 3 projective linear subspaces of P9 intersect. In particular, show that

HL1 ∩ HL2 ∩ HL3 6= ∅,

and conclude that L1,L2,L3 are contained in a quadric Q.

(iv) Explain (briefly) that if L1,L2,L3 are disjoint lines, then Q can be assumed to be irreducible. Answer: (i) Consider a homogeneous degree 3 polynomial 2 2 F (X0,X1,X2,X3) = a0X0 + a1X0X1 + ··· + a10X3 . 4 This has 4 square terms and 2 = 6 mixed terms, i.e. 10 terms in total. We associate the quadric Q := Z(F ) the point point 9 pQ = [a0 : a1 : ... : a10] ∈ P . Note that this association is independent of the defining equation of the quadric. Indeed, if F and G define the same quadric, then by the Nullstellensatz it follows that G = cF for some constant c 6= 0. But then the point pQ does not change, since the coefficients 8

ai are considered projectively. Finally, this association is clearly bijective, since for any 9 pQ ∈ P we can recover the equation of the quadric Q (up to scalars). (ii) Assume that the line L is given by X0 = X1 = 0. Then ∼ L ⊂ Q = F (0, 0,X2,X3) = 0. Since 2 2 F (0, 0,X2,X3) = a8X2 + a9X2X3 + a10X3 , we obtain a8 = a9 = a10 = 0. 9 These equations deﬁne a codimension 3 projective space HL ⊂ P such that

L ⊂ Q if and only if pQ ∈ HL. If L0 is any line and Q0 is a quadric, we can change coordinates linearly so that L0 be- 0 comes the line L: X0 = X1 = 0. After the coordinate change, Q is mapped to another quadric Q and the coefﬁcients of Q are linear combinations of coefﬁcients of Q0. From above L ⊂ Q if and only if pQ ∈ HL. Note that L0 ⊂ Q0 if and only if L ⊂ Q. 9 Therefore there exists a codimension 3 projective space HL0 ⊂ P , the transformation of HL via the change of coordinates, such that 0 0 L ⊂ Q if and only if pQ0 ∈ HL0 .

(iii) The codimension 3 subspaces HL correspond to 7 dimensional linear subspaces AL ⊂ A10 cut out by the same 3 linear equations. Now,

AL1 ∩ AL2 ∩ AL3 is described by 9 linear equations in A10. Such a system of equations must have a non- trivial solution p. This solution p considered projectively satisﬁes

p ∈ HL1 ∩ HL2 ∩ HL3 .

Now, p determines a quadric Q with p = pQ. It follows by (iii) that

L1 ∩ L2 ∩ L3 ⊂ Q.

(iv) If the quardric Q is reducible, then Q is the union of two hyperplanes H1 and H2. If L1, L2 and L3 are 3 disjoint lines and

L1 ∪ L2 ∪ L3 ⊂ Q, ∼ 2 ∼ 2 there should be two lines in the same hyperplane H1 = P or H2 = P . But we have seen on the midterm that two lines on P2 always intersect in 1 point. This is a contradiction. Therefore we can assume Q is irreducible.