div(f ) = D and deg(d) = deg(f ) = d i deg(f i ) (compare this with the definitions for smooth curves). Let:

Transcription

1 Algebraic Curves/Fall 015 Aaron Bertram 4. Projective Plane Curves are hypersurfaces in the plane CP. When nonsingular, they are Riemann surfaces, but we will also consider plane curves with singularities. We will prove Bézout s Theorem and Noether s Theorem together with a standard classical application. This treatment follows Fulton s Algebraic Curves Chapter 5. A hyperplane L CP is a projective line, and every cone in CP is a union of lines through a vertex point p CP. Given finitely many points p 1,..., p n CP, there is a line L 0 that contains none of the points and lines L i that contain the point p i but none of the others. These are simple results that you should check for yourself. C = V (F ) CP is an irreducible curve and: m D = d i C i (for curves C i = V (F i ) and d i 0) i=1 is an effective divisor on CP, with associated homogeneous polynomial F = F d i i that is well-defined up to a scalar multiple. We will write: div(f ) = D and deg(d) = deg(f ) = d i deg(f i ) (compare this with the definitions for smooth curves). Let: P(d) := P(C[x 0, x 1, x ] d ) = CP d(d+3) be the projective space associated to the vector space C[x 0, x 1, x ] d. The last equality is explained by: ( ) d + d(d + 3) dim(c[x 0, x 1, x ] d ) = = + 1 and via the div operation (and Theorem 3.1), we may interpret: P(d) = {effective divisors of degree d on CP } Example 4.1. (Degree divisors) Every conic divisor is projectively equivalent to one of the following: div(x 0 + x 1 + x ) = C, an irreducible smooth conic. div(x 0 + x 1) = div(x 0 + ix 1 ) + div(x 0 ix 1 ) = L 1 + L div(x 0) = div(x 0 ) = L, a double line. 1

2 Exercise 4.1. Every degree 3 divisor on CP is equivalent to one of: (a) An elliptic curve E λ = div(x 0 x x 1 (x 1 x 0 )(x 1 λx 0 )), λ 0, 1. One of the following two singular irreducible curves: (b) E 0 = div(x 0 x x 1(x 1 x 0 )) (the nodal cubic) or (c) E = div(x 0 x x 3 1) (the cuspidal cubic) (d) Or else a reducible divisor L + D where D is a conic divisor. The multiplicity of an effective divisor D at p: mult p (D) is the degree of the polynomial defining the tangent cone to D at p. In projective coordinates chosen so that p 0 = (1 : 0 : 0), mult p0 (D) m if and only if the associated polynomial of degree d: F = c I x i 0 0 x i 1 1 x i I =d has the property that c I = 0 for all I = (i 0, i 1, i ) satisfying i 0 > d m. Definition 4.1. Given a degree d > 0 and integers m i 0 at p i CP, P(d; m 1 p 1,..., m r p r ) P(d) is the linear series of divisors D P(d) with each mult pi (D) m i. Remark. The linear series involving a single point P(d; mp) P(d) are projective linear subspaces of dimension: d(d + 3) (m + 1)m (or else empty) since as above, the linear series P(d; mp 0 ) is defined by: c I = 0 for I = (d, 0, 0), (d 1, 1, 0), (d 1, 0, 1),..., (d m + 1,, ) and changing coordinates in the plane induces a (projective) linear transformation on each of the projective spaces P(d), so the result holds for an arbitrary point p CP. Corollary 4.1. By definition: P(d; m 1 p 1,..., m r p r ) = P(d; m 1 p 1 ) P(d; m r p r ) and so the linear series is either a projective linear space satisfying: d(d + 3) r (m i + 1)m i dim(p(d; m 1 p 1,..., m n p n )) which is the expected dimension or else it is empty, which can only occur when the expected dimension is negative. i=1

3 Example 4.. P(1) = CP is the (dual) projective plane of lines. (a) dim(p(1; p)) = 1 and P(1; p) is empty. (b) P(1; p, q) is the unique line through p and q. (c) P(1; p, q, r) is empty unless the three points are collinear. Thus the dimension of a linear series may depend upon the location of the points. The problem of determining exact dimensions of linear series for a particular configuration of points is very subtle in general. Exercise 4.. P() = CP 5 is the projective space of conic divisors. (a) Show that dim(p(; p, q, r)) = for all points p, q, r. (b) Show that dim(p(; p 1,..., p 5 )) = 0, i.e. there is a unique conic through the five points unless four of the points are collinear. (c) Show that P(; p, q) is a single point for all pairs p and q. Remark. (c) is interesting because the expected dimension is 1, and yet for all pairs p, q CP, the linear series is not empty. Proposition 4.1. Fix m 1,..., m r 0. Then for all sufficiently large d, d(d + 3) r (m i + 1)m i dim(p(d; m 1 p 1,..., m r p r )) = is as expected for all collections of (distinct) points p 1,..., p r CP. Proof. First consider the case m i = 1 for all i. It suffices to show: i=1 P(d) P(d; p 1 ) P(d; p 1,..., p r ) are strict inclusions when d is large, since this forces the linear series to drop in dimension with the introduction of each additional point. Recall the lines L 0 (containing none of the points) and L i (containing one p i but none of the others) from the top of this section. Then: (d i)l 0 + L L i P(d; p 1,..., p i ) P(d; p 1,..., p i+1 ) exhibits the desired proper inclusions for each i = 1,..., r 1, provided that d i 0. Thus, d r 1 is sufficiently large for this argument. We reason similarly in the general case. We need to show: P(d) P(d; m 1 p 1 ) P(d; m 1 p 1,..., m r p r ) each has the expected codimension in the one before. This can be assured by finding projective linear subspaces P(V i ) P(d; m 1 p 1,..., m i p i ) P(d) 3

4 4 with the property that the intersection P(V i ) P(d; m i+1 p i+1 ) P(V i ) has the expected codimension. We do this by letting: P(V i ) = {m 1 L m i L i + D D P(d m 1... m i )} where L j are the fixed lines. Then: P(V i ) P(d; m i+1 p i+1 ) = {m 1 L m i L i + D D P(d m 1... m i ; m i+1 p i+1 )} reduces to the case of a single point linear series for divisors of degree d m 1... m i, which has the expected codimension provided that d m j m i+1 1. For this argument, d m m r 1 is sufficiently large, and is, in fact, optimal for the result. Next, we turn to Bézout s Theorem, which is useful to think of as a comparison between globally and locally defined data coming from a pair of distinct irreducible plane curves C, C CP associated to prime homogeneous polynomials A, A C[x 0, x 1, x ] of degrees a, a. Global Data. The Hilbert function of the projective plane CP is: ( d+ ) for all d 0 h CP (d) = dim C[x 0, x 1, x ] d = 0 for all d < 0 which is a quadratic polynomial in d for all d 0. The Hilbert function of the curve C is the eventually linear function: ( d+ ) ( d+ a ) ( = ad + 1 a 1 ) for all d a ) for all 0 d < a h C (d) = dim(r C ) d = ( d+ 0 for all d < 0 where R C = C[x 0, x 1, x ]/ A is the homogeneous coordinate ring of C. The Hilbert function of C C is the eventually constant function: h C (d) h C (d a ) = aa for all d a + a h C C (d) = dim(r C C ) d = other stuff for d < a + a where R C C = C[x 0, x 1, x ]/ A, A is the homogeneous coordinate ring of the intersection C C of the two curves. Observation. The quotient ring R C is a graded integral domain, and R C C is graded, but usually not a domain. The notation is a little misleading, since the set C C is not enough information, in general, to determine R C C (the scheme structure on C C is required).

5 In 5 we will prove that Hilbert functions are eventually polynomial functions in much greater generality, and the degree will be one of the ways of defining the dimension of a projective variety. Remark. It follows already that the cardinality of the set C C satisfies: C C aa hence in particular that C C is finite. This is explained by Proposition 4.1, since each point p C C supports a homogeneous polynomial F p P(d; C C p) P(d; C C ) that by definition vanishes at all points of C C except p whenever d C D 1. These are evidently linearly independent vectors in (R C C ) d, which has dimension aa. Local Data. With suitable coordinates, x 0 does not divide A or A and the intersection C C is contained in the set U 0 = CP V (x 0 ) which may be identified with C with coordinates y i = x i /x 0. Let: α(y 1, y ) = A/x a 0 and α (y 1, y ) = A/x a 0 and consider the affine coordinate rings: C[y 1, y ]/ α and C[y 1, y ]/ α, α associated to the affine curve and intersection of affine curves: C aff = C U 0 = {(q 1, q ) C α(q 1, q ) = 0} and C aff (C ) aff = C C respectively. This affine data has the advantage that elements of the affine coordinate rings are functions on C aff and C C respectively. We pass to local data via the fields of rational functions, which are the same whether viewed projectively or affinely: { } { } F K(CP ) = G F, G C[x 0, x 1, x ] d f = g f, g C[y 1, y ] = K(C ) (with the usual equivalence relation on fractions) via the isomorphism: F G F/xd 0 and f G/x d 0 g fxn 0 for sufficiently large n gx n 0 Notice in particular that α (but not A) is an element of K(CP ). Definition 4.. To each point p CP, define the ring: { } F O CP,p = G G(p) 0 K(CP ) of local functions defined at p, which has a unique maximal ideal: m p = {φ O CP,p φ(p) = 0} O CP,p K(CP ) of local functions vanishing at p. 5

6 6 Remark. These may alternatively be defined with affine coordinates: m p O C,p = O CP,p K(C ) whenever p U 0 = C, and in these coordinates it is clear that m p is generated by two elements. For example, when p = (1 : 0 : 0) CP, then p = (0, 0) C, and m p is generated by the coordinates y 1, y. Similarly, we may define the fields of rational functions on C: { } { } F K(C) = G F, G (R C) d f = g f, g C[y 1, y ]/ α = K(C aff ) with local ring and maximal ideal for all p C given by: { } F m p = {φ φ(p) = 0} O C,p = G G(p) 0 K(C) and if p C aff, then O C,p = O C aff,p. Remark. The coordinate ring R C (or C[y 1, y ]/ α ) of a plane curve C does not usually have unique factorization, in which case there is no natural lowest terms fraction representing a typical φ K(C). Exercise 4.3. (a) Regarding α m p O CP,p for p C, show that: O CP,p/ α = O C,p (b ) Suppose I = f 1,..., f n C[y 1, y ] is an ideal such that the set: V (I) = {p C f i (p) = 0 f i } C is finite. Show that the natural map: C[y 0, y 1 ]/I p V (I) O CP,p/ f 1,..., f n ; f (f,..., f) is an isomorphism onto the product of local rings. Remark. Exercise 4.3 (b) is starred because it is both challenging and extremely important. The local rings should be thought of as germs of functions, and the Exercise shows that an arbitrary choice of germs can be pieced together to come from a single polynomial (mod I). This isn t even obvious when V (I) is a single point! Proposition 4.. (a) If p C is a non-singular point, then m p O C,p is a principal ideal, i.e. O C,p is a discrete valuation ring. (b) For any p C, the dimensions of the quotient vector spaces: dim C m n p/m n+1 p are constant for large n, equal to the multiplicity of C CP at p C. In particular, if p C is a singular point, then O C,p is not a DVR.

7 Proof. For (a), we may choose coordinates that p = (1 : 0 : 0) and the Zariski tangent line to C at p is x 1 = 0, so that: α(y 1, y ) = y 1 f + y g C[y 1, y ]; with f(0, 0) 0 Since y 1, y generate m p O CP,p and y 1 f = α y g and f is a unit in O CP,p, it follows from Exercise 4.3 (a) that y 1 y O C,p and y 1, y generate m p O C,p, so y by itself is a generator of m p O C,p. (b) From the isomorphism: m n p/m n+1 p = ( ) ( O C,p /m n+1 p / OC,p /mp) n it suffices to prove that dim C O C,p /m n p is eventually linear, of the form mult p (C)d + constant. Choose coordinates so that p = (1 : 0 : 0) and let I = y 0, y 1 C[y 0, y 1 ]. Then the natural map: C[y 1, y ]/ α, I n O C,p /m n p = O CP,p/ α, I n is an isomorphism by Exercise 4.3 (a) and (b). Let m = mult p (C). Then multiplication by α maps I n m into I n, and: 0 C[y 0, y 1 ]/I n m α C[y 0, y 1 ]/I n C[y 0, y n ]/ α, I n 0 is an exact sequence of C[y 0, y 1 ]-modules, hence: dim C O C,p /m n p = dim C C[y 0, y 1 ]/I n C[y 0, y 1 ]/I n m for all n m, and it is a quick check to see that: ( ) n + 1 dim C C[y 0, y 1 ]/I n = n = from which it follows that for all n m, ( ) ( ) n + 1 n + 1 m dim C O C,p /m n p = = mn as desired. ( ) m Remark. This is a second occurrence of a Hilbert function. Here it is attached to the local ring O C,p as the (eventually) linear function: dim C O C,p /m n p n=0 Let us return now to our two distinct, irreducible plane curves C C = C aff (C ) aff Definition 4.3. The intersection number at p C C is: I p (C C ) := dim C O CP,p/ α, α 7

8 8 Bézout s Theorem 4.1. The globally defined Hilbert polynomial aa of the graded ring R = C[x 0, x 1, x ]/ A, A is the sum of locally defined intersection numbers: aa = I p (C C ) p C C Proof. In our coordinates so that the line x 0 = 0 does not contain any of the (finitely many!) intersection points of C and C, I p (C C ) = dim C C[y 1, y ]/ α, α p by Exercise 4.3 (b) with α, α defined as earlier, so the proof amounts to showing that the maps R d C[y 1, y ]/ α, α ; F f = F/x d 0 are isomorphisms for sufficiently large d. First we show that the multiplication maps µ x0 : R d R d+1 ; µ x0 (F ) = F x 0 are injective for all d and therefore isomorphisms for all d a + b. To see this, suppose x 0 H = F A + F A for some F, F. Then: 0 = F 0 A 0 + F 0A 0 C[x 1, x ] = C[x 0, x 1, x ]/ x 0 where F 0 = F (0, x 1, x ), etc. But A 0, A 0 C[x 1, x ] are relatively prime (because C C V (x 0 ) = V (A 0 ) V (A 0) = ), so F 0 = EA 0 and F 0 = EA 0 for some E C[x 1, x ]. Let: G = F + EA and G = F EA Then x 0 H = GA + G A and both G and G are divisible by x 0, by construction, so H = (G/x 0 )A + (G /x 0 )A, proving injectivity. Now fix d a + a and F 1,..., F aa C[x 0, x 1, x ] d whose images in R d are a basis. Let f i = F i /x d 0, as usual. Then we need to show: ( ) The images of f 1,..., f aa C[y 1, y ] in C[y 1, y ]/ α, α are a basis Suppose λ i f i = gα + hα C[y 1, y ] for some λ i C. Multiplying by a sufficiently large power of x 0 gives λ i x r 0F i = x s 0GA + x t 0HA for some r, s, t and homogeneous polynomials G, H, which shows that λi x r 0F i = 0 in R d+r and λ i F i = 0 in R d by the injectivity of µ x0. So the f i are linearly independent. Suppose γ C[y 1, y ] and let Γ = x n 0γ C[x 0, x 1, x ] n for n = d + r. Then Γ = λ i x r 0F i + GA + HA since the x r 0F i are a basis for R d+r, and then γ = λ i f i + gα + hα, so γ = λ i f i in C[y 1, y ]/ α, α. This proves surjectivity, and completes the proof of (*) and therefore of Bézout s Theorem.

9 9 Bézout s theorem begs the following: Question. How do we compute the intersection numbers: I p (C C ) = dim C O CP,p/ α, α? This is answered in many cases by: Proposition 4.3. (a) Suppose p C is a non-singular point. Then: I p (C C ) is the order of vanishing of the element α in the DVR O C,p (b) Suppose p C C is an arbitrary intersection point. Then: I p (C C ) mult p (C) mult p (C ) with equality if and only if the tangent cones to C and C at p are transverse, i.e. they do not share a common line. Proof. (a) is straightforward. O CP,p/ α, α = (O CP,p/ α ) / α = O C,p / α by Exercise 4.3 (a), and the dimension of O C,p / α is the order of α (the value of d so that α = m d p) in the DVR O C,p. For (b), assume as usual that p = (1 : 0 : 0) and let m = mult p (C) and m = mult p (C ) and consider the surjective map: q : O CP,p/ α, α O CP,p/ α, α, m m+m p = C[y 1, y ]/ α, α, I m+m where I m+m is the ideal generated by monomials of degree m + m in y 1, y. (The last equality, as usual, comes from Exercise 4.3 (b).) There is an exact sequence: C[y 1, y ]/I m C[y 1, y ]/I m k C[y 1, y ]/I m+m C[y 1, y ]/ α, α, I m+m 0 where k(f, g) = fα gα, from which we conclude the first part of (b): I p (C C ) = dim C O CP,p/ α, α dim C[y 1, y ]/ α, α, I m+m ( ) ( ) ( ) m + m + 1 m + 1 m + 1 = mm with equality if and only if q, k are both injective. We finish the proof with two observations about q and k. (i) k is injective if and only if C, C do not share a tangent line at p. Suppose, on the contrary, that their tangent cones at (0, 0) C share a common tangent line l = 0. Then the Taylor expansions are: α = lα m 1 + α m α a and α = lα m 1 + α m α a and k(α m 1, α m 1 ) = α m 1α α m 1 α exhibits a non-zero Im+m element of the kernel of k.

10 10 Conversely, if (f, g) ker(k) and f = f d +... and g = g e +..., then f d α m g e α m = 0 and d < m, e < m imply that the polynomials α m, α m defining the tangent cones share a common factor. (ii) If C, C do not share a common tangent, then q is injective. First, notice that because I p (C C ) = dim C O CP,p/ α, α is finite, it follows that the map: q N : O CP,p α, α O CP,p/ α, α, m N p = C[y 1, y ]/ α, α, I N is injective for sufficiently large values of N. Thus, it suffices to show: C[y 1, y ]/ α, α, I N C[y 1, y ]/ α, α, I m+m is injective for all N m + m. But this follows from: Exercise 4.4. If α m, α m C[y 0, y 1 ] are homogeneous forms of degrees m, m > 0 with no common factors, then: C[y 0, y 1 ] m 1 α m + C[y 0, y 1 ] m 1 α m = C[y 0, y 1 ] m+m 1 As an example of how to use the Proposition, consider: Exercise 4.5. Find the intersection points of the singular cubic curves: E 0 E (from Exercise 4.1), compute each multiplicity I p (E 0 E ) and check that they sum to 9. Bézout s Theorem extends as stated to divisors: Bézout s Theorem (Divisor Version) Let D P(d), D P(d ) be effective divisors with associated homogeneous polynomials F and F. Then either the supports D and D share an irreducible curve or else their intersection is finite, and: dd = dim C O CP,p/ f, f p D D where f = F/x d i, f = F /x d i for any coordinate x i not vanishing at p. Moreover, dd = dim C R n = dim C (C[x 0, x 1, x ]/ F, F ) n for n d + d the multiplication maps µ xi : R n R n+1 are all injective and all of the maps R n C[y 1, y ]/ f, f O CP,p/ f, f are isomorphisms when n d+d as in the proof of Theorem 4.1 (provided D D U i = C ) Remark. The support of a divisor D = d i C i is the union: D = i d i 0 of the irreducible curves appearing with nonzero coefficient in D. C i

11 Proof. The equality of numbers follows from Theorem 4.1 and: dim C O CP,p/ f, gh = dim C O CP,p/ f, g + dim C O CP,p/ f, h The rest is proved simply by replacing prime polynomials A and A with relatively prime polynomials F and F in all the earlier proofs. Let D = d i C i and D = d jc j. Corollary 4.. Suppose D D consists of dd distinct points. Then every intersection point is a non-singular point of both D and D, i.e. (a) Every intersection point belongs to a unique pair C i and C j. (b) Each intersection is a nonsingular point of C i and C j. (c) Each of the local intersection rings at p C i C j satisfies: f, f = m p O CP (i.e. the intersections of C i and C j are transverse). (d) Each d i = d j = 1 for all i, j (because each C i and C j intersects!) Proof. If any of these conditions were violated, then there would be an intersection point with intersection number > 1. Noether s Theorem. Let F, F C[x 0, x 1, x ] be homogeneous and relatively prime and suppose G C[x 0, x 1, x ] d maps to zero under: R d C[y 1, y ]/ f, f = p V (F ) V (F )O CP,p/ f, f Then G = HF + H F for homogeneous polynomials H, H. Proof. If d n = deg(f )+deg(f ), then the map in question is an isomorphism, and we are done by definition. Otherwise, by injectivity of the maps µ xi, we see that x n d i H = 0 R n, so H = 0 in R d. Corollary 4.3. Suppose D and D intersect in dd points, and D passes through all of them. Then the associated polynomials satisfy: F = HF + H F for some homogeneous H, H Proof. Using Corollary 4.(c), Noether s Theorem applies! As an application, we study the following: Proposition 4.4. Let D, D be cubic divisors on CP intersecting in 9 distinct points p 1,..., p 9. Then: (a) dim P(3; p 1,..., p 9 ) = 1 (b) P(3; p 1,..., p 8 ) = P(3; p 1,..., p 9 ). Recall that the expected dimension of P(3; p 1,..., p n ) = 9 n, so (b) says that p 1,..., p 8 impose independent conditions on cubic divisors. 11

12 1 Proof. (a) is immediate from the Corollary. If D P(3; p 1,..., p 9 ), then the associated cubic polynomial F vanishes at all the points, and F = af + a F for constants a and b. Thus P(3; p 1,..., p 9 ) = P(V ) where V is the two-dimensional vector space with basis F and F. For (b), suppose D P(3; p 1,..., p 8 ) P(3; p 1,..., p 9 ). Let q be the ninth point of the intersection D D (possibly coming from a point with intersection number ). If q p 9, let L be a line through p 9 but not q, and let L D = p 9 + r + s with r, s D. Then: lf = l 1 F + l F by the Corollary for the associated cubic polynomials F, F, F and linear forms l (defining L) and some forms L 1, L since L+D contains all the points p 1,..., p 9 = D D. Evaluating at the points, we see that l (r) = l (s) = 0, so l = l = l 1 (up to scalar multiples), and F = af + a F, which is a contradiction! Example 4.3. (Pascal s Mystic Hexagon) If a hexagon is inscribed in an irreducible conic, then the three sets of opposite sides meet in three collinear points. Proof. Let C be the conic, L 1, L, L 3 be one set of opposite sides and M 1, M, M 3 the other. Then: (L 1 + L + L 3 ) (M 1 + M + M 3 ) = p p 6 + q 1 + q + q 3 where p 1,..., p 6 are the vertices of the hexagon and q 1, q, q 3 are the other points. But a third cubic divisor (L 1 + L + L 3 ) (C + q 1 q ) contains p 1,..., p 6, q 1, q, so it contains q 3 as well, and q 3 C, so it must be on the line q 1 q. Exercise 4.6. (a) Let E CP be a nonsingular cubic curve, choose an arbitrary point 0 E. If L is a line in CP, define: L E = p 1 + p + p 3 := I p (L, E)p as a divisor on E and the sum p 1 p of points p 1, p E by: (i) φ(p 1, p ) = p 3 if L E = p 1 + p + p 3 and: (ii) p 1 p = φ(0, φ(p 1, p )). This is evidently commutative. Show that it is associative with 0 as an additive identity and that it agrees with the addition on the curves E Λ coming from a lattice Λ C from.