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


 Martina Lang
 4 years ago
 Views:
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 welldefined 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 nonsingular 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 nonsingular 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 nonzero 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 nonsingular 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 twodimensional 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.
Exercises for algebraic curves
Exercises for algebraic curves Christophe Ritzenthaler February 18, 2019 1 Exercise Lecture 1 1.1 Exercise Show that V = {(x, y) C 2 s.t. y = sin x} is not an algebraic set. Solutions. Let us assume that
More informationProjective Varieties. Chapter Projective Space and Algebraic Sets
Chapter 1 Projective Varieties 1.1 Projective Space and Algebraic Sets 1.1.1 Definition. Consider A n+1 = A n+1 (k). The set of all lines in A n+1 passing through the origin 0 = (0,..., 0) is called the
More information10. Smooth Varieties. 82 Andreas Gathmann
82 Andreas Gathmann 10. Smooth Varieties Let a be a point on a variety X. In the last chapter we have introduced the tangent cone C a X as a way to study X locally around a (see Construction 9.20). It
More information12. Hilbert Polynomials and Bézout s Theorem
12. Hilbert Polynomials and Bézout s Theorem 95 12. Hilbert Polynomials and Bézout s Theorem After our study of smooth cubic surfaces in the last chapter, let us now come back to the general theory of
More informationMath 6140 Notes. Spring Codimension One Phenomena. Definition: Examples: Properties:
Math 6140 Notes. Spring 2003. 11. Codimension One Phenomena. A property of the points of a variety X holds in codimension one if the locus of points for which the property fails to hold is contained in
More informationMath 418 Algebraic Geometry Notes
Math 418 Algebraic Geometry Notes 1 Affine Schemes Let R be a commutative ring with 1. Definition 1.1. The prime spectrum of R, denoted Spec(R), is the set of prime ideals of the ring R. Spec(R) = {P R
More informationLocal properties of plane algebraic curves
Chapter 7 Local properties of plane algebraic curves Throughout this chapter let K be an algebraically closed field of characteristic zero, and as usual let A (K) be embedded into P (K) by identifying
More informationAlgebraic Geometry (Math 6130)
Algebraic Geometry (Math 6130) Utah/Fall 2016. 2. Projective Varieties. Classically, projective space was obtained by adding points at infinity to n. Here we start with projective space and remove a hyperplane,
More informationEach is equal to CP 1 minus one point, which is the origin of the other: (C =) U 1 = CP 1 the line λ (1, 0) U 0
Algebraic Curves/Fall 2015 Aaron Bertram 1. Introduction. What is a complex curve? (Geometry) It s a Riemann surface, that is, a compact oriented twodimensional real manifold Σ with a complex structure.
More informationComplex Algebraic Geometry: Smooth Curves Aaron Bertram, First Steps Towards Classifying Curves. The RiemannRoch Theorem is a powerful tool
Complex Algebraic Geometry: Smooth Curves Aaron Bertram, 2010 12. First Steps Towards Classifying Curves. The RiemannRoch Theorem is a powerful tool for classifying smooth projective curves, i.e. giving
More information3. The Sheaf of Regular Functions
24 Andreas Gathmann 3. The Sheaf of Regular Functions After having defined affine varieties, our next goal must be to say what kind of maps between them we want to consider as morphisms, i. e. as nice
More informationSESHADRI CONSTANTS ON SURFACES
SESHADRI CONSTANTS ON SURFACES KRISHNA HANUMANTHU 1. PRELIMINARIES By a surface, we mean a projective nonsingular variety of dimension over C. A curve C on a surface X is an effective divisor. The group
More informationπ X : X Y X and π Y : X Y Y
Math 6130 Notes. Fall 2002. 6. Hausdorffness and Compactness. We would like to be able to say that all quasiprojective varieties are Hausdorff and that projective varieties are the only compact varieties.
More informationwhere m is the maximal ideal of O X,p. Note that m/m 2 is a vector space. Suppose that we are given a morphism
8. Smoothness and the Zariski tangent space We want to give an algebraic notion of the tangent space. In differential geometry, tangent vectors are equivalence classes of maps of intervals in R into the
More informationFOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 48
FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 48 RAVI VAKIL CONTENTS 1. A little more about cubic plane curves 1 2. Line bundles of degree 4, and Poncelet s Porism 1 3. Fun counterexamples using elliptic curves
More informationALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ.
ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. ANDREW SALCH 1. Hilbert s Nullstellensatz. The last lecture left off with the claim that, if J k[x 1,..., x n ] is an ideal, then
More information2. Intersection Multiplicities
2. Intersection Multiplicities 11 2. Intersection Multiplicities Let us start our study of curves by introducing the concept of intersection multiplicity, which will be central throughout these notes.
More informationAN EXPOSITION OF THE RIEMANN ROCH THEOREM FOR CURVES
AN EXPOSITION OF THE RIEMANN ROCH THEOREM FOR CURVES DOMINIC L. WYNTER Abstract. We introduce the concepts of divisors on nonsingular irreducible projective algebraic curves, the genus of such a curve,
More informationIntroduction to Arithmetic Geometry Fall 2013 Lecture #23 11/26/2013
18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #23 11/26/2013 As usual, a curve is a smooth projective (geometrically irreducible) variety of dimension one and k is a perfect field. 23.1
More informationElliptic Curves and Public Key Cryptography (3rd VDS Summer School) Discussion/Problem Session I
Elliptic Curves and Public Key Cryptography (3rd VDS Summer School) Discussion/Problem Session I You are expected to at least read through this document before Wednesday s discussion session. Hopefully,
More informationThis is a closed subset of X Y, by Proposition 6.5(b), since it is equal to the inverse image of the diagonal under the regular map:
Math 6130 Notes. Fall 2002. 7. Basic Maps. Recall from 3 that a regular map of affine varieties is the same as a homomorphism of coordinate rings (going the other way). Here, we look at how algebraic properties
More informationIntroduction to Arithmetic Geometry Fall 2013 Lecture #17 11/05/2013
18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #17 11/05/2013 Throughout this lecture k denotes an algebraically closed field. 17.1 Tangent spaces and hypersurfaces For any polynomial f k[x
More informationABSTRACT NONSINGULAR CURVES
ABSTRACT NONSINGULAR CURVES Affine Varieties Notation. Let k be a field, such as the rational numbers Q or the complex numbers C. We call affine nspace the collection A n k of points P = a 1, a,..., a
More informationCourse 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra
Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................
More informationSPACES OF RATIONAL CURVES IN COMPLETE INTERSECTIONS
SPACES OF RATIONAL CURVES IN COMPLETE INTERSECTIONS ROYA BEHESHTI AND N. MOHAN KUMAR Abstract. We prove that the space of smooth rational curves of degree e in a general complete intersection of multidegree
More information11. Dimension. 96 Andreas Gathmann
96 Andreas Gathmann 11. Dimension We have already met several situations in this course in which it seemed to be desirable to have a notion of dimension (of a variety, or more generally of a ring): for
More informationI(p)/I(p) 2 m p /m 2 p
Math 6130 Notes. Fall 2002. 10. Nonsingular Varieties. In 9 we produced a canonical normalization map Φ : X Y given a variety Y and a finite field extension C(Y ) K. If we forget about Y and only consider
More informationAlgebraic Varieties. Notes by Mateusz Micha lek for the lecture on April 17, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra
Algebraic Varieties Notes by Mateusz Micha lek for the lecture on April 17, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra Algebraic varieties represent solutions of a system of polynomial
More informationMath 203A, Solution Set 6.
Math 203A, Solution Set 6. Problem 1. (Finite maps.) Let f 0,..., f m be homogeneous polynomials of degree d > 0 without common zeros on X P n. Show that gives a finite morphism onto its image. f : X P
More information10. Noether Normalization and Hilbert s Nullstellensatz
10. Noether Normalization and Hilbert s Nullstellensatz 91 10. Noether Normalization and Hilbert s Nullstellensatz In the last chapter we have gained much understanding for integral and finite ring extensions.
More informationLECTURE 10, MONDAY MARCH 15, 2004
LECTURE 10, MONDAY MARCH 15, 2004 FRANZ LEMMERMEYER 1. Minimal Polynomials Let α and β be algebraic numbers, and let f and g denote their minimal polynomials. Consider the resultant R(X) of the polynomials
More informationInstitutionen för matematik, KTH.
Institutionen för matematik, KTH. Contents 4 Curves in the projective plane 1 4.1 Lines................................ 1 4.1.3 The dual projective plane (P 2 )............. 2 4.1.5 Automorphisms of P
More information(1) A frac = b : a, b A, b 0. We can define addition and multiplication of fractions as we normally would. a b + c d
The Algebraic Method 0.1. Integral Domains. Emmy Noether and others quickly realized that the classical algebraic number theory of Dedekind could be abstracted completely. In particular, rings of integers
More informationMATH32062 Notes. 1 Affine algebraic varieties. 1.1 Definition of affine algebraic varieties
MATH32062 Notes 1 Affine algebraic varieties 1.1 Definition of affine algebraic varieties We want to define an algebraic variety as the solution set of a collection of polynomial equations, or equivalently,
More informationResolution of Singularities in Algebraic Varieties
Resolution of Singularities in Algebraic Varieties Emma Whitten Summer 28 Introduction Recall that algebraic geometry is the study of objects which are or locally resemble solution sets of polynomial equations.
More informationSPACES OF RATIONAL CURVES ON COMPLETE INTERSECTIONS
SPACES OF RATIONAL CURVES ON COMPLETE INTERSECTIONS ROYA BEHESHTI AND N. MOHAN KUMAR Abstract. We prove that the space of smooth rational curves of degree e on a general complete intersection of multidegree
More informationAlgebraic Geometry. Andreas Gathmann. Class Notes TU Kaiserslautern 2014
Algebraic Geometry Andreas Gathmann Class Notes TU Kaiserslautern 2014 Contents 0. Introduction......................... 3 1. Affine Varieties........................ 9 2. The Zariski Topology......................
More information9. Birational Maps and Blowing Up
72 Andreas Gathmann 9. Birational Maps and Blowing Up In the course of this class we have already seen many examples of varieties that are almost the same in the sense that they contain isomorphic dense
More informationMath 145. Codimension
Math 145. Codimension 1. Main result and some interesting examples In class we have seen that the dimension theory of an affine variety (irreducible!) is linked to the structure of the function field in
More informationIntroduction to Arithmetic Geometry Fall 2013 Lecture #18 11/07/2013
18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #18 11/07/2013 As usual, all the rings we consider are commutative rings with an identity element. 18.1 Regular local rings Consider a local
More informationAlgebraic Varieties. Chapter Algebraic Varieties
Chapter 12 Algebraic Varieties 12.1 Algebraic Varieties Let K be a field, n 1 a natural number, and let f 1,..., f m K[X 1,..., X n ] be polynomials with coefficients in K. Then V = {(a 1,..., a n ) :
More informationRings and groups. Ya. Sysak
Rings and groups. Ya. Sysak 1 Noetherian rings Let R be a ring. A (right) R module M is called noetherian if it satisfies the maximum condition for its submodules. In other words, if M 1... M i M i+1...
More informationADVANCED TOPICS IN ALGEBRAIC GEOMETRY
ADVANCED TOPICS IN ALGEBRAIC GEOMETRY DAVID WHITE Outline of talk: My goal is to introduce a few more advanced topics in algebraic geometry but not to go into too much detail. This will be a survey of
More informationCHAPTER 1. AFFINE ALGEBRAIC VARIETIES
CHAPTER 1. AFFINE ALGEBRAIC VARIETIES During this first part of the course, we will establish a correspondence between various geometric notions and algebraic ones. Some references for this part of the
More informationMAS 6396 Algebraic Curves Spring Semester 2016 Notes based on Algebraic Curves by Fulton. Timothy J. Ford April 4, 2016
MAS 6396 Algebraic Curves Spring Semester 2016 Notes based on Algebraic Curves by Fulton Timothy J. Ford April 4, 2016 FLORIDA ATLANTIC UNIVERSITY, BOCA RATON, FLORIDA 33431 Email address: ford@fau.edu
More information(1) is an invertible sheaf on X, which is generated by the global sections
7. Linear systems First a word about the base scheme. We would lie to wor in enough generality to cover the general case. On the other hand, it taes some wor to state properly the general results if one
More informationAlgebraic Geometry Spring 2009
MIT OpenCourseWare http://ocw.mit.edu 18.726 Algebraic Geometry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.726: Algebraic Geometry
More informationVector bundles in Algebraic Geometry Enrique Arrondo. 1. The notion of vector bundle
Vector bundles in Algebraic Geometry Enrique Arrondo Notes(* prepared for the First Summer School on Complex Geometry (Villarrica, Chile 79 December 2010 1 The notion of vector bundle In affine geometry,
More informationALGEBRAIC GROUPS. Disclaimer: There are millions of errors in these notes!
ALGEBRAIC GROUPS Disclaimer: There are millions of errors in these notes! 1. Some algebraic geometry The subject of algebraic groups depends on the interaction between algebraic geometry and group theory.
More informationNONSINGULAR CURVES BRIAN OSSERMAN
NONSINGULAR CURVES BRIAN OSSERMAN The primary goal of this note is to prove that every abstract nonsingular curve can be realized as an open subset of a (unique) nonsingular projective curve. Note that
More informationLinear Algebra. Paul Yiu. Department of Mathematics Florida Atlantic University. Fall 2011
Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall 2011 Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall 2011 1A: Vector spaces Fields
More information12. Linear systems Theorem Let X be a scheme over a ring A. (1) If φ: X P n A is an Amorphism then L = φ O P n
12. Linear systems Theorem 12.1. Let X be a scheme over a ring A. (1) If φ: X P n A is an Amorphism then L = φ O P n A (1) is an invertible sheaf on X, which is generated by the global sections s 0, s
More informationDIVISORS ON NONSINGULAR CURVES
DIVISORS ON NONSINGULAR CURVES BRIAN OSSERMAN We now begin a closer study of the behavior of projective nonsingular curves, and morphisms between them, as well as to projective space. To this end, we introduce
More informationExercise Sheet 7  Solutions
Algebraic Geometry DMATH, FS 2016 Prof. Pandharipande Exercise Sheet 7  Solutions 1. Prove that the Zariski tangent space at the point [S] Gr(r, V ) is canonically isomorphic to S V/S (or equivalently
More informationHomework 2  Math 603 Fall 05 Solutions
Homework 2  Math 603 Fall 05 Solutions 1. (a): In the notation of AtiyahMacdonald, Prop. 5.17, we have B n j=1 Av j. Since A is Noetherian, this implies that B is f.g. as an Amodule. (b): By Noether
More informationSecant varieties. Marin Petkovic. November 23, 2015
Secant varieties Marin Petkovic November 23, 2015 Abstract The goal of this talk is to introduce secant varieies and show connections of secant varieties of Veronese variety to the Waring problem. 1 Secant
More informationFactorization in Polynomial Rings
Factorization in Polynomial Rings Throughout these notes, F denotes a field. 1 Long division with remainder We begin with some basic definitions. Definition 1.1. Let f, g F [x]. We say that f divides g,
More information(II.B) Basis and dimension
(II.B) Basis and dimension How would you explain that a plane has two dimensions? Well, you can go in two independent directions, and no more. To make this idea precise, we formulate the DEFINITION 1.
More informationBasic facts and definitions
Synopsis Thursday, September 27 Basic facts and definitions We have one one hand ideals I in the polynomial ring k[x 1,... x n ] and subsets V of k n. There is a natural correspondence. I V (I) = {(k 1,
More informationMATH 326: RINGS AND MODULES STEFAN GILLE
MATH 326: RINGS AND MODULES STEFAN GILLE 1 2 STEFAN GILLE 1. Rings We recall first the definition of a group. 1.1. Definition. Let G be a non empty set. The set G is called a group if there is a map called
More information. HILBERT POLYNOMIALS
CLASSICAL ALGEBRAIC GEOMETRY Daniel Plaumann Universität Konstan Summer HILBERT POLYNOMIALS T An ane variety V A n with vanishing ideal I(V) K[,, n ] is completely determined by its coordinate ring A(V)
More informationBasic Algebraic Geometry 1
Igor R. Shafarevich Basic Algebraic Geometry 1 Second, Revised and Expanded Edition SpringerVerlag Berlin Heidelberg New York London Paris Tokyo HongKong Barcelona Budapest Table of Contents Volume 1
More informationBinomial Exercises A = 1 1 and 1
Lecture I. Toric ideals. Exhibit a point configuration A whose affine semigroup NA does not consist of the intersection of the lattice ZA spanned by the columns of A with the real cone generated by A.
More informationLemma 1.3. The dimension of I d is dimv d D = ( d D+2
BEZOUT THEOREM One of the most fundamental results about the degrees of polynomial surfaces is the Bezout theorem, which bounds the size of the intersection of polynomial surfaces. The simplest version
More informationInstitutionen för matematik, KTH.
Institutionen för matematik, KTH. Contents 7 Affine Varieties 1 7.1 The polynomial ring....................... 1 7.2 Hypersurfaces........................... 1 7.3 Ideals...............................
More informationFormal power series rings, inverse limits, and Iadic completions of rings
Formal power series rings, inverse limits, and Iadic completions of rings Formal semigroup rings and formal power series rings We next want to explore the notion of a (formal) power series ring in finitely
More informationFOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 37
FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 37 RAVI VAKIL CONTENTS 1. Application of cohomology: Hilbert polynomials and functions, Riemann Roch, degrees, and arithmetic genus 1 1. APPLICATION OF COHOMOLOGY:
More informationBEZOUT S THEOREM CHRISTIAN KLEVDAL
BEZOUT S THEOREM CHRISTIAN KLEVDAL A weaker version of Bézout s theorem states that if C, D are projective plane curves of degrees c and d that intersect transversally, then C D = cd. The goal of this
More informationThe most important result in this section is undoubtedly the following theorem.
28 COMMUTATIVE ALGEBRA 6.4. Examples of Noetherian rings. So far the only rings we can easily prove are Noetherian are principal ideal domains, like Z and k[x], or finite. Our goal now is to develop theorems
More informationDraft: January 2, 2017 INTRODUCTION TO ALGEBRAIC GEOMETRY
Draft: January 2, 2017 INTRODUCTION TO ALGEBRAIC GEOMETRY MATTHEW EMERTON Contents 1. Introduction and overview 1 2. Affine space 2 3. Affine plane curves 3 4. Projective space 7 5. Projective plane curves
More informationOral exam practice problems: Algebraic Geometry
Oral exam practice problems: Algebraic Geometry Alberto García Raboso TP1. Let Q 1 and Q 2 be the quadric hypersurfaces in P n given by the equations f 1 x 2 0 + + x 2 n = 0 f 2 a 0 x 2 0 + + a n x 2 n
More informationHolomorphic line bundles
Chapter 2 Holomorphic line bundles In the absence of nonconstant holomorphic functions X! C on a compact complex manifold, we turn to the next best thing, holomorphic sections of line bundles (i.e., rank
More informationAPPENDIX 3: AN OVERVIEW OF CHOW GROUPS
APPENDIX 3: AN OVERVIEW OF CHOW GROUPS We review in this appendix some basic definitions and results that we need about Chow groups. For details and proofs we refer to [Ful98]. In particular, we discuss
More informationExercise Sheet 3  Solutions
Algebraic Geometry DMATH, FS 2016 Prof. Pandharipande Exercise Sheet 3  Solutions 1. Prove the following basic facts about algebraic maps. a) For f : X Y and g : Y Z algebraic morphisms of quasiprojective
More informationOn Maps Taking Lines to Plane Curves
Arnold Math J. (2016) 2:1 20 DOI 10.1007/s4059801500271 RESEARCH CONTRIBUTION On Maps Taking Lines to Plane Curves Vsevolod Petrushchenko 1 Vladlen Timorin 1 Received: 24 March 2015 / Accepted: 16 October
More informationMath 203A  Solution Set 1
Math 203A  Solution Set 1 Problem 1. Show that the Zariski topology on A 2 is not the product of the Zariski topologies on A 1 A 1. Answer: Clearly, the diagonal Z = {(x, y) : x y = 0} A 2 is closed in
More informationDedekind Domains. Mathematics 601
Dedekind Domains Mathematics 601 In this note we prove several facts about Dedekind domains that we will use in the course of proving the RiemannRoch theorem. The main theorem shows that if K/F is a finite
More information12 Hilbert polynomials
12 Hilbert polynomials 12.1 Calibration Let X P n be a (not necessarily irreducible) closed algebraic subset. In this section, we ll look at a device which measures the way X sits inside P n. Throughout
More information(dim Z j dim Z j 1 ) 1 j i
Math 210B. Codimension 1. Main result and some interesting examples Let k be a field, and A a domain finitely generated kalgebra. In class we have seen that the dimension theory of A is linked to the
More informationFILTERED RINGS AND MODULES. GRADINGS AND COMPLETIONS.
FILTERED RINGS AND MODULES. GRADINGS AND COMPLETIONS. Let A be a ring, for simplicity assumed commutative. A filtering, or filtration, of an A module M means a descending sequence of submodules M = M 0
More informationALGEBRAIC GEOMETRY (NMAG401) Contents. 2. Polynomial and rational maps 9 3. Hilbert s Nullstellensatz and consequences 23 References 30
ALGEBRAIC GEOMETRY (NMAG401) JAN ŠŤOVÍČEK Contents 1. Affine varieties 1 2. Polynomial and rational maps 9 3. Hilbert s Nullstellensatz and consequences 23 References 30 1. Affine varieties The basic objects
More informationA course in. Algebraic Geometry. Taught by Prof. Xinwen Zhu. Fall 2011
A course in Algebraic Geometry Taught by Prof. Xinwen Zhu Fall 2011 1 Contents 1. September 1 3 2. September 6 6 3. September 8 11 4. September 20 16 5. September 22 21 6. September 27 25 7. September
More informationGEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS
GEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS JENNY WANG Abstract. In this paper, we study field extensions obtained by polynomial rings and maximal ideals in order to determine whether solutions
More informationALGEBRAIC GEOMETRY I, FALL 2016.
ALGEBRAIC GEOMETRY I, FALL 2016. DIVISORS. 1. Weil and Cartier divisors Let X be an algebraic variety. Define a Weil divisor on X as a formal (finite) linear combination of irreducible subvarieties of
More informationMath 210B. Artin Rees and completions
Math 210B. Artin Rees and completions 1. Definitions and an example Let A be a ring, I an ideal, and M an Amodule. In class we defined the Iadic completion of M to be M = lim M/I n M. We will soon show
More information9. Integral Ring Extensions
80 Andreas Gathmann 9. Integral ing Extensions In this chapter we want to discuss a concept in commutative algebra that has its original motivation in algebra, but turns out to have surprisingly many applications
More informationHARTSHORNE EXERCISES
HARTSHORNE EXERCISES J. WARNER Hartshorne, Exercise I.5.6. Blowing Up Curve Singularities (a) Let Y be the cusp x 3 = y 2 + x 4 + y 4 or the node xy = x 6 + y 6. Show that the curve Ỹ obtained by blowing
More information1 Structures 2. 2 Framework of Riemann surfaces Basic configuration Holomorphic functions... 3
Compact course notes Riemann surfaces Fall 2011 Professor: S. Lvovski transcribed by: J. Lazovskis Independent University of Moscow December 23, 2011 Contents 1 Structures 2 2 Framework of Riemann surfaces
More informationAlgebraic Geometry. Question: What regular polygons can be inscribed in an ellipse?
Algebraic Geometry Question: What regular polygons can be inscribed in an ellipse? 1. Varieties, Ideals, Nullstellensatz Let K be a field. We shall work over K, meaning, our coefficients of polynomials
More information4. Noether normalisation
4. Noether normalisation We shall say that a ring R is an affine ring (or affine kalgebra) if R is isomorphic to a polynomial ring over a field k with finitely many indeterminates modulo an ideal, i.e.,
More information4.5 Hilbert s Nullstellensatz (Zeros Theorem)
4.5 Hilbert s Nullstellensatz (Zeros Theorem) We develop a deep result of Hilbert s, relating solutions of polynomial equations to ideals of polynomial rings in many variables. Notation: Put A = F[x 1,...,x
More informationLecture 2: Elliptic curves
Lecture 2: Elliptic curves This lecture covers the basics of elliptic curves. I begin with a brief review of algebraic curves. I then define elliptic curves, and talk about their group structure and defining
More informationMODULI SPACES OF CURVES
MODULI SPACES OF CURVES SCOTT NOLLET Abstract. My goal is to introduce vocabulary and present examples that will help graduate students to better follow lectures at TAGS 2018. Assuming some background
More informationA finite universal SAGBI basis for the kernel of a derivation. Osaka Journal of Mathematics. 41(4) P.759P.792
Title Author(s) A finite universal SAGBI basis for the kernel of a derivation Kuroda, Shigeru Citation Osaka Journal of Mathematics. 4(4) P.759P.792 Issue Date 20042 Text Version publisher URL https://doi.org/0.890/838
More informationk k would be reducible. But the zero locus of f in A n+1
Math 145. Bezout s Theorem Let be an algebraically closed field. The purpose of this handout is to prove Bezout s Theorem and some related facts of general interest in projective geometry that arise along
More informationAugust 2015 Qualifying Examination Solutions
August 2015 Qualifying Examination Solutions If you have any difficulty with the wording of the following problems please contact the supervisor immediately. All persons responsible for these problems,
More informationMath 203A  Solution Set 1
Math 203A  Solution Set 1 Problem 1. Show that the Zariski topology on A 2 is not the product of the Zariski topologies on A 1 A 1. Answer: Clearly, the diagonal Z = {(x, y) : x y = 0} A 2 is closed in
More informationINTERSECTION NUMBER OF PLANE CURVES
INTERSECTION NUMBER OF PLANE CURVES MARGARET E. NICHOLS 1. Introduction One of the most familiar objects in algebraic geometry is the plane curve. A plane curve is the vanishing set of a polynomial in
More informationARCS IN FINITE PROJECTIVE SPACES. Basic objects and definitions
ARCS IN FINITE PROJECTIVE SPACES SIMEON BALL Abstract. These notes are an outline of a course on arcs given at the Finite Geometry Summer School, University of Sussex, June 2630, 2017. Let K denote an
More informationLecture 2. (1) Every P L A (M) has a maximal element, (2) Every ascending chain of submodules stabilizes (ACC).
Lecture 2 1. Noetherian and Artinian rings and modules Let A be a commutative ring with identity, A M a module, and φ : M N an Alinear map. Then ker φ = {m M : φ(m) = 0} is a submodule of M and im φ is
More information