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Algebraic plane curves: from multiplicities and intersection numbers to Bézout's theorem
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Algebraic plane curves: from multiplicities and intersection numbers to Bézout's theorem
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Algebraic Plane Curves: from Multiplicities and Intersection Numbers to B ezout's Theorem Thesis submitted by Jiajun Yan to satisfy the requirements of the degree of Master of Arts in Mathematics at the University of Southern California August 2017 Guidance Committee: Professor Francis Bonahon, Chair Professor Kenneth Alexander Professor Sheel Ganatra 1 Acknowledgements: With the given chance, I want to acknowledge a few individuals who helped me signicantly along my way of bringing this thesis into completion: First of all, I want to thank Prof. Francis Bonahon for being my thesis advisor. Without his guidance, I could never have been able to nish this thesis. He taught me not only the math, but also LaTex, good writing strategies, and so on. He was patient and understanding when I fell behind; He was funny and cheerful the rest of the time. It was a great pleasure to have Prof. Bonahon as my thesis advisor. I also want to thank Prof. Keith Mayerson, who gave me heartfelt suggestions about my paintings and my future pursuit of art and math, who is always willing to talk and share, and who is simply so kind. He oered me great help and guidance when I was struggling with my important life decisions. His kindness and encouragement will always push me forward when I face diculties in my life. Another person to thank here is Prof. Kenneth Alexander. Prof. Alexander has been important to me as a professor, a mentor and a friend. I am very lucky to get to know him well in the past year. His help and in uence on me cannot be replaced. 2 Last but not the least, I am very grateful to have Prof. Francis Bonahon, Prof. Kenneth Alexander and Prof. Sheel Ganatra to be on my thesis committee. I sincerely appreciate their willingness to spend time reviewing this thesis. I regret for not being able to mention the many more names of the people who have also helped and supported me in various ways. 3 Contents Introduction 7 Chapter 1. Preliminaries 9 1. Ane curves 9 2. Coordinate rings and local rings 9 3. Nullstellensatz 10 Chapter 2. Multiplicities 13 1. Forms 13 2. Tangent lines and multiplicities 13 3. Ideals with a nite number of zeros 14 4. Discrete valuation rings 18 5. Multiplicities and local rings 20 Chapter 3. Intersection Numbers 27 1. Intersection numbers 27 Chapter 4. B ezout's Theorem 37 1. Projective Curves 37 2. B ezout's Theorem 37 Bibliography 41 5 Introduction This thesis is an exposition of material taken from Fulton's book [1]. This thesis focuses on algebraic plane curves, exploring the materi- als through the discussions of multiplicities, interesection numbers and B ezout's theorem. Chapter 1 is a quick review of the materials that will be used fre- quently in the following chapter, but without proofs. Then, in Chapter 2, we give the denitions of tangent lines and study the multiplicity of a plane curve at a given point. In Chapter 3, we introduce the notion of intersection number of two curve and the way to compute it. In these two chapters, we limit our focus on ane curves. Finally, in Chapter 4, we shift the study to projective curves. Since many denitions regarding to projective curves are similar to those of ane curves, they are not explicitly listed out in this chapter. The main result from this chapter is B ezout's theorem, which relates the intersection numbers to the degrees of the curves. 7 CHAPTER 1 Preliminaries 1. Ane curves 1.1. Ane Planes. Letk be an algebraically closed eld, and we should consider it as C in the following discussion. The ane plane A 2 (k) over k is the cartesian product kk. We usually abbreviate A 2 (k) byA 2 . 1.2. Ane curves. The vanishing set of an ideal Ik[X;Y ] is V (I) =fP2A 2 ;F (P ) = 0;8F2Ig: An ane curveV (I(F ))A 2 is the vanishing set of the polynomial ideal I(F )k[X;Y ] generated by F , namely V (I(F )) =fP2A 2 ;F (P ) = 0g: In practice, we simply denote the curve V (I(F )) by F . The polynomial Fk[X;Y ] is irreducible if it cannot be factored. This is equivalent to the property that I(F ) is a prime ideal. 2. Coordinate rings and local rings 2.1. Coordinate rings. LetF be an irreducible ane curve. Be- cause I(F ) is a prime ideal, k[X;Y ]=I(F ) is an integral domain. We let (F ) =k[X;Y ]=I(F ), and call it the coordinate ring of F . 9 The elements of (F ) can be viewed either as equivalence classes of polynomials in k[X;Y ], or as polynomial functions dened on the curve F and valued in k. Let k(F ) denote the quotient eld of (F ). The elements of k(F ) can be interpreted as rational functions dened over all ofF except for a subset of measure zero. 2.2. Local rings. The local ringO P (F ) =f f g : f;g2 (F );g(P )6= 0g of the curveF at the pointP2F is the ring of all rational functions contained ink(F ) that are dened at P. Namely,O P (F ) consists of all rational functions on F whose denominator does not vanish at P . We have the following inclusion: k (F )O P (F )k(F ): Remark 1. An alternative way to look at the local ringO P (F ) for a given pointP is to view it as the germ of rational functions dened at P . Due to continuity, if a rational function is dened atP , it is dened on an arbitrarily small open disk containingP . Since the vanishing set for a rational function, in particular for a polynomial, has dimension at most 1, if two functions agree on an open disk, they must be the same function. Thus each rational function in the germ of P extends to a unique rational function inO P (F ). We get the other way of the inclusion by restricting each function inO P (F ) toP . Thus the two are isomorphic. 3. Nullstellensatz Theorem 1 (Hilbert's Nullstellensatz). LetI be an ideal ink[X 1 ;:::;X n ]. Then Rad(I) =ff2k[X 1 ;:::;X n ]jf(P ) = 0;8P2V (I)g. 10 Corollary 2. Let I be as above. Then V (I) is a nite set if and only ifk[X 1 ;:::;X n ]=I is a nite dimensional vector space over k. If this occurs, the number of points in V (I) is at most dim k (k[X 1 ;:::;X n ]=I). 11 CHAPTER 2 Multiplicities 1. Forms A polynomial f2k[X;Y ] is a form if f is homogeneous, that is if all terms of f have the same degree n, for some positive integer n. Note, for a general polynomialg2k[X;Y ], we can always rearrange the terms so that g is the sum of forms. Proposition 3. Let f2k[X;Y ] be a form. Then f can always be factored into the product of linear terms. Proof. Letf be a form of degreen, and letf =f(X; 1). Consider f =gY r , whereY does not divideg. Then,f =g . Butf can always be factored into linear factors, and thus so can g = c Q (X i ). Hence, f =cY r Q (X i Y ): 2. Tangent lines and multiplicities Let F be an irreducible curve. We want to study the tangent lines of F at a given point P . By a change of coordinates map, we can always assume that P = (0; 0). Let F2k[X;Y ] be a polynomial. Arrange F so that it is the sum of forms. Thus, we can write F =F m + higher terms; 13 whereF m is the lowest degree form of F with degree m. SinceF m can be factored into linear terms, write F m = Q i L i r i . Definition 4. (Tangent lines and multiplicities) Each L i denes a tangent line of F at P = (0; 0), and is counted r i times. We have m P (F ) = P i r i = deg(F m ) =m, call it be the multiplicity of F at P . It is easy to see that m P (F ) is invariant under ane changes of coordinates, so the denition of m P (F ) applies to an arbitrary point P . In section 5, we will also prove thatO P (F ) stays invariant up to isomorphism under ane changes of coordinates, which will give us an alternative denition for m P (F ). We sayP is a simple point if it admits a unique tangent line counted once, or equivalently m P (F ) = 1, and P is a multiple point otherwise, or equivalently m P (F ) 2. 3. Ideals with a nite number of zeros Lemma 5. Let I, J be ideals of a commutative ring R. Suppose I is nitely generated and IRad(J). Then I n J, for some n. Proof. Letf 1 ;:::; m g be the generating set of I, and thus we havef n 1 ; n1 1 2 ;:::; n m g a generating set for I n . Since I Rad(J), for each i , exists r i such that s i i = i 2J. Let k =maxfs 1 ;:::;s m g, then k 2 J, for all i. Let n = (k 1)m + 1, then for all r2 I n , r2J. Lemma 6. Let I 1 , ..., I N be ideals in R such that I i and\ j6=i I j are comaximal, that is, I i +\ i6=j I j = R, for all i. Then, I n 1 \:::\I n N = (I 1 :::I N ) n = (I 1 \:::\I N ) n , for all n. 14 Proof. We apparently have the following inclusion for all n: (I 1 :::I N ) n (I 1 \:::\I N ) n I n 1 \:::\I n N : Thus, if we can showI n 1 \:::\I n N (I 1 :::I N ) n , then we have shown the desired equality. Also notice, if we can show the equality for n = 1, then we show it for all n. Let r 2 I 1 \ :::\ I N = I i \ (\ i6=j I j ). Since I i and\ i6=j I j are comaximal, exist a2 I i and b2\ 6=j I j such that a +b = 1. Thus, r = (a +b)r = ar +br. This implies that ar2\ i6=j I j , br2 I i . But r2 I i \ (\ i6=j I j ), so we have ar2 I i (\ i6=j I j ), br2 I i (\ i6=j I j ). Hence, r2 I i (\ i6=j I j ). Since i is arbitary, r2 I 1 :::I N . Thus by the previous remarks, we showed the desired equality I n 1 \:::\I n N = (I 1 :::I N ) n = (I 1 \:::\I N ) n , for all n. Proposition 7. Let I be an ideal in k[X;Y ], and suppose V (I) = fP 1 ;:::;P N g is nite. LetO i = O P i (A 2 ). Then there is a natural isomorphism of k[X;Y ]=I with Q N i=1 O i =IO i . Proof. Let I i be the maximal ideal corresponding tofP i g, by Nullstellensatz. Let R =k[X;Y ]=I and R i =O i =IO i . Let ' i : R! R i be the natural ring homomorphism dened as r +I7! r +IO i , r2 k[X;Y ]. Thus, we have an induced ring homo- morphism' :R! Q N i=1 R i , wherer +I7! (r +IO 1 ;:::;r +IO N ). We want to show that ' is an isomorphism. 15 Again by Nullstellensatz, Rad(I) =\ N i=1 I i . Thus, by Lemma 5, (\I i ) d I (), for some d. Now, we want to show that I i and\ j6=i I j are comaximal by proving the following lemma: Lemma 8. Let I, J be ideals of k[X;Y ]. Then I and J are comax- imal if and only if V (I)\V (J) =;. Proof. (() AssumeV (I)\V (J) =;. ConsiderI +J, we want to show it isk[X;Y ]. Suppose not, thenI +J is contained in some maxi- mal ideal, and henceV (I +J) is nonempty, that is,fPg2V (I +J) A 2 . But this meansfPg2V (I)\V (J), contradicting the assumption. Thus, I +J =k[X;Y ], that is, I and J are comaximal. ()) Let f2 I, g2 J be such that f +g = 1. Let V (f) be the vanishing set off andV (g) be the vanishing set ofg. Sincef +g = 1, we have V (f)\V (g) =;. But V (I)V (f) and V (J)V (g). Thus, we have shown V (I)\V (J) =;. By the previous lemma, we know I i and\ j6=i I j are comaximal, for all i. Thus, we can apply Lemma 6 to () and get the following: (\I i ) d = (I 1 :::I N ) d =\I d i I. Now, we want to show that for each i, we can nd F i such that F i (P i ) = 1 and F i (P j ) = 0, for all j6= i: Since we can always nd e F i 2 (\ j6=i I j )nI i , we have e F i (P j ) = 0, for all j6= i, and e F i (P i ) = where 6= 0. Let F i = 1 e F i , we have the desired polynomial. Let E i = 1 (1F d i ) d . Observe that E i = F d i D i , for some D i . Since F i 2 I d j , j 6= i, we have E i 2 I d j , j 6= i. Note also that 1 P i E i (P i ) = (1E i (P i )) ( P j6=i E j (P i )) = (1 1) + 0 = 0, for all i. Thus 1 P i E i 2\I d i I. 16 Observe thatE i E 2 i =E i (1E i ) =E i (1F d i ) d . SinceE i 2\ j6=i I d j and (1F d i ) d 2I d i ,E i E 2 i 2 (\ j6=i I d j )I d i = ( Q j6=i I j ) d I d i = (I 1 :::I N ) d I. For i6=j, E i E j = (1 (1F d i ) d )(1 (1F d j ) d ) = 1 (1F d i ) d (1F d j ) d + (1F d i ) d (1F d j ) d =F d i A +F d j A (F d i A + 1)(F d j A + 1) = F d i F d j A 2 2I. Let e i be the residue of E i in k[X;Y ]=I, then e i e 2 i = 0, for all i. For i6=j, e i e j = 0, and P i e i = 1. Lemma 9. If G2 k[X;Y ] and g denotes the residue of G in R = k[X;Y ]=I. Suppose G(P i ) 6= 0, then there exists t 2 R such that gt =e i . Proof. By multiplying a constant, we can always assumeG i (P ) = 1. Let H = 1G, then H(P ) = 0, and thus H 2 I i . Notice that G(E i +HE i +::: +H d1 E i ) = (1H)(E i +HE i +::: +H d1 E i ) = (E i +HE i +::: +H d1 E i ) (HE i +H 2 E i +::: +H d E i ) =E i H d E i . Since H 2 I i , H d E i 2 I, so let t = e i +he i +::: +h d1 e i , we have gt =e i . With the preceding lemma, we are now able to prove that ' is an isomorphism: ' is one-to-one: Let h be in R. Suppose '(h) = 0, then for each i, ' i (h) = 0. Thus, h2 IO i , that is, for each H where its residue in R is h, there exist i 2 I and G i 2 k[X;Y ] with G i (P i )6= 0 such that H = i G i , for each i. Let g i be the residue of G i . Thus by the previous lemma, we can nd t i g i =e i . Since HG i = i 2 I, hg i = 0 2 R. Now write h = P e i h = P t i (g i h) = 0. Hence, ' is one-to-one. 17 ' is onto: Since E i (P i ) =e i (P i ) = 1, ' i (e i )(P i ) = 16= 0, and thus it is invertible inR i , in other words,' i (e i ) is a unit inR i . Notice that since ' i is a ring homomorphism,' i (e i )' i (e j ) =' i (e i e j ) = 0, fori6=j. Since ' i (e i ) is a unit, it is not a zero-divisor, and thus we have ' i (e j ) = 0, for all j6=i. Write ' i (e i ) =' i (e i ) + P j6=i ' i (e j ) =' i (1) = 1. Let z = ( r 1 s 1 ;:::; r N s N )2 Q R i . Again by the previous lemma, we can nd t i such that t i s i =e i , for each i. Thus, ' i (t i r i ) =' i (t i r i )' i (e i ) = ' i (r i t i e i ) =' i (r i e i s i e i ) =' i ( r i s i ) = r i s i 2R i . Hence, '( P t i r i e i ) =z, and ' is onto. This concludes the proof. Corollary 10. With the hypothesis of Proposition 7, dim k (k[X;Y ]=I) = N X i=1 dim k (O i =IO i ): Proof. Follows directly from Proposition 7 by summing up the dimension for eachO i =IO i . Corollary 11. If V (I) from Proposition 7 is equal to a single pointfPg, then k[X;Y ]=I =O P (A 2 )=IO P (A 2 ): Proof. Follows directly from Proposition 7 by taking N = 1. 4. Discrete valuation rings Let R be an integral domain that is not a eld. We say R is a dis- crete valuation ring, written DVR, if R satises either of the following two equivalent conditions. (1) R is Noetherian and local, and the maximal ideal is principal. 18 (2) There is an irreducible elementt2R such that every nonzero z2 R can be written uniquely in the form z = ut n , u a unit in R, t a nonnegative integer. Such t is called a uniformizing parameter for R. Proposition 12. The two conditions are equivalent. Proof. First, we want to show that (1)) (2). Since R is local and the maximal ideal M is principal, we let M = (t), where t2 R is a non-unit that generates M. Suppose9z = ut n = vt m , where u;v are units, and n>m. Then since R is an integral domain, we get the expressionv =ut nm , and thusv =u;n =m. Hence, for allz that can be written asz =ut n , such expression is unique. Now we want to show everyz2R can indeed be written in such form, for appropriate u and n. Suppose not, then we have z a non-unit. Since z is a non-unit, z is in M, and thus we can write z as z =z 1 t, where z 1 is also a non-unit. Again, we have z 1 = z 2 t, for some non-unit z 2 . Thus we can produce an innite ascending chain of ideals: (z) (z 1 ) (z 2 )::: SinceR is Noetherian, this chain terminates, that is,9n2Z such that 8m n; (z m ) = (z n ). Then, z n+1 = vz n , for some v2 R, so z n+1 = vtz n+1 , but that means vt = 1. Since t is not a unit, contradiction. Thus, we have shown that every z 2 R can be written uniquely as z =ut n for some u;n. This shows (1)) (2). Now we want to show (2)) (1). LetM = (t). Then every non-unit element is inM. Thus,M is the unique maximal ideal ofR. This shows R is local, andM is principal. To show thatR is Noetherian, letI be a 19 proper ideal ofR. ThenI is contained inM. SupposeI is not principal, then9z 1 = ut n ;z 2 = vt m ;n > m such that (z 1 ) I; (z 2 ) I, where (z 1 ); (z 2 ) are not contained in each other. But (z 1 ) = (t n ); (z 2 ) = (t m ), since n > m, (z 1 ) (z 2 ), contradiction. So for all I a proper ideal of R,I is principal. Thus,R is a PID. SupposeI = (t n ), we always get a terminating chain: I (t n1 ) (t n2 )::: (t) =M: Hence, we have shown thatR is Noetherian, and we are done with the proof. 5. Multiplicities and local rings Let F be an irreducible curve, write F =F m +higher terms. F has multiplicity m at P , denote m P (F ). Theorem 13. Let F be an irreducible curve, and P a point on F . Then for all suciently large n, m P (F ) = dim k (m P (F ) n =m P (F ) n+1 ); where m P (F ) is the unique maximal ideal ofO P (F ). Proof. Here we will assume Lemma 18 for the moment, so up to an ane change of coordinate, we can assume that P = (0; 0). First, we want to introduce a lemma: Lemma 14. IfO is a local ring, and m the unique maximal ideal, we have the natural short exact sequence: 0! m n =m n+1 !O=m n+1 !O=m n ! 0 20 Proof. Consider the following sequence: 0! m n =m n+1 ' !O=m n+1 !O=m n ! 0; where' :r + m n+1 7!r + m n+1 ,r2 m n , and :s + m n+1 7!s + m n , s2O. Notice clearly that ' is one-to-one, and is onto. Now we need to show that ker( ) = im('). First, we want to show ker( ) im('). Suppose (s + m n+1 ) = 0, then we have s2 m n , and thus s + m n+1 is in the image of '. Hence, ker( ) im('). Now, to show im(') ker( ), let r + m n+1 2 im('). Then, (r + m n+1 ) =r + m n , but r2 m n , so we have (r + m n+1 ) = 02 ker( ). Combining the previous arguments, we have shown the sequence is exact. By the previous lemma, we have the following short exact sequence: 0! m n =m n+1 !O=m n+1 !O=m n ! 0; whereO isO P (F ), and m is the unique maximal ideal. Now, we need another lemma: Lemma 15. If 0!A!B!C! 0 is a short exact sequence of nite dimensional vector spaces, then dimB = dimA + dimC: Proof. This is equivalent to the rank-nullity theorem. By Lemma 15, the theorem will follow if we can show that for all n suciently large, that is, nm P (F ), dim k (O=m n ) =n m P (F ) +s, where s is some constant integer. 21 Lemma 16. LetP = (0; 0)2A 2 . LetI = (X;Y ), thenI r O P (A 2 ) = (m P (A 2 )) r , for all r. Proof. We only need to show the case forr = 1, and the rest will follow. Let f2 IO, whereO denotesO P (A 2 ), then f = th g , where t2 I and g(0; 0)6= 0. In particular, we have t(0; 0) = 0, and so f in not invertible Since for a local ring, all non-invertible elements belong to the maximal ideal, we have IO m, where m denotes m P (A 2 ). Now let f2 m, then f = h g in not invertible, which implies that f(0; 0) = g(0; 0) = 0. Thus, g does not have a constant term, which implies g2I. So we have m2IO. Hence, we have shown the desired equality for all r. By the previous lemma, m n =I n O, where I = (X;Y )k[X;Y ]. Lemma 17. Let I = I(V ) k[X;Y ], P 2 V . J is an ideal that contains I. Let J 0 be the image of J in (V ). Then there is a natural isomorphism ' fromO P (A 2 )=JO P (A 2 ) toO P (V )=J 0 O P (V ). Proof. Consider' :O P (A 2 )=JO P (A 2 )!O P (V )=J 0 O P (V ), where ' : [t]7! e [t] = [t]j V , [t]2O P (A 2 )=JO P (A 2 ). ' is one-to-one: Let [t]2 ker('), then e [t] = [t]j V = 0. This implies that e t2 J 0 O P (V ). Since I J, V (J 0 ) = V (J) V , and thus t2 JO P (A 2 ). Hence, [t] = 02O P (A 2 )=JO P (A 2 ). ' is onto: Let [r]2O P (V )=J 0 O P (V ). We want to nd [t] such that [r] = e [t]. Sincer2O P (V ), we clearly have somet2O P (A 2 ) such that tj V =r. Now notice that '([t]) = [t]j V = [tj V ] = [r]. 22 Thus, ' is an isomorphism. Since V (I n ) =fPg, by the previous lemma and corollary 11 from section 3, we get the following: k[X;Y ]=(I n ;F ) =O P (A 2 )=(I n ;F )O P (A 2 ) =O P (F )=I n O P (F ) =O=m n : Thus, we have dimO=m n = dimk[X;Y ]=(I n ;F ), and we have reduced the problem to computing the dimension of k[X;Y ]=(I n ;F ). Letm =m P (F ),G2I nm , thenFG2I m . Consider the following: 0!k[X;Y ]=I nm !k[X;Y ]=I n ' !k[X;Y ]=(I n ;F )! 0; where : g7! fg, where g denotes the residue of G in k[X;Y ]=I nm , fg is the residue of FG in k[X;Y ]=I n , and ' sends an element that has a factor f to 0, and sends an element that doesn't have a factor f to itself. We want to show the sequence is exact. First observe that since F has multiplicity m, that is, deg(F m ) = m, and that f 6= 0, is injective. It is also clear that ker' = im( ). We also have ' onto. Thus, the sequence is exact. Hence, by Lemma 15, we have dim k (k[X;Y ]=(I n ;F )) = dim k (k[X;Y ]=I n ) dim k (k[X;Y ]=I nm ) = n(n + 1) 2 (nm)(nm + 1) 2 =mn m(m 1) 2 Thus, we have shown that, for nm, dimO=m n = dimk[X;Y ]=(I n ;F ) =mns 23 with s = m(m1) 2 . As indicated before, this proves that for all su- ciently large n, m P (F ) = dim k (m P (F ) n =m P (F ) n+1 ): Now we are going to prove thatO P (F ) remains invariant up to an isomorphism under ane changes of coordinates: Lemma 18. Let T be an ane change of coordinates such that T (P ) = P 0 . Then the induced map e T : O P (V ) ! O P 0(V T ) is an isomorphism, for some subvariety V ofA 2 such that P2V . Proof. Since T is an ane map, we can decompose T as T = T 1 T 2 whereT 1 is a translation such thatT 1 (P 00 ) =P 0 andT 2 is some linear isomorphism onA 2 such that T 2 (P ) =P 00 . First, we want to show that e T 2 is an isomorphism. To show this, we rst need to show that ker( e T 2 ) = 0. Suppose e T 2 (h) = 0, h2O P (V ), then h(T 2 (Q)) = 0, for all Q2A 2 . Since T 2 is an isomorphism onA 2 , it implies that h(Q) = 0, for all Q2A 2 . Thus, h must be 0. Hence, ker( e T 2 ) = 0. Now, to show that e T 2 is onto, let g2O P 00(V T 2 ), then g(Q) =g(T 2 (P )), for someP2A 2 , for allQ. Leth =gT 1 2 2O P (V ), then f is the image of h under e T 2 . Thus, e T 2 is also onto, so it is an isomorphism. So now we haveO P (V ) =O P 00(V T 2 ), if we can show thatO P 00(V T 2 ) = O P 0(V T ), then we will be done with the proof. Since we know that e T 1 :O P 00(V T 2 )!O P 0(V T ) is a map fromO P 00(V T 2 ) toO P 0(V T ), we just want to show that it is an isomorphism. Again, rst we show that the kernel is trivial. Leth2 ker( e T 1 ), then e T 1 (h) =h(Q+~ v) = 0, for all Q2A 2 and for some xed vector~ v. Thus, h must be 0, which shows 24 that the kernel is trivial. Now, to show that the map is also onto, let g2O P 0(V T ), theng(Q) =g(T 1 (Q~ v)). Thus, we can lett =gT 1 1 , then e T 1 (t) =g. Hence, we haveO P 00(V T 2 ) =O P 0(V T ). So we have shown thatO P (V ) =O P 00(V T 2 ) =O 0 P (V T ), or equiva- lently, e T :O P (V )!O P 0(V T ), where e T = e T 1 e T 2 . With the preceding lemma and Theorem 13, we are able to prove the following theorem: Theorem 19. A point P2F is a simple point of F if and only if the local ringO P (F ) is a discrete valuation ring. If L =aX +bY +c is any line passing through P that is not tangent to F , then the image l of L inO P (F ) is a uniforming parameter forO P (F ). Proof. (() IfO P (F ) is a DVR, then m P (F ) n = m n = (t) n = (t n ), for some non-unit t2O P (F ), for all n. Thus, m P (F ) = dim k ((m P (F )) n =(m P (F )) n+1 ) = dim k ((t n )=(t n+1 )) = 1: Hence, P is a simple point. ()) To prove this direction, we need to rst introduce a lemma: Lemma 20. Let P , P 0 be two distinct points in A 2 where L 1 , L 2 are two distinct lines passing through P and L 0 1 , L 0 2 are two distinct lines passing through P 0 . Then we can always nd an ane change of coordinates T such that T (P ) =P 0 and T (L 1 ) =L 0 1 , T (L 2 ) =L 0 2 . Proof. First let T 1 be the translation sending P to 0 and let T 0 1 be the translation sending P 0 to 0. Let T 1 (L 1 ) = l 1 , T 1 (L 2 ) = l 2 , and similarly, T 0 1 (L 0 1 ) = l 0 1 , T 0 1 (L 0 2 ) = l 0 2 , then we have l 1 , l 2 , l 0 1 and l 0 2 lines 25 passing through 0. Thus, we can nd a linear transformation T 2 that sendsl 1 tol 0 1 andl 2 tol 0 2 . Then, letT =T 0 1 1 T 2 T 1 , and we get the desired ane transformation. By an ane change of coordinates, we can assume that P = (0; 0). By the previous lemma, we can also assume that X = L and that Y is the unique tangent line of F passing through P . Thus, it suces to show that m P (F ) = (x), where x is the image of X inO P (F ). First notice that by Lemma 16 and Lemma 17, we haveO P (F ) = O P (A 2 )=(I(F )O P (A 2 )) and so that in any case, m P (F ) is generated by x and y. Recall the denition for tangent lines in Section 2, we can write F =Y +higher terms. Grouping all the terms with aY -factor together, we can write F = YG X 2 H, where G = 1 + higher terms, and H2 k[X]. Thus, yg = x 2 h and so y = x 2 hg 1 , since g(0; 0) = 1 and thus is invertible. Hence, we have shown that y2 (x) so we have the desired result. 26 CHAPTER 3 Intersection Numbers In this chapter, we want to dene the intersection number of two ane plane curves F and G at a point P. Up to applying an ane change of coordinates, we will always assumeP = (0; 0) in the following discussions. 1. Intersection numbers Before explicitly dening the intersection number of F and G at P, we will rst introduce some of the properties it has. Write the intersection number of F and G at P as I P (F;G). Properties of intersection numbers (1) I P (F;G) is a nonnegative integer for any F, G and P such that F and G intersect properly at P, i.e., F and G have no common component that passes through P. I P (F;G) =1 if F and G do not intersect properly at P. (2) I P (F;G) = 0 if and only ifP = 2F\G. I P (F;G) depends only on the components of F and G that pass through P. (3) I P (F;G) is invariant under ane change of coordinates. (4) I P (F;G) =I P (G;F ). (5) I P (F;G)m P (F )m P (G). (6) IfF = F r i i , andG = G s j i , thenI P (F;G) = P i;j r i s j I P (F i ;G j ). (7) I P (F;G) =I P (F; (G +AF )), for all A2k[X;Y ]. 27 Theorem 21. There is a unique intersection number I P (F;G) de- ned for all curves F, G, and all points P 2 A 2 , satisfying the above Properties (1-7). It is given by the following formula I P (F;G) =dim k (O P (A 2 )=(F;G)): Proof. (Uniqueness) To prove uniqueness, we may assume there is a number satisfying properties (1) (7), and calculate the this number explicitly using the properties. We can assumeP = (0; 0) by (3), andI P (F;G) is nite by (1), that is, F and G intersect properly at P . By (2), I P (F;G) = 0 if and only if P = 2F;G, so we can also assume P2F\G. Now we proceed by induction. Since the case for I P (F;G) = 0 is true automatically by (2), we can assume I P (F;G) = n and I P (A;B) can be calculated for all A, B such that I P (A;B)<n. ConsiderF (X; 0),G(X; 0) with degreer,s ink[X], wherer ors is taken to be 0 if F (X; 0) or G(X; 0) is 0. we can assume rs by (4). Case 1. r = 0. Then Y divides F , so F = YH, for some H 2 k[X;Y ]. Thus, by (6),I P (F;G) =I P (Y;G) +I P (H;G). Now if we can show I P (Y;G) > 0, then both I P (Y;G) and I P (H;G) are less than n, and we thus are done by induction. WriteG(X; 0) =X m (a 0 +a 1 X +:::);a 0 6= 0. Note thatm> 0 since P = (0; 0)2G. Now by (5), I P (Y;G)m P (Y )m P (G)m, so we are done by induction. Case 2. r > 0. First by (6), we can multiply constants to make F (X; 0) and G(X; 0) monic. Let H = G X sr F . Then by (7), I P (F;G) = I P (F;H), and the degree of H(X; 0) is less than s, call it 28 t. Repeat the same process (interchange F and H if t < r), we can eventually get two curves A, B that satises Case 1. Thus we have completed the proof of uniqueness. (Existence) Now we need to show existence. Dene the intersection numberI P (F;G) = dim k (O P (A 2 )=(F;G)). We need to show this num- ber satises property (1)-(7). (3) and (4) are immediate. Thus, without loss of generality, we can assume P = (0; 0). Since (F;G) = (F;G +AF ), A2k[X;Y ], (7) is also satised. To show (2), let f g 2O P (A 2 ). Observe ifP = 2F\G, thenF (P )6= 0. Thus F f Fg = f g 2 (F;G). ) I P (F;G) = 0. Conversely, assume I P (F;G) = 0, and thus equivalently,O P (A 2 ) = (F;G). Let f g 2O P (A 2 ) such that f g 6= 0, then exist f 1 g 1 , f 2 g 2 such that F f 1 g 1 +G f 2 g 2 = f g . In particular, f g (P ) =F (P ) f 1 g 1 (P ) +G(P ) f 2 g 2 (P )6= 0. But P2F\G, F (P ) =G(P ) = 0, and henceF (P ) f 1 g 1 (P ) +G(P ) f 2 g 2 (P ) = 0.)P = 2 F\G. Now, we want to show (1). If F and G intersect properly, that is, F , G have no common components, then V ((F;G)) is nite. Thus, by Corollary 2 and Corollary 10, I P (F;G)< dim k (k[X;Y ]=(F;G))<1: Conversely, if F , G do not intersect properly, then they have a com- mon component H with (F;G) (H). Thus, there is a natural homo- morphism sendingO=(F;G) toO=(H), whereO denotesO P (A 2 ). By Lemma 17 and Corollary 2, we know thatO=(H) =O P (H) (H) = 1. 29 We are left to show (5) and (6). For (6), it suces to showI P (F;GH) = I P (F;G) +I P (F;H), whereF has no common components withG,H. By Lemma 10, it suces to show the following sequence is exact: 0!O P (A 2 )=(F;H) ' !O P (A 2 )=(F;GH) !O P (A 2 )=(F;G)! 0; where is the natural onto homomorphism, and ' is dened as ' : z7!Gz, z2O and z the residue of z inO=(F;G). To show ' is one-to-one, let z2 ker('), then Gz = aF +bGH, for some a, b inO. Choose S2k[X;Y ] such that S(P )6= 0, and Sa, Sb and Sz are also in k[X;Y ]. Then we have G(SzSbH) = SaF . Abbreviate the previous equation as G(Z +BH) =AF . Since F and G have not common components, we must haveZ =DFBH. Thus, z = D S FbF , so z = 0. It is also easy to see that im(') = ker( ). Hence, we have shown that the sequence is exact, which implies (6). Finally, to show (5), let m = m P (F ), n = m P (G). Consider the following diagram: k[x;y]=I m k[X;Y ]=I n k[X;Y ]=I m+n ' O P (A 2 )=(F;G) k[X;Y ]=(I m+n ;F;G) // O P (A 2 )=(I m+n ;F;G) 0 0 where is dened to be (A;B) =AF +BG, A;B2k[X;Y ], ' and are natural projections. It is easy to see that ' and are surjective ring homomorphisms. Since V (I m+n ;F;G) = (0; 0) =fPg, is an 30 isomorphism by Corollary 11. Now, observe that im( ) = ker('). Thus we have dim k (k[X;Y ]=I m ) + dim k (k[X;Y ]=I n ) dim k (ker(')) with dim k (ker(')) = dim k (k[X;Y ]=I m+n )dim k (k[X;Y ]=(I m+n ;F;G)); with equality achieved if and only if is one-to-one. Rewrite the above inequality as follows: () dim k (k[X;Y ]=(I m+n ;F;G)) dim k (k[X;Y ]=I m+n ) dim k (k[X;Y ]=I m ) dim k (k[X;Y ]=I n ) =mn: Note also, that ()I P (F;G) = dim k (O P (A 2 )=(F;G)) dim k (O P (A 2 )=(I m+n ;F;G)) =dim k (k[X;Y ]=(I m+n ;F;G)); with equality if and only if is injective, and thus an isomorphism. Combining () and (), we have the following inequality: I P (F;G) = dim k (O P (A 2 )=(F;G)) dim k (O P (A 2 )=(I m+n ;F;G)) = dim k (k[X;Y ]=(I m+n ;F;G)) dim k (k[X;Y ]=I m+n ) dim k (k[X;Y ]=I m ) dim k (k[X;Y ]=I n ) =mn: 31 Thus, we have shown thatI P (F;G) is greater than or equal tomn, with equality only when and are both injective. We will now conclude the proof of (5) with the following lemma: Lemma 22. We have I P (F;G) = mn if and only if F and G have no common tangent lines at P. Proof. We can rephrase the lemma into two parts: Step 1: If F and G have no common tangent lines at P , then I t (F;G)O, for all tm +n 1, that is, is an isomorphism. Proof. If we can show for any large t m +n, we can always reduce t to t 1 such that t 1 still satises the same property, then by Nullstellenstaz, we will be done with the proof. Let such t be chosen. We have earlier given the notation for F = F m + higher terms where F m = Q m i L i , and denedfL i g to be the tangent lines ofF atP . Similarly, for G, we haveG =G n +higher terms where G m = Q n j M j , andfM j g are the tangent lines of G at P . By assumption, L i 6=M j , for all i, j. Now dene the setfA ij j i +j = tg where A ij = L 1 :::L i M 1 :::M j , i;j 0. Note, we let L i = L m , for all i m and M j = M n , for all jn. Lemma 23.fA ij j i +j = tg forms a basis for the vector space of all forms of degree t in k[X;Y ]. Proof. First observe that the size offA ij j i +j = tg is t + 1, which equals to the dimension of the vector space of all forms of degree t in k[X;Y ]. Thus, we only need to show thatfA ij j i +j = tg is linearly independent. 32 To show this, rst notice thatA 0t is the only element in the set that does not have an L-factor, and thus cannot be the linear combination of the rest. Hence, we can eliminate A 0t from the set for the moment, and we need to show the rest of the set is linearly independent, but then A 1(t1) is the only term that has anL 1 -factor, so it cannot be the linear combination of the rest of the terms, and thus we can again eliminate it from the set and only look at the rest of the terms. Proceed in a similar fashion, we can show thatfA ij ji +j =tg is linearly independent. Now, we want to introduce an induction lemma: Lemma 24. If I t+1 (F;G)O, and t m +n 1, then I t (F;G)O. Proof. AssumeI t+1 (F;G)O andtm+n1. If we can show that eachA ij withi+j =t is in (F;G)O, then by the previous lemma, we have I t (F;G)O. Let i +j =t. Since tm +n 1, we have either im or jn. First let's assume that i m. Consider A ij = A m0 B, where B is a form of degree i +jm = tm. Write F = F m + higher terms = A m0 + higher terms =A m0 +F 0 where all term ofF 0 has degree greater than or equal to m + 1. Thus A m0 = FF 0 , so we can write A ij = (FF 0 )B = FBF 0 B. FB is apparently in (F;G)O. Now notice, sinceB only has terms of degree greater thani+jm, andF 0 m+1, we conclude thatF 0 B only has terms of degree greater thani+j+1 =t+1. Thus by assumption, F 0 B2 I t+1 (F;G)O. Hence, A ij 2 (F;G)O, for all i +j = t. The argument goes the same if for the case where jn. We are done with the proof of the induction lemma. 33 Now, we only need to show that for suciently larget,I t (F;G)O, which is a consequence of Nullstellensatz and ideals of nite zeros: Let V (F;G) =fP;Q 1 ;:::;Q r g, and nd some H 2 k[X;Y ] such that H(P )6= 0 andH(Q i ) = 0, for all i, as in the proof of Proposition 7. Hence, HX, HY are both in I(V (F;G)). So, by Nullstenlensatz, (HX) N ; (HY ) N 2 (F;G), for someN. SinceH(P )6= 0,H is invertible inO, and so we have X N ;Y N 2O(F;G). Thus, I 2N O(F;G). Step 2: is injective if and only ifF andG have no common tangent lines. Suppose F and G do not have common tangent lines, we need to show that ker( ) is trivial. Let (A;B)2 ker( ), then AF +BG = 0, that is, AF +BG only has terms of degree greater than or equal to m +n. Write A = A r + higher terms, and B = B s + higher terms. If r < n or s < m, then we must have A r F m =B s G n . But since F m , G n has no common factors, F m must divide B s and G n must divide A r , contradiction. Thus, we must havern andsm, which means AF +BG = 0. Conversely, if F m and G n have a common tangent line L, then we can write F m = F 0 m L and G n = G 0 n L. Then, FG 0 n GF 0 m = F 0 m G 0 n LF 0 m G 0 n L + higher terms, so (G 0 n ;F 0 m ) =FG 0 n GF 0 m = 0. Hence, is not-one-to-one. Now, we have shown thatI P (F;G) exists and satises property (1)- (7). This concludes the proof of existence and hence Theorem 21. 34 Corollary 25. (Property (8) of the intersection numbers) If F and G have no common components, then X P I P (F;G) = dim k (k[X;Y ]=(F;G)): Proof. Recall that from Corollary 10, we have dim k (k[X;Y ]=I) = N X i=1 dim k (O i =IO i ): Thus, let I = (F;G), we have X P I P (F;G) = P X i=1 dim k (O i =(F;G)O i ) = dim k (k[X;Y ]=(F;G)): 35 CHAPTER 4 B ezout's Theorem 1. Projective Curves 1.1. Projective planes. Let k be any eld. The projective plane P 2 (k) over k is dened asP 2 (k) =fPjP2A 3 g=, where P 1 P 2 if 92k;6= 0 such that P 1 =P 2 . We usually abbreviateP 2 (k) asP 2 . IfP2A 3 has coordinates (x 1 ;x 2 ;x 3 ), then the image ofP inP 2 can be written as [x 1 :x 2 :x 3 ]. If x 1 6= 0, this is the same as [1 : x 2 x 1 : x 3 x 1 ]. 1.2. Projective curves. A projective plane curve can be dened in a similar fashion as an ane plane curve: a projective plane curve is the mutual vanishing set of a homogeneous ideal contained ink[X;Y;Z]. 2. B ezout's Theorem Theorem 26. Let F , G be projective plane curves of degree m and n respectively. Suppose in addition that F , G have no common component. Then X P I(P;F\G) =mn: Proof. First notice that since F and G don't have common com- ponents, the setF\G must be nite. Thus, up to a projective change of coordinates, we can assume that none of the points in F\G lies on lines with Z = 0. 37 Remember that we have previously introduced the notationF (X;Y;Z) = F (X;Y; 1). With this notation, and by property (8) of the intersection number, we have P P I P (F;G) = P P I P (F ;G ) = dim k k[X;Y ]=(F ;G ). Now LetR =k[X;Y;Z], =k[X;Y;Z]=(F;G) and =k[X;Y ]=(F ;G ). LetR d , d denote the vector spaces of forms of degreed inR and , re- spectively. In the following proof, we will show that dim k k[X;Y ]=(F ;G ) = dim k = dim k d , for some large d. Step 1: To show dim k d =mn, for all dm +n. Consider the following sequence: 0!R !RR ' !R ! ! 0; where ' : (a;b)7!aF +bG, :h7! (hG;hF ), and is the natural projection. We want to show that the sequence is exact: First notice that is one-to-one and is onto. Now we want to show that ker(') = im( ) and ker() = im('). Let (r;s)2 ker('), then rF +sG = 0. Thus, rF =sG. Since F and G has no common components, we must have some h such that r = hG and s =hF . Hence, ker(') im( ). The other way of the inclusion is obvious, so ker(') = im( ). Let t2 ker(), then t(P ) = 0, for all P2F\G. Thus, t2 (F;G), and equivalently, exist a and b such that t =aF +bG. The other way of inclusion is obvious, so we have ker() = im('). We have shown that the sequence is exact. Now we restrict the sequence as follows, for dm +n, which is also clearly exact: 0!R dmn !R dm R dn ' !R d ! d ! 0 38 We have discussed how to compute the dimension ofR d previously, and thus we have dim k R d = (d+1)(d+2) 2 , dim k R dmn = (dmn+1)(dmn+2) 2 and dim k (R m R n ) = (dm+1)(dm+2) 2 + (dn+1)(dn+2) 2 . Now, we can compute that dim k d = dim k R d dim k ker() = dim k R d dim k im(') = dim k R d (dim k (R m R n ) dim k ker( )) = dim k R d (dim k (R m R n )dim k R dmn ) = (d+1)(d+2) 2 (dm+1)(dm+2) 2 (dn+1)(dn+2) 2 + (dmn+1)(dmn+2) 2 =mn. Thus, we have shown that dim k d =mn, for all dm +n. This completes Step 1. Step 2: To show dim k = dim k d , for dm +n. To prove Step 2, we rst need to introduce a lemma: Lemma 27. The map : ! where :H7!ZH is an isomor- phism. Proof. Since is an endomorphism, to show it is an isomorphism, we only need to show that it is one-to-one, or equivalently to show ker()=0. SupposeZH = 0, that isZH =AF +BG, for someA,B, we need to show that H2 (F;G). For some J 2 k[X;Y;Z], let J 0 denotes J(X;Y; 0). Since we as- sumed that no point in F\G lies on Z = 0, F , G and Z do not have common zeros. Thus, F 0 and G 0 do not have common zeros, and in particular they are relatively prime in k[X;Y ]. Now we have ZH = AF +BH, and so A 0 F 0 =B 0 G 0 . Since F 0 and G 0 are relatively prime, we have A 0 = G 0 C, B 0 =F 0 C, for some C2 k[X;Y ]. Now notice that (ACG) 0 = (B +CF ) 0 = 0, so ACG = ZA 0 and B +CF = ZB 0 , for some A 0 , B 0 . Hence we 39 have (ACG)F + (B +CF )G =AF +BG =ZH =ZA 0 F +ZB 0 G. Canceling Z on both sides, we get H = A 0 F +B 0 G, and thus is in (F;G). Fordm+n, we have shown that dim k d =mn, so we can pick a basisfA 1 ;:::;A mn g for d . We want to show thatfA 1 ;:::;A mn g forms a basis for . Notice that restricting to d is an isomorphism to d+1 , for d m +n. Thus,fZ r A 1 ;:::;Z r A mn g forms a basis for d+r , for all r 0. Now, we are ready to show thatfA 1 ;:::;A mn g is a basis: fA 1 ;:::;A mn g spans: LetH2 ,H2k[X;Y ], and letH denote the form in k[X;Y;Z] achieved by adding Z to each term of H, where deg(H) = deg(H ). Thus, Z N H is a form of degree r, for some suitable N and r. SincefZ r A 1 ;:::;Z r A mn g, Z N H = P mn i=1 i Z r A i , i 2k. Since H = (Z N H ) , we have H = P mn i=1 i A i . fA 1 ;:::;A mn g is independent: Suppose P mn i=1 i A i = 0, then we have P mn i=1 i A i = BF +CG . Thus, for suitable r, s, t, we have Z r P i A i = Z s B F + Z t C G. Hence, P i Z r A i = 0. But since fA 1 ;:::;A mn g forms a basis, all i 's must be 0. Thus,fA 1 ;:::;A mn g is independent. Hence, we have completed the proof of B ezout's theorem. 40 Bibliography [1] William Fulton, Algebraic Curves, An Introduction to Algebraic Geometry, Mathematics Lecture Notes Series, W. A. Benjamin Inc., New York-Amsterdam, 1969. [2] Christopher Hacon, Summer Course in Algebraic Geometry, Lec- ture Notes, University of Utah, 2016. [3] Thomas W. Hungerford, Algebra, Graduate Texts in Mathematics, Springer-Verlag, New York, 1980. 41
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This thesis is an exposition of material taken from Fulton’s book with a focus on multiplicities, intersection numbers, and Bézout's Theorem.
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Algebraic plane curves: from multiplicities and intersection numbers to Bézout's theorem
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