What is the Multipliity of a Base Point

What is the Multipli ity of a Base Point? Outline  Bezout's Theorem  Degree of a Rational Surfa e  Serre's De nition  Hilbert-Samuel De nition  Combinatorial Computation  Algebrai Computation  Proof of Degree Formula  Other Stu (Time Permitting) Bezout's Theorem Suppose that C; D  P2 are urves of degrees n; m with no ommon omponents. Then X nm = 2 \ p C D I (C; D) p where ~ ~gi I (C; D) = dim OP2 =hf; p ;p ~ ~g are lo al equations of and f; C; D near p. Degree of a Rational Surfa e ' : P2 ! P3 is de ned by a; b; ; d of degree n, no ommon fa tors.  Z = fp j a(p) =


= 0g is the base point lo us.  S = '(P2 n Z ) is the image. We assume dim S = 2. Degree Formula: n2 = deg S
deg ' + X m(p): 2 p Z m(p) is the \multipli ity" of p. Naive guess: m(p) = dim OP2 =I = dim R =I ; ;p p p where I = h~a; : : : ; d~i p Counterexample: a = s2u + t3, b = t2u + s3, = stu, d = s2u gives ' with basepoint p = (0; 0; 1) and I = hs2; st; t2i. p The naive guess implies m(p) = dim R =I = 3: p p p However: A Grobner basis al uluation shows that the image surfa e S  P3 has degree 5. Consequen e: Sin e n = 3, the Degree Formula gives 32 = 5
1 + m(p); so that m(p) = 4 in this ase. Serre's De nition R lo al ring, m maximal ideal. Assume R ontains k = R=m. Then R-modules M; N with dim M R k N <1 have interse tion multipli ity (M; N ) de ned by X ( 1) dim Tor (M; N ): 0 i k i R i Example: Bezout Situation Let R = OP2 , M = R=hf~i, N = R=h~g i. Then ;p ~ 0 !R !R !M !0 gives ! Tor1 (R; N ) ! Tor1 (M; N ) ~ !N !N !M N !0 However, no ommon omponent implies ~ N = R=h~g i ! N = R=h~g i is one-to-one. f R R f R f Thus Tor (M; N ) = 0 for i > 0, so that R i (M; N ) = dim M N ~ ~gi: = dim R=hf; R Observations:  f;~ ~g form a regular sequen e in the ring R.  hs2; st; t2i is not generated by a regular sequen e. Hilbert-Samuel De nition Let R, m and k = R=m be as above and let M be a f.g. Rmodule. For `  0, the Hilbert polynomial implies that e +1 dim (M=m M ) = ` + : : : d! where d = dim R and e = e(M ) is the multipli ity of M . d ` k Theorem: If dim R = 0, then e(R) = dim R: k Re nement: Let I be an ideal with m M  IM for some s. Then `  0 implies that ~ e +1 dim (M=I M ) = ` + : : : d! ~e = e(I; M ) is the multipli ity of I in M . s d ` k Main Claim: In the Degree Formula, m(p) = e(I ; R ); for R = OP2 and I = h~a; : : : ; d~i. p p ;p p p Combinatorial Computation Let I  k[x1; : : : ; x ℄h 1 ni = R have nite odimension. Then:  The exponents of a monomial basis of R=I give a nite set E  Z0.  Let C = Conv(Z0 n E ). This gives the multipli ity: n x ;:::;x n n Theorem: e(I; R) = n! Vol (R0 n C ) n n Example: I = hs2; st; t2i  k[s; t℄h i has basis 1; s; t of R=I . Then s;t i i i i i i i i i i i i i  y i i y yi i i i i i i gives e(I; R) = 2
shaded area = 2
2 = 4. Example: I = hs2; t2i  k[s; t℄h i has the same multipli ity. s;t Algebrai Computation Assume m  I  R, where R = regular lo al ring of dimension n. Then:  If m  J  I  R, then s s e(J; R)  e(I; R):  If in addition I J = I +1, then ` ` e(J; R) = e(I; R): J is a redu tion ideal of I .  If I is generated by a regular sequen e, then e(I; R) = dim R=I k I is a omplete interse tion. Two Important Fa ts:  I has a redu tion ideal whi h is generated by a reg. seq.  The reg. seq. an be hosen to be generi linear ombinations of the generators of I . Consequen e of First Fa t: e(I; R) = min dim R=J; k where the minimum is taken over all omplete interse tion ideals J ontained in I . This follows be ause e(I; R)  e(J; R) = dim R=J k holds for any CI ideal ontained in I and be ause (by the rst fa t) I has a CI redu tion ideal. Proof of Degree Formula For ' : P2 ! P3 given by a; b; ; d, we need to prove n2 = deg S
deg ' + X e(I ; R ) 2 p p p Z for R = OP2 and I = h~a; : : : ; d~i. p ;p p Pi k a generi line `  P2. Then deg S = #S \ `: We an assume:  ` meets S at smooth points.  ` meets S transversely.  ' is etale above these points. For oordinates x; y; z; w on P3, let ` = H1 \ H2 for H1 : 1x +


+ 4w = 0 and H2 : 1x +


+ 4w = 0. Then, on P2, onsider the urves C : f = 1a +


+ 4d = 0 D : g = 1 a +


+ 4 d = 0: Sin e Z is the basepoint lo us of a; b; ; d, we have C \ D = ' 1(S \ `) [ Z: By Bezout's Theorem, n2 = deg S
deg ' X ~ ~gi: + dim OP2 =hf; 2 p k ;p Z However, the se ond important ~ ~g generate a fa t implies that f; redu tion ideal for I . Thus ~ ~gi e(I ; R ) = dim OP2 =hf; and the theorem is proved! p p p k ;p Another Proof We will use Fulton's Interse tion Theory. By p. 79, the Segre lass of Z  P2 is the 0- y le s(Z; P2) = X e(I ; R )[p℄: 2 p p p Z Then Prop. 4.4 implies Z deg S
deg ' = P2 1(L)2 Z (1 + 1(L))2 \ s(Z; P2) Z where L = OP2(n). Then we are done sin e Z has dimension 0! The Rees Ring This is the graded ring R Then set 1 M +( ) = f= R I R i=0 +( ) ii I t +( ) I =mR I : This is graded and nitely generated over k = R=m. One an show that dim R = dim R, whi h we denote n. f By graded Noether normalization, there are generi ~s1; : : : ; ~sn 2 I =mI su h that R is a f. g. module over k [~ s1 ; : : : ; ~ sn ℄. f One an show:  +( ) is nitely generated over [1 n℄.  1 n are a regular sequen e. R I R s ;:::;s s ;:::;s We may assume that the si are generi linear ombinations of generators. Now suppose that u1; : : : ; uN generate R+ (I ) over R[s1 ; : : : ; sn ℄. Set ` = max degree of 1 N: u ;:::;u Then one an easily show that I `+1 = hs1; : : : ; sniI `: See Se tion 4.5 of Cohen-Ma aulay Rings by Bruns and Herzog for details.