Introduction to Arithmetic Geometry Fall 2013 Lecture #20 11/14/2013

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1 Introduction to Arithmtic Gomtry Fall 2013 Lctur #20 11/14/ Dgr thorm for morphisms of curvs Lt us rstat th thorm givn at th nd of th last lctur, which w will now prov. Thorm Lt φ: C 1 C 2 b a morphism of curvs dfind ovr k. Thn for ach closd point Q of C 2 /k, dg φ (Q) = dg φ dg Q Bfor bginning th proof, lt us first show that w can assum without loss of gnrality that k is algbraically closd. If th closd point Q is th G k -orbit {Q 1,..., Q d }, with d = dg Q, aftr bas xtnsion to k w hav dg φ (Q) = dg φ (Q Q d ) = dg φ (Q 1 ) + + dg φ (Q d ), sinc both th dgr map and th pullback map φ : Div k(c 2 ) Div k(c 1 ) ar group homomorphisms. If w assum th thorm holds ovr k, thn vry trm on th right is qual to dg φ and th sum is d dg φ = dg φ dg Q. W now prov th thorm assuming k = k, following th approach of [1, III.2]. Proof of Thorm Fix Q C 2, and lt O Q b its local ring of rgular functions. Th st φ 1 (Q) is finit bcaus φ is not constant and C 1 is an irrducibl algbraic st of dimnsion on (so all its propr closd substs ar finit). Lt P 1,..., P n C 1 b th lmnts of φ 1 (Q), lt O 1,..., O n b th corrsponding local rings of rgular functions, and dfin n O = O i. By Lmma 20.4 blow, thr xist uniformizrs t 1,..., t n for O 1,... O n such that { 1 if i = j, ord Pi (t j ) = 0 othrwis. Th maximal idals of O ar (t 1 ),..., (t n ) and ach nonzro f O factors uniquly as i=1 f = ut 1 1 tn n, with u O and i = ord Pi (f). Undr th map φ : k(c 2 ) k(c 1 ), for any f O Q w hav ord Pi (φ f) = ord Pi (f φ) = ord Q (f) 0, thus φ (O Q ) is a subring of O. If w now lt t Q b a uniformizr for O Q, and put t = φ t Q w hav t = φ t = ut 1 t n Q 1 n whr i = ord Pi (φ t Q ) = φ (P i ). Sinc t 1,..., t n ar pairwis rlativly prim (maning that (t i ) + (t j ) = O for all i = j), by th Chins rmaindr thorm w hav n O/(t) i=1 O/( ti i) (1) 1 Andrw V. Suthrland

2 as a dirct sum of rings that ar also k-vctors spacs (hnc k-algbras). To prov th thorm w will comput th dimnsion of O(t) in two diffrnt ways, corrsponding to th two sids of th quality dg φ (Q) = dg φ that w ar trying to prov. On th LHS of th quality w wish to prov, th dgr of th divisor φ (Q) is dg φ (Q) = φ (P i ) dg P i = i, (2) sinc w hav dg P i = 1 for k = k. W claim that this is prcisly th dimnsion of O(t) i O/(t i i ) as a k-vctor spac, which w will prov blow. On th RHS of th quality, aftr idntifying k(c 2 ) with its imag φ (k(c 2 )) w hav dg φ = [k(c 1 ) : k(c 2 )], which w claim is qual to th rank of O as an O Q -modul (O Q is mbddd in O via φ ). Th ring O is an intgral domain that is finitly gnratd as a modul ovr th principal idal domain O Q, so it is torsion fr and isomorphic to O r Q for som intgr r (by th structur thorm for moduls ovr PIDs), hnc it maks sns to spak of its rank r. Th filds k(c 1 ) and φ (k(c 2 )) ar th fraction filds of th rings O and O Q, rspctivly, and it follows that th maximal numbr of lmnts of O that ar linarly indpndnt ovr O Q is xactly th sam as th maximal numbr of lmnts of k(c 1 ) that ar linarly indpndnt ovr k(c 2 ), which is prcisly [k(c 1 ) : k(c 2 )] = dg φ = d. If w choos a basis α 1,..., α d for k(c 1 ) ovr k(c 2 ) and lt = min{ ord Pi (α j ) : 0 i n, 0 j d }, thn th functions α 1 /t,..., α d /t ar rgular at all th P i and thrfor li in O. Thy ar linarly indpndnt ovr O Q, thus r d, and clarly r d, sinc any r lmnts of O k(c 1 ) that ar linarly indpndnt ovr O ar also linarly indpndnt ovr its fraction fild k(c 2 ). W hav O Q /(t) k, sinc (t) is a maximal idal, so dim k O/(t) = r = d = dg φ. To prov dim k O(t) = dg φ i (Q), by (1) and (2) it suffics to show that dimk O/(t i ) = i. W claim that for any positiv intgr n, ach function f O can b writtn uniquly as f a 0 + a 1 t i + + a n i mod t n i, 1 t n 1 with ach a i k. Applying this with n = i will yild th dsird rsult. For n = 1 w lt a 0 = f(p i ) k. W thn hav ord Pi (f a 0 ) = ord Pi (f f(p i )) 1, so f a 0 mod t i as dsird, and clarly a 0 is uniquly dtrmind. W now procd by induction on n, assuming that f g = a 0+a1t i+ an 1t n 1 i mod t n i. Th ord P i (f g) n, so h = t n i (f g) is rgular at P i and thrfor lis in O (sinc ord Pj (t i ) = 0 for j = i). Now lt a n = h(p i) k. Thn ord P (t n (h a )) n + 1 and w hav f g + a t n mod t n+1 i i n n as dsird. Th ky to th proof of Thorm 20.1 is Lmma 20.3, which gav us th indpndnt uniformizrs t 1,..., t n w ndd. In ordr to prov th lmma w nd a tight form of th (nonarchimdan) triangl inquality for valuations. Lmma 20.2 (Triangl quality). Lt v : F Γ b a valuation on a fild F. For any x, y F such that v(x) = v(y) w hav v(x + y) = min((v(x), v(y)). Proof. Assum v(x) < v(y). By th triangl inquality, v(x + y) min(v(x), v(y)). If this is not tight, v(x + y) > v(x), but thn v(x) = v((x + y) y) min(v(x + y), v(y)) > v(x), a contradiction. 2

3 W now prov th main lmma w nd, which is mor gnrally known as th thorm of indpndnc of valuations for function filds. Lmma 20.3 (Indpndnc of valuations). Lt P 1,..., P n b distinct placs of a function fild F. Thn thr xist t 1,..., t n so that v i (t j ) = δ ij (Kronckr dlta), whr v i dnots th valuation for P i. Proof. If n = 1, w can tak t 1 to b any uniformizr for P 1. W now procd by induction, assuming that t 1,..., t n 1 satisfy v i (t j ) = δ ij. It suffics to find tn with v n (t n ) = 1 and v i (t n ) = 0 for 0 i < n. With such a t n, w can thn rplac ach t i with t i /t n, whr = v n (t i ), so that v n (v i ) = 0 and v i (t j ) = δ ij as rquird. If v n (t i ) = 0 for 0 i < n, w can simply pick a uniformizr for P n and multiply it by suitabl powrs of th t i so that this is achivd, so lt us assum othrwis. W now pick s 1,..., s n 1 in O Pn with s i O Pi ; this is possibl bcaus non of th O Pi contain O Pn, by Thorm Thn v n (s i ) 0 and v i (s i ) < 0 for 0 i < n. By rplacing ach s i with si i for som suitably larg i > 0 w can arrang it so that at ach valuation v j, for 0 j < n, th valu min{ v j (s i i ) : i 0 i < n} is achivd by a uniqu si (possibly th sam s i i for diffrnt v j s). For s = si i w thn hav v j (s) < 0 for 0 j < n, by th triangl quality, and v n (s) 0. Now lt t b a uniformizr for O Pn, so v n (t) = 1. If v n (s) = 0 thn w can rplac t by s t for som suitabl so that v i (t) < 0 for 0 i < n and v n (t) = 1, and if v n (s) > 0 w can achiv th sam goal by rplacing t with s + t (again by th triangl quality). Now lt w b th product of t with suitabl powrs of t 1,..., t n 1 so that v i(w) = 0 for 0 i < n. If v n (w) = 0 thn apply th sam procdur to t + t for som suitably chosn > 0 so that this is not th cas (w hav v n (t i ) = 0 for som t i, so this is always possibl). Finally, if v n (w) < 0 thn rplac w with 1/w so v n (w) > 0. W than hav v i (w) = 0 for 0 i < n and v n (w) > 0. Now lt z = w + 1/t. W hav v i (1/t) > 0 for 0 i < n and v n (1/t) = 1, so by th triangl quality, v i (z) = 0 for 0 i < n and v n (z) = 1. For t n = 1/z w thn hav v i (t n ) = 0 for 0 i < n and v n (t n ) = 1 as dsird, and w ar don. Corollary Lt O 1,..., O n b distinct discrt valuation rings of a function fild F/k. Th ring O = i O i has xactly n nonzro prim idals (t 1 ),..., (t n ), ach principal and gnratd by a uniformizr for O i. Evry nonzro f O can b uniquly factord as f = ut 1 1 tn n with u O and i = ord Pi (f) 0. Proof. Th lmnts t 1,..., t n givn by Lmma 20.3 ar uniformizrs for O 1,..., O n, and it follows that vry f F can thn b writtn uniquly in th form x = ut 1 1 tn n with u O and i = ord Pi (x). Th nonzro lmnts of O ar prcisly thos for which th i ar all nonngativ, and th lmma is thn clar. W now not a furthr corollary of th lmma, which is an analog of th wak approximation thorm w provd in Lctur 11. Corollary 20.5 (Wak approximation for function filds). Lt P 1,..., P n b distinct placs of a function fild F/k, and lt f 1,..., f n F b givn. For vry positiv intgr N thr xists f F such that ord Pi (f f i ) > N for 0 i < n. Proof. Lt t 1,..., t n b as in Lmma As in th proof of Thorm 20.1, w can construct Laurnt polynomials g i k((t i )) such that g i f i mod t N i, whr th first nonzro trm of 3

4 g i is a s t s i whr a s = ord Pi (f). W thn hav ord Pi (g i f i ) N, and ord Pj (g i ) 0 for j = i sinc ord Pj (t i ) = 0 for j = i, this follows from th triangl inquality. Multiplying ach g i by (t N 1 t i 1 t i+1 t n ) and summing th rsults yilds th dsird function f. Not that in trms of absolut valus, making th valuation ord Pi (f f i ) larg corrsponds to making th corrsponding absolut valu f f i Pi small. To mak th analogy with Thorm 11.7 mor prcis, w could construct th compltions of F Pi at ach plac P i and thn th f i givn in th thorm would li in F Pi but f would still li in F. Th rlationship btwn F and its compltions F Pi is thn xactly analogous to th rlationship btwn Q and its compltions Q pi Divisors of dgr zro It follows from Thorm 20.1 that th group of principal divisors Princ k C is a subgroup of th group of dgr zro divisors Div 0 k C, th quotint Div0 k C/ Princ k C is dnotd Pic 0 k C. Equivalntly, Pic 0 k C is th krnl of th dgr map Pic C Z. W thn hav th xact squnc 1 k k(c) Div 0 k C Pic0 k C 0. Up to now all th groups of divisors and divisor classs w hav considrd hav bn infinit, but this is not tru of Pic 0 k. Th cas whr Pic0 k is trivial is alrady an intrsting rsult. Thorm Assum k = k. Thn C P 1 if and only if Pic 0 k C = {0}. Proof. Th forward implication is asy. Each point P = (a 0, a 1 ) P 1 is th zro locus of th polynomial f P (x 0, x 1 ) = a 1 x 0 a 0 x 1, and if w hav a divisor D = n P P w can n construct a corrsponding homognous rational function f = fp P. If D has dgr zro thn th numrator and dnominator of f hav th sam dgr and f is an lmnt of k(p 1 ) k(c), so D = div f. Thus Div k C = Princ k C and Pic 0 k C = 0. Now lt P and Q b distinct points in C(k); such P and Q xist bcaus k is algbraically closd. Thn f = f P /f Q is a non-constant function in C(k) that dfins a morphism (f P : f Q ) from C to P 1. Th polynomials f P and f Q hav dgr on, and this implis that th morphism f has dgr on and is an isomorphism. To chck this, w can us Thorm 20.1 with Q = 0 and t 0 = x/y to comput dg f = dg f (0) = f (P ) = ord P (f t 0 ) = ord P (t 0 f) = ord P (f P /f Q ) = 1. Now lt us considr th gnral cas, whr k is not ncssarily algbraically closd. W thn nd to work with closd points, but th forward implication still holds: if C/k is isomorphic to P 1 /k thn Pic 0 k C is trivial; th polynomials f P in th proof ar now irrducibl polynomials that may hav dgr gratr than on, but that dosn t chang th argumnt. But th convrs is mor intrsting. W can always find closd points P and Q on C/k, but for th abov proof to work w nd thm to hav dgr on, othrwis th function f P /f Q will not b an isomorphism. Equivalntly, w nd C/k to hav two distinct rational points P and Q; ths ar closd points of dgr on. W alrady know from arlir in th cours that if C/k has gnus 0 and vn on rational point thn it is isomorphic to P 1 /k (and thn it has mor than two rational points). But if C/k has positiv gnus it can happn that C/k has on rational point and Pic 0 k C = {0}, but C cannot b isomorphic to P 1, bcaus P 1 has gnus zro. Indd, this is xactly what happns for th lliptic curv 4

5 y 2 = x ovr Q, whos only rational point is. So w nd to add th hypothsis that C/k hav two distinct rational points in ordr to gt a thorm that works for gnral k. Corollary Lt C/k b a curv with at last two distinct rational points. Thn C/k is isomorphic to P 1 /k (with th isomorphism dfind ovr k) if and only if Pic 0 k C = {0}. As an intrsting consqunc, if C has gnus gratr than zro and at last two rational points, thn Pic 0 k C cannot b trivial. Th lliptic curv C : y2 = x 3 1 ovr k = Q is such an xampl, with Pic 0 k C of ordr 2. Rfrncs [1] I. R. Shafarvich, Basic algbraic gomtry, 2nd dition, Springr-Vrlag, [2] H. Stichtnoth, Algbraic function filds and cods, Springr,

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