p-adc Galos representatons of G E wth Char(E) = p > 0 and the rng R Gebhard Böckle December 11, 2008 1 A short revew Let E be a feld of characterstc p > 0 and denote by σ : E E the absolute Frobenus endomorphsm x x p. Based on the noton of étale ϕ-module, ntroduced n the last talk, the followng theorem had been proved: Theorem 1.1 There are equvalences of categores Rep Fp (G E ) V M M et ϕ (E), where V assgns to M M et ϕ (E) the mod p Galos representaton (E s E M) ϕ=d of G E and M assgns to V Rep Fp (G E ) the étale ϕ-module ((E s Fp V ) G, σ d) on E. Recall that an étale ϕ-module s a fnte dmensonal E vector space, equpped wth a σ-sem-lnear endomorphsm ϕ such that the lnearzaton of ϕ s an somorphsm. The frst am of ths talk s to present a generalzaton to p-adc Galos representatons of G E. Recall from last tme that O E s a Cohen rng of E. E := Frac(O E ). O be unr s the p-adc completon of lm F/E O F where the lmt s over all fnte unramfed extensons F/E, where unramfed means that the extenson F/E of resdue felds s fnte separable and that p s a unformzer of O F. Êunr := Frac(O be unr). If E s perfect then O E s unque (up to unque somorphsm) and somorphc to W (E) and O be unr = W (E s ). A Frobenus endomorphsm: Usng a basc property of Cohen rngs, there exsts a lft σ : O E O E of σ : E E and we fx one such. It has a unque extenson σ : O be unr O be unr whch reduces to σ : E s E s on resdue felds. [For F over E fnte, the resdue feld F of F contans Eσ(F ), and hence by standard feld theory there exsts a unque extenson of σ to F. The extenson from lm F to the p-adc completon s the unque contnuous F/E one.] Abbrevate G := G E = Gal(E unr /E) = Aut cont (Êunr /E). 1
2 p-adc Galos representatons of G E Theorem 2.1 There are equvalences of categores Rep Zp (G E ) V M M et ϕ (O E ), where V assgns to M M et ϕ (O E ) the Galos representaton V(M) := (O be unr OE M) ϕ=d of G E and M assgns to V Rep Zp (G E ) the étale ϕ-module M(V ) := (O be unr Zp V ) G on O E wth ϕ = σ d and where V and M are quas-nverse to each other. Example 2.2 Perhaps the smplest Cohen rng whch s not a rng of Wtt vectors s the followng: Let k be a perfect feld of characterstc p and W := W (k) ts rng of Wtt vectors. Let E := k((x)). Then a Cohen rng of E s gven by { } O E := a x : a W and lm a = 0. Z One can easly verfy that O E s a complete dscrete valuaton rng wth maxmal deal p O E and resdue feld E. A Frobenus lft s σ : O E O E sendng λ W to σ(λ), where σ : W W s the unque lft of σ restrcted to k, and sendng x to x p. Another possble choce for the mage of x s x p + px. Proposton 2.3 The followng hold: (a) (O be unr) G = O E and (Êunr ) G = E. (b) (O be unr) σ=d = Z p and (Êunr ) σ=d = Q p. Proof: s clear n all cases. (For (b) note that d s a lft to Z p of σ on F p ). : For (b) note that (O be unr) σ=d (W (E s )) σ=d = W (F p ) by drect nspecton of ϕ on Wtt vectors. The assertons of (a) are clear for the uncompleted rngs O E unr and E unr. To prove (a) one can ether use a smlar argument as for (b), or one can consder the followng dagram 0 p n (O be unr) G (O be unr) G (O be unr /p n ) G H 1 (...) 0 p n (O E unr) G (O E unr) G (O E unr/p n ) G H 1 (G, O E unr). The term H 1 (G, O E unr) s zero by the addtve Hlbert 90 theorem, and thus the second row s 0 p n O E O E O E /p n 0. So the dagram yelds that the p-adc completon of (O be unr) G s a subrng of O E = lm O n E /p n. By contnuty of the acton of G, the rng (O be unr) G s p-adcally complete whch completes the proof. 2
For the proof of Theorem 2.1, we need the followng Key Lemma 2.4 (a) Suppose X M et ϕ (O be unr). Then O be unr Zp X ϕ=d = X. (Lang s Thm) (b) Suppose X s a contnuous G-module, fntely generated over O be unr. Then O be unr OE X G = X. (Hlbert 90) Note that (b) (for all X... ) s equvalent to H 1 cont(g, Aut(X)) = {1} (for all such X). Proof: We frst explan why t wll suffce to prove both parts of the lemma for O be unrmodules X of fnte length. So suppose ths s done and consder X = lm X/p n X. n Havng the assertons of the lemma for all X/p n X, one easly deduces that the nverse lmt systems (X/p n X) ϕ=d or (X/p n X) G, respectvely, have surjectve transton maps. Snce G and ϕ act contnuously, t follows that the nverse lmts of these agree wth X ϕ=d or X G, respectvely. The assertons for X n the lemma now drectly follows from the assertons for all X/p n X (of fnte length) n the nverse lmt system. Suppose now that len(x) <. The am s to reduce the proof to assertons proved n the last talk,.e., to the case where X s a vector space over E s. We nduct over n N such that p n X = 0. Defne X := {x X px = 0} and consder the short exact sequence 0 X X X 0. Takng ϕ-fxed ponts, or G nvarants, respectvely, yelds the exact sequences 0 (X ) ϕ=d (X) ϕ=d (X ) ϕ=d X /(ϕ d)x, (1) 0 (X ) G (X) G (X ) G H 1 cont(g, X ). (2) The module X s p-torson and hence a fnte dmensonal vector space over E s. So to t we can apply the results of last tme. In case (a) t wll be an étale ϕ-sheaf over E s. Any such s trval,.e., somorphc to a fnte sum of copes of (E s, σ) (by Lang s Theorem). Snce ϕ d s surjectve on E s, the rght-most term of (1) s zero. Smlarly, n case (b) the results from last tme mply that X s a trval G-module over E s,.e., somorphc to a fnte sum of copes of E s wth the canoncal Galos acton (by Hlbert 90). By the addtve Hlbert 90, the rght-most term of (2) s zero. To fnsh the proof (say, only n case (b)), consder the followng dagram: 0 O be unr OE (X ) G O be unr OE X G O be unr OE (X ) G 0 0 X X X 0. By the result from last tme, the vertcal arrow on the left s an somorphsm, because X s trval. By our nducton hypothess, the vertcal arrow on the rght s an somorphsm. Hence by the Snake Lemma, then central vertcal arrow s an somorphsm. 3
Corollary 2.5 The followng natural maps are somorphsms: (a) For T Rep Zp (G) and M(T ) = (O be unr Zp T ) G the map α T : O be unr OE M(T ) O be unr Zp T. (b) For M M et ϕ (O E ) and V(M) = (O be unr OE M) ϕ=d the map α M : O be unr Zp V(M) O be unr OE M. The njectvty of the maps follows from the Artn trck note that the results of the last talk are not drectly applcable to the d.v.r. O be unr. By applyng the prevous lemma to X = O be unr Zp T and X = O be unr OE M, respectvely, t follows that the α? are somorphsms. Proof of Theorem 2.1: By applyng V to the somorphsm α T and M to the somorphsm α M, Proposton 2.3 yelds that V M and M V are naturally somorphc to the respectve dentty functors. 3 The rng R The am of ths part of the talk s to ntroduce a rng R whch wll be useful when descrbng p-adc Galos representatons of the absolute Galos group G K of a local feld K. Before we come to ts defnton, we try to gve a motvaton. 3.1 A motvaton Let ε (n) denote a prmtve p n -th root of unty n Q p. (Later we wll assume that (ε (n+1) ) p = ε (n) for all n.) Defne K cyc := n K(ε(n) ). Usng that Gal(Q cyc p /Q p ) = Z p = Z p F p ( or = Z 2 Z/(2) for p = 2) one deduces Gal(K cyc /K) = Z p K for a fnte subgroup K F p (or K Z/(2) for p = 2). One defnes K := (K cyc ) K and Γ K := Gal(K /K). Consder K K cyc H K G K K K K tot. ramf. K 0 unr. Γ K Q p. 4
An mportant observaton by Fontane-Wntenberger (?) s that there exsts a feld of characterstc p, the feld E K of norms of K such that H K = Gal(E s K /E K ). One has E K = kk ((π K )) for k K the resdue feld of K and π K an ndetermnate. We wll not prove ths, but only recall the defnton of E K : Defne K n = (K ) pn Γ K and Norm K1/K E K = lm(k 0 Norm K2 /K 0 1 K1 K2...) One can prove that that E K s also somorphc to x x E K = lm(k p x x K p K...) We wll bypass the theory of feld of norms. But the second descrpton of E K s remnscent of the defnton of R we are about to learn. 3.2 The rng R(Ā) Let Ā be a rng of characterstc p and ϕ : Ā Ā : x xp be the Frobenus endomorphsm of Ā. Defnton 3.1 R(Ā) := lm (Ā ϕ Ā ϕ Ā ϕ...) = {(x n ) ĀN n : x p n+1 = x n} The rng R(Ā) s perfect and reduced, because (x n) = (x n+1 ) p and f (x n ) pm = 0, then (x n+m ) = (0). Let θ m : R(Ā) Ā : (x n) x m. The followng lemma, may later be useful: Lemma 3.2 Suppose R R(Ā) s a topologcally closed subrng such that θ m( R) = θ m (R(Ā)) for all m. Then R = R(Ā). Proof: R(Ā) = {(x n)...} = lm Im(θ 1,..., θ m ) = lm (θ 1,..., θ m )( R) = R. Note that f ϕ s njectve, then R(Ā) s smply the ntersecton Ā pn. The cases we wll be nterested n are cases where Ā s hghly non-reduced, such as O K /po K. Suppose A s a separated p-adcally complete topologcal rng,.e., A = lm A/p n. Set Ā := A/p. Proposton 3.3 The map S A := {(x (n) ) A N n : (x (n+1) ) p = x (n) } R(Ā) : (x(n) ) (x (n) mod pa) n s a bjecton. It s a rng somorphsm f on S A one defnes ( (x (n) ) (y (n) ) = (x (n) y (n) ) and (x (n) ) + (y (n) ) = lm m (x(n+m) + y (n+m) ) pm). n 5
Proof: Defne R(Ā) S A : (x n ) (x (n) ) as follows: Lft x n Ā to ˆx n A for all n. Then ˆx p n+1 ˆx n pa because mod p the element s x p n+1 x n = 0. Usng the bnomal theorem for (x + y) p, one deduces for all n: ˆx pm+1 n+1 ˆxpm n p m+1 A. It follows that for all n the sequence (ˆx n+m ) pm s a Cauchy sequence n m. Defnng x (n) as ts lmt, t follows that (x (n) ) s n S A and t obvously reduces to (x n ) modulo pa. The x (n) are ndependent of the choce of the lfts ˆx n. The sequence s also the unque sequence n S A lftng R(Ā). Ths shows that the map s a bjecton. It remans to see that for (x n ) and (y n ) n R(Ā) wth lfts (x(n) ) and (y (n) ) n S A the gven addton formula n S A descrbes the lft of (x n +y n ). But ths s clear: (x (n) +y (n) ) s a lft of x n + y n and by the formula whch gves the canoncal lfts from any sequence of lfts, we obtan the addton formula of the proposton. 3.3 The rng R Let C := K. Defnton 3.4 R := R(O K /po K ) = R(O C /po C ) = {(x (n) ) OC N n : (x n+1 ) p = x n } Theorem 3.5 The followng hold: v R ((x (n) )) := v C (x (0) ) (x (n) ) R. (a) The rng R s perfect of characterstc p. (b) (R, v) s a complete valuaton rng wth valuaton v = v R, wth v(r) = Q 0 { } wth maxmal deal m R = {x R v(x) > 0} and resdue feld k = F p. (c) If the Techmüller lft k O K unr = W ( k) s denoted a â, then k R s the 0 map a (âp n ) n. (d) Frac(R) s algebracally closed. It follows that R s the completon of the algebrac closure of k((x)) for any x R wth strct postve valuaton. (Hence R = Ês K.) Also, note that one has the dentfcaton Frac(R) = {(x (n) ) C N n : (x (n+1) ) p = x (n) }. Proof: Part (a) s clear. Part (c) s straghtforward from the prevous proposton. We frst prove (b): Snce C s algebracally closed t s closed under takng p-th roots and hence any c C can occur as c (0) n a sequence (c (n) ) such that (c (n+1) ) p = (c (n) ) for all n. Hence v R (R) = v C (O C ) s as descrbed. Also v R ((x (n) )) = f and only f x (0) = 0 whch s equvalent to all x (n) beng zero. The condton v(x y) = v(x)+v(y) s straghtforward from the defnton of multplcaton on sequences x = (x (n) ) and y = (y (n) ): v R ((x (n) ) (y (n) )) = v R ((x (n) y (n) )) = v C (x (0) y (0) ) = v C (x (0) ) + v C (y (0) ). 6
The ultrametrc trangle nequalty follows from ( v R (x + y) = v R ((x (n) ) + (y (n) )) = v R lm m (x(n+m) + y (n+m) ) pm) ( = v C lm m (x(m) + y (m) ) pm) ( ) = lm m pm v C (x (m) + y (m) ) }{{} lm nf m ( { mn p m v C (x (m) ) }{{} =v C (x (0) ) mn{v C (x (m) ),v C (y (m) )} }), p m v C (y (m) ) = mn{v }{{} R (x), v R (y)}. =v C (y (0) ) For (b) t remans to show that v defnes the topology on R gven by the nverse lmt topology (of the dscrete sets O C /po C ) and that v s complete. For the topology, note that a neghborhood bass of R s gven by the sets (Ker(θ m )) m. Now for a sequence (x (n) ) we have (x (n) ) Ker(θ m ) (x (m) mod p 0O C v C (x (m) ) 1 v R ((x (n) )) p m. Thus the topologes agree. The completeness follows from the dscreteness of O C /po C : If n r n the r n R tend to zero, then under any θ m the sum becomes statonary and thus t converges. We fnally prove (d): Consder an rreducble polynomal P (x) = x d + a d 1 x d 1 +... + a 1 x + a 0 R[x]. (It suffces to consder coeffcents n R nstead of Frac(R).) Because R s perfect, we may assume that P s separable,.e., that t has no multple roots. Wrte a = (a (n) ) S OC and consder P (n) (x) = x d + a (n) d 1 xd 1 +... + a (n) 1 x + a(n) 0 O C [x]. Because C s algebracally closed, the polynomal P (n) has roots α (n) 1,..., α(n) d n C. We would lke to see that for n 0 these roots are parwse dstnct modulo po C. For ths we show that the dscrmnants of the P (n) have v C valuatons convergng to 0, so that also the valuaton of the dfference of dstnct roots has to converge to zero. The dscrmnant of P (n) s the Resultant of P (n) and (P (n) ). The latter can be computed from a determnant contanng as ts entres the coeffcents of P (n) and of (P (n) ). Determnants can be explctly wrtten n terms of sums and products of matrx entres. Based on ths, one can verfy the followng: If (a j ) M d (R) and a j = (a (n) ), then det(a (n+m) pm m j ) (det(a j )) (n). If one apples ths to the above way of computng the dscrmnant of P, t follows that the sequence n C (n upper ndexng) representng dscr(p ) s gven by dscr(p ) (n) = lm dscr(p (n+m) ) pm. m Snce P s has no multple roots, dscr(p ) s non-zero, and so v C (dscr(p ) (0) ) 0 s non-zero. Clearly v C (dscr(p ) (n) ) = 1 p n v C (dscr(p ) (0) ), and so t follows that dscr(p (n) ) n 0. In partcular, for all n 0 we can order (for fxed n) the α (n) n such a way that they form sequences n O C satsfyng (α (n+1) ) p = α (n) (for all n 0). It easly follows that these sequences defne elements of R whch are roots of P. j 7