THE FUNDAMENTAL EXACT SEQUENCES AND THE SCHEMATIC CURVE

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1 THE FUNDAMENTAL EXACT SEQUENCES AND THE SCHEMATIC CURVE 1. The sace Y ad and X ad 1.1. Notation. In this talk we continue with the following notation from revious talks: (1) Let C be an algebraically closed field of characteristic, comlete under : C R 0. (2) Let ϖ C be a quasi-uniformizer, 0 < ϖ < 1. (3) Let A inf = W (O C ) be Fontaine s eriod ring A inf. 1 (4) Un(C ) denotes the set of characteristic 0 untilts of C Review. We have introduced Y ad = Y ad C = Sa(A inf, A inf )\V ( [ϖ ]) which arametrizes all characteristic 0 untilts of C. As notation suggests, Y ad is an adic sace (though it is not affinoid and the fact the adic structural resheaf is a sheaf is nontrivial). The adic Fargues-Fontaine curve will be a certain quotient X ad = Y ad /ϕ Z which arametrizes Un(C )/ϕ Z. We have already seen the heuristic icture of Y ad : it can be thought of as a unctured oen disc, where C Un(C ) C gives the radial coordinate. The origin of the disc can be thought of as = 0 while the boundary of the disc is [ϖ ] = 0. The untilts of C (which in our setu where C is algebraically closed can be identified with classical oints of Y ad ) tries to degenerate to the characteristic untilt C as they tend to the origin of the disc. Remark. In fact, the global functions on Y ad (which we recall in next subsection) have aeared in -adic Hodge theory before the curve is introduced, but it is Fargues and Fontaine s oint of view that such functions can be regarded as holomorhic functions with variable The ring B. The ring B that Jacob defined in the revious lecture should be regarded as global functions on Y ad, namely B = O(Y ad ). To construct B, we consider a sequence of rings R(k): where R(k) = A inf [ k 1 A inf R(k) R(k-1) B 0 = A inf [ [ϖ ], 1 ] [ϖ ], [ϖ ] k ]. Then we take the Q Banach sace R(k) Q := lim m (R(k)/ m R(k))[ 1 ] 1 In the book of Fargues and Fontaine, they consider more generally Ainf,E = W OE (O F ) where E/Q is totally ramified finite extension of Q and F is a erfectoid field of characteristic. In this talk, we let E = Q and F = C algebraically closed. 1

2 THE FUNDAMENTAL EXACT SEQUENCES AND THE SCHEMATIC CURVE 2 where R(k) = lim (R(k)/ m ) is bounded oen in R(k) Q. Then we take B = lim k R(k) Q. This is the same as taking the comletion of B 0 with resect to the seminorms determined by {R(k)} k 1, and hence a Frechet algebra. Let us further recall the universal roerty of this construction: to give a continuous ring homomorhism from B A where A is a Q Banach algebra, it is the same as giving a homomorhism R(k) A with bounded image for some large enough k. Finally, recall that P := d 0 B ϕ=d and is the schematic curve. X := ProjP 1.4. The ring B +. Now we introduce a variant of the ring B, denoted by B +. For this, consider 2 A inf R(k) + R(k-1) + B + 0 = A inf[ 1 ] with R(k) + := A inf [ [ϖ ] k ] relacing R(k). Then take the inverse limit as before we arrive at Definition 1.1. B + := lim k (R(k) + ) Q. Remark. (1) Consider Un(C ) as oints in the unctured disk described above. Then functions in B 0 and B 0 + have finitely many oles at the origin (since only a bounded ower of can aear in the denominator); but B + and B could have essential singularities. 3 (2) The difference between B and B + is that one should heuristically think of elements in B + as ower exansions in variable where coefficients are integral. Though this descrition does makes sense (recall the next remark), it can be exlained later (and in future talks) when we discuss Newton olygons. Another heuristic to kee in mind is that, if we interret B = O(D (0,1) ) as functions on the oen unctured disc, then B + = O(D (0,1] ) are functions on the closed unctured disc. (3) Elements in B or B + might not have the form of ower series exansions, and when they do have such exansions they might not be unique. 2 The ring B0 and B + 0 are called Bb = B 0 and B b,+ = B + 0 resectively in Fargues and Fontaine s article. 3 The geometric interretation I gave in the lecture was incorrect.

3 THE FUNDAMENTAL EXACT SEQUENCES AND THE SCHEMATIC CURVE Preview of newton olygons. The newton olygons of elements in B 0 are what we exect: for x = n [x n] n B 0, Newt(x) := decreasing convex hull of {(n, v (x n )) n Z } where v is the (fixed) valuation on O C. From definition, one can see that (1) For any x B 0, Newt(x) is bounded from below and Newt(x) = + for x c for some c; (2) B + 0 B 0 are recisely those x such that Newt(x) 0. The ushot is, desite art (3) of the remark above, there will be a way (somewhat indirect, using Legendre transfer) to define Newton olygons for all x B, which satisfy the following roerties: Lemma 1.2. (1) B 0 = {x B : Newt(x) is bounded from below, and c s.t. Newt(x) [,c] = + } (2) B + = {x B : Newt(x) 0} B. (3) B = {x B : Newt(x) has 0 as its only finite sloe.} Proof. The construction of Newt(x) and the roof of the lemma will be a future talk. We will only need the following corollary today: Corollary 1.3. (1) B = (B 0 ) (2) B ϕ=1 = (B + ) ϕ=1 = Q B ϕh = d = (B + ) ϕh = d for h 1, d 0 B ϕh = d = 0 if h 1, d < 0 Sketch. (1) follows from the Lemma. For (2), if x B is such that ϕ(x) = d x, then Newt(x)(T ) = Newt(x)(T d). We rove the first assertion and leave others as exercises. Suose x 0, then ϕ(x) = x imlies Newt(x) = 0, so x (B 0 )ϕ=1, but (B 0 ) B 0 +, so we may write x = [x n ] n where x n O C. Now from ϕ(x) = x we can see that x Q. Another corollary that we will need is the following: Corollary 1.4. Suose t B ϕ= is a nonzero element, then ( B + [ 1 t ]) ϕ=1 = ( B[ 1 t ]) ϕ=1 n Proof. This follows from art (2) of the revious corollary, since for all t 1 (t k B + ) ϕ=1 = t k (B + ) ϕ=k = t k (B) ϕ=k = (t k B) ϕ=1 Now take union on both sides for all k 1.

4 THE FUNDAMENTAL EXACT SEQUENCES AND THE SCHEMATIC CURVE 4 2. Divisors 2.1. Review of B + dr. Let C Un(C ) be an untilt, recall that we have the surjective ma θ C : A inf O C given by n [x n ] n x n. ker(θ C ) is rincially generated, by any element α ker(θ C ) such that α O C α = C. There are many choices of this generator, we will choose ξ = [ɛ] 1 1 [ɛ] 1/ 1 = i=0 [ɛ] i/ where ɛ = (1, ζ, ζ 2,...) O C. Then we form the discrete valuation ring has absolute value B + dr = B+ dr,c := lim A inf [ 1 n ]/(ker θ C[ 1 ])n by ξ-adically comleting A inf [ 1 ]. As usual, let t = log[ɛ], which is a uniformizer for B+ dr Points on Y ad. Definition 2.1. (1) x = n 0 [x n] n A inf is rimitive if x 0 0 and there exists k such that x k O C. (2) If x = n 0 [x n] n A inf is rimitive, then deg(x) := min{k : x k O C }. (3) A rimitive element x is irreducible if deg(x) 1 and x cannot be written as a roduct of two rimitive elements of strictly less degrees. Note that rimitive elements of deg 0 are (A inf ), and we define Y as the set Y := { rimitive irreducible elements of A inf } /(Ainf ) Inside Y, we have Y deg=1 consisting of degree 1 elements. Let = (x) A inf be a rincial ideal generated by a deg 1 element, then A inf / is an untilt of O C. Conversely, if C C, then ker(θ C ) = ([ ] ) for some degree 1 element ([ ] ) A inf. 4 Moreover, we have Theorem 2.2 (Fargues-Fontaine). Suose C is algebraically closed, then rimitive irreducible elements have degree 1. Namely Y deg=1 = Y. Remark. This is really a factorization statement, namely, for any x A inf rimitive of degree d, one can write (in a non-unique(!!) way): where a i O C and a i 0. x = u( [a 1 ]) ( [a d ]) 4 The choice of [ ] in this exression is not unique.

5 THE FUNDAMENTAL EXACT SEQUENCES AND THE SCHEMATIC CURVE 5 To summarize, we now have Un(C ) = Y deg=1 = Y. Remark. Let X be the set of ϕ-orbits in Y deg=1 = Y, which can be identified with Un(C )/ϕ Z. Later we will see that these are recisely the closed oints of the schematic curve X, so the notation X is justified Divisors on Y ad. Let C = C m Un(C ) be an untilt (corresonding to the oint m Y ), then the ma A inf O C C sends [ϖ ] to ϖ := (ϖ ) a quasi-uniformizer of C, therefore for large enough k, the image of R(k) is a bounded subalgebra of C. By the universal roerty of B, θ C [ 1 ] : A inf[ 1 ] C extends to A inf [ 1 ] B θ C C. Remark. ker( θ C ) is again rincial and can be generated by any generator of ker(θ C ), and if we comlete B along ker( θ C ) we get B + dr,m, which is (canonically) isomorhic to B+ dr,c. In articular, we can view Un(C ) as a subset of the maximal ideals of B. 5 Moreover, for each m Y, we have ord m : B + dr,m N { }. Definition 2.3. Define the set (or monoid) of effective divisors of Y ad to be Div + (Y ad ) = { a m [m] ct I (0, 1), {m Y I : a m 0} is finite} m Y where Y I = {m : C m I} Y is the closed annulus consisting of oints of radius r I inside the unctured oen disc Y. Namely for a divisor D Div + (Y ad ), we allow only finitely many nonzero terms on any closed annulus, but they could accumulate on the boundary (or origin). Now for f B\{0}, define div(f) = ord m (f)[m] m Y It remains to be justified 6 in a future lecture that div(f) Div + (Y ad ). Comaring zeroes of functions in B gives a way to factorize in B. More recisely, we state the following theorem, which we need as ablackbox in this talk: Theorem 2.4 (Blackbox 1). Let x, y be two nonzero elements of B, then x = yz for some z B if and only if div(x) div(y). 5 In a latter draft, I will try to exlain closed maximal ideals of B. 6 It is clear that f only has zeroes and no oles on Y, what we need to rove is that f has only finitely many zeroes on each Y I

6 THE FUNDAMENTAL EXACT SEQUENCES AND THE SCHEMATIC CURVE Divisors on X ad. Similarly, we define Div + (X ad ) := {D Div + (Y ad ) ϕ D = D}. Remark. (1) Naturally, we have Un(C )/ϕ Z = X = Y /ϕ Div + (X ad ) C m m n Z[ϕ n (m)] (2) If f 0 B ϕ=d where d 0, then div(f) Div + (X ad ), so we have a morhism of monoids: div : ( d 0 B ϕ=d \{0})/Q Div + (X ad ) Proosition 2.5. The morhism div : ( d 0 B ϕ=d \{0})/Q Div + (X ad ) is an isomorhism. Proof. Injectivity follows from our blackbox theorem 1 (2.4), if x B ϕ=d and y B ϕ=e such that div(x) = div(y) then y = ux where u B is a unit. Moreover u (B ) ϕ=d e, by corollary 1.3, (B ) ϕ=k is either Ø (if d e) or Q (if d = e), hence injectivity follows. Surjectivity if not too difficult, but it was omitted in the talk. I will insert a roof for a second draft. Corollary 2.6. (Assuming C algebraically closed). The graded algebra P := d 0 Bϕ=d is graded factorial, more recisely, if x P d := B ϕ=d is nonzero, then there exists a unique (u to scaling) factorization Q where t i P 1 = B ϕ=. x = t 1 t d Proof. This is clear by induction on d. For x P d nonzero, take any oint [m] Y in the suort of div(x), then div(x) n Z [ϕn m] = divt for some t P 1 (unique u to Q ), now by blackbox theorem 1 (2.4), x = y t where y P d The fundamental exact sequence Our first major goal is to show that X = Proj P is a curve, namely a noetherian integral scheme of dimension 1. The key ingredient is the following fundamental exact sequence in -adic Hodge theory.

7 THE FUNDAMENTAL EXACT SEQUENCES AND THE SCHEMATIC CURVE The statement. Theorem 3.1. Let t B ϕ= be a nonzero element, corresonding to the divisor div(t) = n Z[ϕ n m] for some oint m Y, then we have the following exact sequence Remark. 0 B ϕ=n t k B ϕ=n+k B + dr,m /mk B + dr,m (1) In the statement we do not distinguish the oint m Y and the maximal ideal m = ker(b C m ) (2) div(t) = n [ϕn m], so there Z-ambiguity of choices for m. The ring B + dr,m and the ma B B + dr,m both deend on m. It does not matter which one we choose. Remark. More generally, Fargues and Fontaine rove that, for t 1,..., t r such that t i / Q t j and a 1,..., a r 1, we have r r 0 B ϕ=n t ai i B ϕ=n+k B + dr,m /mai i B+ dr,m i=1 We immediately have the following corollary, which is almost what we want to rove: Corollary 3.2. Let t B ϕ= be a nonzero element, then (P [ 1 t ]) (0) = (B[ 1 t ])ϕ=1 is a PID. Proof. We know already that (B[ 1 t ])ϕ=1 is factorial, so it suffices to show that each irreducible element t /t (B[ 1 t ])ϕ=1 where t / Q t generates a maximal ideal. Take the fundamental exact sequence for t and divide by t we get 0 Q t i=1 t (t 1 B) ϕ=1 B + dr,m /t B + dr,m = C m 0 The surjective ma (t 1 B) ϕ=1 C m factors through (t 1 B) ϕ=1 (B[ 1 t ])ϕ=1 (t k B) ϕ=1 C m, therefore we get a surjection (B[ 1 t ])ϕ=1 C m whose kernel is generated by t /t: suose b/t n lies in this kernel, then alying the fundamental exact sequence again we know that b = b t where b (B) ϕ=n 1, so b/t n = (t /t)(b /t n 1 ) A digression to -adic Hodge theory. Before we rove the result above, let us recall the classical fundamental exact sequence in -adic Hodge theory and a variant. For this we fix an untilt C of C and retain notations from 2.1. First recall Fontaine s eriod rings. A cris := ( A inf [ξ k k! ] k 1) () is the -adic comletion of the PD enveloe of A inf with resect to ker(θ : A inf O C ). This is in fact a comlicated ring, which has the following roerties:

8 THE FUNDAMENTAL EXACT SEQUENCES AND THE SCHEMATIC CURVE 8 (1) There exists a commutative diagram [ A ξ k ] inf k! k 1 A inf [ 1 ] A cris j B + dr (2) The ma j above is injective (which is a nontrivial fact), in articular, A cris is a domain. (3) The image of j has the following descrition: A ξ n cris im(j) = a n a n A inf, a n 0 -adically n! n 0 [Warning: an element in A cris = im(j) may not have a unique such exansion.] One can easily check that t A cris and more generally α k /k! A cris for any α ker(θ C ) and k 1. 7 Finally we make the following definitions B + cris = A cris[ 1 ], B cris = B + cris [1 t ] The oint here is that we have created B cris B dr which admits a Frobenius endomorhism ϕ, extending the Frobenius on A inf [ 1 ]. One first shows that A [ ξ k ] inf k! is stable under the Frobenius k 1 of A inf [ 1 ], which then extends uniquely by continuity to A cris and B + cris. With some care 8 it is not difficult to check that ϕ(t) = t, therefore the Frobenius on B + cris uniquely extends to B cris. Remark (Some further roerties of A cris ). (1) ϕ : A cris A cris is injective but not surjective. (2) The filtration on B cris is given by Fil i B cris = Fil i B dr B cris, note that Fil 0 B cris B + cris. For examle, the element [ɛ]1/2 1 [ɛ] 1/ 1 Fil0 B cris does not lie in B + cris. Now we state the fundamental exact sequences for B cris : Theorem 3.3. (1) (2) 0 Q t k (B + cris )ϕ=k B + dr,c /tk B + dr,c 0 0 Q Fil 0 B cris ϕ 1 B cris 0 Remark. (1) By dividing t k in the first short exact sequence and let k, we easily obtain 0 Q B cris ϕ=1 B dr,c /B + dr,c 0 7 more recisely t k /k! im(j) B + dr, but from now on we will not distinguish these two descritions of A cris. 8 this is not surrising since one would exect ϕ log[ɛ] = log[ɛ ] = log[ɛ], but some additional work is needed to make sense of this.

9 THE FUNDAMENTAL EXACT SEQUENCES AND THE SCHEMATIC CURVE 9 (2) The exactness in the middle of the second exact sequence formally follows from the exactness (in the middle) of the sequence above: (Fil 0 B cris ) ϕ=1 = (B cris ) ϕ=1 B + dr = ker(b cris ϕ 1 B dr /B + dr ) = Q. (3) The crucial observation here is that we can extract Q from B cris via its filtration and ϕ action, and this is the key to rove the Dieudonné functor D cris : Re cris Q (G K ) ((K K 0 B cris) ) GK MF φ K = {Filtered φ-modules}/k is fully faithful A variant of B cris. Due to the technical nature of the ring B cris, in this talk we rove something much less technical to give some ideas of how the arguments work 9, namely we introduce a variant of B cris. ξ k A max := a k k a k A inf, a k 0 -adically k 0 B + max = A max [ 1 ], B max = B + max[ 1 t ] A max A inf [ 1 ] is stable under Frobenius. First we state a lemma which relaces our blackbox theorem in the case of B max Lemma 3.4. (1) If x A max B + dr is such that θ(ϕn (x)) = 0 for all n 0, then ϕ(x) = [ɛ] 1 a for some a A max. (2) If x B + max is an element such that θ(ϕ n (x)) = 0 for all n 0, then t ϕ(x) in B + max (3) (B + max) ϕ=1 = Q. Proof. (1). Note that in A max, Now write x = i 0 a i ξi i ϕ( ξ 1 ) = 1 + i=0 [ɛ i ] 1 1 mod ( [ɛi ] 1 ) where a i A inf, then the above imlies ϕ n (x) ϕ n ( i 0 a i ) mod ( [ɛi ] 1 ) which imlies that ϕ n (x) = ϕ n ( i 0 a i ) + [ɛi ] 1 b n 9 The ring Bmax has a nicer toology and the Frobenius action is easier to describe. It gives the same notion of admissibility as B cris. For arguments of the corresonding statements for B cris, see Fontaine s original aer and his asterisque article

10 THE FUNDAMENTAL EXACT SEQUENCES AND THE SCHEMATIC CURVE 10 where b n A max. By assumtion, θ ( ϕ n ( i 0 a i ) ) = 0 for all n 0. Now the lemma follows from the following exercise 10 : Exercise. If α A inf is such that ϕ n (α) ker(θ) for all n 0, then ϕ(α) ([ɛ] 1) A inf. (2). Without loss of generality assume that x A max, which means ([ɛ] 1) ϕ(x) in B + max, but (1 [ɛ]) n 1 n. ([ɛ] 1)/t is a unit since t/([ɛ] 1) = 1 + n 2 (3). This is the same as (A max ) ϕ=1 = Z, which is an exercise. Now we are ready to rove Theorem 3.5. There is a short exact sequence: 0 Q t k (B + max) ϕ=k B + dr /tk B + dr 0. Proof. (1). We first rove the exactness in the middle by induction on k. The case k = 0 simly says (B + max) ϕ=1 = Q. Now suose x (B + max) ϕ=k t k B + dr, therefore, θ(ϕ n (x)) = θ( nk x) = 0 By lemma 3.4, t ϕ(x) = k x in B + max, therefore x = t y where y (B + max) ϕ=k 1 t k 1 B + dr, so y Q t k 1 by induction. Remark. The ushot here is that we want to factor x as y t, and we can check that x contains t as a factor by showing that ϕ n (x) vanishes in O C, i.e., x has zeroes at ϕ n (m C ), this is recisely why we need the blackbox theorem for the ring B. (2). The surjectivity was essentially roven last time by Jacob. We claim that the following diagram (of sets) commutes log[ ] 1 + m C B + dr log 1 + m C C where on the left, the ma is given by x x. First we check that if x = 1 + y m C, namely y = (y (0), y (1),...) m C (here lower indices means each y (i) O C / and y (i+1) = y (i)), then θ Remark. (1 + y) = lim n (1 + ỹ (n)) n 1 + m C. (1) To check commutativity, we need to justify θ(log[x]) = θ( ([x] 1) n ) = (x 1) n = log(x ). n n n 1 n 1 We need to make sense of log[x], namely it converges to an element. For this we need the assumtion x 1 < 1 and the convergence is shown in the revious lecture. 10 for examle, one get this by looking at Fontaine s I r filtration on A inf

11 THE FUNDAMENTAL EXACT SEQUENCES AND THE SCHEMATIC CURVE 11 θ(log[ ]) (2) Last time we have shown already that 1 + m C C is surjective, this is equivalent to log : 1 + m C C being surjective. This is a standard fact in -adic analysis, usually roven by looking at Newton olygons. Now we induct on k to rove the surjectivity of (B + max) ϕ=k B + dr /tk B + dr. Let x B + dr and let α 1 + m C be an element such that θ(log[α])k = θ(x). It is easy to check (similar to showing that log[ ] converges) that for z 1 + m C, log[z] B + max. Furthermore, since ϕ(log[α]) = log[α], we know that (log[α]) k (B + max) ϕ=k. Now since x (log[α]) k ker(θ) we conclude that x = t y + (log[α]) k where y B + dr, by induction y = t k 1 z + b where z B + dr and (B+ max) ϕ=k 1, therefore, x = t k z + t b + (log[α]) k t k B + dr + (B+ max) ϕ=k Back to the curve. Now we rove the fundamental exact sequence for B (assuming blackbox 2.4). Proof of Theorem 3.1. We rove the exactness of the following sequence from which the general case follows (for examle see the roof of 3.5). 0 B ϕ=1 t B ϕ= C m 0 (1). For the exactness in the middle, we need to factor t out in the ring B. Suose x mb + dr,m, which is another way to say that div(x) [m]. Since div(x) is invariant under ϕ, we know that div(x) n Z[ϕ n (m)] = div(t) Now by theorem 2.4 (this is where we use the blackbox theorem) we know that x = y t where y B ϕ=1. (2). In the revious lecture, we established the following commutative diagram log[ ] = 1 + m C (B) ϕ= log 1 + m C m The surjectivity of log([ ]) (or the surjectivity of log : 1 + m C m which is more classical) forces θ : (B) ϕ= C m to be surjective. θ

12 THE FUNDAMENTAL EXACT SEQUENCES AND THE SCHEMATIC CURVE The schematic curve The goal of this subsection is to show that (1) The nonzero element t B ϕ= corresond to a oint in X, with local ring B + dr,m. (2) There is an isomorhism (B[ 1 t ])ϕ=1 (B cris ) ϕ=1. Note that this means X can be obtained by gluing Sec(B cris ) ϕ=1 and Sec B + dr along Sec B dr, as stated in the first lecture by Jacob Local ring at t. Now we have shown that away from t B ϕ=, the affine scheme (B[ 1 t ])ϕ=1 is a PID, and it follows that X is a regular noetherian integral scheme of dimension 1. We can say a little more about the local ring at t. Exercise. V (t) = Proj(P/tP ) consists of a closed oint in X. Lemma 4.1. Let O X,t be the local ring at the oint defined by t, then there is an isomorhism (deending on the choice of m): Ô X,t B + dr,m Proof. O X,t = { x y Frac(P ) : x P d = B ϕ=d, y P d \tp d 1, where d 1} with uniformizer of the form t/t where t P 1 \Q t. y P d \tp d 1, by the fundamental exact sequence 0 P d 1 t P d B + dr,m /mb+ dr,m 0 we know that y (B + dr,m ), so x/y B + dr,m. This gives a ma O X,t B + dr,m. Consider an element x/y O X,t, where Note that, any uniformizer t/t of O X,t is a uniformizer of B + dr,m since t / mb + dr,m, and it induces an isomorhism on residue fields, so after comletion we get the lemma. Finally, let us summarize what we have shown so far (modulo the theory of Newton olygons and the blackboxed theorem): (1) We have shown that X is a regular noetherian integral scheme of dim = 1. (2) For a closed oint V (t) defined by t P 1 \{0} in X, the comleted local ring is (noncanonically) isomorhic to B + dr ; removing V (t) we get an affine scheme (B[ 1 t ])ϕ=1 which is a PID. (3) It is not hard to show that Q t V (t) gives a bijection between (P 1 \{0})/Q {closed oints in X}. Combined with 2.5 and the discussions above it, we have identifications Un(C )/ϕ Z = X = (P 1 \{0})/Q = {closed oints in X}, where X = Y /ϕ Z by definition is the ϕ-orbits of (degree 1 rimitive elements) Y.

13 THE FUNDAMENTAL EXACT SEQUENCES AND THE SCHEMATIC CURVE (B cris ) ϕ=1. Finally we connect this talk with lecture 1. Consider the quasi-uniformizer ϖ, and let ϖ = (ϖ ) O C, note that k 1, one has ([ϖ ] ϖ) k A cris k! In articular, for k large enough, [ϖ ] k A cris. From the discussion in 3.2, we have a natural ma A inf [ 1 ] B+ cris, and for k large enough, the image of R(k)+ = A inf [ [ϖ ] k ] lies in A cris, therefore it extends to a continuous ma B + B + cris which is comatible with Frobenius. Inverting t 11, this induces a ma B + [ 1 t ] B cris Lemma 4.2. (B + [ 1 t ])ϕ=1 (B cris ) ϕ=1 Remark. Once we know B + B + cris (in fact B+ B + cris is a subring), one can in fact show that B + = ϕ n (B + cris ) n 0 then the lemma also follows. We give the following amusing argument. Sketch. By the remark immediately after theorem 3.3, we have 0 Q t (B[ 1 t ])ϕ=1 B dr /B + dr 0 0 Q t (B cris ) ϕ=1 B dr /B + dr 0 Finally, by corollary 1.4, we know that (B[ 1 t ])ϕ=1 (B cris ) ϕ=1 Therefore, indeed, the schematic curve X can be regarded as Sec(B cris ) ϕ=1 and Sec B + dr glued together along Sec B dr. 11 We need to be careful about what t means on the right hand side. I will make this recise in a later draft