Lecture notes for a class on perfectoid spaces

Transcription

1 Lecture notes for a class on perfectoid spaces Bharav Bhatt April 23, 2017

2 Contents Preface 3 1 Conventions on non-archimedean fields 4 2 Perfections and tiltin 6 3 Perfectoid fields Definition and basic properties Tiltin Almost mathematics Constructin the cateory of almost modules Almost commutative alebra Almost étale extensions Some more almost commutative alebra Non-archimedean Banach alebras via commutative alebra Commutative alebras: completions and closures The dictionary Perfectoid alebras Reminders on the cotanent complex Perfectoid alebras Adic spaces Tate rins Affinoid Tate rins Affinoid adic spaces: definition and basic properties Spectrality of affinoid adic spaces The structure presheaf and adic spaces The adic spectrum via alebraic eometry Topoloical spaces Comparin sheaves

3 8.3 A fully faithful embeddin of affinoid Tate rins into a eometric cateory Perfectoid spaces Perfectoid affinoid alebras Tiltin rational subsets Tate acyclicity and other sheaf-theoretic properties Zariski closed subsets The almost purity theorem The direct summand conjecture DSC for maps which are unramified in characteristic Quantitative Riemann Hebbarkeitssatz Almost-pro-isomorphisms Proof of DSC

4 Preface These are notes for a class on perfectoid spaces tauht in Winter The oal of the class was to develop the theory of perfectoid spaces up throuh a proof of the almost purity theorem, and then explain the proof of the direct summand conjecture. I have tried to make these notes self-contained, and hopefully accessible to anyone with a backround alebraic eometry. In particular, I have not assumed any familiarity with riid eometry, so the relevant theory of adic spaces is developed from scratch. Likewise, I have not assumed any familiarity with Hochster s network of homoloical conjectures, so the direct summand conjecture is proven directly, and not via (a non-trivial) reduction to some other statement. Two exceptions: (a) I have used the lanuae of derived cateories in a couple of spots where I think it brins out the essence of the arument faster, and (b) I have used some results in point set topoloy (of spectral spaces) without proof. Disclaimers. There are surely many errors, so please use at your own risk. The notes are unstable, and bein constantly revised. Also, essentially all references and attributions are missin, and will be added later. 3

5 Chapter 1 Conventions on non-archimedean fields We establish some standard notation about non-archimedean fields. Definition A (complete) non-archimedean or NA field is a field K equipped with a multiplicative valuation : K R >0 such that K is complete for the valuation topoloy. The roup K R >0 is called the value roup of K. Remark Some comments are in order: 1. It is convenient to extend the valuation to a map : K R 0 by settin 0 = 0. With this extension, the valuation topoloy on K is the unique roup topoloy with a basis of open subroups iven by 1 ((0, γ)) for γ R >0. This topoloy defined by a metric on K: d(x, y) = x y. 2. Most authors do not impose a completeness hypothesis. However, our later constructions with adic spaces work best for complete rins, so we impose completeness riht away. 3. We shall typically work in a mixed/positive characteristic settin, i.e., p < 1 for some prime p. 4. We shall always assume that K is nontrivially valued, i.e., there exists a nonzero element t K with 0 < t < 1. A NA field comes naturally with some associated rins and ideals: Definition Let K be a NA field. The subset K := {x K x 1} is called the valuation rin of K; this is an open valuation subrin of K with maximal ideal K := {x K x < 1}. The quotient k := K /K is called the residue field of K. Any nonzero element t K is called a pseudo-uniformizer. The next exercise shows that iven the field K, specifyin the topoloy on K is equivalent to specifyin the valuation. In particular, it is meaninful to ask if a topoloical field K is NA. Exercise Fix a NA field K. 4

6 1. A subset S K is bounded if there exists a nonzero t K such that t K K ; equivalently, S is bounded for the metric topoloy on K. An element t K is power bounded if the set t N := {t n n 0} K is bounded. Check that K K is exactly the set of power bounded elements. 2. Check that K K is exactly the topoloically nilpotent elements of K, i.e., those t K such that t n 0 as n. 3. Fix a pseudouniformizer t K. Show that the t-adic topoloy on K coincides with the valuation topoloy. 4. Show that the iven NA valuation on K can be reconstructed from the valuation rin K. table:na my-label The next table records some examples the concepts introduced above. K K pu Value roup Z p Q p p p Z O K K/Q p finite π π Z if π is a uniformizer F p t F p ((t)) t t Z Z p [p 1 p ] Q p (p 1 p ) p p Z[ 1 p ] Z p C p := Q p p p Q K perfect K perfect t p-divisible Table 1.1: NA fields 5

7 Chapter 2 Perfections and tiltin Recall that a characteristic p rin R is perfect is the Frobenius φ : R R is an isomorphism; if instead φ is merely assumed to be surjective, we say that R is semiperfect. In this chapter, we introduce and study Fontaine s tiltin functor: it attaches a perfect rin of characteristic p any commutative rin (and is typically of interest when the latter has mixed/positive characteristic). def:tilt Definition (The tiltin functor). Let R be a rin. 1. If R has characteristic p, set R perf := lim R and R perf φ denotes the Frobenius. := lim φ R, where φ : R R 2. (Fontaine) For any rin R, set R := (R/p) perf := lim φ R/p. Unless otherwise specified, this rin is endowed with the inverse limit topoloy, with each R/p bein iven the discrete topoloy. Remark (Universal properties of perfections). When R has characteristic p, both R perf and R perf are perfect. The canonical map R R perf (resp. R perf R) is universal for maps into (resp. from) perfect rins. Moreover, the projection R perf R is surjective exactly when R is semiperfect. inrins Example We record some examples of these concepts. 1. F p [t] perf = F p [t 1 p ] and F p [t] perf = F p. 2. F p [t] F p. 3. Say R is a finite type alebra over an alebraically closed field k of characteristic p. Then R k π 0(Spec(R)) is the alebra of k-valued continuous functions on Spec(R). To see this, we may assume Spec(R) is connected and reduced (see Exercise 2.0.4). We must show k R perf. Assume first that Spec(R) is irreducible. Pickin a closed point x Spec(R) ives a map R R x, where R x is the completion of the local rin at x; this map is injective as R is a domain. It is therefore enouh to show that R x perf k. We have Rx = lim n R x /m n x, and thus R x perf = limn (R x /m n x) perf. Usin Exercise 2.0.4, it is easy to see that (R x /m n x) perf 6

8 (R x /m x ) perf k perf for all n, which ives the claim. The eneralization to the case where R is not a domain is left to the reader. 4. (F p [t 1 p ]/(t)) perf F p [t 1 p ]. More enerally, if R is a perfect rin of characteristic p and f R is a nonzerodivisor, then (R/f) perf is the f-adic completion of R. 5. (Z p ) F p. 6. ( Z p [p 1 p ]) F p [t 1 p ] The perfection functors kill nilextensions. F ( perf. p [t] perf F p [t] perf /(t)) tionnilp Exercise Let f : R S be a map of characteristic p rins that is surjective with nilpotent kernel. Then R perf S perf and R perf S perf. More enerally, the same holds if f factors a power of Frobenius on either rin. We repeatedly use the followin elementary lemma. Binomial Lemma Let R be a rin, and let t R be an element such that p (t). Given a, b R with a = b mod t, we have a pn = b pn mod t n+1 for all n.. Proof. We prove this by induction on n. If n = 0, there is nothin to show. Assume inductively that a pn = b pn + t n+1 c for some c R. Raisin both sides to the p-th power, and usin that p ( ) p i for 1 i p 1, we et a pn+1 = b pn+1 + p t n+1 d + t p (n+1) c p for some d R. As p (t) and p 2, the claim follows. The next lemma is critical in future applications. It ives a strict description of the ( ) functor. Abstract Lemma Assume R is p-adically complete. The projection map R R/p induces a bijection lim x x p R lim φ R/p =: R of multiplicative monoids. Proof. We first check injectivity. Fix (a n ), (b n ) lim x x p R with a n = b n mod p for all n. Then a pk n+k = a n for all n, k, and similarly for the b s. Applyin Lemma to both sides of a n+k = b n+k mod p then shows that a n = b n mod p k+1 for all n and k. As R is p-adically separated, it follows that a n = b n for all n, as wanted. For surjectivity, fix (a n ) lim x x p R/p. Choose arbitrary lifts a n R of a n. Then a p n+k+1 = a n+k mod p for all n, k. Lemma shows that for each n, the sequence k a pk n+k is Cauchy for the p-adic topoloy, and thus has a limit b n. Then one checks that b p n+1 = b n for all n, and that b n lifts a n, provin surjectivity. 7

9 Topoloy SharpMap tionrin Remark Note that the construction of (b n ) from (a n ) in the second half of the proof above is well-defined (i.e., independent of auxiliary choices), and ives an explicit inverse to the projection lim x x p R lim φ R/p =: R. Remark (Exercise) In Lemma 2.0.6, if one topoloizes R with the p-adic topoloy and R/p with the discrete topoloy, then the bijection of Lemma is a homeomorphism. Indeed, the map is clearly continuous. For continuity in the other direction, note that we have a homeomorphism ( ) lim R lim lim x x p x x p n R/pn lim x xp R/p 3 x x p R/p 2 x x p R/p. Thus, for each k 0, a basic open subroup U k lim x x p R is iven by those (a n ) with a i (p i ) for i k. Now a p n+1 = a n for all n, so U k is exactly those (a n ) with a i (p k ). It suffices to show that the imae of U k in lim φ R/p contains those (b n ) with b i = 0 for all i 2k; this follows from the explicit inverse constructed in the proof above. Remark (Sharp map). In Lemma 2.0.6, via projection to the last term, we et multiplicative map : R R denoted f f. Its imae is exactly those f R that admit a compatible system {f 1 p k } of p-power roots. We shall sometimes call such elements perfect. Usin the map, we can understand valuation rins under tiltin; this will be useful when discussin adic spaces later. Lemma If a p-adically complete rin R is a domain (resp. a valuation rin), the same is true for its tilt R. In fact, if : R Γ {0} is the valuation on R, then the map R R Γ {0} ives the valuation on R. In particular, the rank of R is bounded above 1 by the rank of R. Proof. We first check that R is a domain whenever R is a domain. We have R lim x x p R as a multiplicative monoid. Fix elements (a n ), (b n ) lim x x p R with a n b n = 0 for all n. Then either a 0 or b 0 vanishes as R is a domain. By symmetry, assume a 0 = 0. Then, as the transition maps involve raisin to powers, we et a n = 0 for all n, and thus (a n ) = 0, so R has no zero divisors. Now assume R is a valuation rin. Fix a, b R, correspondin to (a n ), (b n ) lim x x p R. As R is a valuation rin, we have a 0 b 0, or vice versa. Assume a 0 b 0 by symmetry. Then, for valuation reasons, we must have a n b n for each n 1: the element an b n in the fraction field of R must lie in R (as its p n -th power does). Thus, (a n ) (b n ) in lim x x p R, and thus a b in R, provin that R is a domain. Explicitly, this construction shows that if : R Γ {0} is the valuation, then := : R Γ {0} ives a valuation on R : indeed, iven a = (a n ), b = (b n ) R, we checked above that a b exactly when a 0 b 0, which happens exactly when a b as a = a 0 (and similarly for b). The assertion about ranks is automatic. 1 In this enerality, the rank can indeed o down under tiltin. For example, if R = Z p, then R = F p. We shall check later that this does not happen for perfectoids. 8

10 Chapter 3 Perfectoid fields In this chapter, we introduce and study perfectoid fields. These are NA fields that contain lots of p-power roots. The main result is that the tilt of (the rin of inteers of) a perfectoid field K is a perfectoid field K of characteristic p that reflects the alebraic properties of K. In particular, we formulate (and prove the key special case of) the almost purity theorem for perfectoid fields, equatin the Galois theory of K and K. 3.1 Definition and basic properties Fix a prime number p. Definition A perfectoid field K is a NA field with residue characteristic p such that: The value roup K R >0 is not discrete. K /p is semiperfect, i.e., the Frobenius map K /p K /p is surjective. oidfield Example The first condition rules out fields like Q p itself. Interestin examples are: 1. Let K = Q p (p 1 p ). Then the value roup is Z[ 1 ]. To calculate the valuation rin, observe p that both completions and filtered colimits of valuation rins are valuation rins. It follows that K = Z p [p 1 p ]: the natural map from the riht to the left is an extension of rank 1 valuation rins with the same field of fractions, and must thus be an isomorphism. It is then easy to see that K is perfectoid. A similar analysis applies to Q p (µ p ). 2. Let K = C p = Q p. Then the value roup Q. As K is alebraically closed, every t K admits a p-th root, so the perfectoidness is clear. Alternately, one may arue directly without usin the alebraic closedness of K by observin that Z p /p is semiperfect (as Q p is alebraically closed), and that Z p K is an isomorphism modulo any power of p. 9

11 3. Let K be a NA field of characteristic p. Then K is perfectoid if and only if K is perfect. In this case, semiperfectness of K implies its perfectness, and hence the nondiscreteness of the value roup (as lon as the valuation is not trivial). As these examples illustrate, the perfectoid world is very non-noetherian: Lemma Let K be a perfectoid field. 1. The value roup K is p-divisible. 2. We have (K ) 2 = K. Moreover, K is flat. 3. The rin K is not noetherian. Proof. For (1). We temporarily call x K small if p < x 1. We first check the p-divisibility of x K for small x. The perfectoidness of K ives a y, z K such that y p = x + p z for some z K. Takin absolute values and usin the NA property shows that y p = y p = x, so x K is divisible by p. In eneral, as K is not discrete, the containment p Z K must be strict, so we can choose an x K with x / p Z. After rescalin by a suitable power of p, we can assume x is small. By the total orderin of ideals in K, we must have p = xy for small y. But then p = x y, so p K is divisible by p. A similar arument shows p Z and x for x small enerate K, so we are done. For (2). Pick some f K. By perfectoidness, we can write f = p + ph for K and h K. The previous proof shows that p (K ) 2, so this formula proves that f (K ) 2. For flatness, we simply note that any torsionfree module over a valuation rin is flat. (2) implies (3) by Nakayama s lemma. Alternately, K is not finitely enerated as it has elements of arbitrarily small valuation. rfectoid Remark The proof above shows that K R >0 is enerated by x for x K with p < x < 1. This observation will be useful later in analyzin the value roup under tiltin. We shall see later that differential forms tend to vanish in the perfectoid world. An elementary instance of this is: Exercise Let K be a perfectoid field. Show that Ω 1 K /Z p completion always vanishes. is never 0, and yet its p-adic 3.2 Tiltin Fix a perfectoid field K. Our oal is to attach to K a perfectoid field K of characteristic p. We shall do so by first constructin the rin of inteers of K as K, and then constructin K as 10

12 a suitable localization. For the rest of the section, we fix 1 a pseudouniformizer π K with p π < 1, so p (π). We obtain a commutative diaram lim x x p K pr 0 K K := lim x x p K /p pr 0 K /p (3.1) eq:tiltpe lim x x p K /π pr 0 K /π, where all the vertical maps are the canonical reduction maps, the top half comes from Lemma when K has characteristic 0 (and is trivial in characteristic p), the bottom half comes from Exercise when K has characteristic 0 (and is trivial in characteristic p), and the bottom two horizontal maps are surjective by the semiperfectness of K /p. We can then topoloize K as follows: Exercise Check that the followin 3 topoloies on K are equivalent: The inverse limit topoloy arisin via K lim φ K /π. The inverse limit topoloy arisin via K lim x x p K. The inverse limit topoloy arisin via K lim φ K /p, where the topoloy on K /p is the one induced from K (and is thus discrete when K has characteristic 0, but not so in characteristic p). Recall that we want to show that K is the valuation rin of a perfectoid field K. To this end, we first find a pseudouniformizer: m:tiltpu Lemma There exists some element t K such that t = π. Moreover, t maps to 0 in K /π, and this ives an isomorphism K /t K /π. Proof. As p (π), the canonical projections ive surjective maps K K /p K /π. By p-divisibility of the value roup, we may choose some f K such that f p = π, and hence f > π. In particular, f K /π is nonzero. Choose some K liftin f under the canonical map K K /π. Then = f mod π by the diaram (3.1). By the NA property, this ives = f since f > π. Settin t = p and usin the multiplicativity of shows that t = f p = π. For the second part, note that t K maps to f p = π = 0 in K /π, thus ivin a map K /t K /π. To show this map is an isomorphism, consider the diaram (3.1) aain. Choose some K such that maps to in K /π. We must show that (t). The diaram shows that K maps to 0 in K /π, and hence (π). But π = t, so (π) = (t ). Hence, we can write 1 For practical purposes, we may take π = p in characteristic 0, and any pseudouniformizer in characteristic p. 11

13 = at for suitable a K. It is then easy to see that a lifts to an element ã = (a n ) lim x x p K alon the projection pr 0 : simply set a n = ( 1 p n ) (t 1 p n ) K, and then observe that a pn n = a K, so a n K for all n. By construction, we have = ã t as elements in lim x x p K. Goin down the left vertical arrow, we learn that (t), as wanted. Usin elements constructed above, the topoloy on K is seen to be alebraic: Topoloy Corollary With t as above, K is t-adically complete, and that the t-adic topoloy coincides with the iven topoloy. Proof. Usin Lemma 3.2.2, we have a map of inverse systems of rins... K /(t pn ) std... K /(t p ) std K /(t)... K /π std φ n φ... K /π std φ 1 φ std K /π with all vertical maps bein isomorphisms. Comparin inverse limits as topoloical rins proves the claim. oidfield Proposition Fix an element t as in Lemma The rin K is a valuation rin, and the rin 2 K := K [ 1 ] is a (necessarily perfect) field. t 2. The ideal (t 1 p ) is maximal, and the Krull dimension of K is The valuation topoloy on K comin from (1) coincides with the one induced by the t-adic topoloy on K. In this topoloy, K is a perfectoid field, and K, = K. 4. The value roups and residue fields of K and K are canonically identified. We shall repeatedly use the followin: usin the presentation K lim x x p K as multiplicative monoids, we learn that an element shows that a K is a unit if and only if a is a unit. Proof. 1. Specializin Lemma to R = K, we learn that K is a valuation rin of rank 1. In fact, the rank is exactly 1 since we know that t := t = π is non-trivial. In particular, invertin any nonzero nonunit in K, such as t, produces the fraction field K. 2 There is abuse of notation here: when K has characteristic 0, the rin K is not the tilt of K in the sense of Definition

14 2. The assertion about Krull dimensions follows from (1) as the rank of a valuation rin equals its Krull dimension. For the rest, we already know that K /t K /π by Lemma As the maximal ideal of K /π is its nilradical (since K is a rank 1 valuation rin), the same must be true for K /t. But the nilradical of K /t is just the imae of (t 1 p ) (as the latter clearly lies in the nilradical, and the quotient K /(t 1 p ) is perfect, whence reduced), so (t 1 p ) must indeed be maximal. 3. As K is a rank 1 valuation rin, the valuation topoloy coincides with the f-adic topoloy for any nonzero nonunit f. Takin f = t ives the first part of the claim. Corollary shows that K is t-adically complete, and thus K is a NA field. Finally, it is clear that K is perfect, and thus K is perfectoid. 4. The claim about residue fields follows from (2) usin the identification K /π K /t from Lemma For value roups, usin the notation above, we trivially have K K. To show equality, note that K is enerated by x for x K with p < x < 1 (see Remark 3.1.4). We must show x K for any such x. But this is immediate from Lemma tcontval Remark (Tiltin Continuous Valuations). Proposition (1) is a special case of the followin: Proposition For any continuous valuation : K Γ (of any rank), the function = : K, Γ is also a continuous valuation. This construction identifies the space of continuous valuations on either field. Proof. Fix a continuous valuation on K. It is clear that is multiplicative. Moreover, as : K K has trivial fiber over 0, it is clear that f = 0 if and only if f = 0. To check the NA property, fix f := (f n ), := ( n ) lim x x p K K, so f = f 0 and = 0. We must check that f + max( f, ). But this follows from f+ := (f+) = lim n (f n + n ) pn = lim n f n + n pn lim n max( f n, n ) pn = lim n max( f 0, 0 ) = max( f where the second equality is obtained by chasin the behaviour of addition across the isomorphism lim x x p R lim x x p R/p for a p-adically complete rin R, and the third equality uses the continuity of on K. For the second part, write std for the iven NA valuation on K. Observe that a valuation : K Γ is continuous if and only if for one (or, equivalently, any) pseudouniformizer f K, we have f n 0 as n. Usin this remark, one checks the followin about the valuation rin R K attached to : 1. R contains K inside its maximal ideal as f < 1 for any f K. 13

15 2. We have R K. Indeed, if not, then R has an element from K[ 1 f ] \ K, i.e., an element of the form a/f n with a K and a std > f n std. But then f n /a std < 1, so f n /a K, so f n /a lies in the maximal ideal of R. As a/f n R as well, we find an element of the maximal ideal of R that is invertible, which is absurd. Conversely, one may check any valuation subrin R K that satisfies (1) and (2) defines a continuous valuation on K: the key point is that the map R K is a localization of R (at its unique heiht 1 prime, by the classification of rins between a valuation rin and its fraction field), so K must lie in all primes of R, and thus t n 0 for t K. Passin to the quotient, we learn that continuous valuations on K identify bijectively with valuation rins in K /K. Repeatin the same arument for K, we conclude usin the identification of K /K with K, /K,. We ive an explicit example of a rank 2 valuation on a perfectoid field. tionrin Example Let k be a perfect field of characteristic p, and let K = W (k)[p 1 p ]. Then K = K [ 1] is a perfectoid field with p K /K k. In particular, iven any valuation rin R k, the preimae R K of R is a valuation rin of K whose attached valuation : K Γ is continuous. For an explicit example, set k = F p ((t)) perf. Consider its valuation rin R = F p t perf k. Then the preimae R K of R defines a rank 2 valuation on R with value roup Γ := K t Z[ 1 p ] Z[ 1] Z[ 1 ] ordered lexicoraphically. p p The main theorem about perfectoid fields is: Theorem (Almost purity in dimension 0). Let L/K be a finite (necessarily separable) extension. Endow L with its natural topoloy as a finite dimensional K-vector space. Then 1. L is perfectoid. 2. The field extension L /K is finite of the same deree as L/K. 3. The association L L defines an equivalence K fet K fet. Example Let K = arument shows that if L = K( p), then L = the isomorphism K Q p (p 1 p ). We explained in Example that K = Z p [p 1 p ]. A similar Z p [p 1 2p ]. It is then easy to calculate that if we fix F p ((t)) perf by requestin t = p, then L = K ( t). We shall prove the full result later, once the lanuae of almost mathematics has been introduced. In fact, the latter will allow us to explicitly construct an inverse to the operation L L. Grantin the existence of this construction, the result will follow from the followin consequence, which we prove directly: APTField Proposition Assume that K is alebraically closed. Then K is alebraically closed. The proof below is due to Kedlaya. 14

16 Proof. We assume that K has characteristic 0. Set x 0 = 0. We shall inductively construct a sequence {x n K } such that the followin hold for each n: 1. P (x n ) p n. 2. x n+1 x n p n d. Then (2) shows that {x n } converes to some x K, and (1) shows that P (x) = 0, and thus P (x) = 0. As x 0 = 0 is already defined, assume by induction we have constructed x 0, x 1,..., x n satisfyin the above two properties. Write d P (T + x n ) = b i T i, so b d = 1. If b 0 = 0, then P (x n ) = 0, so we may simply take x i = x n for each i n. Assume from now on that b 0 0. Consider the quantity i=0 c = min{ b 0 b j 1 j j > 0, bj 0}. Considerin j = d shows that c b 0 1 d 1. By Proposition (4) and the alebraic closedness of K, we know that K is a Q-vector space, so c = u for some u K; in fact, we have u K as u = c 1. We have b i b 0 u i K by construction. Moreover, as the minimum definin c is achieved, there exists i > 0 such that b i b 0 u i is a unit. Usin Lemma 3.2.2, choose t K with t = p. Consider any polynomial Q(T ) K [T ] liftin d b i i=0 b 0 u i T i K /p[t ] under the identification K /t K /p. By construction, this is a polynomial of deree > 0 whose constant coefficient and (at least) one non-constant coefficient is a unit. Lemma shows that there exists some unit y K, such that Q(y) = 0. We shall check that x n+1 = x n + u y K satisfies the analos of (1) and (2). First, note that P (x n+1 ) = P (u y + x n ) = d b i u i (y i ) = b 0 ( d b i u i (y ) i). b 0 i=0 i=0 Now the parenthesized term is conruent to Q(y) modulo p, and thus 0 modulo p as y is a root of Q. It follows that P (x n+1 ) b 0 p p n p = p n+1, where we use induction to et b 0 = P (x n ) p n. This ives (1), and for (2) we observe that x n+1 x n = u y = u = c b 0 1 d = P (xn ) 1 d p n d, where we use that y is a unit in the second equality, and the inductive hypothesis in the last one. The followin lemma was used above. 15

17 utnewton Lemma Let V be the valuation rin of a complete and alebraically closed NA field. Let P (T ) V [T ] be a polynomial of deree 1 such that the constant coefficient and (at least) one non-constant coefficient are units in V. Then P vanishes at a unit of V. This lemma can be proven usin Newton polyons, but we ive a direct alebraic proof. Proof. Write m V for the maximal ideal with residue field k = V/m. Our hypothesis ensures that reduction of P (T ) modulo m is a non-constant polynomial with unit constant and leadin coefficients. We may then choose a pseudouniformizer t V such that the reduction modulo t of P (T ) is also a non-constant polynomial with unit constant and leadin coefficients. We view P as a map f P : V [T ] V [T ]. This map realizes the taret V [T ] as a torsionfree module over the source V [T ] that is finite free when reduced modulo t by our choice of t. Writin V T for the t-adic completion of V [T ], it follows that the t-adic completion f P : V T V T is a finite free morphism as well. In particular, the rin A := V T /P (T ) is a finite free V -alebra. As V is henselian, we can decompose A i A i as a finite product with each A i bein a finite free local V -alebra. Reducin modulo m, this ives k[t ]/P (T ) i A i/m with each factor bein local. Now the assumption on P (T ) ensures that one of the roots of P (T ) over k is a unit. It follows that T maps to a unit in one of the residue fields of k[t ]/P (T ). As each A i is local, it follows that T maps to a unit in some A i. Fix one such index i. As A i is finite free over V and K is alebraically closed, the rin A i,red [ 1 ] decomposes as a non-empty product of copies of K. Pickin one such t copy ives a map A i K. This map has imae contained inside V K as A i is interal over V. Thus, we obtain a map A i V. As T mapped to a unit in A i, the same must be true for its imae in V. Puttin everythin toether, we have produced a map V [T ]/P (T ) V that carries T to a unit, as wanted. clotomic Example Let K = Q p (µ p ), so K = as the p-adic completion of Z p [µ p ]. We have an explicit presentation of K Z p [ɛ 1 p ]/( ɛ 1 ɛ 1 p 1 ) = Z p [ɛ 1 p ]/(1 + ɛ 1 p + ɛ 2 p ɛ p 1 p ), iven by choosin a compatible system ɛ n µ p n of p-power roots of 1, and sendin ɛ 1 p n to ɛ n. Reducin modulo p, and usin that xp 1 = (x x 1 1)p 1 in characteristic p, we learn that K /p F p [ɛ 1 p ]/(ɛ 1 p 1) p 1 F p [t 1 p ]/(t p 1 ), where we use the substitution t ɛ 1 (and similarly for p-power roots). By Exercise 2.0.4, we learn that K = (K /p) perf identifies with the t-adic completion of F p [t 1 p ], and hence K F p ((t)) perf. Remark Examples and show that K = Q p (µ p ) and L = Q p (p 1 p ) have isomorphic tilts, i.e., K L. In particular, the tiltin functor K K is not fully faithful on perfectoid fields K/Q p. We shall see later that this is a consequence of workin over the nonperfectoid base Q p : the functor R R will be fully faithful on perfectoid fields (in fact, alebras) over a perfectoid base field K. 16

18 Chapter 4 Almost mathematics In this chapter, we introduce Faltins theory of almost mathematics. This theory is essentially a softenin of commutative alebra that is possible when one works over a R equipped with an ideal I such that I = I 2 ; the basic idea is to redevelop the basic notions of commutative alebra whilst systematically inorin I-torsion modules. This idea was inspired by work of Tate (who observed that inorin K -torsion modules when workin over the perfectoid field K = Q p (µ p ) was a sensible and useful idea), and is crucial to a finer study of Fontaine s tiltin functor. We follow the treatment of Gabber-Ramero [GR]. 4.1 Constructin the cateory of almost modules Let R be a rin equipped with an ideal I. In this situation, we have the followin standard pair of adjoints: Construction Restriction of scalars alon R R/I ives a fully faithful functor This functor has a left adjoint i iven by and a riht adjoint i! iven by We now specialize to the case of interest: i : Mod R/I Mod R. i (M) = M R R/I, i! (M) = Hom R (R/I, M) = M[I]. ostsetup Assumption (The setup of almost mathematics). Assume I R is a flat ideal and satisfies I 2 = I. This implies I R I I 2 I. The precedin assumption will be in place for the rest of this section. 17

19 ostsetup Example Two classes of examples that will be relevant to us are: Let K be a perfectoid field. Set R = K for a perfectoid field K and set I = K to be the maximal ideal. As torsionfree modules over valuation rins are flat, it is easy to see that I is flat. In fact, we can be more explicit: if t K, is a pseudo-uniformizer, then a = t has a compatible system of p-power roots, and I = (a 1 p ). In particular, I = colim n (a 1 p n ) is a countable union of free K -modules, so it is flat of projective dimension 1. Let R be a perfect rin of characteristic p, and let I = (f 1 p ) for f R. It is easy to see that I 2 = I in this case. Moreover, I is clearly flat if f R is a nonzerodivisor. To verify flatness of I in eneral, set M i = R for i 0, and consider the inductive system M 0 f p p f 1 p M 2 1 M 2... M n 1 p f n 1 p n+1 M n+1... Write M = colim M n, so M is a flat. There is an obvious map M I iven by sendin 1 M n to f 1 p n I. This map is surjective by construction, so it suffices to show it is also injective. If α M n = R oes to 0 in I, then αf 1 p n = 0. But then α pm f = 0 for all m n. By perfectness of R, we learn that αf 1 p m = 0 for all m 0. In particular, the transition map M n M n+1 kills α, so α = 0 in M, provin injectivity of M I. Exercise Usin the second example in Example 4.1.3, show the followin: if R is a perfect rin of characteristic p and I R is the radical of a finitely enerated ideal, then R/I has finite flat dimension over R. In the situation above, one can construct an interestin localization Mod a R of the cateory Mod R of R-modules. This will be the cateory of almost R-modules. It can defined directly as an abstract cateory. However, in order to have a tiht relationship between this cateory and Mod R, we will need the followin construction. Construction (The almost cateory in disuise). Let A Mod R be the full subcateory spanned by all R-modules M such that the action map I R M M is an isomorphism; equivalently, as I R I I via the multiplication map, we can also describe A as the essential imae of the idempotent functor M I R M on Mod R ; this functor is exact by the flatness of I. Usin flatness of I, one checks that A is abelian subcateory of Mod R that is closed under takin kernels, cokernels, and imaes inside Mod R. We will construct a series of auxiliary functors relatin A with Mod R, eventually allowin us to realize A as a quotient of Mod R. Write j! : A Mod R for the resultin exact inclusion. The inclusion j! has an exact riht adjoint j : Mod R A iven by the formula j (M) = I R M. The unit map N j j! N is an isomorphism for any N A. 18

20 Proof. We first note that I R M A as I R I I, so we have a well-defined functor. The exactness is clear from the flatness of I. For adjointness, fix some N A and M Mod R. We must show that Hom A (N, I R M) Hom ModR (N, M). Usin the exact trianle I R M M M L R R/I, it is enouh to show that RHom R (N, M L R R/I) 0. By adjointness, we have RHom R (N, M L R R/I) RHom R/I (N L R R/I, M L R R/I). But our hypothesis on N tells us that N R I N, and that this tensor product is derived (by flatness of I). By associativity of (derived) tensor products, it is enouh to show that I L R R/I 0. This follows from the flatness of I and the hypothesis I = I 2. Grantin adjointness, the assertion about the unit map results from the isomorphism I I R I. The riht adjoint j has a further riht adjoint j iven by the formula j (M) = Hom R (I, M). The counit map j j M M is an isomorphism for any M A. Proof. Fix some N Mod R. Then Hom A (j N, M) = Hom R (I N, M) Hom R (N, Hom R (I, M)) = Hom R (N, j (M)), which proves the adjointness; here we use that Hom- -adjunction the second isomorphism. For the rest, fix some M A. We must show that I R Hom R (I, M) M via the natural evaluation map. As tensorin with I is exact, it suffices to show the stroner statement that I L R RHom R(I, M) M. As M A, we have I L R M M, so it is enouh to check that the natural map M RHom R (I, M) induces an isomorphism after tensorin with I. Usin the exact trianle RHom R (R/I, M) M RHom R (I, M), this reduces to showin that tensorin with I kills the term on the left. But the term on the left admits the structure of an R/I-complex, so we can write I L R RHom R (R/I, M) I L R R/I L R RHom R (R/I, M). We now conclude usin I L R R/I 0, as before. 19

21 The composition i j! is 0: we must show that if M I R M, then M R R/I 0. This follows by observin that I R R/I 0. The composition i! j is 0: we must show that if M I R M, then Hom R (R/I, Hom R (I, M)) 0. But the adjointness of and Hom identifies this with Hom R (R/I R I, M), so we conclude usin R/I R I 0. The composition j i is 0: we must show that M R I 0 if M is I-torsion, but this follows from M R I M R R/I R/I I 0, where the last equality uses R/I R I 0. The kernel of j is exactly Mod R/I : iven M Mod R with j (M) := I R M 0, we must check that M is I-torsion. Tensorin M with the standard exact sequence 0 I R R/I 0 shows that M M/IM, so M is I-torsion. The followin remark explains why a more naive definition of A runs into trouble. Warnin For eneral M Mod R, the action map I R M M has imae inside IM M, and thus if M A, then IM = M. However, the converse need not be true. Consider R = k[t] perf for a perfect field k of characteristic p. Let I = (t 1 p ), so R/I k is the residue field at the oriin. Let M R/t be the maximal ideal inside the local rin R/t; this is the imae of I/t inside R/t, but does not coincide with I/t. As I = I 2, it is easy to see that M = IM. However, the action map I R M M need not be injective. The kernel of this map is Tor R 1 (R/I, M). To calculate this, we use the definin exact sequence 0 M R/t R/I 0. Tensorin with R/I, and usin that 1 Tor R i (R/I, R/I) = 0 for i > 0, we learn that Tor R 1 (M, R/I) Tor R 1 (R/t, R/I). The second roup is computed to be nonzero usin the standard resolution the claim follows. We can now construct the promised cateory of almost R-modules. Proposition In the above situation, we have: ( ) R t R of R/t, so 1. The imae of i is closed under extensions. In particular, i realizes Mod R/I as an abelian Serre subcateory of Mod R, so the quotient Mod a R := Mod R /Mod R/I exists by eneral nonsense (see [SP, Ta 02MS]). 1 This is a eneral fact about perfect rins. In our case, we may prove this as follows. Writin I = n I n with I n := (t 1 p n ), we et Tor R i (R/I, R/I) colim Tor R i (R/I n, R/I n ); here we use that Tor commutes with direct limits in either variable. Usin the standard resolution for (R t p 1 ) n R of R/I n, we see that the Tor s vanish for i > 1. For i = 1, we have a canonical identification Tor R 1 (R/I n, R/I n ) I n /I 2 n. We now observe that the transition maps R/I n R/I n+1 induce the 0 map I n /I 2 n I n+1 /I 2 n+1 as I n I 2 n+1. It follows that colim n Tor R 1 (R/I n, R/I n ) = 0, as wanted. 20

22 2. The quotient functor q : Mod R Mod a R admits fully faithful left and riht adjoints. Thus, q commutes with all limits and colimits. 3. The imae of i is a tensor ideal of Mod R, so the quotient Mod a R inherits a symmetric monoidal -product from Mod R. 4. The -structure on Mod a R is closed, i.e., to X, Y Mod a R, one can functorially an object alhom(x, Y ) Mod a R equipped with a functorial isomorphism Hom(Z X, Y ) Hom(Z, alhom(x, Y )). Proof. 1. If an R-module M can be realized as an extension of two R-modules killed by I, then M is itself killed by I 2. But I = I 2, so M is also killed by I. Thus, the imae of i is closed under extensions. The rest is by cateory theory, but we shall construct an explicit candidate for Mod a R in the proof of (2). 2. We claim that the functor j : Mod R A introduced above provides an explicit realization of the quotient functor q : Mod R Mod a R. To check this, we must show that: j (Mod R/I ) = 0: this amounts to show that I R M = 0 for an R-module M killed by I. But, for such M, we have I R M = I R R/I R/I M = I/I 2 R M = 0 as I = I 2. j is exact: this follows from the description j (M) = I R M and the flatness of I. j is universal with the previous two properties: let q : Mod R B be an exact functor of abelian cateories such that q (Mod R/I ) = 0. Fix some M Mod R. We then have the canonical action map I R M M. The kernel and cokernel of this map are identified with Tor R i (R/I, I) for i = 1, 0. In particular, the kernel and cokernel are killed by I, and thus also by q. As q is exact, we learn that q (I R M) q (M). But I R M =: j! j (M), so we have shown that q q j! j, so q factors throuh j, as wanted. 3. If M is killed by I, so M R N for any R-module N. By cateory theory, this implies that the symmetric monoidal structure on Mod R passes to the quotient A. In particular, j is symmetric monoidal. 4. Given M, N Mod R, we simply set alhom(j (M), j (N)) = j Hom R (M, N); this is well-defined by (3) as Hom R (M, N) is I-torsion if either M or N is so. Checkin the rest is left to the reader. Remark (The topoloical analo). The construction above can be summarized in the followin diaram: i Mod R/I i ModR i! 21 j! j Mod a R = A. j

23 Here we have the adjoint pairs (i, i ), (i, i! ), (j!, j ) and (j, j ). Moreover, each composition in a straiht line in the above diaram is the 0 functor: i j! 0, i! j 0, and i j 0. The notation is meant to remind the reader of the analoous situation in topoloy: if X is a topoloical space with an open subset j : U X with complement i : Z X, then we have a similar diaram of topoi: Shv(Z) i i i! Shv(X) j! j j Shv(U). Moreover, the pairs of functors ivin adjoint pairs and the pairs composin to 0 are the same as before. In fact, the analoy can be stretched a bit further. Recall that we may view Mod R as quasicoherent sheaves on X := Spec(R), and Mod R/I as quasi-coherent sheaves on Z := Spec(R/I). However, the cateory Mod a R is not the cateory of quasi-coherent sheaves on U := X Z (or any other open subset of X). Instead, we think of Mod a R as quasi-coherent sheaves on some nonexistent open U X that contains U (as restriction to U factors throuh j ). Exercise Let R = k[t 1 p ] for a perfect field k of characteristic p, and let I = (t 1 p ). Show that the extension of scalars functor Mod R Mod R[t 1 ] iven by M M R R[t 1 ] factors throuh Mod a R, and that the resultin functor Mod a R Mod R[t 1 ] is not an equivalence. 4.2 Almost commutative alebra We continue in the setup of Assumption For notational ease and compatibility with Gabber- Ramero, we rewrite some of the functors introduced in the previous section as: Definition Fix an R-module M. We say that an element f M is almost zero if I f = 0. We say that M is almost zero if all its elements are almost zero, i.e., M is I-torsion. In eneral, write M a := j M Mod a R, M := j M a := Hom R (I, M), M! := j! M a = I R M, We refer to elements of M as almost elements of M. A map f : M N of R-modules is almost surjective (resp. almost injective, almost isomorphism) if f a is surjective (resp. injective, isomorphism). We sometimes refer to an almost R-module as an R a -module, and likewise for alebras. With the notation above, we have canonical maps M! M M, and they both become isomorphisms on almostification, i.e., after applyin ( ) a. We record some useful examples of the notion of almost elements. Elements Example Let R = K for a perfectoid field K and I = K. If M is I-torsion, then M = 0. If M is a torsionfree R-module, then M {m M K K ɛ m M for all ɛ I}. 22

24 R = R and I = R. More enerally, iven an ideal J R, the ideal J is principal exactly when c := sup( x x J) 1 lies in K. This is clear if J is principal: we have c = x for a enerator x J, and J = J. If J is not principal, then J = {a K a < c} for c = sup( x x J) by valuation yoa 2. It then follows that J = {a K a c}, which is principal exactly when c K. If t R is a pseudouniformizer, then R/t (R/t) is injective, but need not be surjective: the obstruction lies in Ext 1 R(I, R), which may be nonzero. Let K = Q p [p 1 p ] and let L := (W (F p (t) perf )[p 1 p ]) [ 1 ]; these are both perfectoid fields, p and K L. Let A L be the rank 2 valuation rin in Example Then A is a K -alebra, and A = L : the cokernel of A L is killed by K by construction. The functor ( )! is exact, but the functor ( ) is only left exact. Its hiher derived functors will appear (at least implicitly) in some aruments that follow, so we explain how to calculate them. Elements Exercise (Derived functors of ( ) ). The bifunctor Hom R a(m a, N a ) can be derived in any variable to convert short exact sequences to lon exact sequences (in the usual fashion). In fact, we et these derivatives by the formula Ext i R a(m a, N a ) := Ext i R(I M, N) Ext i R(M, RHom R (I, N)). In particular, if 0 M M M 0 be a short exact sequence of R a -modules, we have a lon exact sequence... Ext i R a(ra, M ) Ext i R a(ra, M) Ext i R a(ra, M ) Ext i+1 R a (Ra, M )..., derivin the functor of almost elements. An explicit nonzero example of a hiher Ext-roup is iven in Remark We now extend some basic notions of commutative alebra to the almost world: Definition Let M Mod R with imae M a Mod a R. Then 1. We say that M or M a is almost flat if M a ( ) is exact on Mod a R; equivalently, Tor R >0(M, N) is almost zero for any R-module N. 2. We say that M or M a is almost projective if alhom(m, ) is exact; equivalently, Ext >0 R (M, N) is almost zero for any R-module N. 2 The containment is clear as J is non-principal, and the reverse follows by notin that if a < c with a / J, then a > x for any x J (as J K is closed under takin smaller elements, via scalar multiplication), whence a c, contradictin the assumption a < c. 23

25 ojective njective 3. We say that M or M a is almost finitely enerated (resp. almost finitely presented) if for each ɛ I, there exists a finitely enerated (resp. finitely presented) R-module N ɛ and a map N ɛ N with kernel and cokernel killed by ɛ. If the number of enerators of N ɛ can be bounded independently of ɛ, we say that M is uniformly almost finitely enerated. The last definition iven above depends a priori on the choice of the R-module M ivin M a on almostification. However, one checks readily that, in fact, the definition is independent of this choice. Remark (Cateorical projectivity ives the incorrect notion). The notion of almost projectivity is distinct from the cateorical notion of projectivity in the abelian cateory Mod a R: the latter is far more restrictive. Indeed, the rin R is almost projective with the above definition. However, R a need not be a projective object of Mod a R. In fact, usin Exercise 4.2.3, one can show: with R = K and I = K for a perfectoid field K with residue field k, the roup Ext 1 R a(ra, R a ) identifies with Ext 2 R(k, R), and is nonzero if K is not spherically complete (such as K = Q p ). Remark (Injectives behave well). The cateory Mod a R has enouh injectives. In fact, if I is an injective R-module, then I a is injective (as ( ) a has an exact left-adjoint ( )! ). Conversely, if J Mod a R is injective,then J is an injective R-module (as ( ) has an exact left-adjoint ( ) a ) such that (J ) a J. Thus, we may construct injective resolutions in Mod a R by simply computin them in Mod R (under either ( ) or ( )! ) and applyin almostification. We ive an example illustratin why the notion of almost finite eneration is defined as above (instead of a stroner condition). raticroi Example (A quadratic extension of a perfectoid field). Let K = Q p [p 1 p ], and L = K( p) with p 2. Then we claim that L is a uniformly almost finitely presented projective K -module. For this, it suffices to show: for each n, there exists a finite free K -module R n of rank 2 and an injective map R n L with cokernel annihilated by p 1 p n. (Indeed, then p 1 p n on either module will factor over this map, showin Ext >0 K (L, ) is killed by p 1 p n for all n, and thus almost zero.) Set R n = K K p 1 2p n L, so we have L colim n R n. We first claim that the cokernel of R n N n+1 is killed by p 1 p n. To see this, observe that p 1 p n p 1 2p n+1 = p (p+1)/2 p n+1 p 1 2p n. It follows that the cokernel of R n colim m R m is killed by p m n 1 p n = p p p n (p 1). Strictly speakin, the element written above does not make sense: its norm does not live in K = p Z[ 1 p ]. However, any element that is more divisible (i.e., has smaller norm) also works by the same reasonin. In particular, the cokernel of R n colim m R m is killed by p 1 p n 1. This shows that, as R 0 is p-adically complete, so is colim n R n, and thus L = colim n R n. But then each R n is finite projective of rank 2, and the cokernel of the injective map R n L is killed by p 1 p n 1. In particular, L is an uniformly almost finitely presented projective K -module. 24

26 We also pause to explain the construction of honest alebras from almost alebras in a fashion that respects faithful flatness. Strictly speakin, the rest of these notes can be developed without reference to this functor. However, this functor is quite convenient in extractin honest consequences out of an almost mathematical statement, so we shall use it. Alebras Remark (Left adjoint to almostification for alebras). As Mod a R has a symmetric monoidal structure, there is an evident notion of a commutative alebra object in this cateory: it is iven by an object S Mod a R with maps R a S and S S S satisfyin some natural natural axioms). Write CAl(Mod a R) for the cateory of commutative alebras in almost R-modules. The almostification functor M M a commutes with tensor products, and hence takes commutative alebras to commutative alebras, inducin a functor ( ) a : CAl(Mod R ) CAl(Mod a R). The functor M M is lax symmetric monoidal (i.e., there is a canonical map M R N (M N) for M, N Mod a R), and hence takes commutative alebras to commutative alebras. In fact, the resultin functor ( ) : CAl(Mod a R) CAl(Mod R ) is easily seen to be a riht adjoint to the almostification functor ( ) a. In contrast, the left adjoint M M! to almostification does not preserve commutative alebras: if A is an almost R-alebra, then A! does carry a multiplication as ( )! commutes with non-empty tensor products, but, as R! a = I does not coincide with R, so one does not have a unit for A!. To fix this, iven such an A, define A!! as pushout of R I R! a A!. There is then a unique way to make A!! into a commutative rin such that the definin map R A!! is a unit and the definin map A! A!! is compatible with the multiplication. This construction ives a functor ( )!! : CAl(Mod a R) CAl(Mod R ) that is left-adjoint to the almostification functor ( ) a, see [GR, ]. The crucial properties are: 1. ( )!! commutes with colimits (clear by adjointness). 2. (Exercise) ( )!! preserves faithful flatness. However, it may not preserve flatness; see [GR, Remark 3.1.3]. Note that there is a natural map A!! A of almost alebras which is an almost isomorphism; this can often be used to move completeness properties between the two rins. 4.3 Almost étale extensions The main point of makin the definitions earlier is to enable discussion of the followin key notion: Definition A map A B of R a -alebras is almost finite etale if 25