PERFECTOID SPACES NOTES - University of Chicagohaolee/Perfectoid Spaces Notes...PERFECTOID SPACES...

PERFECTOID SPACES NOTES

HAO (BILLY) LEE

Abstract. These are notes I took in class, taught by Akhil Matthews. I claim no credit to the originality of the contents

of these notes. Nor do I claim that they are without errors, nor readable.

Course website: math.uchicago.edu/ amathew/perfcourse.html.

Let X/C is a smooth projective variety. Then by viewing it as a compact manifold, we can associate to it, singularcohomology H?sing (X;C) . We also have Hodge decomposition

Hn (X,C) ∼= ⊕p+q=nHp(X,ΩqX/C

).

Additionally,

Hn (X,C) ∼= Hn (X,Q)⊗Q C.

In the direct sum decomposition, if we let V p,q = Hp (X,Ωq) then V p,q = V q,p.

Goal of p-adic Hodge theory is to get an analog of this formula with X/E with E/Qp a �nite extension. The right handside of this Hodge decomposition makes sense over any �eld. The question is what is supposed to replace the left hand

side. This is étale cohomlogy.

1. Review of `-adic étale cohomology

Let K be an algebraically closed �eld of characteristic not `. Suppose X/K is an algebraic variety. This gives us

H?et (X,Q`) which are �nite dimesional Q`-vector spaces. If K = C, then this is isomorphic to singular cohomology withQ` coe�cients.

• H?et (X,Q`) = 0 for all ? > 2 dimX.• This cohomology depends on the choise of the prime `. That is, if K has positive characteristic, then it's not evenknown that dimH?et (X,Q`) varies in `.• This construction does not depend on the structure map X → Spec K (only depends on the scheme). Therefore,if K = F̄ for some �eld F , and X/K is de�ned as X = Y ⊗F F̄ for Y/F variety, then H?et (X,Q`) is a continuousrepresentation of GF .

Example 1.1. Take F = Q and Y/Q a variety. Then H?sing (Y (C),Q`) has a natural action of GQ. This is remarkablebecause there is not GQ action on Y (C) (discontinuous).

De�nition 1.2. Let F be a �eld of characteristic not `. Consider

µ`∞(F̄ ) ={`-power roots of unity in F̄

} ∼= Q`/Z`.The Galois group GF naturally acts on this. De�ne the Z`(1) to be

Z`(1) = lim← µ`n(F̄ ) = hom(Q`/Z`, µ`∞(F̄ )

)is called the Tate-twist. De�ne

Q`(1) = Z`(1)[1

`] and Q`(n) = Q`(1)⊗n

for all n ∈ Z.

Example 1.3. We know that Q`(−1) ∼= H2et(P1F ⊗F F̄ ,Q`

).1

PERFECTOID SPACES NOTES 2

Example 1.4. Suppose E is an elliptic curve over F of characteristic not `. Then H1et(E ⊗F F̄ ,Q`

)has an explicit

description. This is lim←E(F̄ )[`n].

Abstractly,

E(F̄ )[`∞] ∼= (Q`/Z`)2 .

Can take

T` = hom(Q`/Z`, E(F̄ )[`∞]

) ∼= Z2` .Then H1et

(E ⊗F F̄ ,Q`

) ∼= T∨` ⊗ Z[ 1` ].Let F be a �nite extension of Qp. SupposeX/F is a proper smooth variety, then we can associate to itH?et

(X ⊗F F̄ ,Qp

)which has a continuous action of GF .

De�nition 1.5. De�ne Cp = Q̂p is the completion of the algebraic closure of Qp, which is itself algebraically closed, witha norm extending the p-adic norm, and a continuous action of GQp .

Construction: suppose V is a continuous representation of GF on a Qp-vector space. Consider

V ⊗Qp Cp

which has an action of GF (acting diagonally). This will then be a semi-linear representation of GF .

Example 1.6. We have Cp(1) = Qp(1)⊗Qp Cp.

Theorem 1.7. (Faltings) Hodge-Tate decomposition. Suppose F is a �nite extension of Qp. Suppose X/F is a propersmooth variety. Then

Hnet(X ⊗F F̄ ,Qp

)⊗Qp Cp ∼= ⊕i+j=nHi

(X,ΩjX/F

)⊗ Cp(−j).

This isomorphism is in the category of Cp-vector spaces with semi-linear Galois action.

In particular,

H?et (XF̄ ,Qp)⊗Qp Cp ∼= ⊕jCp(−j)nj

in semilinear representations.

Consequences:

• The Galois representation on étale cohomology determines Hi(Ωj)(the Hodge numbers). This is by the following

theorem.

Theorem 1.8. (Ax-Sen-Tate) For F/Qp �nite,

Cp(i)GF =

F i = 00 i 6= 0 .Then

(H?et (X)⊗Qp Cp(j)

)GF ∼= Hi (X,Ωj) .Idea of perfectoid spaces. Suppose Y/C is a smooth algebraic variety. Then as a complex manifold, it is locally analytic

and contractible. Can compute de Rham cohomology and singular cohomology locally.

Question: how much of this carries over to the p-adic setting. Consider the n-ball in Cp, BnCp . The theory of rigidanalytic geometry enables us to work analytically locally and produces a locally ringed space. Here, functions on BnCp is

the Tate-algebra

Cp 〈z1, ..., zn〉 =

∑i1,...,in≥0

ai1...inzi11 ...z

inn : convergent on unit ball

.However, things are not locally contractible. This is why we get to perfectoid spaces.


Theorem 1.9. (Scholze) If X/Cp is a proper smooth rigid analytic space, then there exists a natural spectral sequencewhere

E2 = Hi(X,Ωj

)(−j) =⇒ Hi+jet (X,Cp) .

This is proved by working locally in the analytic topology. If A is an a�noid algebra over Cp (quotient of Tate-algebra),we associate it to F (A) where H? (F (A)) ∼= Ω?A/Cp and

RΓ (X,F (OX)) ∼= H?et (X,Cp) .

2. Perfect Fp-Algebras

De�nition 2.1. Let R be an Fp-algebra. It is said to be perfect if the Frobenius map ϕ : R → R given by z 7→ zp is anisomorphism.

Example 2.2. Consider Fp[z1/p

∞]= ∪n≥1Fp[z1/p

n

] is called the perfect polynomial ring. It has the property that if S

is any perfect ring,

hom(Fp[z1/p

∞], S)∼= S.

That is, it is the free perfect ring on 1 generator.

Remark 2.3. Perfect rings are reduced.

De�nition 2.4. (Perfection) Let R be any Fp-algebra.

(1) Let Rperf = lim→R is called the direct limit perfection (the map R→ R is by ϕ)

Example 2.5. R = Fp[x] then Rperf = Fp[x1/p∞

].

(2) Rperf = lim←R where R→ R is still by ϕ. This is called the inverse limit perfection

A standard element is of the form (..., zn, zn−1, ..., z0) where zpn = zn−1 ∈ R

Remark 2.6. If R is any Fp-algebra, and S is any perfect ring, then R→ S extends uniquely over Rperf

R S

Rperf

Similarly,

Rperf R

S

De�nition 2.7. A ring R is semiperfect if ϕ : R→ R is surjective, which implies Rperf � R is a surjective.

Example 2.8. The ring R = Fp[x1/p∞

]/(x) is semiperfect. Additionally, Rperf = Fp[x1/p∞

](x) is semiperfect.

Proposition 2.9. (Bhatt-Scholze) Let A,B,C be perfect Fp-algebras. If A→ B and A→ C are maps then

TorAi (B,C) = 0

for i > 0. Additionally, B ⊗A C is a perfect ring.

Proof. We will give two groups.


(1) We can form R ⊗LA C (relative derived tensor product) and Hi(B ⊗LA C) = Tori. We can use the topologicalcommutative rings to form

B ⊗LA C πi(B ⊗LA C

)= Tori.

Since B ⊗LA C is a topological Fp-algebra, it comes with a Frobenius ϕ. We can construct this so that ϕ is a weakhomotopy equivalence. Therefore, ϕ induces an isomorphism on Tori.

Observe that if R is a topological commutative Fp-algebra, ϕ : R → R induces zero on πi for i > 0. This isbecause Frobenius factors as a map R→ Rp → R (breaks into p-copies then multiplies). The multiplication mapRp → R is zero on higher homotopy groups. This is because it induces a map

πi(R)p ∼= πi(Rp)→ πi(R)

and the multiplication map on each {0} × ...×R× ...× {0} → R is zero on the homotopy group.(2) Suppose B = A/I then I has to be a radical ideal. The claim is that TorAi (A/I,C) = 0 for i > 0, I ⊆ A radical

ideal and C prefect. Without loss of generality, assume I = rad (�nitely generated ideal). Can even assume

I = (f1/p∞

) = rad(f) for f ∈ A. (This is done by some Tor spectral sequence to show that you can make thesereductions). Here,

rad(f) = (f1/p∞

) = ∪(f1/pn

)

because if z ∈ rad(f), then zn ∈ (f). Assume zpn ∈ (f) and so z ∈ (f1/p∞) by perfect-ness of A.

Lemma 2.10. If A,C are perfect rings, I = (f1/p∞

) then TorAi (A/I,C) = 0 for i > 0.

Proof. The ideal I is actually �at as an A-module and I ⊗A C = IC, then

0→ I → A→ A/I → 0

gives long exact sequence

...→ TorA1 (A/I,C)→ I ⊗A C → C → A/I ⊗A C → 0

Everything in the long exact sequence degenerates.

Now, I is �at because of the following. If f is a non-zero divisor in A then I is �at. If A is also non-zero divisor

in C then I ⊗A C ∼= IC. In general, write I as a �ltered colimit of free A-modules, which means it is �at.

A A A ...

I

f1− 1

p

f

f1p− 1p2

f1p

f1/p2

Claim that this lim→A ∼= I. This direct limit surjects onto I, so just need injective. Suppose z ∈ A that is in the

�rst stage of the diagram is in the kernel. That is fz = 0. Then

f1/pz1/p = 0→ f1−1p z1/p = 0→ ... = 0.

Therefore, we have injectivity. �

�

Remark 2.11. The key to this proof is that for A a perfect ring, f ∈ A, we can consider f -power torsion elements z ∈ Asuch that zfN = 0 for N � 0. That is zf

1

pN = 0 for n > 0. Then zfq = 0 for all q ∈ Z[ 1p ]>0.

De�nition 2.12. Suppose R is a Fp-algebra. R is said to be p-complete if R→ lim← R/pn is an isomorphism.

De�nition 2.13. If R is any Fp-algebra, a lift to characteristic 0 of R is a ring R̃ which is p-complete, p-torsion free suchthat R̃/p ∼= R.


Example 2.14. If R/Fp is smooth, then it has a lift. However, the lift is only unique up to non-unique isomorphism.

De�nition 2.15. A strict p-ring is a p-complete, p-torsion free ring R̃ such that R̃/p is perfect.

Example 2.16. If E/Qp is a �nite extension which is unrami�ed, then OE is a strict p-ring.

Example 2.17. The ring ̂Zp[x1/p∞ ], the p-completion of ∪n≥0Zp[x1/pn

], is a strictly p-ring. Mod p, this is Fp[x1/p∞

].

Theorem 2.18. There is an equivalence of categories

strict p-rings↔ perfect Fp-algebra

by R̃ 7→ R̃/p. The inverse is the Witt vector functor W .More generally, if A is a strict p-ring, and B is any p-complete ring, then hom (A,B) ∼= hom (A/p,B/p).

Remark 2.19. If R is any ring, the ring of Witt vectors W (R) is de�ned. However, if R is perfect, then W (R)/p ∼= R.The construction R 7→ W (R) is analogous to A 7→ AJtK, but the analogy is not complete because W (R)/p ∼= R only

makes sense for perfect rings.

Example 2.20. Let R̃ = ̂Zp[x1/p∞ ]. Then R̃/p = Fp[x1/p∞

]. Then if S is any p-complete ring,

hom (R,S) ∼= hom (R/p, S/p) .

The left hand side is just

lim←

x 7→xp

S,

because we specify where x goes, then x1/p, x1/p2

,... in a compatible way. Similarly, the right hand side is

lim←

x 7→xp

S/p.

The theorem says that for any p-complete ring S,

lim←

x 7→xp

S ∼= lim←

x 7→xp

S/p ∼= (S/p)perf .

We will prove this.

Lemma 2.21. Suppose S is any ring, x, y ∈ S. If x ≡ y mod pn then xp ≡ yp mod pn+1.

Proof. We write y = x+ (y − x) and use binomial theorem to get yp =∑(p

i

)xi (y − x)p−i . �

Proof. of the isomorphism.

For injectivity: Suppose (..., x2, x1, x0) is such that xpn+1 = xn and similarly for (..., y2, y1, y0) with yi ≡ xi mod p for

all i.

Since xn ≡ yn mod p, this implies that x0 ≡ y0 mod pn by the lemma. Let n→∞, x0 = y0.Now for surjectivity: Given an element of (..., ȳ2, ȳ1, ȳ0) with ȳ

pn+1 = ȳn. Choose any limit y

′n of ȳn for all n. This is

not compatible under p-powers, but we can de�ne a new sequence

yn = lim←m

(yn+m)pm.

�

De�nition 2.22. If S is any p-complete ring, then de�ne S[ (the tilt of S) to be

lim←

x7→xp

S ∼= (S/p)perf .

This is a functor from p-complete rings to perfect Fp-algebras.


There is a natural map # : S[ → S sending (..., y3, y2, y1, y0) 7→ y0. NOTE! this is a map on monoids (it is multiplicative,but not additive).

Let R be a strict p-ring. Then R[ = R/p because it is perfect. However, we still have

# : R/p→ R

which is multiplicative. Explicitly, the map is given x̄, we choose a lift yn of x̄1/pn . Then

x# = lim←yp

n

n .

Note that

R/p#→ R→ R/p

is the identity. It follows that every x ∈ R can be written unqiuely as

x =∑i≥0

(a#i )pi

for unique ai ∈ R/p. Note that since # is not additive, we can't add these power series the usual way, but this can becalculated.

Claim 2.23. Let A be a perfect Fp-algebra. There is a strict p-ring A such that R/p ∼= A. Additionally, for all p-completerings S, we have

hom (R,S) = hom (Rn/p, S/p) .

Proof. We already proved this for A = Fp[x1/p∞

]. Formally follows that we can does this for

A = Fp[x1/p

∞

α

]α∈I

and perfect polynomial algebra. We want to reduce to this case.

Let A be a perfect ring. Then we can �nd a surjective P0 = Fp[x

1/p∞

α

]α∈A→ A. Suppose I is the kernel, we can take

generators for I ⊆ P0. We can take in image of a map of perfect polynomial rings P1 → P0 and so A = P0 ⊗P1 Fp.There exists strict p-rings

P̃1 → P̃0↓Zp

such that mod p, we have

P1 → P0↓Fp

.

We want to set

R = ̂P̃0 ⊗P̃1 Zp.

We need to check that R is a strict p-ring. Then

hom (R,S) = hom (R/p, S/p)

by diagram chasing.

The only thing we need to check is p-torsion free. Consider the derived tensor product P̃0 ⊗LP̃1 Zp. We want H0 is

p-torsion free. If there were p-torision then (P̃0 ⊗LP̃1 Zp

)⊗LZp Fp

will have non-trivial H1. However, this derived tensor product has no higher homology, which are Tor of Fp-algebras (andby theorem of earlier). �


Suppose S1 and S2 are p-complete rings. Suppose S1 is a strict p-ring. Then

hom (S1, S2) ∼= homFp (S1/p, S2/p) ∼= homFp(S1/p, (S2/p)

perf)

∼= hom(S1, S

[2

).

Corollary 2.24. There is a pair of adjoint functors

Perfect Fp-algebras→ p-complete ring

where the left adjoint is the Witt vector functor, and right adjoint is the tilt. That is, suppose A is a perfect Fp-algebra,S is any p-complete ring then

hom (W (A), S) ∼= hom (A,S/p) ∼= hom(A, (S/p)perf

).

3. Topological Rings

De�nition 3.1. A topological ring R is a non-archimedean or linearly topologized if it has a basis at 0 consisting of

subgroup.

Example 3.2. Let R be any ring, and I ⊆ R be an ideal. We can give R the I-adic topology, so the basis of neighbourhoodat 0 is {In} . We call these adic rings.

Let R0 be any ring. Let ω ∈ R0 non-zero divisor. Let

R = R0

[1

ω

],

which is a topological ring such that a basis at 0 is given by ωnR0. Observe that the restricted topology on R0 ⊆ R is theω-adic topology, and R0 ⊆ R is an open subring. Note that ω ∈ R is a topologically nilpotent unit. These are called Taterings.

Example 3.3. This is the same as the topology on Qp = Zp[ 1p ].

More formally, a Tate ring is a topological ring R, which contains an open subring R0 ⊆ R and an element ω ∈ R0which is both topologically nilpotent in R and a unit in R. Additionally, the topology on R0 is the ω-adic topology.

In general, R0 is not unique. Suppose given R0 ⊆ R and ω as above. Suppose I have a slightly larger R0 ⊆ R′0 ⊆ Rin the sense that there exists n such that ωnR′0 ⊆ R0. Then R can be recovered as a Tate ring from (R0, ω) or (R′0, ω).Similarly, we can use Z + ωR0 ⊆ R0.

De�nition 3.4. A non-archimedean �eld K is a �eld K with an absolute value |·| : K → R≥0 such that

• |x| = 0 i� x = 0• |xy| = |x| · |y|• |x+ y| ≤ max {|x| , |y|}

Then K is a metric space, and a topological ring.

Example 3.5. For any �eld K, consider |·| : K → R≥0, there's a trivial absolute value which is |x| = 1 if x 6= 0 and|x| = 0 else. We will ignore this from now on.

Example 3.6. Fp((t)) with the t-adic absolute value. That is,∣∣∣∣∣ ∑t�−∞

aiti

∣∣∣∣∣ = p−jwhere j is the smallest i such that ai 6= 0.


Remark 3.7. Given a non-archimedean �eld, we can complete it as a metric space K̂.

A non-archimedean �eld is a Tate ring by

K0 = {x ∈ K : |x| ≤ 1} .

This is a local ring, with maximal ideal

K◦◦ = {x ∈ K : |x| < 1} .

If we choose w ∈ K∞\ {0} , 0 < |w| < 1 then ω is called a pseudo-uniformizer. It is not hard to see that K0 has the ω-adictopology and K = K◦

[1ω

].

In cases such as Qp,Fp((t)) the maximal ideal K◦◦ is principal. This is if and only if the image of |·| is discrete. In thiscase, K◦ is a discrete valuation ring.

De�nition 3.8. Let R be a topological ring. We say S ⊆ R is bounded if for all neighbourhood U of 0, there exists aneighbourhood V such that SV ⊆ U .

Example 3.9. If R is a non-archimedean �eld. Then S is bounded i� |·| is bounded on S.

De�nition 3.10. Given a non-archimdean topological ring, we say x ∈ R is power bounded if {xi}i≥0 forms a boundedset. Let R0 ⊆ R be the set of power bounded elements. We say x ∈ R is topologically nilpotent if xi → 0 as i→ 0.

Proposition 3.11. If R is a non-archimedean topological ring, then R0 ⊆ R is a subring and R◦◦ ⊆ R0 is an ideal.

Example 3.12. Let R be a Tate ring, R = R0[1ω ]. Then

(1) R0 is an open bounded (a subset S ⊆ R is bounded i� S ⊆ ω−NR for some N � 0)(2) R0 ⊆ R0 (that is, R0 is union of all possible choices of R0).

Remark 3.13. A Tate ring is a topological ring which contains

(1) an open bounded subring

(2) topological nilpotent unit

In general, R0 is not bounded.

Example 3.14. Consider the topological ring Qp[�]/�2 =(Zp[�]/�2

)[ 1p ]. Since � is nilpotent, any multiple of it is power

bounded. Therefore,

R0 = Zp ⊕Qp�

is not power bounded itself.

De�nition 3.15. If R0 is bounded, it is called uniform. (The previous example suggests that a uniform R0 must be

reduced).

Proposition 3.16. If R is a non-archimedean topological ring, R0 ⊆ R is integrally closed. That is, if x ∈ R is such thatf(x) = 0 for some f ∈ R0[X] monic, then x ∈ R0.

Proof. That means the module generated by x is �nitely generated, which is bounded. �

Return to the non-archimedean �eld example. Look at K0 more closely. If |·| has discrete image, then K0 is a discretevaluation ring. In general, K0 is always a valuation ring.

De�nition 3.17. A valuation ring V is an integral domain such that given x, y ∈ V either x | y or y | x.

Example 3.18. In K0, x | y i� |y| ≤ |x|, so it is a valuation ring.


In a valuation ring, the set of ideals form a totally ordered set. That is, Spec V is a totally ordered set.

Spec K0 = {(0),K◦◦}

because K◦ is a rank 1 valuation ring i� x, y ∈ K0 non-zero, non-units, then x | yn and y | xm.

Proof. If the property holds, R has a nonzero prime ideal and p contains a non-zero non-unit x and so every other y by

the property. �

Claim 3.19. To give an non-archimedean �eld K is the same as giving a rank 1 valuation ring V .

Remark 3.20. K is complete i� V is x-adically complete for any (or all) pseudo-uniformizers x ∈ V .

To go in reverse, suppose V is a valuation ring. Let K be its fraction �eld. K×/V × is a totally ordered group under

multiplication and divisibility. Then K×/V × is archimedean in that given γ1, γ2 ∈ K×/V × there exists n ∈ Z\{0} suchthat nγ1 > γ2. Given an ordered group, it maps to R>0 which is order absolute value. Note, implicit in this, is that tohave this archimedean property, my group can't be too large. So this valuation map is injective.

4. Perfectoid Fields

Fix a prime number p. Let K be a complete non-archimedean �eld such that |p| < 1.

De�nition 4.1. Say K is perfectoid if

(1) K is not discretely valued, that is, there exists elements x ∈ K× such that |x| is arbitrarily close to, but not equalto 1.

(2) K0 = {x ∈ K : |x| ≤ 1} has the property that

Frob : K◦/p→ K◦/p

is surjective (semi-perfect). Note, K◦ is a rank one valuation ring, but its maximal ideal K◦◦ is not principal, so

not Noetherian

Example 4.2. Qp and any unrami�ed extension of this, satis�es 2 but not 1. Cp is a perfectoid �eld. In fact, anyalgebraically closed (complete non-archimedean �eld) is perfectoid �eld, because we can take p-th roots on K◦ already.

De�nition 4.3. A perfectoid rank 1 valuation ring V is a rank 1 valuation ring such that

(1) p ∈ rad(V )(2) V is x-adically complete for all x pseudo-uniformizer (that is V ∼= lim

←V/xn).

(3) V is not a DVR, so the maximal ideal of V is not principal

(4) Frob : V/p→ V/p is surjective

Example 4.4. In characteristic p, a perfectoid �eld is the same as a complete non-archimedean �eld that is perfect, which

is the same as a perfect rank 1 valuation ring which is complete.

Example 4.5. Fp is not allowed, but Fp((t1/p∞

))∧ is a perfectoid �eld, with valuation ring Fp[t1/p∞

]∧.

Example 4.6. Consider Qp(p1/p∞

)∧. This absolute value is not discretely valued, because∣∣p1/pn ∣∣ = |p|1/pn .

Claim 4.7. K◦ = Zp[p1/p∞

]∧ = Zp[x1/p∞

]∧/x = p.

Proof. This is because

K = ∪̂n≥0Qp(p1/pn

).

Let Kn = Qp(p1/pn

). Then K◦n = Zp[un]/upn

= p because we can adding a root of an Eisenstein polynomial. Let n→∞and p-adically complete. �


Then K◦/p = Fp[x1/p

∞]/x, and Frob is surjective.

Example 4.8. L = Qp (ζp∞)∧ is a perfectoid �eld. Then L0 = Zp[ζp∞ ]∧, so L0/p has surjective Frobenius, becauseζpn � ζpn−1 .

Proposition 4.9. TFAE: for a (complete non-archimdean) �eld K of characteristic 0

(1) K perfectoid

(2) there exists ω ∈ K◦, which is a pseudo-uniformizer with(a) |ω|p = |p| (that is, ωp = p·unit)(b) ω admits a system of p-power roots

(c) K◦/ωFrob∼= K◦/ωp = K◦/p

Lemma 4.10. If K is perfectoid, then im |·| is uniquely p-divisible.

Proof. If K is not discretely valued, the image of absolute value is generated by |x| such that 0 < |x| < 1. Assume wehave such an x. Then we want to �nd some y such that |y|p = |x|. Since x ∈ K◦, we can wirte x = yp + pz for y, p ∈ K◦

(by the surjective Frobenius condition). Then |y|p = |x|. �

Proof. of proposition. By the lemma, we can �nd ω0 ∈ K◦ such that |ω0|p = |p|. By the surjectivity condition, we can�nd a compatible sequence ωi ∈ K◦ such that ωpi+1 ≡ ωi mod p. This gives an element in

lim←

x 7→xp

K◦/p ∼= lim←

x7→xp

K◦.

By the isomorphism, there exists (w′n) ∈ K◦ with ω′n = wn mod p and ω′pn+1 = ω

′n for all n. Then w

′0 ≡ ω0 mod p and

has p-power roots. Take ω = ω′0, then

K◦/ωFrob→ K◦/ωp = K0/p.

By assumption, surjective since K is perfectoid. It is also injective by divisibility reasons. �

De�nition 4.11. If K is a perfectoid space of characteristic 0, say ω ∈ K◦ a is a perfectoid pseudo-uniformizer if thefollowing holds.

(1) ω admits a compatible p-power root ω1/pn

(2) ω | p

Proposition 4.12. If K is a perfectoid �eld of characteristic 0, ω ∈ K0 such that |ωp| = |p|. Every element of K◦/pω(not an Fp-algebra) is a p-th power.

Proof. Let x ∈ K0, then x = yp + pz = yp + ωpz for y, z ∈ K0. Let z = zp1 + ωpz2 and continue. Then

x = yp + ωpzp1 mod ω2p.

Additionally, x = yp + ωpzp1 = (y + ωz1)p

mod pω. �

4.1. Tilting. Tilting is a functor from perfectoid �elds in characteristic 0 to perfectoids in characteristic p. Recall we

have map

R→ R[

from a p-adically complete ring to perfect Fp-algebra. Consider K◦[ and its fraction �eld K[. We claim that this is a rank1 valuation ring.

Theorem 4.13.

(1) Let V be a perfectoid rank 1 valuation ring, then V [ is a rank 1 perfectoid valuation ring (in characteristic p)


(2) Say K = Frac(V ) has characteristic 0. Choose a pseudo-uniformizer ω of V , with ω = p(unit) in V and ω admits

a compatible system of p-power roots {ω1/pn}, which determines ω[ ∈ V [. Therefore, V/ω = V [/ω[.

Proof. For x, y ∈ V [, we can identify them with x = (x0, x1, ...) and y = (y0, y1, ...) sequence in V such that xpi+1 = x andypi+1 = y. Since V is a domain, so is V

b. Similarly, this expression as a monoid shows that V [ is a rank 1 valuation ring.

There is a ring homomorphism V [ → V/p by projection to the last coordinate. Since V/p is semiperfect, this issurjective. Letωb =

(ω, ω1/p, ...

)∈ V [. Notice that proj

(ω[)

= ω = 0 in V/p. Therefore, we get a natural map

V [/ω[ → V/p.

Need that the kernel is exactly (ω[). Again, we can write x ∈ V [ as (x0, x1, ...) a p-power sequence in V . If x→ 0 in V/pthen p | x0 and so ω | x0. Similarly, ω1/p

n | xn. That is, ω[ | x. �

Example 4.14. Let K1 = Qp(p1/p∞

)∧. Then

K◦1 = Zp[p1/p

∞]∧

= Zp[x1/p∞

]∧/(x− p).

Then

K◦1/p = Fp[x1/p∞

]/x, and K◦[1 = FpJx1/p∞K∧.

Then

K[1 = Fp((x1/p∞

))∧.

Here, x =(p, p1/p, ...

).

Remark 4.15. If K is perfectoid, then

K[ = lim←

x→xp

K.

Example 4.16. Let K2 = Qp(ζp∞)∧. Then

K◦2 = Zp[ζp∞ ]∧ = Zp[x1/p∞

]∧/

(xp − 1x− 1

).

Then

K◦2/p = Fp[x1/p∞

]∧/ (x− 1)p−1

and

K◦[2 = Fp[y1/p∞

]∧

where y = x− 1 andK[2 = Fp((y1/p

∞)),

and y =[(ζp, ζp2 , ...)

]− 1.

Remark 4.17. Note that K[1 = K[2. This means that tilting loses some information.

Proposition 4.18. For a perfectoid �eld K, write K[ = lim←

x→xp

K. De�ne # : K[ → K to be projective to the last coordinate

which is a multiplicative map. We can de�ne |x| =∣∣x]∣∣

K.

Proof. Clearly a multiplicative construction. The set{x ∈ K[ : |x| ≤ 1

}is exactly K◦[ and has the non-archimedean property. �

Theorem 4.19. (Fontaine-Winterberger, Scholze, Kedlaya-Lin) If K is a perfectoid �eld, then there is a canonical

isomorphism

Gal (K) ∼= Gal(K[).


More precisely, if L/K is a �nite extension, then L[/K[ is a �nite extension. This gives an equivalence of categories

{�nite extensions of K} ↔{�nite extensions of K[

}.

Example 4.20. We have equalities

Gal(Qp(p1/p

∞))

Gal (Fp((t)))

Gal(Qp(p1/p

∞)∧)

Gal (Fp((t))∧)

Henselian perfection

Recall for any R a p-complete ring, we have θ : W (R[) → R which comes from adjunctions. An element in W (R[) isof the form x =

∑∞i=0 [xi] p

i where [·] : R[ →W (R[) is the Teichmuller lift. Then

θ

( ∞∑i=0

[xi]pi

)=

∞∑i=0

pix]i .

Theorem 4.21. For V a perfectoid (rank 1) valuation ring, consider θ : W (V [) → V . Then θ is surjective and kernelgenerated by a non-zero divisor.

Proof. To see that it's surjective, notice that mod p, V [ → V/p is surjective, so θ is surjective.Choose perfectoid pseudouniformizer ω ∈ V such hat ω = p(unit) in V and ω1/pn . This determines ω[ ∈ V [. Then

under θ,[ω[]∈W (V [) maps to ω. Consider an element ũ ∈W (V [) unit such that θ(ũ) = u. Now[

ω[]− pũ ∈W (V [)

projects to zero under θ. Want to show that this generates the kernel. We will check that

W (V [)/(

[ω[]− pũ)

is p-torsion free. Since both sides are p-torsion free and p-complte, su�ces to be isomorphisms mod p. Which is to say

V [/ω[ ∼= V/p,

which we have. �

De�nition 4.22. For V a perfectoid (rank 1) valuation ring of characteristic p. An element x =∑i≥0 [xi] p

i ∈ W (V ) iscalled primitive of degree 1 i�

(1) x0 is in the maximal ideal of V

(2) x1 is a unit in V

The element constructed in the proof is a primitive element of degree 1. The degree 0 term is ω a pseudo-uniformizer.

The degree 1 term is u which is a unit.

Remark 4.23. Given x ∈ W (V ), how do re recognize that it is primitive of degree 1. Suppose V has maximal ideal mVand residue �eld kV .

Claim 4.24. x ∈W (V ) is primitive of degree 1 i� x maps to p(unit) in W (kV ).

Proof. If x is primitive, x maps to p [x1] + p2 + ... is p(unit). If x is not primitive, then either x0 is a unit, or x1is a

non-unit. These are not possible. �

Example 4.25. Recall that K1 = Qp(p1/p∞

)∧ and K2 = Qp(ζp∞)∧ have the same tilt Fp((t1/p∞

))∧. In the �rst case, t

corresponds to (p, p1/p, ...) and in the second, to(1, ζp, ζp2 , ...

)− 1. By convention, p[ = (p, p1/p, ...) ∈ Qp(p1/p

∞)∧[ and

� = (1, ζp, ...) ∈ Qp(ζp∞)∧[. What are the associated primitive elements.

Case 1. p−[p[]∈W (K◦[1 ). This is because θ

([p[])

= p and so clearly a primitive element of degree 1.


Case 2. θ([�]) = 1, and

θ(

[�1/p])

= θ([(ζp, ζp2 , ..., )]

)= ζp.

A primitive element is then

[�]− 1[�1/p

]− 1

= 1 +[�1/p

]+[�2/p

]+ ...+

[�p−1p

]and generates ker θ. It is clear that it is in the kernel. To check primitive, su�ces to map to W (residue �eld).

Since K2 is totally rami�ed, mapping to W (Fp) = Zp. Moreover, � = 1 mod maximal ideal. Under the map

W (K◦[)→W(K◦[/max ideal

)the elements [�]−1

[�1/p]−1is mapped to p.

Proposition 4.26. Suppose V is a perfectoid rank 1 valuation ring of characteristic p, and x ∈ W (V ) is a primitivedegree 1 element. Then

V # = W (V )/x

is a perfectoid valuation ring such that (V ])[ = V .

Example 4.27. Let x = p, then V ] = V .

Proof. Suppose x =∑i≥0[xi]p

i = [x0] + pu where u is a unit in W (kV ) and assume x0 6= 0. First, V ] is p-torsion free.Let z ∈ V ] such that pz = 0. Then we can lift to z̃ ∈W (V ) such that

pz̃ = ([x0] + pu) y.

Then y must be divisible by p. Therefore, z̃ = ([x0] + p) (yp ) is in (x).

Note that p is a uniformizer in W (V )/x. Therefore, W (V )/p = V and x acts as x0 in V . Note that V]/p =

W (V )/(p, x) = V/x0. Therefore,

V ][ = (V/x0)perf

= V ∧x0 = V.

Now, we want to show V # is a valuation ring. Need to show that if a, b ∈ V # then either a | b or b | a. To prove this,want a = [a′0] (unit) and b = [b

′0] (unit) for a

′0, b′0 ∈ V . This would prove the claim, because V is a valuation ring.

Fix a ∈ V # and assume a not divisible by p = [x0](unit) mod x. Lift a to ã =∑i≥0 [ã] p

i ∈ W (V )s. Since it's notdivisible by p in V ], ã is not divisible by x0. Therefore, ã = [ã0] + p(various). In V

],

a = [ã0] + pω = [ã0]

(1 +

p

[ã0]ω

).

By our choices, this p[ã0] is topologically nilpotent in V] because p = [x0](unit). �

4.2. Perfectoid Algebra. Let K be a complete non-archimedean �eld with respect to |·| : K → R≥0.

De�nition 4.28. A K-Banach algebra is a commutative K-algebra A equipped with an absolute value satisfying

• |x| ≥ 0 with equality i� x = 0 and |1| = 1• |x+ y| ≤ max (|x| , |y|)• |xy| ≤ |x| · |y| with equality if x ∈ K

and A is complete with respect to |·|.

This makes A into a topological ring and a map of K-Banach algebras is continuous homomorphisms.

Example 4.29. Construction: Suppose K◦ ⊆ K is the ring of integers. A0 is a K0-algebra such that for all x ∈ K0

pseudo-uniformizer, A0 is x-adically complete, torsion free. Let A = A0[

1x

], which ahs the structure of a K-Banach


algebra,

|a| = inft∈K×, ta∈A0

1

|t|.

Given a K-Banach algebra A, let

A0 = {x ∈ A : |x| ≤ 1} ⊆ A

is a subring which is a K0-algebra complete and torsion free over K0. Conversely, given this, we get K−Banach algebra.

Warning! Given a map A→ B of K-Bananch algebras, it is not true that it carries

{a ∈ A : |a| ≤ 1} → {b ∈ B : |b| ≤ 1} .

Additionally, there are di�erent constructions of A0. We can �nd A′0 such that A0, A

′0 di�er by some bounded power of x.

Example 4.30. Topologically �nite type algebras. Consider K0-algebra K0 [T1, ..., Tn] . Let A0 = K0 [T1, ..., Tn]

∧x then

An = A0

[1

x

]=

∑i1,...,in≥0

ai1...inTi11 ...T

inn

where ai1,...,in ∈ K with a· → 0. This is called a Tate algebra.A K-algebra of topologically �nite type is a quotient of a Tate algebra.

Fact 4.31.

• An is a Noetherian ring, where all ideals are closed• Any topologically �nite type K-algebra canonically has the structure of a K-Banach algebra, any map is continuous(of K-algebras)

Suppose A is a K-Banach algebra. De�ne A0 ⊆ A to be the set of power bounded elements. That is, x ∈ A0 i�supn |xn| < ∞. De�ne A◦◦ be the topologically nilpotent elements, which are x ∈ A, |xn| → 0 as n → ∞. These arefunctors from K-Banach algebra to K◦-algebras.

De�nition 4.32. A K-Banach algebra A is uniform if A◦ is bounded.

Example 4.33. Consider K[�]/�2. Then A◦ = K◦ +K� is not uniform.

Example 4.34. Consider R = K◦ [x1, x2, ...] viewed as a graded ring with |xi| = 1. De�ne R′ ⊆ R to be the subringgenerated by all deg ≥ 2 elements, and pixi for all i ≥ 1. Let A =

(R′∧p

)[ 1p ] with respect to the ring of de�nition R

′∧p .

Observe that xi ∈ A, not in R′ (disk of radius 1) because pixi ∈ R′. Additionally, x2i ∈ R′ for all i. All the xi's are powerbounded, but xi is not bounded.

Any topologically �nite type K-Banach algebra is uniform i� it is reduced. This can fail in general.

De�nition 4.35. Let A be a uniform Banach algebra over K (K perfectoid). De�ne the spectral norm to be

|a|sp = limn→∞ |an|1/n = inf

n≥1|an|1/n .

Note that |a|sp < 1 i� a is topologically nilpotent.

Proposition 4.36. Suppose A is a uniform K-Banach algebra. Since K is perfectoid, we have ω pseudouniformizer with{ω1/p

n}n≥0. Then for x ∈ A, TFAE:

(1) x ∈ A0

(2) |x|sp ≤ 1(3) for all n, ω1/p

n

x is topologically nilpotent

Proof. 1 =⇒ 2 =⇒ 3 is easy to see, and without the use of uniformity.


Now, for 3 =⇒ 1. For x such that ω1/pnx is topologically nilpotent for all n, this implies that there are in A◦. Thenfor all i ≤ pn, ωi/pnxi ∈ A◦ and so xi ∈ ω−1/pnA◦ ⊆ ω−1A0. Therefore, ωxi ∈ A◦ for all i. Since A◦is bounded (byuniformity) the powers of x are bounded. That is, x ∈ A◦. �

Remark 4.37. A uniform, |·|sp is a norm on A such that the unit disk is A◦. Any A→ B (uniform) is contracting for |·|sp.Note that for uniform A, A◦ is canonical and is a K◦-algebra, ω-complete and torsion free.

Proposition 4.38. R = A◦ for some uniform K-Banach algebra A i�

(1) R is ω-complete, torsion free K◦-algebra

(2) R is saturated in the sense that if x ∈ R[ 1ω ] with ω1/pnx ∈ R then x ∈ R

(3) R is p-root closed. That is, for x ∈ R[ 1ω ] such that xp ∈ R then x ∈ R

Proof. Given R, let A = R[ 1ω ] and make it into a K-Banach algebra such that R is the unit disk. Need to show that A is

uniform and R = A◦.

Clearly, R ⊆ A◦. Let a ∈ A◦. Since ω1/pna is topologically nilpotent, for some N large enough,(ω1/p

n

a)N ∈ R. Now

take N = phigh power then ω1/pn

a ∈ R (by property 3). Therefore, ω1/pna ∈ R. �

Let K be a perfectoid �eld, ω ∈ K◦ a pseudouniformizer with

• ω | p and |p| ≤ |ω| < 1• Have ω1/pn for all n

De�nition 4.39. Say A a K-Banach algebra is perfectoid if

(1) A is uniform (A◦ is bounded)

(2) A◦/ω is semiperfect (ie. Frobenius surjective)

Proposition 4.40. The second condition is equivalent to A◦/p has surjective Frobenius

Proof. A◦ is p-adically complete and ω-adically complete. For a ∈ A◦, we can write a = bp + ωz and z = cp + ωy. Then

a = bp + ωcp +O(ω2)

and so

a ≡(b+ ω1/pc+ ...

)pmod p.

�

Example 4.41. (Perfectoid disc) Let A = K〈T 1/p

∞〉is K-Banach algebra obtained by K◦

[T 1/p

∞]∧ω

[ 1ω ]. We claim that

this is perfectoid.

Proof. A◦ =(K◦[T 1/p

∞])∧

and A◦/p is semiperfect. �

Observe that A is a perfectoid K-algebra,

ϕ : A◦/ω1/p ∼= A◦/ω

Proof. Just need to show that ϕ is injective. Suppose z ∈ A◦, zp ∈ ωA◦ then z is divisible by ω1/p and so zω1/p

∈ A.Hence, ( z

ω1/p

)p=zp

ω∈ A◦

and so ω1/p | z in A◦. �

De�nition 4.42. A K◦-algebra R is called integral perfectoid if

(1) R is ω-adically complete and ω torsion free

(2) ϕ : R/ω1/p ∼= R/ω

Example 4.43. R = K◦[T 1/p

∞]∧0is integral perfectoid


Claim 4.44. If R is integral perfectoid, A = R[ 1ω ] perfectoid Banach algebra.

Proposition 4.45. R is integral perfecoid, then R is p-root closed in R[ 1ω ].

Proof. Let x ∈ R[ 1ω ] with xp ∈ R. Want x ∈ R. There exists some i > 0 such that ωi/px ∈ R. Now, (ωi/px)p = ωixp ∈ ωR.

This means that

ω1/p | ωi/px in R and so x ∈ ω(i−1)/pR.

Induction, to show that x ∈ R. �

R is not necessarily saturated.

Example 4.46. Let k be the residue �eld of K. Let

R = K◦ ×k K◦ ={

(x, y) ∈ (K◦)2 : x = y mod K◦◦}.

This is not saturated because (1, 0) ∈ R[ 1ω ] and so not saturated.

De�nition 4.47. Let M be a torsion free K◦-module,

M? =

{x ∈M [ 1

ω] : ω1/p

n

x ∈M for all n > 0}.

Proposition 4.48. R integral perfectoid K◦-algebra, then A = R[ 1ω ] is a perfectoid K-algebra and A◦ = R? (which is

integral perfectoid).

Remark 4.49. Consider A◦/A◦◦ = T0. If T1 ⊆ T0 is a perfect subalgebra, then A◦ ×T0 T1 is integral perfectoid.

Lemma 4.50. R integral perfectoid, then so is R?.

Proof. Need to check R?/ω1/p → R?/ω is an isomorphism. Can check that ω admits p-power roots. �

Proof. To prove the proposition, enough to show if R is saturated and integral perfectoid. Apply the criterion from last

time. Then R = A◦ and A = R[ 1ω ] with A uniform. �

Theorem 4.51. (Tilting equivalence) There is an equivalence of categories

{perfectoid K-algebras} ↔{perfectoid K[-algebras

}.

Also true that integral perfectoid K◦-algebras correspond to integral perfectoid K◦[-algebras.

Proof. The construction is the following. We send an integral perfectoid K-algebra R to R[ = (R/p)perf

= (R/ω)perf

.

The proposition shows that if R is perfectoid K◦-algebra, then R[ is a perfectoid K◦[ algebra, and

R/ω ∼= R[/ω[.

Also, this correspondence preserves saturation, and gives correspondence for perfectoid Banach algebras.

Construction: if R is a (integral) perfectoid K◦-algebra, then

θ : W (R[)→ F

is surjective and R ∼= W (R[)⊗W (K◦[). �

Proposition 4.52. Let K be a perfectoid �eld of characteristic p.

(1) An (integral) perfectoid K◦-algebra is a ω-complete , ω-torsion free K◦-algebra R which is perfect

(2) (André) A K-Banach space A is perfectoid i� it is perfect.

Proof. We will prove 2 �rst. Need to show A is uniform. Let A0 ⊆ A be a ring of de�nition in A. If A0 is perfect (henceintegral perfectoid), A = A0[

1ω ] is perfectoid. Consider ϕ : A → A is an isomorphism. By the Banach open mapping


theorem, it is an open linear homeomorphism. Hence, ϕ(A0) ⊇ ωNA0 for N su�ciently large (so it's an open set). Thenϕ−1 (A0)ω

N/p ⊆ A0 and so ϕ−1(A0) ⊆ ω−N/pA0.Repeat this argument to get ϕ−2(A0) ⊇ ω−(N/p+N/p

2)A0. Inductively,

ϕi (A0) ⊆ ω−(N/p+N/p2+...+N/pi)A0 ⊆ ω−MA0

for M independent of i.

∪iϕi(A0) = (A0)perf ⊆ ω−MA0.

Therefore, A admits an open bounded subring (A0)perf . �

5. Equivalence of category of étale modules

Theorem 5.1. Let K be a perfectoid �eld, R perfectoid K-algebra, then if S/R is a �nite étale algebra, then S is a also

perfectoid as a K-algebra. The construction S → S[ establishes an equivalence between the categories

{�nite étale R-algebras} ↔{�nite étale R[-algebras

}.

Proof goes through a reduction to perfectoid �elds.

Theorem 5.2. (Special case) If L/K is a �nite extension, then L is also perfectoid, giving an isomorphism


Strategy is to reduce to the algebraically closed case.

Proposition 5.3. Let K be a perfectoid �eld. If K[ is algebraically closed, then so is K.

Proof. Use a successive approximation argument. Claim that the image of the absolute values on K and K[ in R≥0 agree.

Note that for x ∈ K[,|x|K[ =

∣∣x#∣∣K,

so we just need the other way.

Recall that we can choose ω and ω1/pn

in K where ω[ ∈ K[ and K◦/ω ∼= K◦[/ω[. This implies that the value group ofK and K[ are the same, because the value group of K are generated by |x| where |w| < |x| < 1 and similarly for K[.

Since K[ is algebraically closed, its value group is a Q-vector space (because you can take roots). Goal is to show thatany monic f(x) ∈ K◦[x] has a root in K◦. We know that f has a root mod w. We know that f has a root mod ω. Thefollowing proposition then �nishes o� the proof. �

Proposition 5.4. Suppose K is a complete non-archimedean �eld with

(1) The image of |·| : K× → R>0 is a Q-vector space(2) There exists a pseudo-uniformizer ω such that any f(x) ∈ K◦[x] monic has a root mod ω.

Then K is algebraically closed.

Proof. Let f(x) ∈ K◦[x] is an irreducible monic polynomial of degree d. Suppose we have an approximate root, that is,we have α ∈ K◦ such that |f(α)| < �. We claim that there exists α′ ∈ K◦ such that

(1) |α− α′| ≤ �1/d

(2) |f(α′)| ≤ � |ω|

Starting with any α, we get a sequence α, α′, α′′, ... which comverges to a root. First, assume α = 0 (by replacing f with

f(x+ α)). f is an irreducible polynomial, f(0) is constant term of absolute value �. That is,

f(x) =

d∑i=0

aixid−i


and |ad| = �. Since f is irreducible, all of its roots have the same absolute value, and must be �1/d. Choose an elementof K with |v| = �1/d. De�ne f̃(x) = v−df(vx). Then f̃(x) ∈ K◦[x] by construction. By 2, there exists β ∈ K◦ such that∣∣∣f̃(β)∣∣∣ ≤ |ω| and so |f(vβ)| ≤ � |ω| and |vβ| ≤ �1/d. Take α′ = vβ. �Lemma 5.5. (Artin) Let L be a �eld, G a �nite group acting faithfully on L. Then if K = LG then L/K is Galois with

group G.

Lemma 5.6. (Krasner) If K is a complete non-archimedean (perfect) �eld, α, β ∈ K̄ and |β − α| < |α− ᾱ| for eachGalois conjugation ᾱ of α, then K(α) ⊆ K(β).

Proof. of the tilting equivalence.

Recall that if L/K[ is a �nite extension, L# =(W (L◦)⊗W (K[) K◦

)[ 1p ] is a perfectoid �eld over K. Suppose L/K

[ is

Galois with Galois group G. Then G acts naturally on L◦ and L#. Additionally,(L])G

=(W (K◦[)⊗W (K◦[) K◦

)[1p

]= K.

We can take G-invariants inside the brackets, because cohomology disappears since we are in characteristic 0.

It follows that if L/K[ is a G-Galois extension then so is L]/K by Artin's lemma. Get a correspondence under tilting.

To complete the proof, we need to show there are enough of these sharp extensions.

To complete the proof, it's enough to show that any �nite extension of K is contained in some L] or that E =

∪L/K[ �nite L] is algebraically closed. E is not necessarily complete, but Ê is perfectoid.(Ê)[

= ∪̂L/K[ �niteL =ˆK[

is algebraically closed. Therefore, E is algebraically closed. �

6. Almost mathematics

Let R be a ring, I ⊆ R an ideal, such that

(1) I2 = I

(2) I is �at as an R-module.

Example 6.1. Let V be a perfectoid valuation ring V = K◦, K perfectoid. V is local with maximal ideal m that is

torsion free and m2 = m. We can take (R, I) = (V,m).

Example 6.2. Let R be any ring. Let t ∈ R, with p-power roots t1/p∞ . Suppose t is a uniformizer, then we can takeI =

(t1/p

∞)= ∪n≥0

(t1/p

n).

Example 6.3. If R is a perfect Fp-algebra, t ∈ R, can take I = (t1/p∞

) = rad(t). Then I is �at, even if t is a zero divisor.

Remark 6.4. R/I ⊗LR R/I ∼= R/I. (That is, TorRi (R/I,R/I) = 0 for i > 0).

Proof. We have

0→ I → R→ R/I → 0.

Take Tor with R/I. Note that

TorRi (I,R/I) = 0 and TorRi (R,R/I) = 0 for i > 0.

So the long exact sequence degenerates

0→ Tor1 (R/I,R/I)→ I ⊗R/I → R⊗R/I → R/I ⊗R/I → 0.

By I ⊗R/I ∼= I/I2 = 0, so the Tor is zero. �


De�nition 6.5. An R-module M is almost zero, if IM = 0.

Remark 6.6. Note, this is equivalent to the condition I ⊗RM = 0. This is because we have a surjection I ⊗RM → IM .On the other hand, I ⊗R R/I = 0.

Observe that almost zero R-modules is closed under extensions, that is, given a short exact sequence

0→M ′ →M →M ′′ → 0

and if 2/3 of the above are almost zero, then so is the third.

Loose analogy: Consider S, J = (Z, (p)) (not a set-up admissible as above). Analog is to consider

Fp vector spaces ⊆ abelian groups.

Note that we have a short exact sequence

0→ Z/p→ Z/p2 → Z/p→ 0,

and things are not closed under extension this way.

De�nition 6.7. A an abelian category B ⊆ A a full subcategory is same on beat, if

(1) B is an abelian subcategory of A(2) B is closed under extensions. That is, if

0→M ′ →M →M ′′ → 0

short exact sequence, and if 2 of the three are in B, the third is as well

(3) B is closed under subobjects, quotient and contains 0

Remark 6.8. Don't need 1, lol.

Example 6.9. Non-example. Fp-vector spaces ⊆abelian groups

Example 6.10. p-power torsion abelian groups ⊆abelian groups.

Example 6.11. Almost zero objects (R/I-modules) ⊆ R-modules.

Remark 6.12. p-power torsion abelian groups are not stable under products but almost zero modules are.

Construction: (Quotient category). B ⊆ A is an Serre subcategory, then one forms a quotient A/B with an exactfunction A → A/B which send objects in B to 0. If f : A → C is exact, and f(B) = 0, f canonically factors throughA/B. Explicitly, one constructs A/B such that the objects are those of A,

homA/B(X,Y ) = lim→x′→x

homA (X′, Y )

all maps f : X ′ → X such that ker f, cokerf ∈ B.

Example 6.13. For p-power torsion abelian groups ⊆ abelian groups, the quotient is category of Z[ 1p ]-modules.

De�nition 6.14. Let (R, I) be as before. The category of Moda(R) or Ra-modules as the quotient category

Mod(R)

almost zero modules.

Example 6.15. R = Zp[p1/p∞

]∧, and I = ∪(p1/p∞).

There is a functor Mod(R)→Moda(R) by M 7→Ma. Now,

hommoda (Ma, Na) = lim

→M′→M

homR (M′, N) = homR (I ⊗M,N) .


Analogy: if X is a topological space, L ⊆ X open,

Sh(X)j?→ Sh(U)

is exact. It has

(1) a right adjoint j?

(2) a left adjoint j! (exact functor)

The almosti�cation function Mod(R) → Moda(R) is analogous, and says adjoints (·)i and (·)? : Moda(R) → Mod(R).Here, (·)i is left adjoint, and the other right adjoint.

(1) Mi = I ⊗RM(2) M? = homR (I,M)

Remark 6.16. Consider the category of all R-modules M such that I ⊗RM ∼= M . That's a full subcategory of R-modules

equivalent to Moda(R). (ie. Moda(R)j!⊆Mod(R)).

Remark 6.17. Moda(R) has a tensor product. one has the notion of an Ra-module A (if so, A? is an actual A-algebra).

Example 6.18. If M ∈ModR, then M1 →M →M? these are almost isomorphisms.

Example 6.19. R! = I.

Example 6.20. Consider M?. I = (t1/p∞) with t non-zero divisor. If M is t-torsion free, then M? ⊆M [ 1t ] is exactly{

x ∈M [ 1t] : t1/p

n

x ∈M for all n > 0}.

In general, M 7→M? is not exact.

Example 6.21. I? = R and R? = R.

Example 6.22. Consider (R/t)?, which has elements such as∑n>0 t

1−1/pn .

Let R be a commutative right, I ⊆ R an ideal such that I2 = I.

De�nition 6.23. Ma ∈Moda(M) is almost �nitely generated with for all � ∈ I, there exists a �nitely generated A-moduleM ′ and a map M ′ →M whose cokernel is annihilated by � (WLOG M ′ ⊆M).

Example 6.24. R = Fp[t1/p∞

] and I =(t1/p

∞). Take M = ⊕i≥0R/t1/p

i

.

De�nition 6.25. M is uniformly almost �nitely generated, if the number of generators of M ′ = M ′� is indepdent of �.

Similarly, we say M is almost �nitely presented if for all � ∈ I, there exists M ′ = M ′� and maps

f : M ′� →M and g : M →M ′�

such that

(1) M ′ is �nitely presented

(2) f ◦ g = g ◦ f = �

Example 6.26. R = Fp[t1/p∞

] with p 6= 2, S = Fp[t1/2p∞

]. There is a map R → S by R → R[√t] ↪→ S, but S is not a

�nitely generated module over R (don't have t1/2p2

).

Remark 6.27. If we replace everything with t-adic completions, then we can consider K = Fp((t1/p∞

))∧, L = K(√t). The

map R̂t → S∧t is K◦ → L◦.

Example. Back to the example. S is (uniformly) almost �nitely generated over R. For each n, R[√

t1/pn]⊆ S and this

has cokernel annihilated by t1/pn

. We can approximate S by degree 2 �nitely generated projective R-modules.


Let V = K◦, for K a perfectoid �eld. Let R = V and I = mV .

Example 6.28. If I ⊆ V any ideal, determined by γ ∈ [0, 1], such that I = {x ∈ V : |x| ≤ γ} (or |x| < γ), call I = Iγ .Iγ is principal i� γ ∈value group of K. Iγ is always almost �nitely generated.

De�nition 6.29. If M,N are V -modules, M ≈ N if there are maps f� : M → N and g� : N → M such thatf� ◦ g� = g� ◦ f� = �.

Theorem 6.30. (Scholze) If M is an almost �nitely generated V -module, then M ≈ V/Iγ1 ⊕ V/Iγ2 ⊕ ... such that0 ≤ γ1 ≤ γ2 ≤ ... and lim γi = 1.

De�nition 6.31. Let A be an Ra-algebra, Ma ∈ Moda(A) is almost �at if TorAi (M,N)a= 0 for all N , i > 0. (That is,

tensoring with M is exact in the Moda(A) category).

Ma is almost projective if ExtiA(M,N)a= 0 for all N .

Example 6.32. In Moda(R), R is generally not projective. Let R = Fp[t1/p∞

]

0→ R t→ R→ R/t→ 0

and apply the function (·)? = homModa(R) (R, ·) to get

0→ R t→ R→ (R/tR)? .

However,∑n>0 t

1− 1pn ∈ (R/tR)?, but does not come from R.

Note that if V = K◦, K is spherically complete (every descending sequence of disks has non-empty ∩), then the (·)?has no higher derived functors on R itself.

Proposition 6.33. Given an R-module M , M is �nitely generated i� for all inductive systems {Nα}α∈A, the map

lim→

homR (M,Nα)→ homR(M, lim

→Nα

)is injective. M is �nitely presented if it is always an isomorphism.

De�nition 6.34. For M,N ∈Moda(R), de�ne alHomR(M,N) as the almosti�cation of HomR(M,N).

Proposition 6.35. M ∈ Moda(R) is almost �nitely generated (resp. almost �nitely presented) if for all {Nα}α∈A themaps

lim→alHom (M,Nα)→ alHom

(M, lim

→Nα

)is injective (resp. isom).

Proposition 6.36. M is almost projective if alHom(M, ·) is exact on Homa(R).

The principle is that in perfectoid settings, rational statements hold almost integrally.

Proposition 6.37. R perfect Fp-algebra, I = (t1/p∞

). Suppose f : S → S′ is a map of perfect R-algebra, such that

(1) f [ 1t ] is an isomorphism

(2) S, S′ are integral over R

Then f is almost isomorphism.

Proof. WLOG, S, S′ have no t-torsion. This is because if J ⊆ S are t-power tors, then for x ∈ J , tNx = 0. ThentN/p

R

x1/pR

= 0 and so tN/pR

x = 0.

Can assume that S ⊆ S′ ⊆ S′[ 1t ] − S[1t ]. Need to see almost equal. Let x ∈ S

′, The powers of x generate a �nitely

generated R-module. For each xM , there exists N = N(M) such that tNxM ∈ S. Can choose N uniformly in M , becausepowers of x form a �nitely generated R-modules (x is integral). Then tNxi ∈ S for all i. Take i = pR for R large, thentNxp

R ∈ S . By perfection, tN/pRx ∈ S. �


De�nition 6.38. For A a commutative ring, then an A-algebra B is called �nite étale if

(1) B is a �nitely generated and projective

(2) m : B ⊗A B → B exists B ⊗A B ∼= C ×B where C is some other A-algebra, and m is the projectionEquivalently, this multiplication map m has a section in B ⊗A B-module, so there exists an idempotent elemente ∈ B ⊗A B such that m(e) = 1 and e (kerm) = 0

Example 6.39. Geometrically, Y = Spec B, X = Spec A, Y → A is analogous to a �nite sheeted cover. Y → Y ×X Yis an open embedding when it's a �nite sheeted cover (always a closed embedding) and so it is a union of connected

components.

Remark 6.40. Instead, could assume that B is �at as an A-algebra, and �nitely presented as an algebra (étale morphism,

but not �nite étale) (replacing 1 and keeping 2).

Could assume instead, that B is �at over A is �at, and m is �at as well. This gives the de�nition of a weakly étale

morphism.

Example 6.41. Any �ltered colimit of étale A-algebras is weakly étale.

Example 6.42. E ⊆ F �nite separable, it is �nite étale.

Example 6.43. A a ring, f(x) is a monic polynomial, then A→ (A[x]/f(x)) [ 1f ′(x) ] is an étale map.

In �nite étale algebra, have a natural idempotent e ∈ B ⊗A B. This correponds to the factor B under m.Construction: B/A is an algebra such that B is �nitely generated and projective as A-module. De�ne Tr : B → A (by

b 7→ TrA(b : B → B)). This de�nes a bilinear pairing B ⊗A B → A via b1 ⊗ b2 7→ TrA(b1b2). If B/A is �nite étale, thenTr is non-degenerate and induces B ∼= B∨. We can de�ne a Casimir element in B ⊗A B, in the following way.

If B is free over A, choose basis bi ∈ B and a dual basis b′i ∈ B, then the Casimir is∑bi ⊗ b′i ∈ B ⊗A B. In general,

we can do this Zariski locally.

Claim 6.44. The Casimir element is exactly the idempotent e.

Suppose B/A is just projective. Suppose e =∑nj=1 cj ⊗ dj ∈ B⊗AB. Then we have a map B → An by b 7→ {Tr(bcj)}

and An → B by {aj} 7→∑ajdj . This Casimir element allows us to say that this

B → An → B

is the identity map (from being an idempotent).

Example 6.45. Suppose A is a Z[ 12 ]-algebra, x ∈ A×. B = A [

√x] then e = 12

(1 +√x⊗ 1√

x

).

Theorem 6.46. Suppose A is I-adically complete. There is an equivalence of categories

�nite étale A-algebras↔ �nite étale A/I-algebra

by B 7→ B/IB.

Example 6.47. If Fq/Fp is a �nite extension, then W (Fq)/Zp is �nite étale.

Theorem 6.48. Suppose (A, I) is a pair where I ⊆ A is nilpotent, then

weakly étale A-algebra ↔ wealy étale A/I-algebras

by B 7→ B/IB.

Now, we consider the almost version, (R, I) setting for almost ring theory. (Here, I2 = I with I �at).

De�nition 6.49. Suppose A is Ra-algebra. Then an A-algebra B is almost �nite étale if


(1) B almost �nitely generated projective as A-module

(2) The map m : B ⊗A B → B has a splitting in the almost category (ie, B ⊗A B ∼= B × C in the almost category)

This gives the idempotent e ∈ (B ⊗A B)? which comes from the splitting.

Example 6.50. For p 6= 2, Fp[t1/p∞

] ⊆ Fp[t1/2p∞

] (comes from Fp[t] ⊆ Fp[t1/2] which is a rami�ed cover, but afterapplying perfection, we get almost �nite étale).

Recall that Fp[t1/2p∞

] = ∪Fp[t1/p∞

][√

t1/pn

]and Fp[t±1/p

∞] ⊆ Fp[t±1/2p

∞]. Then e = 12

(1 +√t⊗ 1√

t

).

Let A = Fp[t]perf and B = Fp[√t]perf . e ∈ B ⊗A B[ 1t ] . But we can also write

e =1

2

(1 +

√t1/pn ⊗ 1√

t1/pn

)for all n. This means we can make the denominator as small as possible. Therefore, e ∈ (B ⊗A B)?. Consequently, A/Bis almost �nite étale.

Theorem 6.51. (Gabber-Romero) Suppose A is an Ra-algebra, I ⊆ A nilpotent. One has a notion of almost weakly étaleA-algebras (B such that B/A is almost �at, and B/B ⊗A B is also almost �at). This implies

weakly étale A-algebra ↔ weakly étale A/I-algebra.

Proof. Relies on the cotangent complex on the cotangent category. �

Theorem 6.52. Suppose A is a perfectoid algebra over a perfectoid �eld K. If B/A is �nite étale, then B is perfecoid.

Additionally, B◦/A◦ is almost �nite étale. There is an equivalence of categories

�nite étale A-algebras ↔ (almost) �nite étale A◦a-algebras.

By B 7→ B◦.

Theorem 6.53. (Almost purity in characteristic p) Let R be a perfect Fp-algebra, I = (t1/p∞

), then

(1) S is a perfect R-algebra such that S/R is integral, and S[ 1t ] is �nite étale over R[1t ]. Then S is automatically

almost �nite étale over R.

(2) There is an equivalence of categories between

{almost �nite étale Ra-algebras} ↔{�nite étale R[

1

t]-algebra

}.

Example 6.54. (From last time) R = Fp[t1/p∞

], R[ 1t ] = Fp[t±1/p∞ ] → Fp[t±1/2p

∞] is �nite étale. Therefore, R →

Fp[t1/2p∞

] is �nite étale.

Proof. Recall that there is an equivalence of categories from

{perfect R-algebras integral over R} ↔{perfect R[

1

t]-algebras integral over R[

1

t]

}up to almost isomorphism. The forward direction is by inverting t, the over is taking integral closure of.

In the situation of the theorem, in 1, need S ⊗R S → S decomposes S ⊗R S ∼= S × C in the almost category. Thisdecomposition exists after inverting t, by assumption, hence exists in almost category. Still need to show S is almost

�nitely generated projective. Any t-power torsion in R,S, S ⊗R S is automatically almost zero (can take p-th roots, alsoannihlated by t means the same for powers of t).

Let's reprove the splitting. Need to write down an idempotent element in (S ⊗R S)?.

S[1

t]⊗R[ 1t ] S[

1

t]→ S[ 1

t]

splits so we have a corresponding idempotent e ∈ S[ 1t ]⊗R[ 1t ] S[1t ]. t

Ne ∈ (S ⊗R S) /torsion for N su�ciently large. ApplyFrob−r, then tN/p

r

e ∈ (S ⊗R S) /torsion and so e ∈ (S ⊗R S)?.


Now, need to see S is almost �nitely generated projective as R-module. Here, use S[ 1t ] is �nite étale as an R[1t ]-algebra,

we have a trace pairing

Tr : S[1

t]× S[ 1

t]→ R[ 1

t].

Recall that with respect to this pairing, e is a Casimir element for this pairing. Write e =∑ai ⊗ bi, then

S[1

t]→ Rn[ 1

t]→ S[ 1

t]

is the identity as a whole, and s 7→ {Tr(sai)}ni=1 for the �rst map.We can do these constructions without denomintaors, Tr : S → R can imitate above constructions over S. This relies

on e, which has denominators. However, for any � > 0, t�e ∈ S ⊗R S. Can use this to de�ne S → Rn and Rn → S suchthat the composite is t�id. Therefore, S is almost �nitely generated projective.

For 2, we use the fact that A→ B is a map of Fp-algebras which is étale, and if A is perfect, then so is B. This is alsotrue in the almost category.

If S′/R [1/t] is �nite étale, then S′ is perfect. Consider S ⊆ S′ which is the integral closure of R. �

Theorem 6.55. K perfectoid �eld, A/K perfectoid K-Banach algebra. Then there is an equivalence of categoryies

{�nite étale A-algebra} ↔{almost �nite étale A0-algebra

}by B/A to B◦/A◦.

Remark 6.56. We already knew this for K in characteristic p.

If we choose a perfectoid pseudo-uniformizer ω with ω ∈ K◦, and ω1/p∞ ∈ K◦. Then almost �nite étale A◦-algebras isthe same as almost �nite étale A◦/ω-algebras. Recall, we also can associated A[,K[ (perfectoid K[-algebras),

A[◦/ω[ ∼= A◦/ω.

Can continue the equivalence

almost �nite étale A◦/ω-algebras ↔ almost �nite étale A[◦/ω[-algebras

↔ almost �nite étale A[◦-algebra

(almost purity) ↔ �nite étale A[-algebras.

Combining, we get �nite étale A-algebras are equivalent to �nite étale A[-algebras. This is a generalization of


Strategy for the proof of the theorem.

• characteristic p is okay by Frobenius• �elds of characteristic 0 is okay by tilting• reduce everything to the case of �elds

Proposition 6.57. K perfectoid, L/K �nite extension. Then L◦/K◦ is almost �nite étale.

Proof. Fix ω the pseudo-uniformizer. Observe that for L◦/K◦ to be almost �nite étale, su�ces to work mod ω (by

completeness, �atness).

L◦/ω ∼= L[◦/ω[ and K◦/ω ∼= K[◦/ω[.

However, we already know that K[◦ → L[◦ is almost �nite étale. �

Example 6.58. Qp(p1/p∞

)∧ ⊆ Qp(p1/2p∞

)∧. On rings of integers, Zp[p1/p∞

]∧ ⊆ Zp[p1/2p∞

]∧ is almost �nite étale.

Example 6.59. The map Tr : mL◦ → mK◦ is surjective, if L/K is a �nite extension.


7. Berkovich Spectrum

Let K be a complete non-archimedean �eld. A a K-Banach algebra, with || · ||.

De�nition 7.1. A multiplicative semi-norm on A is a function |·|? : A→ R≥0 such that

(1) |f + g|? ⊆ max (|f |? , |g|?)(2) |0|? = 0, |1|? = 1(3) |fg|? = |f |? |g|?(4) |·|? restricts to given |·| on K(5) |·|? ≤ C|| · || for some constant C

De�nition 7.2. The Berkovich spectrum M(A) is the set of multiplicative semi-norms on A. This is a topological space

in the following way. Given a ∈ A, it de�nes a function M(A) → R≥0 (evaluation). Give M(A) the weakest topologymaking these functions continuous.

Construction: Given a Banach space A, de�ne

A 〈T 〉 =

∑i≥0

aiTi : ||ai|| → 0, i→∞

.For A = R, this is the Tate-algebra. Here, A 〈T 〉 is a Banach space such that ||

∑i aiT

i|| = supi≥0 ||ai||.Assume the image of |·| : K → R>0 is non-discrete.

Proposition 7.3. (Berkovich) A a non-zero Banach algebra, then M(A) 6= ∅.

Proof. If I ⊆ A is closed, then A/I is a K-Banach alegbra. All maximal ideals are closed. Choose a maximal ideal m ⊆ A,and consider A/m. Replacing A with A/m, we can assume that A is a �eld.

Consider all seminorms on A, which are submultiplicative. Consider it as a poset (≤ everywhere). By Zron's lemma,there is a minimal submultiplicative seminorm |·|min over A.

Claim. |·|min is multiplicative.

Proof. A→ Ã =completion of A with respect to |·|min. Ã is a Banach algebra.Let f ∈ A. By assumtion,

∣∣f2∣∣min≤ |f |2min. Suppose we have strict inequality, ie. |f |min > 1 but

∣∣f2∣∣min

< 1. We will

get a contradiction.

Consider A→ Ã→ Ã 〈T 〉 → Ã 〈T 〉 /(T − f). This composition sends f to an element of norm ≤ 1 (T ) .Idea: take composite + Banach norm on the target, to get a smaller seminorm. To do that, enough to show that

T − f = −f−1(

1− Tf)∈ Ã 〈T 〉 is non-invertible. It is enough to show that 1 − Tf is non-invertible in Ã 〈T 〉. But the

inverse would be ∑i≥0

T i

f i

and∣∣f2i∣∣

min< 1 so

∣∣f−2i∣∣min

> 1.

By a similar argument, we can show that∣∣f−1∣∣

min= |f |−1min, and so |·|min is strictly multiplicative. �

�

Theorem 7.4. If A is a non-zero K-Banach algebra, then

(1) M(A) 6= ∅ is a compact Hausdor� space(2) For f ∈ A, f is invertible i� |f |? 6= 0 for all x ∈M(A)

Recall that if R is a ring,

Spec R = {equivalence classes of maps R→ K for K a �eld}


where two maps

K

R �

L

are equivalent. We can give a similar de�nition for the Berkovich spectrum.

If A→ L is a continuous map, L non-archimedean �eld, get a point of M(A) via the norm on L. If |·|? ∈M(A), de�nep? = {a ∈ A : |a|x = 0} which is a prime ideal. We get an absolute value K (A/p?) and complete.

Example 7.5. (The unit disk) A = K 〈T 〉 for K a compelte algebraically closed non-archimedean �eld. One can classifypoints in M(A).

(1) x ∈ K◦, get a multiplicative semi-norm f 7→ |f(x)|K(2) Given a disk D(a, r) ⊆ K◦, we can de�ne a multiplicative seminorm f 7→ supx∈D(a,r) |f(x)|.(3) For a descending sequence of disks

K◦ ⊇ D1 ⊇ D2 ⊇ ...

such that ∩Di = ∅, we can de�ne a seminorm f 7→ limn

supx∈Dn |f(x)|.

7.1. Spectral radis formula.

De�nition 7.6. Given a K-Banach algebra A, f ∈ A

ρ(f) = limn→∞

||fn||1/n

is called the spectral radius.

Theorem 7.7. (Spectral radius formula) ρ(f) = supx∈M(A) |f |x.

Proof. Enough to show that f ∈ A with |f |x < 1 for all x ∈M(A) then ρ(f) < 1.Consider 1− fT ∈ A 〈T 〉. For any multiplicative seminorm |·|y ∈M (A 〈T 〉), |1− fT |y = 1 and so 1− fT is invertible

in A 〈T 〉 with inverse ∑f iT i ∈ A 〈T 〉

which has to converge. This means taht f i → 0 and so |f |x < 1. �

Fact 7.8.

• ρ(g + g) ≤ max (ρ(f), ρ(g))• ρ(fg) ≤ ρ(f)ρ(g)• ρ(fn) = ρ(f)n

ρ de�nes an equivalent Banach seminorm i� A is uniform as a K-Banach algebra.

Suppose A is a quotient of K 〈T1, ..., Tn〉 locally path connected topological space.

Example 7.9. De�ne a path xt, t ∈ [0, 1] where

|f(xt)| = supD(0,t)

|f(u)| = supi≥0

∣∣citi∣∣where f(x) =

∑cix

i.

De�nition 7.10. Let R be a ring. Then a valuation on R consists of a totally ordered abelian group Γ (written

multiplicatively) and a function |·| : R→ Γ ∪ {0} satisfying:


(1) |0| = 0, |1| = 1(2) |x+ y| ≤ max (|x|, |y|)(3) |xy| = |x| |y|

The only di�erence this time, is that we replaced R>0 with Γ.

Remark 7.11. If |·| is a valuation on R, then de�ne support of |·| to be

supp (|·|) = {x ∈ R : |x| = 0} ∈ Spec R.

If p = supp (|·|), then |·| de�nes a valuation on R/p, and a valuation on k(p).

Remark 7.12. If K is a �eld, then to give a valuation on K (up to equivalence) it is enough to specify {x ∈ K : |x| ≤ 1}(the valuation ring).

De�nition 7.13. A valuation ring inside a �eld K is a subring R ⊆ K such that, for all x ∈ K\{0}, either x or x−1 ∈ R.

Given R ⊆ K a valuation ring, we can de�ne a valuation on K, by K → K×/R× ∪ {0} (so Γ = K×/R×).If |·| and |·|′ are valuations on R, such that value groups are generated by absolute values. Say two are equivalent if we

have ordered preserving isomorphism Γ ∼= Γ′ which carries |·| to |·|′.

De�nition 7.14. R a topological ring, then |·| is continuous if for all γ ∈ Γ, {x : |x| < γ} is open. If π is a topologicalnilpotent element, |π|n → 0 in valuation topology.

De�nition 7.15. An a�noid K-algebra is a pair (A,A+) where A is a K-Banach algebra and A+ ⊆ A◦ is an openintegrally closed subring.

Remark 7.16. Note that A+ ⊇ A◦◦ (use huge powers to get it in A+, then use integrally closed).

De�nition 7.17. De�ne

Spa(A,A+

)={equivalent classes of cont. valuations |·| : A→ Γ ∪ {0} such that |f | ≤ 1 if f ∈ A+

}.

Remark 7.18. Continuity means that π ∈ K pseudouniformizer, then |π|n → 0.

Example 7.19. M(A) ⊆ Spa (A,A+). Any bounded multiplicative seminorm A → R≥0 is a a continuous valuation onA. If f ∈ A◦, then |f | ≤ 1 (because powers are bounded). A+ really plays no roles here.

Notation: for x ∈ Spa (A,A+), and f ∈ A, write |f(x)| (lives in a totally ordered abelian group).

Proposition 7.20.

(1) If f ∈ A is such that |f(x)| 6= 0 for all x ∈ Spa (A,A+) then f is invertible(2) If f ∈ A is such that |f(x)| ≤ 1 for all x ∈ Spa (A,A+) then f ∈ A+

Fact 7.21. If R is a domain, with fraction �eld L, then consider all valuations on L, |·|, such that |f | ≤ 1 for all f ∈ R.If x ∈ L is such that |x| ≤ 1 for all |·|, then x is in the integral closure of R in L.

Example 7.22. (The unit disk) A = K 〈T 〉, A+ = OK 〈T 〉. Last time, we saw

M(A) = {equiv. classes of nested sequences of disks} .

In Spa (A,A+), we get some additional rank 2 points.

Consider Γ = {R>0, γ} where 1− � < γ < 1 for all � > 0. De�ne∣∣∣∣∣∣∑i≥0

ciTi

∣∣∣∣∣∣ = supi≥0 |ci| γi.It is important that γ > 1 is not allowed because |T | ≤ 1.


Can also de�ne if 0 < r < 1, de�ne a totally ordered abelian group R>0× γZ where r < γ < r+ � for all � > 0 (can alsodo r − � < γ < r) and de�ne ∣∣∣∑ ciT i∣∣∣ = sup

i≥0|ci| γi.

Theorem 7.23. All the points of the adic space arise in this way.

K◦ considered naively, D (0, 1) ⊆ K◦ is totally disconnected.

Example 7.24. Consider

{x : |x| = 1} q ∪r


De�nition 7.34. Let f1, ..., fn, g as above. De�ne a K-Banach algebra A〈f1,...,fn

g

〉to be the completion of A[ 1g ] with

ring of de�nition A0[f1g , ...,

fng ], which is (

A

[f1g, ...,

fng

])∧π

[1

π].

Proof. We claim that this is independent of all the choices. �

Proposition 7.35. If B is a K-Banach algebra, then ϕ : A→ B factor overs A〈f1,...,fn

g

〉i�

(1) g maps to an invertible element in B

(2) ϕ(fi)/ϕ(g) ∈ B◦

This factorization is unique if it exists.

Remark 7.36. If (A,A+) is an a�noid K-algebra, then one forms the corresponding construction for a�noid algebras(A

〈f1, ..., fn

g

〉, smallest open integrally closed subring containing A+ and

fig

).

Theorem 7.37. If (A,A+) is an a�noid algebra, U = U(f1,...,fn

g

)then

ϕ :(A,A+

)→(B,B+

)factors through (

A

〈f1, ..., fn

g

〉, A+

〈f1, ..., fn

g

〉)i� ϕ? : Spa (B,B+)→ Spa (A,A+) has image contained in U . The factorization is unique if it exists.

In particular,

{rational open subsets}op → rings, OX ,O+Xgiven by

U = U

(f1, ..., fn

g

)→(A

〈f1, ..., fn

g

〉, A+

〈f1, ..., fn

g

〉).

In general, this does not de�ne a sheaf, so we do not get a locally ringed space. In good situations, OX and O+X de�nesheaves.

Example 7.38. (Tate) If A = K 〈T1, .., Tn〉 /I with A+ = A◦ then OX ,O+X are sheaves.

Example 7.39. (Scholze) If A is perfectoid, OX ,O+X are sheaves.Locally ringed topological spaces X = Spa (A,A+) , then x ∈ X de�nes a valuation on OX,x.

Remark 7.40. Suppose x ∈ Spa (A,A+). Then Supp(x) = {f ∈ A : |f(x)| = 0} and get a valuation on k(x) of A at x suchthat A+ are elements of norm ≤ 1 and π ∈ K◦ is topologically nilpotent for valuation topology.

De�nition 7.41. An a�noid �eld (L,L+) is a pair such that L is a complete non-archimedean �eld containing K and

L+ is an open valuation subring of L◦.

Given x ∈ Spa (A,A+) , consider k(x) the residue �eld of A at x. Consider the associated valuation ring k(x)+. Takek(x)∧ and k(x)+∧ with respect to the π-adic topology. Then these form an a�noid �eld.

Given an a�noid �eld, (L,L+) we have a natural map to (L,L◦). There is always an associated rank one valuation

and valuation ring, by L◦ which gives the same topology.

Remark 7.42. If V is a valuation ring containing K◦, then V ∧π is also a valuation ring. The valuation topology is the

π-adic topology. If K = Frac (V ∧π ), considering the ring K◦ which is a rank 1 valuation ring which contains V ∧π but can

be slightly bigger. These are called microbial valuation rings.


Example 7.43. Suppose L is a non-archimedean �eld. Start with L◦ which has a residue �eld k(L◦). Suppose we have

a non-trivial valuation subring V ⊆ k(L◦). De�ne L+ = L◦ ×k(L◦) V . Then (L,L+) is still an a�noid �eld.

Theorem 7.44. (Huber) Spa (A,A+) is a spectral space.

De�nition 7.45. (Hockster) A topological space X is spectral, if

(1) quasi-compact and T 0

(2) the collection of all quasi-compact open subsets in X are stable under �nite intersection and forms a basis of

topology on X

(3) if Z ⊆ X is an irreducible subset, then Z has a generic point

Example 7.46. R commutative ring, then Spec R is a spectral space

De�nition 7.47. A continuous map f : Y → X of spectral if f−1(U) is quasi-compact for all U quasi-compact.

Example 7.48. If X is quasi-compact, quasi-separated scheme, then its space is spectral.

Theorem 7.49. (Hochster) If X is spectral, there exists R such that X ∼= Spec (R).

Example 7.50. A compact Hausdor� topological space is a spectral space (inverse limit of �nite sets). These are examples

of Hausdor� spectral spaces.

Example 7.51. Any �nite poset de�nes a T0 topological space (order topology) which is a �nite spectral space.

If X is a spectral space, then we can de�ne a specialization y x if x ∈ {y}. This de�nes a partial ordering on X.

Theorem 7.52. (Hochster) Spectral space ∼=Pro(�nite T0-spaces)× Pro (�nite posets).

Inside spectral spaces, we have the category of totally disconnected compact Hausdor� spaces which comes from

Pro(�nite sets).

Remark 7.53. Spa (A,A+) is almost never Noetherian. This is because for f ∈ A,

{x : |f(x)| 6= 0} = ∪n≥1 {x : |f(x)| ≥ |π|n} .

What are specializations in Spa (A,A+).

Proposition 7.54. If x, y ∈ Spa (A,A+) such that x ∈ {y} then

(1) x and y have the same support, given by valuation subrings Vx and Vy ⊆ k(x)(2) Vx ⊆ Vy

The converse is also true.

Proof. Suppose x ∈ {y}. Then every rational subset containing x also contains y. Suppose f ∈ supp(x) so |f(x)| = 0 ≤ |π|n

for all n. Then

x ∈ {z : |f(z)| ≤ |πn|}

also contains y. Therefore, |f(y)| = 0. Can also do a reverse argument. �

Corollary 7.55. If x ∈ Spa (A,A+), then the generalization of x forms a totally ordered subset S. All maximal elementis the rank 1 valuation.

Remark 7.56. For f ∈ A, {x : |f(x)| = 0} is a closed set of the adic spectral. This is a closed subset is ∩ of open subsets{x : |f(x)| ≤ |π|n} for each n. This is surprising for the scheme-theoretic point of view.

Huber's construction takes as input (A,A+) and associates it with

• Spa (A,A+) spectral space


• rational subsets• associated a presheaf of Banach algebras

(OX ,O+X

)on the poset of rational subsets

In good situations,(OX ,O+X

)are sheaves on basis of rational subsets and get a ringed space (X,OX).

Proposition 7.57. If (A,A+) is an a�noid K-algebra, then x ∈ Spa (A,A+), can form

OX,x = lim→x∈U,U rational

OX(U)

and

O+X,x = lim→ O+X(U).

These are local rings with

mOX,x = {f : |f(x)| = 0}

mO+X,x= {f : |f(x)| < 1} .

To each U(f1,...,fn

g

), we associate an a�noid algebra(

A

〈f1, ..., fn

g

〉, A+

〈f1, ..., fn

g

〉)through some universal property. This has the property that

Spa

(A

〈f1, ..., fn

g

〉, A+

〈f1, ..., fn

g

〉)∼= U

(f1, ..., fn

g

).

On Spa (A,A+) we have presheaves(OX ,O+X

)of K-algebras. They are not sheaves in general.

Example 7.58. (Buzzard-Verbmoles). One can have a K-Banach algebra A and element t ∈ A such that A → A〈t1

〉×

A〈

1t

〉has non-trivial kernel. Consider the graded ring Zp

[T±1, U

]/U2 = 0. The grading is where |T | = 1 and |U | = 0.

Consider the subring R′ = Zp[p|i|T i, p−|i|T iU

]i∈Z (not �nitely generated). De�ne a Qp-Banach algebra

A = (R′)∧p [

1

p].

Note that U ∈ A and is a non-zero nilpotent element.

Claim 7.59. U maps to 0 in A〈T1

〉or A

〈1T

〉.

Proof. Suppose T is power bounded, so there exists a ring of de�nition B0 containing T and R′. In the ring B0,(

p−NT−NU)

(TN ) = p−NU

and so p−NU ∈ B0. Then U → 0 in the completion. �

Theorem 7.60. (Huber, Tate) If (A,A+) is strongly Noetherian (if A 〈T1, ..., Tn〉 are Noetherian) then OX , O+X aresheaves. In addition, Hi (X,OX) = 0 for i > 0 (where X = Spa (A,A+)). This is not true for O+X .

Theorem 7.61. (Buzzard, Verberkmoes) If A is a K-Banach algebra, which is stabily uniform (ie. all rational localizations

A〈f1,...,fn

g

〉are uniform) then OX and O+X are sheaves. Additionally, Hi (X,OX) = 0 for i > 0.

Proposition 7.62. X = Spa (A,A+), given x ∈ X, OX,x and O+X,x are local rings and

mOX,x = {f : |f(x)| = 0} , mO+X,x = {f : |f(x)| < 1} .

Note that x de�nes a valuation on the residue �eld k(x).

If (A,A+) is sheafy, we get locally rings space and gives valuation |·|x which is slightly more than schemes.


De�nition 7.63. An adic space is an object of category V where it is locally isomorphic to Spa (A,A+).

Let K be perfectoid �eld. A perfectoid a�noid K-algebra is an a�noid K-algebra (A,A+) such that A is perfectoid.

Example 7.64.(K〈x1/p

∞〉,OK

〈x1/p

∞〉).

Remark 7.65. The notion of a perfectoid K-Banach algebra A can be for unrelated purely algebraically. It is the same as

giving a K◦a-algebra Ba such that

(1) Ba is �at over K◦a

(2) π ∈ K◦ is a perfectoid pseudouniformizer, Ba/π ∼= Ba/πp by the Frobenius.

De�nition 7.66. A perfectoid K◦a/π-algebra is a K◦a/π-algebra B̄ where

(1) B̄ is �at over K◦a/π

(2) B̄/π1/p ∼= B̄/π in almost category

Remark 7.67. A+ is an integral perfectoid K◦-algebra, A+ �at, A+/π1/p ∼= A+/π and A+ is π-adically complete.

Theorem 7.68. (Scholze) (A,A+) is a perfectoid a�noid algebra.

(1) OX is a sheaf on Spa (A,A+) = X(2) Hi (X,OX) = 0 for i > 0(3) Hi

(X,O+X

) a= 0 for i > 0

(4) For every rational open subset U ⊆ X,(OX(U),O+X(U)

)is a perfectoid a�noid K-algebra

(5) Suppose f1, .., fn, g de�ne a rational subset. Suppose f1, ..., fn, g admits compatible p-power roots of unity, fn = πN

and fi, g ∈ A+. Let U = U(f1,...,fn

g

). Then

O+X(U)a= A+

〈(f1g

)1/p∞, ...,

(fng

)1/p∞〉.

We can also form K[, A[, and A+[. Then

(1) Spa (A,A+) ∼= Spa(A[, A[+

)identifying rational subsets

(2) U,U [ be corresponding rational subsets, then(OX(U),O+X(U)

)[ ∼= (OX(U [),O+X (U [))Let K be a perfectoid �eld. (A,A+) a�noid perfectoid algebra.

De�nition 7.69. A perfectoid space is an adic space which is locally isomorphic to Spa (A,A+).

Remark 7.70. A perfectoid a�noid �eld is a pair (L,L+) where L is a perfectoid �eld, L+ ⊆ L open valuation subring.Then the points of Spa (A,A+) are equivalent classes of maps(

A,A+)→(L,L+

).

Under tilting, (A,A+

)→(A[, A[+

)gives

Spa(A,A+

) ∼= Spa(A[, A[+)corresponding structure sheafs go to each other. We have already seen this on the level of points. If |·| is a continuousvaluation on A, then we get a continuous valuation on A[ by x 7→

∣∣x]∣∣.When char(K) = p, the de�nition of a perfectoid algebra simpli�es, as just a perfect K-Banach algebra A and A+ is

also perfect (π-complete and π-torsion free), integrally closed in A = A+[ 1p ].

Proposition 7.71. Let (A,A+) be a�noid perfectoid of characteristic p. Suppose f1, ..., fn, g ∈ A+ and suppose fn = πN .Let U = U

(f1,...,fn

g

)then

O+X(U)a= A+

[(fig

)1/p∞]∧π


ie. form A[ 1g ] ⊇ A+

[(fig

)1/p∞].

Proof. How to construct OX(U). We construct a Banach ring from A+[f1g , ...,

fng

]the π-adically complete, then invert

π. In this setting,

A+[fig

]⊆ A+

[(fig

)1/p∞]but this is an isogeny (its cokernel is annhilated by a bounded power of π). In particular, we can take πnN . Consider(

f1g

)1/pπN(f1g

)1/p=

(fng

)g

(f1g

)1/p=

(fng

)f

1/p1 g

1− 1p ∈ A+[fig

].

This means that the rings di�er only by a bounded power of π. It follows that

A+

[(fig

)1/p∞]∧π

is also a ring of de�nition for OX(U). Note that this is a perfect ring, so OX(U) is perfect. Additionally,

A+

[(fig

)1/p∞]a= OX(U)

a= O+X(U).

�

Key point: given a K-Banach algebra A, it is di�cult to recognize A◦. If A is perfect(oid) then if R ⊆ A is a ring ofde�nition, and perfect then R

a= A◦.

Proof. πNA◦ ⊆ R ⊆ A◦. By perfect, apply p-th roots to get πN/pMA◦ ⊆ R ⊆ A. �

Theorem 7.72. If (A,A+) has characteristic p, perfectoid, then Hi(X,O+X

) a= 0 for i > 0. (Note, this is not true for

rigid analytic geometry).

Idea is to write (A,A+) as a �ltered colimit of algebras(Ai, A

+i

)which are topologically of �nite type. Then, on(

Ai, A+i

)the cohomology groups are bounded torsion, by Tate's theorem. Use a perfection process to turn bounded

torsion to almost zero. Can be described as colimit of uniform Banach algberas

A+ =(

lim→A+i

)∧t.

The category of a�noid perfectoids (A,A+) has �ltered colimits

lim→

(Ai, A

+i

)=

((lim→A+)∧t

[1

t],(

lim→A+i

)∧t

).

We say an a�noid perfecoid (A,A+) is p-�nite if

A+ = completed perfection of a tft K-algebra integral closure.

Also follows from Buzzard-Verberkmoes (for OX), A is perfectoid (in characteristic p) then A is stably perfectoid (allrational local localizations are perfectoids, in particular uniform). Therefore, Hi

(X,O+X

) a= 0 by a Frobenius argument.

Lemma 7.73. Suppose given a complex of K◦-modules P? such that each Pi is π-torsion free and π-complete. Suppose

H?(P )[1π ] = 0. Then for each i, there exists N = N(i) such that π

N(i)Hi(P ) = 0.

Proof. Invert π to get a complex of Banach spaces

Pi+1 ⊗Kdi→ Pi ⊗K

di−1→ Pi−1 ⊗K.

The map di : Pi ⊗K → ker(di−1)⊗K is a surjection by assumption. Since this is a map of Banach spaces, it is an openmap (open mapping theorem). Therefore, Pi+1 → ker (di−1) is open, which is exactly what we want. �


Lemma 7.74. Perfectoid algebras in characteristic p are stably uniform.

Proof. Recall that A,A+ are perfect. Will check that if U = U(f1,...,fn

g

)is rational open subset, then

(OX(U),O+X(U)

)is perfect a�noid.

WLOG, fi, g ∈ A+ and fn = πN . Then

O+X(U)a= A+

[(fig

)1/p∞, ...,

(fng

)1/p∞]∧π

and OX(U) is perfect K-Banach algebra. Since we are in characteristic p, this means that it is perfectoid and uniform.Then, (A,A+) is stably uniform. By Buzzard-Verberkmoes, OX , O+X are sheaves and Hi (X,OX) = 0 for i > 0. �

Lemma 7.75. Hi(X,O+X

) a= 0 for i > 0 in characteristic p.

Proof. Su�ce to show that suitable Cech complexes of O+X are almost exact. Suppose U1, ..., Um is a covering of X byrational open subsets. Then we have a cochain complex

O+X(X)→∏O+X(Ui)→

∏i 0, there existsg ∈ A[ such that ∣∣f(x)− g#(x)∣∣ ≤ |π|1−� max (|f(x)| , |π|c)for x ∈ Spa (A,A+).

Example 7.79. max (|f(x)| , |π|c) = max(∣∣g](x)∣∣ , |π|c)


Example 7.80. |f(x)| ≥ |π|c/2 i�∣∣g](x)∣∣ ≥ |π|c/2 . The rational open subsets

U

(f

πc/2

)= U

(g#

πc/2

).

Using this, can re-express any rational open subset in Spa (A,A+) can be expressed as U(f#1 ,...,f

#n

g#

)and f ]n = π

N . It

follows that OX(U) is perfectoid, hence uniform. Therefore, A is stably uniform, so(OX ,O+X

)are sheaves of rings.

Recall that

O+X/π ∼= O+X[/π[

and use almost vanishing in characteristic p to prove the Hi = 0 result.

8. Étale Cohomology

De�nition 8.1. (A,A+) be an a�noid K-algebra. A map (A,A+)→ (B,B+) is �nite étale if

(1) A→ B is �nite étale(2) B+ is the integral closure of A+ in B

De�nition 8.2. A map of adic spaces f

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