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De Rham cohomology of smooth proper pairs

3. An explicit description of Frobenius-stable lattices

3.1. De Rham cohomology of smooth proper pairs

such that

OX(−Z)(U) =t1· · ·trOX(U),

for some 0≤r≤ndepending onU. We always assume thatrin the above formula is minimal in the sense that notj,j= 1, . . . , r, can be left out. For simplicity, we callU together with thet1, . . . , tn apleasant open affine ofX and apleasant neighbourhood of any pointP ofU.

For the remainder of this chapter, let(X, Z)be a smooth proper pair over W(k)whereX is of relative dimension nandZ nontrivial.

3.1. De Rham cohomology of smooth proper pairs

This section is divided into three subsections. In the first we define differentials with logarithmic poles and the de Rham cohomology of (X, Z). The middle subsection contains Theorem B and its proof. The third part specialises to the case of curves.

Differentials with logarithmic poles. — To define the de Rham cohomology of (X, Z) one introduces the sheaf Ω(X,Z) of differentials on X with logarithmic poles alongZ: LetU,t1, . . . , tn, be a pleasant open affine. SinceU is ´etale overAnW(k), the differentialsdt1, . . . , dtn form anOX(U)-basis of ΩX(U). Set

dt˜j df=

(dtj/tj if j ≤r dtj otherwise.

The sections of Ω(X,Z) overU are defined as Ω(X,Z)(U)df=

n

M

j=1

OX(U) ˜dtj (3.4)

⊆ΩX(Z)(U).

This definition glues and gives rise to a subsheaf Ω(X,Z)of ΩX(Z). We let

∂˜j: OX(U)→OX(U) (3.5)

be the map sending an elementf to the coefficient of ˜dtj in df∈Ω(X,Z)(U).

As usual, we set Ωi(X,Z)df=Vi

(X,Z). The Ωi(X,Z) together with the universal deriva-tionddefine a complex Ω(X,Z). Thede Rham cohomlogy of (X, Z) is

HdR(X, Z)df=H(Ω(X,Z)).

Main theorem and proof. — To compute zeta functions one is interested in the Frobenius action on the lattices HidR(X, Z)/(tor) by virtue of the isomorphisms of (3.2). In many cases, the only interesting action is the one on HndR(X, Z)/(tor). For geometrically integral curves, this was already remarked in the introduction. Simi-larly, when X is a hypersurface inPn+1k , one is usually interested in the Frobenius action on Hn+1dR (Pn+1W(k), X) [Ger07, AKR11] or HndR(X, Z) [Wal09] withZa smooth hyperplane section.

To find a description of HndR(X, Z)/(tor) akin to (3.1), let us compare the hypercoho-mology of Ω(X,Z)to the hypercohomology of the complex

(X,Z)(E)df= Ω(X,Z)⊗OX(E), whereE is an effective divisor onX such that

E≤`Z and Ω(X,Z)(E) is acyclic.

We say that a complex of quasi-coherent sheaves is acyclic if every sheaf in this complex is acyclic. In this case

Hn(Ω(X,Z)(E)) = H0(Ωn(X,Z)(E))

dH0(Ωn−1(X,Z)(E)).

For example, if X is a geometrically integral curve of genus g, then Ω(X,Z)(E) is acyclic if and only ifOX(E) is. By Riemann-Roch the latter happens wheneverE is of degree at least 2g−1. Let

Q= Coker

(X,Z),→Ω(X,Z)(E) .

Lemma 3.1. — The complexH(Q) vanishes upon multiplication bypblogp`c. Proof. — There is a spectral sequence E2rs = RrΓ(Hs(·)) ⇒ Hr+s(·) [GM03, Para. III.7.14] so it suffices to show that the cohomology sheaves Hi(Q) vanish upon multiplication bypblogp`c. Taking cohomology commutes with cokernels (in fact with any direct limit) and so Hi(Q) equals the cokernel of Hi(Ω(X,Z))→Hi(Ω(X,Z)(E)).

Thus the lemma is a theorem by Abott et al. [AKR11, Thm. 2.2.5].

One can now describe lattices that are “close” to HidR(X, Z)/(tor) and “almost”

Frobenius-stable: The short exact sequence

0→Ω(X,Z)→Ω(X,Z)(E)→Q→0 induces the long exact sequence

. . .→Hi−1(Q)→HidR(X, Z)→Hi(Ω(X,Z)(E))→Hi(Q)−→. . . . (3.6) Since pblogp`cHi(Q) vanishes, both kernel and cokernel of the map HidR(X, Z) → Hi(Ω(X,Z)(E)) are torsion. Thus, there is a natural embedding

HidR(X, Z)

(tor),→Hi(Ω(X,Z)(E))

(tor), (3.7)

3.1. DE RHAM COHOMOLOGY OF SMOOTH PROPER PAIRS 35

whose cokernel is killed by pblogp`c, see also [vdB08, Prop. 3.5.4], [Wal09, Prop. 5.1.11], or [Tui11, Cor. 3.7.6]. By (3.2), the sequence of maps HndR(X, Z)/(tor) ,→Hn(Ω(X,Z)(E))/(tor),→HndR(XK\ZK) is Frobenius-equivariant. Consequently,

pblogp`cF

Hi(Ω(X,Z)(E)) (tor)

⊆Hi(Ω(X,Z)(E))

(tor). (3.8) Let us now give a description of the Frobenius-stable lattice HndR(X, Z)/(tor) itself.

Since Ωn(X,Z)(E) andQn are the last nonzero sheaves in their respective complexes, there are maps H0(Ωn(X,Z)(E))→Hn(Ω(X,Z)(E)) and H0(Qn)→Hn(Q). These sit in the following commutative square in which ϕis defined as the diagonal.

H0(Ωn(X,Z)(E))

//

ϕ '' H0(Qn)

Hn(Ω(X,Z)(E)) //Hn(Q)

(3.9)

The kernel ofϕis closely related to the de Rham cohomology of (X, Z). To describe this relation, it is useful to consider the stalks of Ωn(X,Z)(E) at points ofX. In partic-ular, the completion of a stalk along the intersection of the irreducible components of Z passing through the point is of interest. To this end, for each pointP ofX write

iP df= X

p⊇OX(−Z)P, minimal

p,

where the sum is over the minimal primes of OX,P that contain OX(−Z)P. For an OX,P-moduleM, set

df= lim

←−M/ii

PM.

Lemma 3.2. — Let U,t1, . . . , tn be a pleasant open affine. The minimal primes of OX(U) that contain OX(−Z)(U) are exactly the ideals tjOZ(U), j = 1, . . . , r. In particular, ifP is a point ofU andsthe number of irreducible components ofZ that pass through P, thens≤r and the tj can be ordered such that

iP =

s

X

j=1

tjOX,P.

This observation is certainly not original to this thesis, but lacking a reference we give our own proof.

Proof. — If p is a minimal prime of OX(U) containing OX(−Z)(U), then its inverse image q in W(k)[x1, . . . , xn] is a minimal prime containing the ideal x1· · ·xrW(k)[x1, . . . , xn] [Mil80, Prop. I.3.17]. Hence q equals xjW(k)[x1, . . . , xn] for an appropriate 1≤j≤r. SinceU is unramified overAnW(k), tj generatesp.

Conversely, by minimality of r (see the definition of a pleasant open affine) each tjOX(U), j = 1, . . . , r, is a proper ideal. Moreover, it cannot properly contain a minimal prime p over OX(−Z)(U): Otherwise, p would equal some tiOX(U). But thenj must equali sinceZ is reduced. Thus any minimal prime pover tjOX(U) is also minimal overOX(−Z)(U). ThenpequalstiOX(U) for somei. Again, sinceZ is reduced, we deduce thatj equalsi. HencetjOX(U) is prime.

Similar to the Cohen structure theorem for regular local rings containing a field [Bou06, Prop. VIII.5.6 & Thm. VIII.5.2], there is

Proposition 3.3. — Let P be a point of X. There is a section OX,P/iP ,→OX,P

to the canonical epimorphism OX,P OX,P/iP. If s is the number of irreducible components ofZ that pass throughP, then there is anOX,P/iP-algebra isomorphism

OX,P/iP

[[T1, . . . , Ts]]−→X,P

Tj 7→tj,

where thetj are part of thet1, . . . , tn of a pleasant neighbourhood ofP ordered in such a way that t1, . . . , ts generateiP.

Abott et al. use a similar power series representation for certain rational functions in their proof of the central ingredient to our Lemma 3.1.

Proof. — Write i = iP. Let us prove the first claim first. By assumption on U, there is a morphism U → AnW(k) of finite type (´etale even). The closed immer-sion V df= V(t1, . . . , ts) ,→ U is induced by the closed immersion An−sW(k) ,→ AnW(k) (corresponding toW(k)[x1, . . . , xn]W(k)[xs+1, . . . , xn]) and base change to along U → AnW(k). The latter immersion has an obvious retraction and hence, so does V ,→U. In particular, there is a retraction locally at P and this corresponds to a section ofOX,P OX,P/i.

The image of the sectionOX,P/i,→OX,P has trivial intersection withi. Thei-adic filtration onOX,P is exhaustive by definition and separated sinceOX,P is a Noetherian domain [Bou89a, Prop. III.3.5]. Hence, the proposition is proven if we can show that the images of the tj, j = 1, . . . , s, in the graded algebra gr ˆOX,P associated to OˆX,P and i form an algebraically free system of generators over OX,P/i [Bou89a, Prop. III.2.11].

To this end, let us adapt a proof of Bourbaki [Bou06, Proof of Thm. VIII.5.1]: The OX,P/i-modulei/i2is free of ranks[Liu06, Lem. 6.3.6] and so the symmetric algebra

3.1. DE RHAM COHOMOLOGY OF SMOOTH PROPER PAIRS 37

S of i/i2 over OX,P/i is the polynomial ring

OX,P/i

[T1, . . . , Ts]. Let grOX,P be the graded algebra associated toOX,P andi. Note that it coincides with gr ˆOX,P. To prove the proposition, let us show that the canonical morphismγ:S →grOX,P is an isomorphism.

It is certainly epi (so the map in the proposition is epi, too). If it is not iso, then there is a homogeneous polynomial h(T1, . . . , Tr) of degree dat least 1 that gets mapped to 0 by γ, i.e. h(t1, . . . , ts) ∈ Id+1. In other words, there is a polynomial relation 0 =h(t1, . . . , ts) +“terms of higher degree in thetj” overOX,P. But thedtj are part of anOX,P-basis of ΩX,P sinceU is ´etale overAnW(k), so this cannot happen. Hence γ is iso.

LetP be a point ofX. Then

Ωˆn(X,Z)(E)P ∼= Ωn(X,Z),P ⊗OX(E)P ⊗OˆX,P.

To describe ˆΩn(X,Z)(E)P we still need to determine the stalkOX(E)P. IfU,t1, . . . , tn

are a pleasant neighbourhood of P as in Lemma 3.2, then OX(E)P is the ideal t−v1 P1(E)· · ·t−vs Ps(E)OX,P. Let us show that the numbers vPj(E) are already deter-mined by data from the special fibre of (X, Z): On the one hand, OX(E)P equals t−v1 P1(E)· · ·t−vs Ps(E)OX,P. On the other hand, going through the proof of Lemma 3.2 again, this time overkinstead ofW(k), we deduce thatvPj(E) equalsvP

j(E), where Pj is the irreducible component ofZ that corresponds totj. Hence,

OX(E)P =t−v1 P1(E)· · ·t−vs P s(E)OX,P. (3.10) Now specialise to the case where P is a generic point of Z. Then the number s of irreducible components ofZ that pass through P equals 1. HenceiP is generated by t1, andOX,P/iP andOZ,P =OX,P/t1OX,P coincide. We write

tP ,jdf=tj and tP df=tP ,1. By the proposition, there is an isomorphism

t−vP(E)

P OZ,P[[tP]] ˜dtP −→ Ωˆn(X,Z),P(E), with ˜dtP = ∧jdt˜P ,j. For an element ωP of ˆΩn

(X,Z),P(E), let ω(i)

P ∈ OZ,P be the coefficient ofti

P

dt˜P. Finally, let

∂˜P ,j:OX,P →OX,P be the germ of the map ˜∂j:OX(U)→OX(U) of (3.5).

Theorem B (Generalised van den Bogaart’s Lemma) Let ϕbe as in(3.9). There is a natural isomorphism

Kerϕ

dH0(Ωn−1(X,Z)(E))−→ HndR(X, Z)

(tor). (3.11)

Furthermore with the above notation, Kerϕ=

ω∈H0(ΩnX(Z+E))|For all generic pointsP ofZ and i=−vP(E), . . . ,−1, there existf(i)

P ,1, . . . , f(i)

P ,n∈OZ,P

such that ω(i)

P =if(i)

P ,1+

n

X

j=2

∂˜P ,jf(i)

P ,j .

IfX is a geometrically integral curve andE=`Z, then tensoring(3.11)withK yields the left-hand isomorphism in(3.1).

One can also describe HndR(X) by comparing the hypercohomologies of the complexes ΩX and Ω(X,Z)(E). ForX a geometrically integral curve of genusg,Z a closed point of degree 1, andE= (2g−1)Z, this was done by van den Bogaart in his thesis [vdB08, Sec. 3.3]. He defines a map that is exactly the map ϕ from above and shows that there exists a natural isomorphism Kerϕ/dH0(OX((2g−1)Z))−→ H1dR(X) [vdB08, Lem. 3.3.10]. His proof was the inspiration for this section and gives Theorem B its name.

While in theory it might suffice to describe the cohomology of X to determine the zeta function of X, in actual computations it can prove advantageous to use the cohomology of a pair (X, Z). In the case of hypersurfaces in Pn+1k , as noted before, one usually considers (Pn+1W(k), X) or (X, Z) withZ a smooth hyperplane section. For curves there are further examples where it is particularly easy to describe a basis for the cohomology of (X, Z) (Remark 3.8).

Proof. — On the one hand, the hypercohomology of the cokernel complex is torsion (Lemma 3.1) and thus the long exact sequence (3.6) induces 0→HndR(X, Z)/(tor)→ Hn(Ω(X,Z)(E))→Hn(Q). On the other hand, Ω(X,Z)(E) is acyclic and so its hy-percohomology is given by the cohomology of the complex of global sections. In par-ticular, Hn(Ω(X,Z)(E)) is the quotient of H0(Ωn(X,Z)(E)) bydH0(Ωn−1(X,Z)(E)). Hence

3.1. DE RHAM COHOMOLOGY OF SMOOTH PROPER PAIRS 39

there is the following diagram whose row and column are exact.

H0(Ωn−1(X,Z)(E))

d

H0(Ωn(X,Z)(E))

ϕ

&&

0 //HndR(X, Z)

(tor) //Hn(Ω(X,Z)(E))

//Hn(Q)

0

This gives (3.11). Since tensoring withK is flat, this isomorphism induces the left-hand isomorphism in (3.1) ifX is a geometrically integral curve andE=`Z. Turning to the kernel of ϕ, fix an open affine covering U = {Ui}i∈I of X. Let us writeC(F)=dfC(U,F) for the ˇCech complex of a sheafF after applying the global sections functor, i.e. Cr(F) = ⊕#J=r+1F(UJ) where J ⊆ I and UJ = ∩j∈JUj. The hypercohomology of a complexF of quasi-coherent sheaves onX is thus given by the cohomology of the total complexSC(F) associated to the double complex C(F) [GM03, Sec. III.7]. Recall that SCn(F) = ⊕r+s=nCr(Fs). With this identification, the mapϕ(thought of as concatenation of the top and the right-hand map in (3.9)) becomes the concatenation of the maps in the following diagram.

Ker

C0(Ωn(X,Z)(E))→C1(Ωn(X,Z)(E))

//Ker C0(Qn)→C1(Qn)

 _

Ker SCn(Q)→SCn+1(Q)

Ker SCn(Q)→SCn+1(Q)

/Im SCn−1(Q)→SCn(Q) An element (ωU)U∈Uof Ker

C0(Ωn(X,Z)(E))→C1(Ωn(X,Z)(E))

is in the kernel ofϕ if and only ifωU + Ωn(X,Z)(U) lies in the image ofQn−1(U)→Qn(U) for allU ∈U.

SinceQis supported onZ, it suffices to check this at the stalks of the closed points of Z. In fact, it suffices to check this at the stalks of the generic points ofZ (Lemma 3.4) of which there are only finitely many. LetP be such a point. We set

dt˜(j)

P =∧k6=jdt˜P ,k.

Then by (3.4) and Proposition 3.3, QPn−1∼= Ωn−1(X,Z),P(E)/Ωn−1

(X,Z),P

∼= ˆΩn−1

(X,Z),P(E)/Ωˆn−1

(X,Z),P

∼=

n

M

j=1

−1

M

i=−vP(E)

OZ,PtiPdt˜(j)

P .

and similarly,

QnP ∼=

−1

M

i=−vP(E)

OZ,PtiPdt˜P.

With respect to the bases ˜dt(j)

P , j= 1, . . . , n, and ˜dtP, the mapQPn−1→QnP is X

i

f(i)

P ,1tiP,X

i

f(i)

P ,2tiP, . . . ,X

i

f(i)

P ,ntiP

! 7→X

i

if(i)

P ,1+

n

X

j=2

∂˜P ,jf(i)

P ,j

tiP.

The calculation is straightforward: d(f(i)

P ,jti

P

dt˜(j)

P ) = ∂˜P ,jf(i)

P ,jti

P

dt˜P whenever j is at least 2. Only the case j equals 1 is interesting: Here, d(f(i)

P ,1ti

P

dt˜(1)

P )

=

if(i)

P ,1+ ˜∂P ,1f(i)

P ,1

ti

P

dt˜P. But f(i)

P ,1 belongs to OZ,P = OX,P/tP ,1OX,P, so

∂˜P ,1f(i)

P ,1 vanishes. Noting that H0(Ωn(X,Z)(E)) equals H0(ΩnX(Z+E)) concludes the proof.

The following lemma was used in the proof.

Lemma 3.4. — With the notation of Theorem B and its proof, let P be a generic point of Z and Qbe a closed point in the closure of P. If ωP+ Ωn

(X,Z),P lies in the image ofQPn−1→QnP, thenωQ+ Ωn

(X,Z),Q lies in the image of Qn−1Q →QQn. Proof. — Without loss of generality, we assume that the pleasant neighbourhoodU, t1, . . . , tn ofP that is used in Theorem B is also a pleasant neighbourhood ofQ, and thatPj=V(tj),j= 1, . . . , s, are the irreducible components ofZ that pass through Q. AsP is fixed, we drop the indexP for thetjand ˜∂j in this proof. As before, write P =P1 andt=t1. By Proposition 3.3 and (3.10),

QnQ=

(−1,...,−1)

M

α=(−vP

1(E),...,−vP s(E))

(OX,Q/iQ)tα11· · ·tαssdt.˜

Thus,

ωQ+ Ωn(X,Z),Q=X

α

f(α)

Q tα11· · ·tαssdt˜

3.1. DE RHAM COHOMOLOGY OF SMOOTH PROPER PAIRS 41

for appropriatef(α)

Q ∈OX,Q/iQ. SinceωP+ Ωn

(X,Z),P is in the image ofQPn−1→QnP, there existf(i)

P ,j inOX,P/iP such that ωP+ Ωn(X,Z),P =

−1

X

i=−vP(E)

if(i)

P ,1+

n

X

j=2

∂˜jf(i)

P ,j

ti1dt˜

=

n

X

j=1

∂˜jf(i)

P ,j

dt˜ + Ωn(X,Z),P,

withfP ,j=dfP

if(i)

P ,jti1. The mapU →AnW(k)is ´etale and so the ideal ofP inOX(U) is generated by{p, t1}. In particular,

OX,P = OX,Q

(p,t1)

.

Therefore, each fP ,j lies inOX(F)Q for some suitable effective divisorF supported onZ. By Proposition 3.3, there is an expansion

fP ,j = X

α≥(−vP

1(F),...,−vP s(F))

f(α)

P ,jtα11· · ·tαss

for suitablef(α)

P ,j ∈OX,Q/iQ. Hence there is an equality f(α)

Q =

s

X

j=1

αjf(α)

P ,j +

n

X

j=s+1

∂˜jf(α)

P ,j ∈OX,Q/iQ for all (−vP

1(F), . . . ,−vP

s(F)) ≤ α ≤ (−1, . . . ,−1). The element f(β)

Q vanishes wheneverβi is less than−vP

i(E) for somei, so the same holds for the right hand side of the equation. We may therefore assume that the summands vanish individually so that

(−1,...,−1)

X

α=(−vP

1(E),...,−vP s(E))

f(α)

P ,1tα11· · ·tαssdt˜(1), . . . ,

(−1,...,−1)

X

α=(−vP

1(E),...,−vP s(E))

f(α)

P ,ntα11· · ·tαssdt˜(n)

defines the desired preimage ofωQ+ Ωn(X,Z),Qin

Qn−1Q =

n

M

j=1

(−1,...,−1)

M

α=(−vP

1(E),...,−vP s(E))

(OX,Q/iQ)tα11· · ·tαssdt˜(j).

The case of curves. — In this subsection, X is a geometrically integral curve of genusg. For such anX, Theorem B takes a particularly nice form: The generic points P of Z coincide with the closed points of Z. Let P be such a point. Its local ring OZ,P is an ´etale extension ofW(k) of degree [k(P) :k] whose completion (with respect to the maximal ideal) is canonically isomorphic toW(k(P)) [Gro64, Rem. 0.19.8.7].

Moreover,nequals 1, so to check whether a differentialωlies in the kernel, it suffices to check that the O(degE) many coefficients ω(i)

P , i=−vP(E), . . . ,−1 lie iniOZ,P. In particular, for anyZ andE there is a natural isomorphism

Kerϕ

dH0(OX(E))−→ H0(ΩX(Z))⊕M, (3.12) whereMis a freeW(k)-module of rankg. (A quick argument for the rank is as follows:

The rank of Kerϕ

dH0(OX(E)) equals the dimension of HdR(XK\ZK). The latter space is the colimit of the spaces H0(ΩXK((r+ 1)ZK))/dH0(OX(rZK)) ordered by inclusion. Riemann-Roch says that the dimension of the latter spaces stabilises forr large enough and that it equalsg+ degZK−1. This gives the desired formula since degZK equals degZ.)

Proposition 3.5. — Let P be a point ofZ of degree1,P be the irreducible compo-nent of Z that reduces to P, and choose E = `P in Theorem B where ` is at least the maximal gap number of P (as point on X). For each gap number r of P, let ωr∈Kerϕbe such that its germ atP lies in

−r−1

X

i=−vP(E)

ω(i)r tiPdt˜P(r)r t−r

P dt˜P+ Ω(X,Z)((r−1)Z)P

(tP as in Theorem B) with v(ωr(i)) > −i for i = −vP(E), . . . ,−r − 1 and v(ωr(r)) = v(r). Then the classes of the ωr form a basis of the quotient of Kerϕ

dH0(OX(E))byH0(ΩX(Z)).

Remark 3.6. — First note thatOX(E) is acyclic so that it makes sense to useEin Theorem B: Indeed,OX(E) is acyclic because` is at least the maximal gap number of P. Since h1(OXK(EK)) is at most h1(OX(E)) [Liu06, Thm. 5.3.20], the claim follows. Second observe that the basis in the above proposition is determined using the gap numbers ofP as point onX. No knowledge of the gap numbers of the point PK onXK (which may differ from those ofP [Sch39,§5]) is required.

Proof. — For each pole numberrofP between 1 andvP(E), choose an elementf of H0(OX(E)) such that its reductionf modulophas pole orderr. This is possible since OX(E) is acyclic and so H0(OX(E))⊗k→H0(OX(E)) is surjective. Setωr=df. We are going to show that theωr generate Kerϕmodulo H0(ΩX(Z)). A rank argument then shows that the classes of theωrgiven in the proposition form a basis as desired.

3.1. DE RHAM COHOMOLOGY OF SMOOTH PROPER PAIRS 43

Let ωr = P−1

i=−vP(E)ωr(i)ti

P

dt˜P be the (beginning of the) expansion of ωr with re-spect to tPdt˜P according to Proposition 3.3. Given the beginning of the expansion P−1

i=−vP(E)ϑ(i)ti

P

dt˜P of some element ϑ ∈Kerϕ, we must show that there exists a solution λto the linear system

ω(−1)−1 . . . ω(−1)

−vP(E)

... ...

ω(−v−1P(E)) . . . ω(−vP(E))

−vP(E)

 λ=

 ϑ(−1)

... ϑ(−vP(E))

.

Since theωrandϑbelong to Kerϕ, thei-th row of the system is divisible byi. After carrying out these divisions, the elements on the diagonal of the matrix become units.

If a column has nonzero entries below the diagonal, then those elements must be a multiple of p (even after the aforementioned divisions). Hence the matrix can be brought into an upper-triangular form with units on the diagonal. A solution λ to the linear system can be found using reverse substitution.

The following lemma is not essential to the remainder of the present thesis. It says that the cohomology of (X, Z) (withX a curve) does not have any torsion. So from now on we will not have to kill any torsion of HidR(X, Z) or Hicr(X, Z).

Lemma 3.7. — The moduleHidR(X, Z) is a lattice inHidR(XK\ZK)for alli∈N.

Lacking a reference, we give a full proof that mimics the proof of [vdB08, Prop. 3.3.2].

Proof. — By (3.2), HidR(X, Z) ⊗K is isomorphic to HidR(XK \ ZK), so it re-mains to show that HidR(X, Z) is torsion-free. To this end, consider the long exact sequence in hypercohomology that is associated to the short exact sequence 0→Ω(X,Z)[1]→Ω(X,Z)→OX[0]→0.(3) The part in degreeireads

Hi−1(Ω(X,Z))→Hi(Ω(X,Z))→Hi(OX).

The modules Hi(OX) are free since OX is cohomologically flat. The mod-ules Hi(Ω(X,Z)) vanish unless i equals 0 (and then the module is free) since Ω(X,Z)= ΩX(Z) is acyclic. Hence, apart from degree 1, the module Hi(Ω(X,Z)) is a submodule of a free module and thus free itself.

Regarding degree 1, note that the boundary map from H0(OX) to H0(Ω(X,Z)) is 0.

Indeed, this is equivalent to surjectivity of H0(Ω(X,Z)) ,→ H0(OX). So let h be an element of H0(OX). Since H0(Ω(X,Z)) and H0(OX) have the same rank (their tensor products withKhave dimension 1 overKsinceX is geometrically integral), a

(3)As usual, when given a sheaf F we denote byF[i] the complex withF in degree i and 0 everywhere else.

multipleλhis in the image. Thus the image ofλhin H0(Ω(X,Z)) is 0, but this module is free, so the image ofhitself is 0. This meanshis in the image of the inclusion.

So the degree 1 part of the long exact sequence is a short exact sequence. Assume h∈H1(Ω(X,Z)) is such thatλhequals 0. Since H1(OX) is free, the image ofhin that module is 0, i.e. his in the image of the map H0(Ω(X,Z))→H1(Ω(X,Z)). Let h0 be a preimage. Then λh0 lies in the kernel and thus, since the coboundary map is 0,λh0 is 0. Since H0(Ω(X,Z)) is free,h0 itself and thereforehmust be 0.

Remark 3.8 (Regarding the choice of E). — Different choices of E influence how easy it is to compute H0(Ω1X(Z +E)). This in turn affects how easy it is to computeH1(Ω(X,Z)(E))/(tor) and H1dR(X, Z)/(tor). There are at least three factors to consider.

(1) If E is of small degree, then H0(Ω1(X,Z)(E)) is a module of small rank and thus (hopefully) fast to compute. A lower bound is given by Riemann-Roch: The sheaf OX(E) is acyclic whenever the degree of E is at least 2g−1, and it cannot be acyclic if the degree is less than g. However, degree g is often possible: IfX is hyperelliptic orpat least 2g−1, then (up to a constant field extension) there exists a strict relative normal crossing divisor Z such that Z is of degree 1 and has gap sequence 1, . . . , g [Sch39, S¨atze 6 & 8]. ThusOX(gZ) is acyclic and

H1(Ω(X,Z)(gZ))

(tor) = H0(ΩX((g+ 1)Z)).

In the higher-dimensional case when X is a hypersurface of degree din PrW(k) and Z a smooth hyperplane section, Walker shows that Ω(X,Z)(`Z) is acyclic whenever

` is at least max{(r−2)d+ 1,(r−1)d−r} [Wal09, Thm. 5.1.14]. Specialising to smooth plane curves of genus g, this bound becomesd−2. Therefore, it says that the degree of`Z should be at least 2g−1 (observe that the degree ofZ isdand that the genus of the curve is (d−1)(d−2)/2). This coincides with the bound obtained from Riemann-Roch.

(2) If E is bounded by `Z where ` is small compared to p, then already H1(Ω(X,Z)(E))/(tor) is Frobenius-stable, see (3.8). For example, if the degree ofZ is at least 2g−1, then the choiceE=Z yields

H1dR(X, Z)∼=H1(Ω(X,Z)(Z))

(tor) = H0(ΩX(2Z))

dH0(OX(Z)).

(3) If E is of “special form”, then a basis for H1dR(X, Z)/(tor) or a generating set of H0(Ω(X,Z)(E)) may be especially easy to compute. An example of the for-mer is Proposition 4.2. For an example of the latter, let X be nondegenerate. This means that there exists a torus T2W(k) such that X ∩T2W(k) can be defined by a Laurent polynomial f(x, y) that is nondegenerate with respect to its Newton poly-tope ∆ [CV09, Def. 1.9]. In his thesis, Tuitman considers the pair (X, Z) where Z =X\(X∩T2W(k)) [Tui11]. Depending on the Newton polytope ∆, one defines a