dieudonné modules, part i: a study of primitively ...€¦ · a study of primitively generated...

Dieudonneacute modules part Ia study of primitively generated Hopf algebras

Alan Koch

Agnes Scott College

May 23 2016

Alan Koch (Agnes Scott College) 1 58

Outline

1 Overview

2 Primitive Elements

3 A Dieudonneacute Correspondence

4 ApplicationsClassification of Hopf algebrasEndomorphisms of Hopf algebrasExtensions of Hopf algebrasForms of Hopf algebrasHopf orders

5 Want more


Let R be an Fp-algebra for exampleR = Fq q = pn

R = Fq[T1T2 Tr ]

R = Fq(T1T2 Tr )

R = Fq[[T ]]

R = Fq((T ))

Objective Classify finite abelian (commutative cocommutative)R-Hopf algebras of p-power rank free as R-modules

Approach Can we construct a category of modules which areequivalent to this category of Hopf algebras

ldquoIt is probably impossible to do sordquo ndash AJ de Jong describingthe geometric statement of this problem


R an Fp-algebra

Let C be a subcategory of finite abelian R-Hopf algebras of p-powerrank free as R-modules

Objective Classify the objects of C using a category of modules

This can be done for certain C using Dieudonneacute modules


Some legal disclaimers

Dieudonneacutersquos original works (Lie groups and hyperalgebras over a ofcharacteristic p gt 0 parts I - VI) focus on abelian Lie groups Heappears to be much more interested in infinite dimensional structures(formal groups p-divisible groups)

Dieudonneacute modules have typically been used by algebraic geometerstrying to describe geometric objects

In many ways the development of Dieudonneacute module theory is betterexpressed geometrically but here (ldquoOmahardquo) algebraicconstructionsexplanations will be a priority


Assumptions

Throughout H is an R Hopf algebra with comultiplication ∆ count εand antipode λ

All Hopf algebras are assumed to be finite abelian free over R and ofp-power rank unless otherwise specified


Possible readings for this talk

A Grothendieck Seacuteminaire de Geacuteomeacutetrie Algeacutebrique duBois-Marie 1967ndash1969 (SGA 7) - Groupes de monodromie engeacuteomeacutetrie algeacutebrique Lecture notes in mathematics 288 BerlinNew York Springer-Verlag (Apparently)AJ de Jong Finite locally free group schemes in characteristic pand Dieudonneacute modules Invent mathematicae 1993 Volume114 Issue 1 pp 89ndash137M Rapoport Formal moduli spaces in equal characteristichttpsmathberkeleyedu aaronlivetexfmsiecpdf

Listed in decreasing order of the amount of algebraic geometryneeded as background


Outline

1 Overview




5 Want more


What is a Hopf algebra

A structure which is both an algebra and a coalgebra where thesestructures get along in a compatible way

Idea what if we focus on Hopf algebras with a particularly simplecoalgebra structure

Let H be finite (ie finitely generated and free) R-Hopf algebra Anelement t isin H is primitive if

∆(t) = t otimes 1 + 1 otimes t

Necessarily if t isin P(H) then ε(t) = 0 and λ(t) = minust

Let P(H) be the set of all primitive elements of H


ExampleLet H = R[t ](tp) ∆(t) = t otimes 1 + 1 otimes t

Exercise 1 Show P(H) = Rt

Example

Let H = R[t ](tpn) ∆(t) = t otimes 1 + 1 otimes t

Exercise 2 Show P(H) = Rt + Rtp + middot middot middot+ Rtpnminus1

Example

Let H = R[t1 t2 tn](tp1 t

p2 t

pn ) ∆(ti) = ti otimes 1 + 1 otimes ti

Exercise 3 Describe P(H) as an R-module


ExampleLet H = RΓ Γ an abelian p-group (or any finite group)


ExampleLet R be a domain and H = RClowast

p = HomR(RCpR) Cp = 〈σ〉Let

t =pminus1983142

i=1

i983185i

where 983185i(σj) = δij

Exercise 5 Show that H = R[t ](tp minus t) and P(H) = Rt


Properties of P(H)

1 P(H) cap R = 02 t u isin P(H) rArr t + u isin P(H)3 t isin P(H) r isin R rArr rt isin P(H)4 P(H) is an R-submodule of H5 If R is a PID then P(H) is free over R6 t isin P(H) rArr tp isin P(H)

Exercise 6 Prove any subset of the above


Consider

HomR-Hopf(R[x ]H)

where x isin R[x ] is primitive

There is a map HomR-Hopf(R[x ]H) rarr P(H) given by

f 983347rarr f (x)

which is an isomorphism

Define a map F HomR-Hopf(R[x ]H) rarr HomR-Hopf(R[x ]H) by

F (f )(x) = f (xp) = (f (x))p

This induces a map t 983347rarr tp in P(H) which we will also denote F andcall the Frobenius map

Note F (rx) = rpF (x) for all r isin R


ExampleLet H = R[t ](tp) ∆(t) = t otimes 1 + 1 otimes t Then P(H) = Rt and F (t) = 0

Example


Then P(H) = Rt + Rtp + middot middot middot+ Rtpnminus1and F (tpi

) = tpi+1

ExampleLet H = RClowast

p = R[t ](tp minus t) ∆(t) = t otimes 1 + 1 otimes t Then P(H) = Rt and F (t) = tp = t


For any H the group P(H) has an R-module structure and is actedupon by F

Thus H is an R[F ]-module where R[F ] is the non-commutative ring ofpolynomials with

Fr = rpF

for all r isin R

Now suppose φ H1 rarr H2 is a Hopf algebra homomorphism Thenφ(P(H1)) sube φ(P(H2)) Furthermore for t isin P(H1)

Fφ(t) = φ(Ft)

so φ commutes with F

Thus φ983066983066983066P(H1)

is an R[F ]-module map


Therefore we have a functor

R-Hopf alg rarr R[F ]-modulesH rarr P(H)

which is not a categorical equivalence

Exercise 7 Show that RC2p and RCp2 correspond to the same

R[F ]-module

Exercise 8 Show that the R[F ]-module (R[F ])[X ] does notcorrespond to any Hopf algebra H

We will get a categorical equivalence by restricting the objects in bothcategories


Outline

1 Overview




5 Want more


We say H is primitively generated if no proper subalgebra of Hcontains P(H)Famous examples

1 H = R[t ](tp) t isin P(H)2 H = R[t ](tpn

) t isin P(H)3 H = R[t1 tn](t

pr1

1 tprn

n ) ti isin P(H)4 H = RClowast

p

5 H = R(Cnp )

lowast

Exercise 9 Let H = R[t ](tp2) with

∆(t) = t otimes 1 + 1 otimes t +pminus1983142

i=1

1i(p minus i)

tpi otimes tp(pminusi)

This is a Hopf algebra (not the exercise) Show that H is not primitivelygenerated (yes the exercise)


From here on we suppose H is primitively generated and R is a PID

Then P(H) is free over R and of finite type over R[F ]

So the functor from before can restrict to983080

primitively generatedR-Hopf algebras

983081minusrarr

983080R[F ]-modules of finite type

free over R

983081

We claim this is a categorical equivalence


H 983347rarr P(H)

We will construct an inverse

Let M be an R[F ]-module of finite type free over R

Let e1 e2 en be an R-basis for M

For 1 le i j le n pick aij isin R such that

Fei =n983142

j=1

ajiej

Let

H = R[t1 t2 tn](tpi minus

n983142

j=1

aji tj)

with ti primitive for all i

Then after identifying ti with ei P(H) = M


We will call any R[F ]-module of finite type over R a Dieudonneacute module

We will write

Dlowast(H) = P(H)

Question Why not simply call the functor P(minus)

For clarity so we may better differentiate H from its DieudonneacutemoduleTo get the audience used to Dlowast which will be used extensivelyover the next few days (and usually wonrsquot mean P(minus))


Properties of Dlowast

1 Dlowast is covariant (although this is typically referred to as an exampleof a ldquocontravariant Dieudonneacute module theoryrdquo)

2 If G = Spec(H) then the Dieudonneacute module isM = HomR-gr(GGa) where Ga is the additive group schemeSo Spec(H) 983347rarr Dlowast(H) is a contravariant functor

3 Dlowast is an exact functor4 Dlowast(H1 otimes H2) sim= Dlowast(H1)times Dlowast(H2)5 rankR H = prankR M

6 Dlowast respects base change (explained later)


Some examples

ExampleLet H = R[t ](tp) t primitiveThen Dlowast(H) = Re with Fe = 0

Example

Let H = R[t ](tpn) t primitive

Then P(H) is generated as a free R-module by t tp tp2 tpnminus1

If we write ei = tpi 0 le i le n minus 1 then

Dlowast(H) =nminus1983131

i=0

Rei Fei = ei+1 Fenminus1 = 0

where i lt n minus 1


Some more examples


p = R[t ](tp minus t) t primitiveThen Dlowast(H) = Re with Fe = e

ExampleLet M = Re1 oplus Re2 with

Fe1 = e2 Fe2 = e1

Then the corresponding Hopf algebra is

H = R[t1 t2](tp1 minus t2 t

p2 minus t1) sim= R[t ](tp2 minus t)

with t primitiveNote that if Fp2 sube R then H = RClowast

p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more


Consider the following problem for a fixed n find all primitivelygenerated R-Hopf algebras of rank pn

Equivalently find all R[F ] modules which are free rank n over R

These can be described with matrices

Let M be a free R-module basis e1 en Pick a matrix A isin Mn(R)and use it to define an F -action on M by

Fei = Aei

Then M corresponds to a Hopf algebra which can be quicklycomputed


Each element of Mn(R) gives a Hopf algebra

Conversely each Hopf algebra gives a matrix

However different choices of matrices can give the same Hopf algebra

It can be shown (as it was in Exeter) that AB isin Mn(R) give the sameHopf algebra if any only if there is a Θ isin Mn(R)times such that

B = Θminus1AΘ(p) where Θ = (θij) rArr Θp = (θpij)

Thus a parameter space for primitively generated Hopf algebras ofrank pn is Mn(R) sim where sim is the relation above

Presumably this allows for a count of Hopf algebras for certain R egR is a finite field


Outline

1 Overview




5 Want more


Let H be a primitively generated R-Hopf algebra and let End(H) bethe R-Hopf algebra endomorphisms of H

If φ H rarr H is a map of Hopf algebras it induces a map Θ M rarr M ofR[F ]-modules where M = Dlowast(H) We view Θ as a matrix with respectto a choice of some basis for M

Again as M is a free R-module for the same choice of basis theaction of F is determined by a matrix A

ThenΘA = AΘ(p)

Thus the elements in End(H) are in 1-1 correspondence with matricesin Mn(R) such that the equality above holds

The ones with Θ isin Mn(R)times are of course the automorphisms


Example

Let H = R[t ](tp2) t primitive Then assuming Dlowast(H) = Re1 oplus Re2

where e1 corresponds to t and e2 to tp

A =

9830720 01 0

983073

Then if Θ =

983072a bc d

983073is an R[F ]-linear map

ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073

so d = ap b = 0The corresponding φ H rarr H is given by φ(t) = at + ctp for a c isin Rclearly an isomorphism iff a isin Rtimes


Exercises

Exercise 10 Use Dieudonneacute modules to describe End(RClowastp)

Exercise 11 Use Dieudonneacute modules to describe Aut(RClowastp)

Exercise 12 Use Dieudonneacute modules to describe End(R(Cp times Cp)lowast)

Exercise 13 Use Dieudonneacute modules to describe Aut(R(Cp times Cp)lowast)


Outline

1 Overview




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The theory of Dieudonneacute modules can be extended beyond finiteR-Hopf algebras

For example H = R[t ]∆(t) = t otimes 1 + 1 otimes t corresponds to a freeR[F ]-module of rank 1

This allows us to construct projective resolutions for Dieudonneacutemodules and extensions of Dieudonneacute modules


An example

Let H = R[t ](tp) M its Dieudonneacute module

Let Dlowast(R[t ]) = R[F ]e (a free R[F ] module of rank 1)

Multiplication by F R[F ]e rarr R[F ]e gives the projective resolution

0 rarr R[F ]e rarr R[F ]e rarr M rarr 0

for M

This gives rise to

0 rarr HomR[F ](R[F ]eM) rarr HomR[F ](R[F ]eM) rarr Ext1R[F ](MM) rarr 0



We have HomR[F ](R[F ]eM) sim= M by f 983347rarr f (e) hence

0 rarr M rarr M rarr Ext1R[F ](MM) rarr 0

The map M rarr M induced from middotF is trivial (since Fm = 0 for allm isin M) giving

Ext1R[F ](MM) sim= M

Explicitly for [f ] isin Ext1R[F ](MM) arising from f isin HomR[F ](R[F ]eM)the extension Mf obtained is (R[F ]e times M)I where

I = (Fx f (x)) x isin R[F ]e


Does Ext1R[F ](MM) give all the Hopf algebra extensions

Exercise 14 Prove that the answer is no Hint considerH = R[t ](tp2

) with


i=1

1i(p minus i)


This H is a Hopf algebra (still not the exercise)

Note That the H above is in fact a Hopf algebra will be obvious aftertomorrow


Outline

1 Overview




5 Want more


Base change

Let S be a flat extension of R

Then we have two Dieudonneacute module theories1 R-Hopf algebras to R[F ]-modules Denote this functor DlowastR2 S-Hopf algebras to S[F ]-modules Denote this functor DlowastS

How are they related


Very closely

PropositionLet H be a primitively generated R-Hopf algebra Then H otimesR S is aprimitively generated S-Hopf algebra and

DlowastS(H otimesR S) = DlowastR(H)otimesR S

ProofObvious or

Exercise 15 Prove the above proposition


Let R = K a field (still char p) and let H1H2 be primitively generatedK -Hopf algebras

Let S = L be an extension of K

Then H1 and H2 are L-forms of each other if and only ifH1 otimesK L sim= H2 otimesK L as L-Hopf algebras

Alternatively H1 and H2 are L-forms of each other if and only ifDlowastL(H1 otimesK L) sim= DlowastL(H2 otimesK L)

If Mi = DlowastL(Hi) i = 1 2 then H1 and H2 are L-forms of each other if

LM1sim= LM2

as L[F ]-modules


ExampleLet H = Fp[t ](tp minus 2t)Then H ∕sim= FpClowast

p since this would imply the existence of an invertible1 times 1 matrix Θ = (θ) θ isin Fp such that

θ middot 2 = θp ( ldquoΘA = BΘ(p)rsquorsquo)

in which case θpminus1 = 2However let L = Fppminus1 Then such a θ isin L existsSo H and FpClowast

p are L-forms of each other


More exercises

Exercise 16 Show that K [t ](tpn) has no non-trivial L forms for any L

Exercise 17 Let H = Fp[t1 t2](tp1 minus t2 t

p2 minus t1) Determine if possible

the smallest field L such that H and (KCp2)lowast are L-forms

Exercise 18 Let M = Dlowast(H) for H a K -Hopf algebra of rank pnSuppose F acts freely on M Show that H and (KΓ)lowast are K sep-formsfor some p-group Γ

Exercise 19 Let M = Dlowast(H) for H a K -Hopf algebra of rank pnSuppose F r M = 0 for some r gt 0 Show that H and (KΓ)lowast are notK sep-forms for any p-group Γ


Outline

1 Overview




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Let R = Fq[[T ]] K = Fq((T )) p | q

(This is more structure than necessary For much of what follows Rcould be a DVR of characteristic p)

Let vK be the valuation on K with vK (T ) = 1

We have Dlowast compatible for the extension R rarr K

Pick AB isin Mn(R) and construct R-Dieudonneacute modules MAMB withrankR MA = rankR MB = n where the action of F is given by ABrespectively for some fixed R-basis for each

Write MA = Dlowast(HA) and MB = Dlowast(HB)

Now KHAsim= KHB if and only if there is a Θ isin GLn(K ) such that

ΘA = BΘ(p) for some Θ isin GLn(K )



Write A = (aij)B = (bij)Θ = (θij) Then

HA is viewed as an R-Hopf algebra using A ie

HA = R[u1 un](upi minus

983142ajiuj)

HB is viewed as an R-Hopf algebra using B ie

HB = R[t1 tn](tpi minus

983142bji tj)

KHB is viewed as a K -Hopf algebra in the obvious wayHB is viewed as an order in KHB in the obvious wayHA is viewed as an order in KHB through Θ ie

HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB

(apologies for the abuses of language amp the re-use of an old slide)


rank p

Recall that K -Hopf algebras of rank p correspond to rank 1 Dieudonneacutemodules over K

Fix b isin K and let M = Ke with Fe = be

The corresponding Hopf algebra is Hb = K [t ](tp minus bt)

Furthermore Ha sim= Hb if and only if there is a θ isin Ktimes with θa = bθp

Case b = 0 This is the algebra K [t ](tp)

Case b ∕= 0 From the exercises Hb is an K sep-form of KClowastp


Case b ∕= 0

Since K [t ](tp minus bt) sim= K [t ](tp minus T pminus1bt) by the map t 983347rarr Tminus1t wemay assume

0 le vK (b) lt p minus 1

Pick θ isin Ktimes and let a = bθpminus1

Provided a isin R (which holds iff θ isin R) we have

Hθ = R[θt ] sub K [t ](tp minus bt)

If u = θt thenup = (θt)p = θptp = θpminus1bθt = au

and so Hθ = R[u](up minus au)Here Hθ depends only on vK (θ) so a complete list is

Hi = R983158T i t

983159 i ge 0


Case b = 0

Clearly θa = bθp can occur only if a = 0

So any two R-Hopf orders in K [t ](tp) are isomorphic as Hopfalgebras

However They are not necessarily the same Hopf order It dependson the chosen embedding Θ = (θ) isin GL1(K )

Let θ isin Ktimes

Then R[θt ] is a Hopf order in K [t ](tp)

As R[θt ] = R[(rθ)t ] r isin Rtimes the complete list is

Hi = R[T i t ] i isin Z


Overview for all n

Pick a K -Hopf algebra H and find the B isin Mn(K ) which is used inthe construction of its K -Dieudonneacute moduleFind A isin Mn(R) such that ΘA = BΘ(p) for some Θ isin GLn(K )(One such example A = B Θ = I )Construct the R-Dieudonneacute module corresponding to AConstruct the R-Hopf algebra HA corresponding to this DieudonneacutemoduleThe algebra relations on HA are given by the matrix AHA can be viewed as a Hopf order in H using ΘHA1 = HA2 if and only if Θminus1Θprime is an invertible matrix in R where

ΘA1 = BΘ(p) and ΘprimeA2 = B(Θprime)(p)

Alternatively HA1 = HA2 if and only if Θprime = ΘU for some U isin Mn(R)times


ΘA = BΘ(p)

One strategy Given B set A = Θminus1BΘ(p)

This will generate a Hopf order iff A has coefficients in R

But we can replace Θ with ΘU for U isin M2(R)times

Notice that

(ΘU)minus1B(ΘU)(p) = Uminus1983070Θminus1BΘ(p)

983071U(p)

and so (ΘU)minus1B(ΘU)(p) isin Mn(R) iff Θminus1BΘ(p) isin Mn(R)



Theorem (New)Let H be a primitively generated K -Hopf algebra of rank pn and let Bbe the matrix associated to H Let H0 be an R-Hopf order in H and letA be the matrix associated to H0 Then there existsΘ = (θij) isin GLn(K ) such that

1 Θ is lower triangular2 vK (θii) ge vK (θij) j le i le n3 θii = T vK (θii )4 ΘA = BΘ(p)

In the case n = 2 the Hopf orders are of the form

Hijθ = R983158T i t1 + θt2T j t2

983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j

Q When is Hijθ = Hi primej primeθprime

Precisely when there is a U isin Mn(R)times such that983072

T i 0θ T j

983073=

983072T i prime 0θprime T j

983073U

Such a U exists if and only if

i = i prime

j = j prime

vK (θ minus θprime) ge j


An example H = K [t ](tp2) = K [t1 t2](t

p1 t

p2 minus t1)

A =

983072θpTminusi T pjminusi

minusθp+1Tminus(i+j) θT (pminus1)jminusi

983073

To give a Hopf order we require pj ge i and

vK (θ) ge minip (i + j)(p + 1) i minus (p minus 1)j

which can be simplified significantly giving


983159= R

983158T i tp + θt T j t

983159

pj ge i i minus (p minus 1)j le vK (θ) le j

Note Hijθ is monogenic if and only if pj = i and vK (θ) = j


More exercises All Hopf algebras below are primitively generated

Exercise 20 Find all Hopf orders in H = K [t1 t2](tp1 t

p2 )


p2 minus t2)

Exercise 22 Find all Hopf orders in H = (KC2p)

lowast

Exercise 23 Determine which of the Hopf orders in (KC2p)

lowast aremonogenic

Exercise 24 Find all Hopf orders in H = K [t1 t2](tp1 minus t2 t

p2 minus t1)

Exercise 25 Determine which of the Hopf orders in the previousproblem are monogenic


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1 Overview




5 Want more


Letrsquos assume you do

There is a theory which is dual to that of primitively generated Hopfalgebras

In other words a theory to classify H where Hlowast is primitivelygenerated

This would include elementary abelian group rings

As far as I can figure out given such an H define

Dlowast(H) = H+(H+)2 H+ = ker ε

It turns out that Dlowast(H) = (Dlowast(Hlowast))lowast = HomR(Dlowast(Hlowast)R)

I donrsquot know how the correspondence works without passing to duals


Thank you


Outline

1 Overview




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R = Fq[T1T2 Tr ]

R = Fq(T1T2 Tr )

R = Fq[[T ]]

R = Fq((T ))





R an Fp-algebra










Assumptions








Outline

1 Overview




5 Want more












Example



Example


p2 t








t =pminus1983142

i=1

i983185i




Properties of P(H)




Consider

HomR-Hopf(R[x ]H)



f 983347rarr f (x)



F (f )(x) = f (xp) = (f (x))p





Example



) = tpi+1






Fr = rpF

for all r isin R


Fφ(t) = φ(Ft)


Thus φ983066983066983066P(H1)







R[F ]-module




Outline

1 Overview




5 Want more





pr1

1 tprn


p

5 H = R(Cnp )

lowast



i=1

1i(p minus i)








983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




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Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you



R = Fq[T1T2 Tr ]

R = Fq(T1T2 Tr )

R = Fq[[T ]]

R = Fq((T ))





R an Fp-algebra










Assumptions








Outline

1 Overview




5 Want more












Example



Example


p2 t








t =pminus1983142

i=1

i983185i




Properties of P(H)




Consider

HomR-Hopf(R[x ]H)



f 983347rarr f (x)



F (f )(x) = f (xp) = (f (x))p





Example



) = tpi+1






Fr = rpF

for all r isin R


Fφ(t) = φ(Ft)


Thus φ983066983066983066P(H1)







R[F ]-module




Outline

1 Overview




5 Want more





pr1

1 tprn


p

5 H = R(Cnp )

lowast



i=1

1i(p minus i)








983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


R an Fp-algebra










Assumptions








Outline

1 Overview




5 Want more












Example



Example


p2 t








t =pminus1983142

i=1

i983185i




Properties of P(H)




Consider

HomR-Hopf(R[x ]H)



f 983347rarr f (x)



F (f )(x) = f (xp) = (f (x))p





Example



) = tpi+1






Fr = rpF

for all r isin R


Fφ(t) = φ(Ft)


Thus φ983066983066983066P(H1)







R[F ]-module




Outline

1 Overview




5 Want more





pr1

1 tprn


p

5 H = R(Cnp )

lowast



i=1

1i(p minus i)








983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you







Assumptions








Outline

1 Overview




5 Want more












Example



Example


p2 t








t =pminus1983142

i=1

i983185i




Properties of P(H)




Consider

HomR-Hopf(R[x ]H)



f 983347rarr f (x)



F (f )(x) = f (xp) = (f (x))p





Example



) = tpi+1






Fr = rpF

for all r isin R


Fφ(t) = φ(Ft)


Thus φ983066983066983066P(H1)







R[F ]-module




Outline

1 Overview




5 Want more





pr1

1 tprn


p

5 H = R(Cnp )

lowast



i=1

1i(p minus i)








983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you






Outline

1 Overview




5 Want more












Example



Example


p2 t








t =pminus1983142

i=1

i983185i




Properties of P(H)




Consider

HomR-Hopf(R[x ]H)



f 983347rarr f (x)



F (f )(x) = f (xp) = (f (x))p





Example



) = tpi+1






Fr = rpF

for all r isin R


Fφ(t) = φ(Ft)


Thus φ983066983066983066P(H1)







R[F ]-module




Outline

1 Overview




5 Want more





pr1

1 tprn


p

5 H = R(Cnp )

lowast



i=1

1i(p minus i)








983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you












Example



Example


p2 t








t =pminus1983142

i=1

i983185i




Properties of P(H)




Consider

HomR-Hopf(R[x ]H)



f 983347rarr f (x)



F (f )(x) = f (xp) = (f (x))p





Example



) = tpi+1






Fr = rpF

for all r isin R


Fφ(t) = φ(Ft)


Thus φ983066983066983066P(H1)







R[F ]-module




Outline

1 Overview




5 Want more





pr1

1 tprn


p

5 H = R(Cnp )

lowast



i=1

1i(p minus i)








983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you






t =pminus1983142

i=1

i983185i




Properties of P(H)




Consider

HomR-Hopf(R[x ]H)



f 983347rarr f (x)



F (f )(x) = f (xp) = (f (x))p





Example



) = tpi+1






Fr = rpF

for all r isin R


Fφ(t) = φ(Ft)


Thus φ983066983066983066P(H1)







R[F ]-module




Outline

1 Overview




5 Want more





pr1

1 tprn


p

5 H = R(Cnp )

lowast



i=1

1i(p minus i)








983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Properties of P(H)




Consider

HomR-Hopf(R[x ]H)



f 983347rarr f (x)



F (f )(x) = f (xp) = (f (x))p





Example



) = tpi+1






Fr = rpF

for all r isin R


Fφ(t) = φ(Ft)


Thus φ983066983066983066P(H1)







R[F ]-module




Outline

1 Overview




5 Want more





pr1

1 tprn


p

5 H = R(Cnp )

lowast



i=1

1i(p minus i)








983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Consider

HomR-Hopf(R[x ]H)



f 983347rarr f (x)



F (f )(x) = f (xp) = (f (x))p





Example



) = tpi+1






Fr = rpF

for all r isin R


Fφ(t) = φ(Ft)


Thus φ983066983066983066P(H1)







R[F ]-module




Outline

1 Overview




5 Want more





pr1

1 tprn


p

5 H = R(Cnp )

lowast



i=1

1i(p minus i)








983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you



Example



) = tpi+1






Fr = rpF

for all r isin R


Fφ(t) = φ(Ft)


Thus φ983066983066983066P(H1)







R[F ]-module




Outline

1 Overview




5 Want more





pr1

1 tprn


p

5 H = R(Cnp )

lowast



i=1

1i(p minus i)








983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you




Fr = rpF

for all r isin R


Fφ(t) = φ(Ft)


Thus φ983066983066983066P(H1)







R[F ]-module




Outline

1 Overview




5 Want more





pr1

1 tprn


p

5 H = R(Cnp )

lowast



i=1

1i(p minus i)








983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you






R[F ]-module




Outline

1 Overview




5 Want more





pr1

1 tprn


p

5 H = R(Cnp )

lowast



i=1

1i(p minus i)








983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Outline

1 Overview




5 Want more





pr1

1 tprn


p

5 H = R(Cnp )

lowast



i=1

1i(p minus i)








983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you





pr1

1 tprn


p

5 H = R(Cnp )

lowast



i=1

1i(p minus i)








983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you






983081minusrarr


free over R

983081



H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


H 983347rarr P(H)





Fei =n983142

j=1

ajiej

Let


n983142

j=1

aji tj)





We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you



We will write

Dlowast(H) = P(H)










Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you








Some examples


Example





i=0




Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Some more examples




Fe1 = e2 Fe2 = e1





p2


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Outline

1 Overview




5 Want more


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Outline

1 Overview




5 Want more






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you






Fei = Aei











Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you










Outline

1 Overview




5 Want more





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you





ThenΘA = AΘ(p)




Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Example



A =

9830720 01 0

983073

Then if Θ =

983072a bc d


ΘA =

983072a bc d

9830739830720 01 0

983073=

983072b 0d 0

983073

AΘ(p) =

9830720 01 0

983073983072ap bp

cp dp

983073=

9830720 0ap bp

983073



Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Exercises






Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Outline

1 Overview




5 Want more






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you






An example





for M

This gives rise to













) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you












) with


i=1

1i(p minus i)





Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Outline

1 Overview




5 Want more


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Base change





Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Very closely



ProofObvious or








LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you







LM1sim= LM2

as L[F ]-modules








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you








More exercises








Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Outline

1 Overview




5 Want more


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Let R = Fq[[T ]] K = Fq((T )) p | q













983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you






983142ajiuj)



983142bji tj)


HA = R

983104

983106

983110983114

983112

n983142

j=1

θji tj 1 le i le n

983111983115

983113

983105

983107 sub KHB



rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


rank p








Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Case b ∕= 0








Hi = R983158T i t

983159 i ge 0


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Case b = 0




Let θ isin Ktimes





Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Overview for all n





ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


ΘA = BΘ(p)




Notice that


983071U(p)








983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you







983159

with vK (θ) le j


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Hi j θ = R983056T i t1 + θt2T j t2

983057 vK (θ) le j



T i 0θ T j

983073=


983073U


i = i prime

j = j prime




p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you



p1 t

p2 minus t1)

A =



983073





983159= R

983158T i tp + θt T j t

983159






p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you




p2 )


p2 minus t2)


lowast


lowast aremonogenic


p2 minus t1)



Outline

1 Overview




5 Want more











Thank you


Outline

1 Overview




5 Want more











Thank you











Thank you


dieudonné modules, part i: a study of primitively ...€¦ · a study of primitively generated...

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