On linear relations in algebraic groupsGeneralities about abelian varieties History of the problem Main Theorem Basic ingredients of proof of Theorem A G. Banaszak, P. Krason´ On

Generalities about abelian varietiesHistory of the problem

Main TheoremBasic ingredients of proof of Theorem A

On linear relations in algebraic groups

P.Krason, Szczecin University

P.Krason On linear .....



G. Banaszak, P. Krason

On arithmetic in Mordell-Weil groups,

Acta Arithmetica 150.4 2011 pp.315-337




Outline1 Generalities about abelian varieties

Basic facts and notationClassification of endomorphism algebras

2 History of the problemnumber field caseStatement of the problem for abelian varietiesResults in this direction

3 Main TheoremCorollariesA counterexampleCase of algebraic tori

4 Basic ingredients of proof of Theorem ASome easy reductionsTheorems about reductionsome semisimple algebras and modulesidea of the proof P.Krason On linear .....












Abelian variety - projective algebraic variety that is also analgebraic group, i.e., has a group law that can be defined byregular functions.

m : A× A→ A

inv : A→ A

An affine group variety is called a linear algebraic group

Each such variety can be realized as closed subgroup of GLnfor some n

In particular Gm = GL1 is a linear algebraic group.






m : A× A→ A

inv : A→ A









m : A× A→ A

inv : A→ A









m : A× A→ A

inv : A→ A









m : A× A→ A

inv : A→ A








A(C) = Cg/Λ Λ- a lattice in R2g

TheoremA torus Cg/Λ is the set of complex points A(C) of an abelianvariety iff there exists an R-bilinear form

E : Cg × Cg → R

such that(1) E(iv , iw) = E(v ,w)

(2) (v ,w)→ E(v , iw) is a symmetric positive defined bilinearform

(3) E(v ,w) ∈ Z for v ,w ∈ Λ

If such a form exists then A → Pn (by means of θ-functions)







E : Cg × Cg → R



(3) E(v ,w) ∈ Z for v ,w ∈ Λ








E : Cg × Cg → R



(3) E(v ,w) ∈ Z for v ,w ∈ Λ








E : Cg × Cg → R



(3) E(v ,w) ∈ Z for v ,w ∈ Λ








E : Cg × Cg → R



(3) E(v ,w) ∈ Z for v ,w ∈ Λ








E : Cg × Cg → R



(3) E(v ,w) ∈ Z for v ,w ∈ Λ








E : Cg × Cg → R



(3) E(v ,w) ∈ Z for v ,w ∈ Λ








E : Cg × Cg → R



(3) E(v ,w) ∈ Z for v ,w ∈ Λ








E : Cg × Cg → R



(3) E(v ,w) ∈ Z for v ,w ∈ Λ






f : A→ B is called an isogeny if f is surjective and has finitekernel.

The product of an abelian variety A of dimension m, and anabelian variety B of dimension n, over the same field, is anabelian variety of dimension m + n.

An abelian variety is simple if it is not isogenous to a product ofabelian varieties of lower dimension.

Any abelian variety is isogenous to a product of simple abelianvarieties.

A ∼= Ae11 × · · · × Aer

r









A ∼= Ae11 × · · · × Aer

r









A ∼= Ae11 × · · · × Aer

r









A ∼= Ae11 × · · · × Aer

r





there exists A∨ such that A = A∨∨

polarisation - an isogeny λ : A→ A∨

principal polarisation - degλ = 1

over C polarisation is equivalent to the choice of skew-symmetic form

ψ : Λ× Λ→ Z

detψ = 1 - polarisation is principal









ψ : Λ× Λ→ Z










ψ : Λ× Λ→ Z










ψ : Λ× Λ→ Z










ψ : Λ× Λ→ Z










ψ : Λ× Λ→ Z






Since λ is an isogeny there exists λ−1 in Hom(A∨,A)⊗Q

Rosati involution on End0(A) = End(A)⊗Q

Rinv : α→ α′ = λ−1 α∨ λ





Since λ is an isogeny there exists λ−1 in Hom(A∨,A)⊗Q

Rosati involution on End0(A) = End(A)⊗Q

Rinv : α→ α′ = λ−1 α∨ λ













Albert’s classification

A/F - simple; E = Z (End0(A)) = Z (End(A)⊗Q);E0 = ERinv

1 (I) End0(A) = E = E0 is a totally real number field Rinv = id2 (II) E = E0 is a totally real number field; End0(A) is a

division algebra over E such that every component ofEnd0(A)⊗QR is isomorphic to M2(R);β ∈ End0(A), tβ = −β; Rinv (α) = β−1 (tα)β

3 (III) E = E0 is totally real; End0(A) is a division algebraover E such that every component of End0(A)⊗QR isisomorphic to H Rinv (α) = tα

4 (IV) E0 is totally real; E - totally imaginary quadraticextension of E0; Rinv |E = cc|E ; End0(A) is a divisionalgebra over E


































number field caseStatement of the problem for abelian varietiesResults in this direction









K - a number field

Theorem (A.Schinzel 1973)If α1, . . . αk , β are non-zero elements of K and the congruence

αx11 α

x22 . . . αxk

k ≡ β mod p

is soluble for almost all prime ideals p of K then thecorresponding equation is soluble in rational integers. i.e. thereexist n1 . . . nk ∈ Z such that β = α1

n1 . . . αknk

proved again by C.Khare by different methods





K - a number field


αx11 α

x22 . . . αxk

k ≡ β mod p


n1 . . . αknk






K - a number field


αx11 α

x22 . . . αxk

k ≡ β mod p


n1 . . . αknk






K - a number field


αx11 α

x22 . . . αxk

k ≡ β mod p


n1 . . . αknk






K - a number field


αx11 α

x22 . . . αxk

k ≡ β mod p


n1 . . . αknk






Theorem of A.Schinzel does not extend in full generality to thesystem of congruences.

Theorem (A. Schinzel 1973)

Let αi,j , βi (i = 1, . . . ,h, j = 1, . . . k) be non-zero elements ofK , D a positive integer.If the system of congruences

Πkj=1α

xjij ≡ βi mod m (i = 1, . . . ,h)

is soluble for all moduli m prime to D then the correspondingsystem of equations is soluble in integers.








Πkj=1α

xjij ≡ βi mod m (i = 1, . . . ,h)









Πkj=1α

xjij ≡ βi mod m (i = 1, . . . ,h)









Πkj=1α

xjij ≡ βi mod m (i = 1, . . . ,h)






Counterexample: [A.Schinzel 1973]

2x3y ≡ 1mod p

2y3z ≡ 4 mod p

for p = 2, 3 (x , y , z) = (0,1,0) resp. (0,0,0)

For other p, fix a primitive root mod p

xind2 + yind3 ≡ 0 mod p − 1,

yind2 + zind3 ≡ 2ind2 mod p − 1






2x3y ≡ 1mod p

2y3z ≡ 4 mod p

for p = 2, 3 (x , y , z) = (0,1,0) resp. (0,0,0)









2x3y ≡ 1mod p

2y3z ≡ 4 mod p

for p = 2, 3 (x , y , z) = (0,1,0) resp. (0,0,0)









2x3y ≡ 1mod p

2y3z ≡ 4 mod p

for p = 2, 3 (x , y , z) = (0,1,0) resp. (0,0,0)









2x3y ≡ 1mod p

2y3z ≡ 4 mod p

for p = 2, 3 (x , y , z) = (0,1,0) resp. (0,0,0)









2x3y ≡ 1mod p

2y3z ≡ 4 mod p

for p = 2, 3 (x , y , z) = (0,1,0) resp. (0,0,0)








gcd((ind2)2

gcd(ind2, ind3), ind3) | ind2

t(ind2)2

gcd(ind2, ind3)+ zind3 = 2ind2

x =−tind3

gcd(ind2, ind3), y =

tind2gcd(ind2, ind3)

z





gcd((ind2)2


t(ind2)2


x =−tind3



z





gcd((ind2)2


t(ind2)2


x =−tind3



z













A/F redv : A(F )→ A(kv )

QUESTION (W.GAJDA 2002)

Let Σ be a subgroup of A(F ). Suppose that x is a point of A(F )such that redv x lies in redV Σ for almost all places v of F . Doesit then follow that x lies in Σ?



























Theorem (Weston 2003)Let A be an abelian variety over a number field F and assumethat EndF A is commutative. Let Σ be a subgroup of A(F ) andsuppose that x ∈ A(F ) is such that redv x ∈ redv Σ for almost allplaces v of F . Then x ∈ Σ + A(F )tors

this covers product of non-isogenous elliptic curves.























Theorem (G.Banaszak, W.Gajda, P.K 2005)Let A be a principally polarized abelian variety of dimension gdefined over the number field F such that EndF (A) = Z anddim(A) = g is either odd or g = 2 or 6. Let P and P1 . . .Pr benon-torsion elements of A(F ) such that P1 . . .Pr are linearlyindependent over Z. Denote by Λ the subgroup of A(F )generated by P1 . . .Pr . Then the following are equivalent:

(1) P ∈ Λ

(2) rv (P) ∈ rv (Λ)






(1) P ∈ Λ

(2) rv (P) ∈ rv (Λ)






(1) P ∈ Λ

(2) rv (P) ∈ rv (Λ)






(1) P ∈ Λ

(2) rv (P) ∈ rv (Λ)






(1) P ∈ Λ

(2) rv (P) ∈ rv (Λ)





Theorem (BGK 2005)

Let P and P1 . . .Pr be non-torsion elements of A(F ) such thatRP is a free R-module and P1 . . .Pr are linearly independentover R. Denote by Λ the R-submodule of A(F ) generated byP1 . . .Pr . Assume that rv (P) ∈ rv (Λ) for almost all primes v ofF . Then there exists a natural number a such that aP ∈ Λ.

W. Gajda and K.Górnisiewicz refined this theorem to a = 1





Theorem (BGK 2005)







Theorem (BGK 2005)







Theorem (BGK 2005)







Theorem (G.Banaszak 2008)

Let P1, . . . ,Pr be elements of A(F ) linearly independent overR = EndF (A). Let P be a point of A(F ) such that RP is a freeR-module. The following conditions are equivalent:(1) P ∈ Σr

i=1ZPi

(2) rv (P) ∈ Σri=1Zrv (Pi)

A. Perucca generalized this theorem to semiabelian varietiesand removed the hypotheses that RP is a free R-module







i=1ZPi









i=1ZPi









i=1ZPi






CorollariesA counterexampleCase of algebraic tori

THEOREM A (G.BANASZAK, P.KRASON)

Let A/F be an abelian variety defined over a number field F .Assume that A is isogeneous to Ae1

1 × · · · × Aett with Ai simple,

pairwise nonisogenous abelian varieties such thatdimEndF ′ (Ai )0 H1(Ai(C); Q) ≥ ei for each 1 ≤ i ≤ t , whereEndF ′(Ai)

0 := EndF ′(Ai)⊗Q and F ′/F is a finite extensionsuch that the isogeny is defined over F ′. Let P ∈ A(F ) and let Λbe a subgroup of A(F ). If rv (P) ∈ rv (Λ) for almost all v of OFthen P ∈ Λ + A(F )tor . Moreover if A(F )tor ⊂ Λ, then thefollowing conditions are equivalent:

1 P ∈ Λ

2 rv (P) ∈ rv (Λ) for almost all v of OF .










1 P ∈ Λ











1 P ∈ Λ











1 P ∈ Λ











1 P ∈ Λ











1 P ∈ Λ











1 P ∈ Λ











1 P ∈ Λ






PROBLEM

Let A/F be an abelian variety over a number field F and letP ∈ A(F ) and let Λ ⊂ A(F ) be a subgroup. Is there aneffectively computable finite set Seff of primes v of OF ,depending only on A, P and Λ such that the following conditionsare equivalent? :

(1) P ∈ Λ

(2) rv (P) ∈ rv (Λ) for every v ∈ Seff





PROBLEM


(1) P ∈ Λ






PROBLEM


(1) P ∈ Λ






THEOREM B ( G.BANASZAK, P.KRASON)

Let A/F satisfy the hypotheses of Theorem A. Let P ∈ A(F )and let Λ be a subgroup of A(F ).There is a finite set of primes vof OF , such that the condition: rv (P) ∈ rv (Λ) for all v ∈ Sfin

implies P ∈ Λ + A(F )tor . Moreover if A(F )tor ⊂ Λ, then thefollowing conditions are equivalent:

(1) P ∈ Λ

(2) rv (P) ∈ rv (Λ) for v ∈ Sfin.








(1) P ∈ Λ









(1) P ∈ Λ














Corollary (Weston )Let A be an abelian variety defined over a number field suchthat EndF (A) is commutative. Then Theorem A holds for A.





Since EndF (A) is commutative,

A is isogeneous to A1 × · · · × At

with Ai simple, pairwise nonisogenous.

In this case the assumption of our theorem

dimEndF ′ (Ai )0 H1(Ai(C); Q) ≥ 1

for each 1 ≤ i ≤ t always holds.





Since EndF (A) is commutative,

A is isogeneous to A1 × · · · × At

with Ai simple, pairwise nonisogenous.

In this case the assumption of our theorem

dimEndF ′ (Ai )0 H1(Ai(C); Q) ≥ 1

for each 1 ≤ i ≤ t always holds.





Corollary

Let A = Ee11 × · · · × Eet

t , where E1, . . . ,Et are pairwisenonisogenous elliptic curves defined over F .Assume that1 ≤ ei ≤ 2 if EndF (Ei) = Z and ei = 1 if EndF (Ei) 6= Z. ThenTheorem A holds for A.





Corollary

Let A = Ee11 × · · · × Eet






Corollary

Let A = Ee11 × · · · × Eet






Observe that for an elliptic curve E/F we have

dimEndF (E)0 H1(E(C); Q) = 2 if EndF (E) = Z

dimEndF (E)0 H1(E(C); Q) = 1 if EndF (E) 6= Z



























E := Ed defined over Q y2 = x3 − d2x

Ed has CM by Z[i]

K.Rubin, A.Silverberg showed that rankEd (Q) can reach 6.

d = 34 rankEd (Q) = 2

d = 1254 rankEd (Q) = 3

d = 29274 rankEd (Q) = 4

d = 205015206 rankEd (Q) = 5

d = 61471349610 rankEd (Q) = 6






Ed has CM by Z[i]


d = 34 rankEd (Q) = 2

d = 1254 rankEd (Q) = 3

d = 29274 rankEd (Q) = 4

d = 205015206 rankEd (Q) = 5

d = 61471349610 rankEd (Q) = 6






Ed has CM by Z[i]


d = 34 rankEd (Q) = 2

d = 1254 rankEd (Q) = 3

d = 29274 rankEd (Q) = 4

d = 205015206 rankEd (Q) = 5

d = 61471349610 rankEd (Q) = 6






Ed has CM by Z[i]


d = 34 rankEd (Q) = 2

d = 1254 rankEd (Q) = 3

d = 29274 rankEd (Q) = 4

d = 205015206 rankEd (Q) = 5

d = 61471349610 rankEd (Q) = 6






Ed has CM by Z[i]


d = 34 rankEd (Q) = 2

d = 1254 rankEd (Q) = 3

d = 29274 rankEd (Q) = 4

d = 205015206 rankEd (Q) = 5

d = 61471349610 rankEd (Q) = 6






Ed has CM by Z[i]


d = 34 rankEd (Q) = 2

d = 1254 rankEd (Q) = 3

d = 29274 rankEd (Q) = 4

d = 205015206 rankEd (Q) = 5

d = 61471349610 rankEd (Q) = 6






Ed has CM by Z[i]


d = 34 rankEd (Q) = 2

d = 1254 rankEd (Q) = 3

d = 29274 rankEd (Q) = 4

d = 205015206 rankEd (Q) = 5

d = 61471349610 rankEd (Q) = 6






Ed has CM by Z[i]


d = 34 rankEd (Q) = 2

d = 1254 rankEd (Q) = 3

d = 29274 rankEd (Q) = 4

d = 205015206 rankEd (Q) = 5

d = 61471349610 rankEd (Q) = 6





Let Ad := Ed × Ed defined over Q(i)

PROPOSITION

There is a nontorsion point P ∈ Ad (Q(i)) and a free Z[i]-moduleΛ ⊂ Ad (Q(i)) such that P /∈ Λ and rv (P) ∈ rv (Λ) for all primesv 6 |2d in Z[i]





Let Ad := Ed × Ed defined over Q(i)

PROPOSITION

There is a nontorsion point P ∈ Ad (Q(i)) and a free Z[i]-moduleΛ ⊂ Ad (Q(i)) such that P /∈ Λ and rv (P) ∈ rv (Λ) for all primesv 6 |2d in Z[i]





Since rankEd (Q) ≥ 2 we can find Q1,Q2 ∈ Ed (Q(i)) such thatthey are linearly independent over Z[i]

P :=

[0

Q1

], P1 :=

[Q10

], P2 :=

[Q2Q1

], P3 :=

[0

Q2

].

Λ := Z[i]P1 + Z[i]P2 + Z[i]P3 ⊂ A(Q(i))

Λ is free over Z[i] hence also free over Z.

Λ is not free over EndQ(i) A = M2 (Z[i]).

P /∈ Λ






P :=

[0

Q1

], P1 :=

[Q10

], P2 :=

[Q2Q1

], P3 :=

[0

Q2

].

Λ := Z[i]P1 + Z[i]P2 + Z[i]P3 ⊂ A(Q(i))



P /∈ Λ






P :=

[0

Q1

], P1 :=

[Q10

], P2 :=

[Q2Q1

], P3 :=

[0

Q2

].

Λ := Z[i]P1 + Z[i]P2 + Z[i]P3 ⊂ A(Q(i))



P /∈ Λ






P :=

[0

Q1

], P1 :=

[Q10

], P2 :=

[Q2Q1

], P3 :=

[0

Q2

].

Λ := Z[i]P1 + Z[i]P2 + Z[i]P3 ⊂ A(Q(i))



P /∈ Λ






P :=

[0

Q1

], P1 :=

[Q10

], P2 :=

[Q2Q1

], P3 :=

[0

Q2

].

Λ := Z[i]P1 + Z[i]P2 + Z[i]P3 ⊂ A(Q(i))



P /∈ Λ






P :=

[0

Q1

], P1 :=

[Q10

], P2 :=

[Q2Q1

], P3 :=

[0

Q2

].

Λ := Z[i]P1 + Z[i]P2 + Z[i]P3 ⊂ A(Q(i))



P /∈ Λ





Let Qi := rv (Qi) for i = 1,2,

Pi := rv (Pi) for i = 1,2,3 and P := rv (P).

We will prove that rv (P) ∈ rv (Λ) for all v of Z[i] over a primep 6 | 2d .

The equationP = r1P1 + r2P2 + r3P3.

in Ev (kv )× Ev (kv ) with r1, r2, r3 ∈ Z[i] is equivalent to a systemof equations in Ev (kv ) :

r1Q1 + r2Q2 = 0

r2Q1 + r3Q2 = Q1










r1Q1 + r2Q2 = 0

r2Q1 + r3Q2 = Q1










r1Q1 + r2Q2 = 0

r2Q1 + r3Q2 = Q1










r1Q1 + r2Q2 = 0

r2Q1 + r3Q2 = Q1










r1Q1 + r2Q2 = 0

r2Q1 + r3Q2 = Q1





Ev (kv ) ∼= Z[i]/γ(v),

c1, c2 ∈ Z[i]Q1 = c1 mod γ(v)Q2 = c2 mod γ(v).the above system of equations is equivalent to the system ofcongruences in Z[i]/γ(v) :

r1c1 + r2c2 ≡ 0 mod γ(v)

r2c1 + r3c2 ≡ c1 mod γ(v).

If c1 ≡ 0 mod γ(v) or c2 ≡ 0 mod γ(v) then the last system ofcongruences trivially has a solution.





Ev (kv ) ∼= Z[i]/γ(v),


r1c1 + r2c2 ≡ 0 mod γ(v)

r2c1 + r3c2 ≡ c1 mod γ(v).






Ev (kv ) ∼= Z[i]/γ(v),


r1c1 + r2c2 ≡ 0 mod γ(v)

r2c1 + r3c2 ≡ c1 mod γ(v).






Ev (kv ) ∼= Z[i]/γ(v),


r1c1 + r2c2 ≡ 0 mod γ(v)

r2c1 + r3c2 ≡ c1 mod γ(v).






Ev (kv ) ∼= Z[i]/γ(v),


r1c1 + r2c2 ≡ 0 mod γ(v)

r2c1 + r3c2 ≡ c1 mod γ(v).






c1 6≡ 0 mod γ(v) and c2 6≡ 0 mod γ(v).

D := gcd(c1, c2). Then

gcd (c21/D, c2) = D

and since D | c1

the equation r c21/D + r3c2 = c1

has a solution in r , r3 ∈ Z[i].

r1 :=−rc2

D, r2 :=

rc1

D







gcd (c21/D, c2) = D

and since D | c1



r1 :=−rc2

D, r2 :=

rc1

D







gcd (c21/D, c2) = D

and since D | c1



r1 :=−rc2

D, r2 :=

rc1

D







gcd (c21/D, c2) = D

and since D | c1



r1 :=−rc2

D, r2 :=

rc1

D







gcd (c21/D, c2) = D

and since D | c1



r1 :=−rc2

D, r2 :=

rc1

D













The methods of the proof of Main Theorem work for somealgebraic tori over a number field F .

Let T/F be an algebraic torus and let F ′/F be a finiteextension that splits T .

Hence T ⊗F F ′ ∼= Gem := Gm × · · · ×Gm︸︷︷︸

e−times

where

Gm := spec F ′[t , t−1].








e−times

where

Gm := spec F ′[t , t−1].








e−times

where

Gm := spec F ′[t , t−1].








e−times

where

Gm := spec F ′[t , t−1].





For any field extension F ′ ⊂ M ⊂ F we have EndM (Gm) = Zand H1(Gm(C); Z) = Z.

Hence the condition e ≤ dimEndF ′ (Gm)0 H1(Gm(C); Q) = 1,analogous to the corresponding condition of our Theorem ,means that e = 1.

Hence we can prove the analogue of the Main Theorem for onedimensional tori which is basically the A. Schinzel’s Theorem .

The torsion ambiguity that appears in our Theorem can be inthis case removed.





































On the other hand A. Schinzel showed that his theorem doesnot extend in full generality to Gm/F × Gm/F , hence it doesnot extend in full generality to algebraic tori T with dim T > 1.

The phrase full generality in the last sentence means for anyP ∈ T (F ) and any subgroup Λ ⊂ T (F ).
















Some easy reductionsTheorems about reductionsome semisimple algebras and modulesidea of the proof









Since the theorem is up to torsion we can assumeA = Ae1

1 × · · · × Aett

Put c := |A(F )tor | and Ω := c A(F ).

Ω is torsion free.

we can assume P ∈ Ω, P 6= 0, Λ ⊂ Ω and Λ 6= 0,

Let P1, . . . ,Pr , . . . ,Ps be such a Z-basis of Ω that:

Λ = Zd1P1 + · · ·+ Zdr Pr + · · ·+ ZdsPs.

where di ∈ Z\0 for 1 ≤ i ≤ r and di = 0 for i > r .






1 × · · · × Aett


Ω is torsion free.



Λ = Zd1P1 + · · ·+ Zdr Pr + · · ·+ ZdsPs.







1 × · · · × Aett


Ω is torsion free.



Λ = Zd1P1 + · · ·+ Zdr Pr + · · ·+ ZdsPs.







1 × · · · × Aett


Ω is torsion free.



Λ = Zd1P1 + · · ·+ Zdr Pr + · · ·+ ZdsPs.







1 × · · · × Aett


Ω is torsion free.



Λ = Zd1P1 + · · ·+ Zdr Pr + · · ·+ ZdsPs.







1 × · · · × Aett


Ω is torsion free.



Λ = Zd1P1 + · · ·+ Zdr Pr + · · ·+ ZdsPs.






put Ωj := c Aj(F ). Note that Ω =⊕t

j=1 Ωejj .

For P ∈ Ω =∑s

i=1ZPi we write

P = n1P1 + · · ·+ nr Pr + · · ·+ nsPs

where ni ∈ Z

Let K/Q be a finite extension such that Di ⊗Q K ∼= Mdi (K ) foreach 1 ≤ i ≤ t .Di := Ri ⊗Z Q






j=1 Ωejj .

For P ∈ Ω =∑s

i=1ZPi we write

P = n1P1 + · · ·+ nr Pr + · · ·+ nsPs

where ni ∈ Z







j=1 Ωejj .

For P ∈ Ω =∑s

i=1ZPi we write

P = n1P1 + · · ·+ nr Pr + · · ·+ nsPs

where ni ∈ Z







j=1 Ωejj .

For P ∈ Ω =∑s

i=1ZPi we write

P = n1P1 + · · ·+ nr Pr + · · ·+ nsPs

where ni ∈ Z






Since Λ ⊂ Ω is a free subgroup of the free finitely generatedabelian group Ω,

observe that P ∈ Λ if and only if P ⊗ 1 ∈ Λ⊗Z OK .

The latter is equivalent to P ⊗ 1 ∈ Λ⊗Z Oλ for all prime idealsλ | l in OK and all prime numbers l .



























Using Kummer maps for abelian varieties we prove thefollowing:

TheoremLet Qij ∈ Ai(L) for 1 ≤ j ≤ ri be independent over Ri for each1 ≤ i ≤ t . There is a family of primes w of OL of positive densitysuch that rw (Qij) = 0 in Ai w (kw )l for all 1 ≤ j ≤ ri and 1 ≤ i ≤ t .





Using Kummer maps for abelian varieties we prove thefollowing:

TheoremLet Qij ∈ Ai(L) for 1 ≤ j ≤ ri be independent over Ri for each1 ≤ i ≤ t . There is a family of primes w of OL of positive densitysuch that rw (Qij) = 0 in Ai w (kw )l for all 1 ≤ j ≤ ri and 1 ≤ i ≤ t .





TheoremLet l be a prime number. Let m ∈ N ∪ 0 for all 1 ≤ j ≤ ri and1 ≤ i ≤ t .Let L/F be a finite extension and let Pij ∈ Ai(L) beindependent over Ri and let Tij ∈ Ai [lm] be aribitrary torsionelements for all 1 ≤ j ≤ ri and 1 ≤ i ≤ t .There is a family ofprimes w of OL of positive density such that

rw ′(Tij) = rw (Pij) in Ai,w (kw )l

for all 1 ≤ j ≤ ri and 1 ≤ i ≤ s, where w ′ is a prime in OL(Ai [lm])

over w and rw ′ : Ai(L(Ai [lm]))→ Ai,w (kw ′) is the correspondingreduction map.





































Let D be a division algebra and let Ki ⊂ Me(D) denote the leftideal of Me(D) which consists of i-the column matrices of theform

αi :=

0 . . . a1i . . . 00 . . . a2i . . . 0...

... . . ....

0 . . . ae i . . . 0

∈ Ki





For ω ∈W put

ω :=

ω0...0

∈W e,





TheoremEvery nonzero simple submodule of the Me(D)-module W e hasthe following form

K1ω = α1ω, α1 ∈ K1 =

a11 ωa21 ω

...ae1 ω

, ai1 ∈ D, 1 ≤ i ≤ e

for some ω ∈W .





Let Di be a finite dimensional division algebra over Q for every1 ≤ i ≤ t .

The trace homomorphisms: tri : Mei (Di)→ Q, for all 1 ≤ i ≤ t ,give the trace homomorphism tr : Me(D)→ Q, wheretr :=

∑ti=1 tri .

Let Wi be a finite dimensional Di -vector space for each1 ≤ i ≤ t . Then W is a finitely generated Me(D)-module.

The homomorphism tr gives a natural map of Q-vector spaces

tr : HomMe(D)(W , Me(D))→ HomQ(W , Q).







∑ti=1 tri .










∑ti=1 tri .










∑ti=1 tri .








LemmaThe map tr is an isomorphism.













Assume P /∈ Λ P ⊗ 1 /∈ Λ⊗Z Oλ for some λ | l for some primenumber l .

Hence nj 6= 0 for some 1 ≤ j ≤ s in the expression

P = n1P1 + · · ·+ nr Pr + · · ·+ nsPs

Consider the equality in Ω⊗Z OK .

Since P /∈ Λ⊗Z Oλ then there is 1 ≤ j0 ≤ s such that λm1 ||nj0and λm2 |dj0 for natural numbers m1 < m2.

Consider the map of Z-modules

π : Ω→ Z

π(R) := µj0

for R =∑s

i=1 µiPi with µi ∈ Z for all 1 ≤ i ≤ s.P.Krason On linear .....






P = n1P1 + · · ·+ nr Pr + · · ·+ nsPs




π : Ω→ Z

π(R) := µj0

for R =∑s







P = n1P1 + · · ·+ nr Pr + · · ·+ nsPs




π : Ω→ Z

π(R) := µj0

for R =∑s







P = n1P1 + · · ·+ nr Pr + · · ·+ nsPs




π : Ω→ Z

π(R) := µj0

for R =∑s







P = n1P1 + · · ·+ nr Pr + · · ·+ nsPs




π : Ω→ Z

π(R) := µj0

for R =∑s





Lift π ⊗Q (by last Lemma) to

π ∈ HomMe(D)(Ω⊗Z Q, Me(D))

s ∈ HomMe(D)(Imπ, Ω⊗Z Q) such that π s = Id .





Lift π ⊗Q (by last Lemma) to

π ∈ HomMe(D)(Ω⊗Z Q, Me(D))

s ∈ HomMe(D)(Imπ, Ω⊗Z Q) such that π s = Id .





Im s(i) =

ki⊕k=1

K (i)1ωk (i),

Ker π(i) =

ui⊕k=ki+1

K (i)1ωk (i).





Choose, for each 1 ≤ i ≤ t , a lattice L′i ⊂ Li such that L′i is afree Ri -module.

Li - Riemann lattice (Ai(C) = Cg/Li)

Put L :=⊕t

i=1 Li and L′ :=⊕t

i=1 L′i .







Put L :=⊕t


i=1 L′i .







Put L :=⊕t


i=1 L′i .







Put L :=⊕t


i=1 L′i .





Analyze

z(n, λ) : L′ ⊗Z Oλ / λε n L′ ⊗Z Oλ → L⊗Z Oλ / λε n L ⊗Z Oλ,

l OK =∏λ | l λ

ε,

Here we need dimension condition

Use the reduction theorem





Analyze


l OK =∏λ | l λ

ε,







Analyze


l OK =∏λ | l λ

ε,







Analyze


l OK =∏λ | l λ

ε,







Let Tk (i) be the image of ηk (i) via the map z(n, λ) for all1 ≤ i ≤ t and 1 ≤ k ≤ pi .





By Reduction Theorem there is a set of primes w of OLof positive density such that rw (ωk (i)) = 0 for 1 ≤ i ≤ t ,ki + 1 ≤ k ≤ uiand rw (ωk (i)) = rw (Tk (i)) for all 1 ≤ i ≤ t , 1 ≤ k ≤ ki .





Choose such a prime w . Since rw (P) ∈ rw (Λ) we take Q ∈ Λsuch that rw (P) = rw (Q). Applying the reduction map rw to theequation

M2(P −Q) =t∑

i=1

ki∑k=1

(αk (i)1 − βk (i)1)M0ωk (i)

+t∑

i=1

ui∑k=ki+1

(αk (i)1 − βk (i)1)ωk (i),

we obtain

0 =t∑

i=1

ki∑k=1

(αk (i)1 − βk (i)1)M0 rw (T k (i)).





Since rw is injective on torsion we have

0 =t∑

i=1

ki∑k=1

(αk (i)1 − βk (i)1)M0 T k (i).

and thusαk (i)− βk (i) ∈ λan+bK (i)1,λ

On the other hand

αk (i)− βk (i) /∈ λm2K (i)1,λ





This part is P.K. Yuval Flicker





Theorem (Schinzel)Let F be a number field, and D > 0 a rational integer. Let T bethe torus Gn

m. Fix t1, . . . , tr , t0 in T (F ). Suppose for each idealm in the ring O of integers of F , that is prime to D, there arex1,m . . . , xr ,m in Z such that tx1,m

1 · · · txr,mr ≡ t0(modm). Then

there are x1, . . . , xr ∈ Z with tx11 · · · t

xrr = t0.





Conjecture

Let F be a number field. Let G be a linear algebraic group overO, viewed as a subgroup of some matrix group GL(n). Fix g1,. . . , gr , g0 in G(F ). Let D > 0 a rational integer, depending ong1, . . . , gr . Suppose for each ideal m in the ring O of integers ofF , that is prime to D, there are x1,m . . . , xr ,m in Z such thatgx1,m

1 · · · gxr,mr ≡ g0(modm). Then there are x1, . . . , xr ∈ Z with

gx11 · · · g

xrr = g0.





If gi =(

1 ui0 1

), when F = Q Conjecture states the following.

PROPOSITION

Suppose u0, u1, . . . , ur are nonzero rational integers. Let D > 0be an integer prime to the g.c.d. u = (u1, . . . ,ur ). Suppose foreach m > 1 prime to D there are integers xi,m (1 ≤ i ≤ r) withx1,mu1 + · · ·+ xr ,mur ≡ u0(mod m). Then there are integers xi(1 ≤ i ≤ r) with x1u1 + · · ·+ xr ur = u0.





If gi =(

1 ui0 1

), when F = Q Conjecture states the following.

PROPOSITION

Suppose u0, u1, . . . , ur are nonzero rational integers. Let D > 0be an integer prime to the g.c.d. u = (u1, . . . ,ur ). Suppose foreach m > 1 prime to D there are integers xi,m (1 ≤ i ≤ r) withx1,mu1 + · · ·+ xr ,mur ≡ u0(mod m). Then there are integers xi(1 ≤ i ≤ r) with x1u1 + · · ·+ xr ur = u0.





When G is a Heisenberg group, say the unipotent radical of theupper triangular subgroup of SL(3), the following questioncomes up, first again in the context of integers. Suppose,instead of u0, u1, . . . , ur , say we have two sequences ofintegers u0, u1, . . . , ur and v0, v1, . . . , vr ; and for D prime to(u1, . . . ,ur ) and (v1, . . . , vr ), for each m prime to D theequations x1,mu1 + · · ·+ xr ,mur ≡ u0(mod m) andx1,mv1 + · · ·+ xr ,mvr ≡ v0(mod m) are solvable, with the sameintegral x ’s.





Thus xi,m are the same for the u’s and for the v ’s. Then there isa solution to x1u1 + · · ·+ xr ur = u0 and to y1v1 + · · ·+ yr vr = v0.We should be able to choose yi = xi . Is this true? Yes it is, evenmore generally, in the context of G being the unipotent radicalU of the upper triangular subgroup of SL(n).





Regarding the above diagonal, that is, the derived groupU/[U,U] of U, we have, in the context of F = Q:

PROPOSITION

Suppose u0,j , u1,j , . . . , ur ,j (1 ≤ j < n) are nonzero integers. LetD > 0 be an integer prime to the g.c.d. uj = (u1,j , . . . ,ur ,j), all j .Suppose for each m > 1 prime to D there are integers xi,m(1 ≤ i ≤ r) with x1,mu1,j + · · ·+ xr ,mur ,j ≡ u0,j(mod m) for all j .Then there are integers xi (1 ≤ i ≤ r) withx1u1,j + · · ·+ xr ur ,j = u0,j .





Regarding the above diagonal, that is, the derived groupU/[U,U] of U, we have, in the context of F = Q:

PROPOSITION

Suppose u0,j , u1,j , . . . , ur ,j (1 ≤ j < n) are nonzero integers. LetD > 0 be an integer prime to the g.c.d. uj = (u1,j , . . . ,ur ,j), all j .Suppose for each m > 1 prime to D there are integers xi,m(1 ≤ i ≤ r) with x1,mu1,j + · · ·+ xr ,mur ,j ≡ u0,j(mod m) for all j .Then there are integers xi (1 ≤ i ≤ r) withx1u1,j + · · ·+ xr ur ,j = u0,j .





TheoremLet Γ be a subgroup of SL(n,O) of finite index. Let φ be anontrivial homomorphism from Γ to Γ. Suppose there is aninfinite set S of prime ideals p of O with the following property.For all p in S, the homomorphism φ factors to give anhomomorphism φp : SL(n,O/p)→ SL(n,O/p), thus thefollowing diagram exists and is commutative:





Γφ //

mod p

Γ

mod p

SL(n,O/p)

φp // SL(n,O/p)





Moreover, suppose φp is inner, thus φp(g) = Int(x)g := xgx−1,for some x = x(φp) in GL(n,O/p), for all p ∈ S. Then φ is anautomorphism of Γ which is the restriction to Γ of theinner-conjugation action by an element of GL(n,F ).





Proof:Put B =

∏p∈S O/p. The ring O embeds in the ring B. Hence

there is an injection SL(n,O) → SL(n,B). So φ lies in acommutative diagram

Γφ //

Γ

SL(n,B)

∏p φp // SL(n,B).





and it is locally inner. Hence the representationφ : Γ→ GL(n,F ) and the identity – natural embedding –representation id : Γ→ GL(n,F ), have equal traces.

But id is irreducible, hence φ and id are conjugate by anelement of GL(n,F ), namely φ is the restriction to Γ of Int(x),for some x ∈ GL(n,F ), and Int(x) takes Γ to itself.

But the index [SL(n,O) : Γ] equals [SL(n,O) : Int(x)Γ], henceφ = Int(x) is an automorphism of Γ.



















Lemma (J.-P.Serre)Let F be an algebraically closed field. Let A be an algebra overF . Consider two A-modules M and N which are finitedimensional over F , with the same dimension n. Suppose thatthey have the same character (trace of the elements) and thatone of them is irreducible. Then they are equivalent.





Note first that we may assume that A has finite dimension overF , by replacing it by its image in End(M)× End(N). Suppose Nis irreducible. Let M ′ be the semisimplification of M (direct sumof its simple Jordan-Hölder subquotients). The character of M ′

is equal to that of M. By linear independence of characters(assume that A is finite dimensional, and has trivial radical.Hence it is a product of matrix algebras, the representations areobvious, and so is the linear independence of their characters),this implies that M ′ is isomorphic to the sum of N and of othercharacters with multiplicities multiples of the characteristic of F .Because N and M ′ have the same dimension, thesemultiplicities are all 0. Hence M ′ is isomorphic to N. Hence Mis irreducible, and isomorphic to N.


Documents

On linear relations in algebraic groupsGeneralities about abelian varieties History of the problem Main Theorem Basic ingredients of proof of Theorem A G. Banaszak, P. Krason´ On