SLIDE 1
Linear algebra over Zp[[u]]
Xavier Caruso, David Lubicz
IRMAR — University Rennes 1
June 20, 2011
SLIDE 2 Notations
Let W be a discrete valuation ring, that is a ring equipped with a surjective map vW : W → N ∪ {+∞} such that
- for all x ∈ W, vW(x) = +∞ iff x = 0;
- for all x, y ∈ W, vW(xy) = vW(x) + vW(y);
- for all x, y ∈ W, vW(x + y) ≥ min(vW(x), vW(y));
- for all x ∈ W, vW(x) = 0 iff x is invertible.
SLIDE 3
Notations
Let W be a discrete valuation ring. Examples: the ring of p-adic integers Zp; vW, here, is the usual p-adic valuation the ring k[[X]] where k is a field; vW, here, is the usual valuation of a serie (that is the smallest power of X having a nonzero coefficient) the ring of integers of a finite extension of Qp Let π be a uniformizer of W, that is an element such that vW(π) = 1.
SLIDE 4
Notations
Let W be a discrete valuation ring. Examples: the ring of p-adic integers Zp; vW, here, is the usual p-adic valuation the ring k[[X]] where k is a field; vW, here, is the usual valuation of a serie (that is the smallest power of X having a nonzero coefficient) the ring of integers of a finite extension of Qp Let π be a uniformizer of W, and set S = W[[u]].
Aim of the talk
Describe efficient algorithms to compute with S-modules.
SLIDE 5 Motivations
Well, it is certainly interesting for itself, but concretely we expect applications to : Iwasawa theory Certain abelian Galois groups inherit a structure of Zp[[u]]-module (and Iwasawa used this fact to study them). p-adic Hodge theory
representations of GQp
− →
+ additional structures
- Example: (restrictions to GQp of) p-adic Galois representations
associated to a modular form of level prime to p are semi-stable (and even crystalline).
SLIDE 6
Precise set-up
Recall that we want to describe efficient algorithms to manipulate S-modules, e.g. compute intersections, sums, kernels, images, etc. Basic assumption: We restrict ourselves to finitely generated modules without torsion. All these modules can be realized as submodules of Sd for a suitable d. In the sequel, we will always assume that our modules are embedded in some Sd and have full rank (i.e. contain a family of d vectors linearly independant).
SLIDE 7
Preliminary problems
Theoretical problem S is not a very nice ring (e.g. it is not a principal domain).
Example: for all n, the ideal (πn, πn−1u, . . . , un) cannot be gen- erated by less than n + 1 elements.
SLIDE 8
Preliminary problems
Theoretical problem S is not a very nice ring (e.g. it is not a principal domain). Problem of representation We do not have a Hermite Normal Form (HNF) for matrices in Mn×m(S).
Example: (πn πn−1u · · · un) = (⋆ · · · 0) · P with P invertible
SLIDE 9 Preliminary problems
Theoretical problem S is not a very nice ring (e.g. it is not a principal domain). Problem of representation We do not have a Hermite Normal Form (NHF) for matrices in Md×d(S) with non-vanishing determinant.
Example: Assume that W = k[[v]] (k is a field) and
v v u
b d
with P invertible.
- a, b and d belong to the ideal (u, v)
- ad = unit · (u − v) · (u + v)
= ⇒ a = unit · (u ± v) (since S is factorial)
SLIDE 10 Preliminary problems
Theoretical problem S is not a very nice ring (e.g. it is not a principal domain). Problem of representation We do not have a Hermite Normal Form (NHF) for matrices in Md×d(S) with non-vanishing determinant.
Example: Assume that W = k[[v]] (k is a field) and
v v u
b d
with P invertible. = ⇒ ∃λ, µ ∈ S, λu + µv = u ± v and λv + µu = 0 By the second equation, u divides λ and v divides µ. Therefore u ± v ∈ u2S + v2S. Contradiction.
SLIDE 11 Preliminary problems
Theoretical problem S is not a very nice ring (e.g. it is not a principal domain). Problem of representation We do not have a Hermite Normal Form (NHF) for matrices in Md×d(S) with non-vanishing determinant. Problem of precision Find a good notion of precision
- f course, for elements in S
Basic idea: truncate series mod uN and coefficients mod πn. In other words, we describe a serie f by the values of f(0) mod pn0, f ′(0) mod pn1, . . . , f (N)(0) mod pnN But why not f(x) for another x? Itself not fully determined? Or even, something more involved?
SLIDE 12 Preliminary problems
Theoretical problem S is not a very nice ring (e.g. it is not a principal domain). Problem of representation We do not have a Hermite Normal Form (NHF) for matrices in Md×d(S) with non-vanishing determinant. Problem of precision Find a good notion of precision
- f course, for elements in S,
but also, for submodules of Sd (of full rank).
NB: It is definitely not the same than for matrices. Example (over Zp): The matrix
O(p2) 1 + O(p2) p + O(p2)
- determines a well-defined lattice in Q2
p.
SLIDE 13 Preliminary problems
Theoretical problem S is not a very nice ring (e.g. it is not a principal domain). Problem of representation We do not have a Hermite Normal Form (NHF) for matrices in Md×d(S) with non-vanishing determinant. Problem of precision Find a good notion of precision
- f course, for elements in S,
but also, for submodules of Sd (of full rank). In the sequel, we will give complete solutions to the first two problems, and just hints for the third one.
SLIDE 14
Solution to the theoretical problem
SLIDE 15 Iwasawa’s Theorem
Let M be a submodule of Sd. We define: Max (M) =
- x ∈ Sd
- ∃n ∈ N, unx ∈ M and πnx ∈ M
- Examples:
if M = (u, π) ⊂ S, then Max (M) = S if M is free, then Max (M) = M.
Proof: Let (e1, . . . , eh) be a basis of M. Let x ∈ Max (M). un · πnx = unα1e1 + · · · + unαheh = πn · unx = πnβ1e1 + · · · + πnβheh Therefore unαi = πnβi for all i. Hence un divides all βi’s and: x = β1 un e1 + · · · + βh un eh ∈ M.
SLIDE 16 Iwasawa’s Theorem
Let M be a submodule of Sd. We define: Max (M) =
- x ∈ Sd
- ∃n ∈ N, unx ∈ M and πnx ∈ M
- Examples:
if M = (u, π) ⊂ S, then Max (M) = S if M is free, then Max (M) = M.
Theorem
The module Max (M) is always free over S.
Corollary
A module M is free iff M = Max (M).
SLIDE 17 Application to algorithmics
Max (M) =
- x ∈ Sd
- ∃n ∈ N, unx ∈ M and πnx ∈ M
- Theorem
The module Max (M) is always free over S. For algorithmic purpose, we want to work only with free modules. For that, we will replace systematically all modules by their Max . In particular, we define (and will work with) a “free sum”: M1 +free M2 = Max (M1 + M2) Example: uS +free pS = Max (u, p) = S NB: We do not need a special “free intersection”
SLIDE 18
Solution to the problem of representation
SLIDE 19 More rings
Sπ =
aiui
vW(ai) bounded below
=
aiui
limi→−∞ vW(ai) = +∞
=
aiui
vW(ai) bounded below limi→−∞ vW(ai) = +∞
- Sπ and Su contain S and are contained in E.
Moreover, Sπ ∩ Su = S.
SLIDE 20 Sπ is an euclidean ring
Sπ =
aiui
vW(ai) bounded below
- Let A =
- i∈N
- aiui. Define:
the Gauss valuation: vG(A) = min
i∈N vW(ai)
- vG(A) = +∞ ⇔ A = 0
- vG(AB) = vG(A) + vG(B)
- vG(A + B) ≥ min(vG(A), vG(B))
- But we do not have vG(A) = 0 ⇔ A is invertible
SLIDE 21 Sπ is an euclidean ring
Sπ =
aiui
vW(ai) bounded below
- Let A =
- i∈N
- aiui. Define:
the Gauss “valuation”: vG(A) = min
i∈N vW(ai)
the Weierstrass degree: degW(A) = min
- i ∈ N | vW(ai) = vG(A)
- degW(AB) = degW(A) + degW(B)
SLIDE 22 Sπ is an euclidean ring
Sπ =
aiui
vW(ai) bounded below
- Let A =
- i∈N
- aiui. Define:
the Gauss “valuation”: vG(A) = min
i∈N vW(ai)
the Weierstrass degree: degW(A) = min
- i ∈ N | vW(ai) = vG(A)
- Proposition (Euclidean division)
Let A, B ∈ Sπ with B = 0. There exists a unique pair (Q, R) with Q ∈ Sπ, R ∈ K[u] such that A = BQ + R and deg(R) < degW(B). Moreover vG(Q) = vG(A) − vG(B) and vG(R) ≥ vG(A) − vG(B).
SLIDE 23 Hermite Normal Form for matrices over Sπ
Every matrix A ∈ Md×d(Sπ) of full rank can be decomposed as a product: A = T1 ⋆ T2 ⋆ ⋆ Td · P P is in GLd(Sπ) Ti = udi + Ri where Ri ∈ πW[u] has degree < di
Lemma: X ∈ Sπ ⇒ X = UT with U ∈ S×
π and T as before.
Proof: Assume vG(X) = 0. Let d = degW(X). Write: T = − R + ud = XQ with deg(R) < d
- d = degW(T) and degW(Q) = 0
- R ∈ πW[u]
- vG(Q) = vG(ud) − vG(X) = 0 ⇒ Q is invertible
SLIDE 24
Hermite Normal Form for matrices over Sπ
Every matrix A ∈ Md×d(Sπ) of full rank can be decomposed as a product: A = T1 ⋆ T2 ⋆ ⋆ Td · P P is in GLd(Sπ) Ti = udi + Ri where Ri ∈ πW[u] has degree < di all entries of the i-th line before Ti are polynomials with coefficients in K of degree < di. Moreover, this writing is unique. It is called the Hermite Normal Form (HNF) of A
SLIDE 25 Su is a discrete valuation ring
Su =
aiui
limi→−∞ vW(ai) = +∞
The Gauss valuation vG turns Su into a discrete valuation ring.
Proof (of the last property in the definition): Let A =
i∈Z aiui ∈ Su with vG(A) = 0. Assume degW(A) = 0.
Write A = a0 +
aiui +
aiui = a0 − uX − πY 1 A = 1 a0 − uX − πY = 1 a0
∞
uX + πY a0 n ∈ Su
SLIDE 26 Su is a discrete valuation ring
Su =
aiui
limi→−∞ vW(ai) = +∞
The Gauss valuation vG turns Su into a discrete valuation ring. We have also a kind of Hermite Normal Form: every matrix B ∈ Md×d(Su) of full rank can be written B = πn1 ⋆ πn2 ⋆ ⋆ πnd · P, P ∈ GLd(Su) with all entries at the left of πni uniquely determined modulo πni. (If W is nice, one can choose a canonical representant.)
SLIDE 27
A useful bijection
Notation: Free d
A is the set of submodules of Ad of full rank.
For M ∈ Free d
S, define:
Mπ ∈ Free d
Sπ to be the submodule of Sd π generated by M
Mu ∈ Free d
Su to be the submodule of Sd u generated by M
Theorem
The maps Free d
S
− → Free d
Sπ × Free d Su
M → (Mπ, Mu) A ∩ B ← (A, B) are bijection inverse one to the other. We will represent M by the pair (Mπ, Mu), each component being itself represented by a matrix in HNF .
SLIDE 28
Few words about precision
An element M ∈ Free d
S is represented by a pair of matrices
(Mπ, Mu) with: Mπ = T1 ⋆ T2 ⋆ ⋆ Td all entries are polynomials in u (with coefficients in K) ⇒ exact with respect to u Mu = πn1 ⋆ πn2 ⋆ ⋆ πnd all entries below the diagonal are determined modulo πn for some n ⇒ exact with respect to π Conclusion: With our representation, the precision splits in two parts (one π-adic, and one u-adic). It helps.
SLIDE 29
Algorithms
SLIDE 30
Usual operations
Membership test x ∈ M ⇐ ⇒ x ∈ Mπ and x ∈ Mu
Proof: Remember that M = Mπ ∩ Mu
SLIDE 31 Usual operations
Membership test x ∈ M ⇐ ⇒ x ∈ Mπ and x ∈ Mu “Free sums” (M +free M′)π = Mπ + M′
π
; (M +free M′)u = Mu + M′
u
Proof (of the first equality): It is clear that the submodule generated by M + M′ is Mπ + M′
π.
Hence, it is enough to prove, for N ⊂ Sd, N and Max (N) generates the same module over Sπ. It is obvious from the definition of Max (N): Max (N) =
- x ∈ Sd
- ∃n ∈ N, unx ∈ N and πnx ∈ N
SLIDE 32 Usual operations
Membership test x ∈ M ⇐ ⇒ x ∈ Mπ and x ∈ Mu “Free sums” (M +free M′)π = Mπ + M′
π
; (M +free M′)u = Mu + M′
u
Intersections (M ∩ M′)π = Mπ ∩ M′
π
; (M ∩ M′)u = Mu ∩ M′
u
Proof: Write M ∩ M′ = (Mπ ∩ Mu) ∩ (M′
π ∩ M′ u)
= (Mπ ∩ M′
π) ∩ (Mu ∩ M′ u)
SLIDE 33
Usual operations
Membership test x ∈ M ⇐ ⇒ x ∈ Mπ and x ∈ Mu “Free sums” (M +free M′)π = Mπ + M′
π
; (M +free M′)u = Mu + M′
u
Intersections (M ∩ M′)π = Mπ ∩ M′
π
; (M ∩ M′)u = Mu ∩ M′
u
Kernels and images Let f : Sd → Sd′ be a morphism. Let fπ : Sd
π → Sd′ π and fu : Sd u → Sd′ u be the induced morphisms.
(ker f)π = ker fπ ; (ker f)u = ker fu (Max (im f))π = im fπ ; (Max (im f))u = im fu
SLIDE 34 Other natural operations
Saturation For M ∈ Free d
S, define Msat =
∃n ∈ N, πnx ∈ M
(Msat)π = Sd
π
; (Msat)u = Mu Intersection with Sπ-modules Let M ∈ Free d
S and M′ π ∈ Free d Sπ. Then M ∩ M′ ∈ Freed S and:
(M ∩ M′)π = Mπ ∩ M′
π
; (M ∩ M′)u = Mu Intersection with Su-modules Let M ∈ Free d
S and M′ u ∈ Free d
S and:
(M ∩ M′)π = Mπ ; (M ∩ M′)u = Mu ∩ M′
u
So, with our representation, everything is very easy, except...
SLIDE 35 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
SLIDE 36 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
SLIDE 37 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
1 π(u + π)
SLIDE 38 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
SLIDE 39 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
SLIDE 40 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π−1
SLIDE 41 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π−1
u
u−2
π2u
SLIDE 42 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
SLIDE 43 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
SLIDE 44 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
SLIDE 45 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
π(u + π)
SLIDE 46 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
π2u π(u + π) π
SLIDE 47 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
π2u 1 π(u + π) π u−2
SLIDE 48 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
π2u u2 π(u + π) π 1
SLIDE 49 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
π2u u2 × det Mπ π(u + π) π 1 × det Mπ
SLIDE 50 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
π2u u3(u + π) π(u + π) π u(u + π)
SLIDE 51 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
π2u u3(u + π) π(u + π) π u(u + π) πu(u + π)
SLIDE 52 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
π2u u3(u + π) π(u + π) π u(u + π) πu(u + π)
Lemma
These vectors generate a module M′ s.t. M′
π = Mπ and M′ u = Mu.
SLIDE 53 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
Generators:
π2u u3(u + π) π(u + π) π u(u + π) πu(u + π)
Lemma
These vectors generate a module M′ s.t. M′
π = Mπ and M′ u = Mu.
SLIDE 54 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
Generators:
π2u u3(u + π) π(u + π) π u(u + π) πu(u + π)
Relations:
SLIDE 55 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
Generators:
π2u u3(u + π) π(u + π) π u(u + π) πu(u + π)
Relations:
u −1
SLIDE 56 Computing a basis of a free S-module
Example: Take Mπ =
π−1 u + π
u−2 π
π
(u + π)
π2
π−1
π2u
u−2
u
Generators:
π2u u3(u + π) π(u + π) π u(u + π) πu(u + π)
Relations:
u −1 u4(u + π) −π2u2 −u2(u − π)
SLIDE 57
Computing a basis of a free S-module
Generators:
π2u u3(u + π) π(u + π) π u(u + π) πu(u + π)
Relations:
u −1 u4(u + π) −π2u2 −u2(u − π)
SLIDE 58
Computing a basis of a free S-module
Generators:
π2u u3(u + π) π(u + π) π u(u + π) πu(u + π)
Relations:
u −1 u4(u + π) −π2u2 −u2(u − π)
SLIDE 59 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj
Generators:
π2u u3(u + π) π(u + π) π u(u + π) πu(u + π)
Relations:
u −1 u4(u + π) −π2u2 −u2(u − π)
SLIDE 60 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj
Generators:
π2u u3(u + π) π(u + π) π u(u + π)
Relations:
−1 −u3(u − π) u4(u + π) −π2u2 −u2(u − π)
SLIDE 61 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj
Generators:
π2u u3(u + π) π(u + π) π u(u + π)
Relations:
−1 −u3(u − π) u4(u + π) −π2u2 −u2(u − π)
SLIDE 62 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj
Generators:
π2u u3(u + π) π(u + π) π u(u + π)
Relations:
−1 −u3(u − π) u4(u + π) −π2u2 −u2(u − π)
SLIDE 63 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj
Generators:
−π2u2 π2u u3(u + π) π2 π u(u + π)
Relations:
−1 −u3(u − π) 2πu4 −π2u2 −u2(u − π)
SLIDE 64 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj
Generators:
−π2u2 π2u u3(u + π) π2 π u(u + π)
Relations:
−1 −u3(u − π) 2πu4 −π2u2 −u2(u − π)
SLIDE 65 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj Second operation Assumption: vG(ai) = 0 and πn divides all other entries in the line. Gen: Vi
1 πn Vi
ai
(to be written)
Generators:
−π2u2 π2u u3(u + π) π2 π u(u + π)
Relations:
−1 −u3(u − π) 2πu4 −π2u2 −u2(u − π)
SLIDE 66 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj Second operation Assumption: vG(ai) = 0 and πn divides all other entries in the line. Gen: Vi
1 πn Vi
ai
(to be written)
Generators:
−πu2 π2u u3(u + π) π π u(u + π)
Relations:
−1 −u3(u − π) 2u4 −πu2 −u2(u − π)
SLIDE 67 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj Second operation Assumption: vG(ai) = 0 and πn divides all other entries in the line. Gen: Vi
1 πn Vi
ai
(to be written)
Generators:
−πu2 π2u u3(u + π) π π u(u + π)
Relations:
−1 −u3(u − π) 2u4 −πu2 −u2(u − π)
SLIDE 68 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj Second operation Assumption: vG(ai) = 0 and πn divides all other entries in the line. Gen: Vi
1 πn Vi
ai
(to be written)
Generators:
−πu(u + 2π) π2u u3(u + π) −π π u(u + π)
Relations:
−1 −u3(u − π) 2πu3 −πu2 −u2(u − π)
SLIDE 69 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj Second operation Assumption: vG(ai) = 0 and πn divides all other entries in the line. Gen: Vi
1 πn Vi
ai
(to be written)
Generators:
−πu(u + 2π) π2u u3(u + π) −π π u(u + π)
Relations:
−1 −u3(u − π) 2πu3 −πu2 −u2(u − π)
SLIDE 70 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj Second operation Assumption: vG(ai) = 0 and πn divides all other entries in the line. Gen: Vi
1 πn Vi
ai
(to be written)
Generators:
−u(u + 2π) π2u u3(u + π) −1 π u(u + π)
Relations:
−1 −u3(u − π) 2u3 −u2 −u2(u − π)
SLIDE 71 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj Second operation Assumption: vG(ai) = 0 and πn divides all other entries in the line. Gen: Vi
1 πn Vi
ai
(to be written)
Generators:
−u(u + 2π) π2u u3(u + π) −1 π u(u + π)
Relations:
−1 −u3(u − π) 2u3 −u2 −u2(u − π)
SLIDE 72 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj Second operation Assumption: vG(ai) = 0 and πn divides all other entries in the line. Gen: Vi
1 πn Vi
ai
(to be written)
Generators:
−u(u + 2π) π2u
u2(u2 +uπ−2π2)
−1 π u(u − π)
Relations:
−1 −u3(u − π) −u2 −u2(u − π)
SLIDE 73 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj Second operation Assumption: vG(ai) = 0 and πn divides all other entries in the line. Gen: Vi
1 πn Vi
ai
(to be written)
Generators:
−u(u + 2π) π2u
u2(u2 +uπ−2π2)
−1 π u(u − π)
Relations:
−1 −u3(u − π) −u2 −u2(u − π)
SLIDE 74 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj Second operation Assumption: vG(ai) = 0 and πn divides all other entries in the line. Gen: Vi
1 πn Vi
ai
(to be written)
Generators:
−u(u + 2π) π2u −1 π
Relations:
−1 −u2 −u2(u − π)
SLIDE 75 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj Second operation Assumption: vG(ai) = 0 and πn divides all other entries in the line. Gen: Vi
1 πn Vi
ai
(to be written)
Generators:
−u(u + 2π) π2u −1 π
Relations:
−1 −u2 −u2(u − π)
SLIDE 76 Computing a basis of a free S-module
First operation Assumption: vG(ai) = vG(aj) = 0. Gen: Vi Vj Vi Vj+qVi
ai=ajq+r
ai aj Ri−qRj Rj Second operation Assumption: vG(ai) = 0 and πn divides all other entries in the line. Gen: Vi
1 πn Vi
ai
(to be written)
Generators:
−u(u + 2π) π2u −1 π
Proposition
The two reminding vectors form a basis of M.
SLIDE 77 Computing a basis of a free S-module
Generators:
−u(u + 2π) π2u −1 π
Proposition
The two reminding vectors form a basis of M.
Proof: Let e1 and e2 be these vectors. Let M′′ be the submodule generated by e1, e2. From the construction, it follows that M′′
π = Mπ and M′′ u = Mu.
Now note that e1 and e2 generate M′′
π, which has dimension 2 over
Sπ. Therefore, they are linearly independant over Sπ, and then also
- ver S. As a consequence, M′′ is free.
Hence M′′ = M as claimed.
SLIDE 78
Implementation
SLIDE 79
Implementation
We have implemented all these algorithms in Magma. The code will be soon avaiable on the web site of the CETHop project : http://cethop.math.cnrs.fr/ Based on this framework, we are currently writing and coding (in Magma) algorithms to compute lattices in semi-stable representations (ask questions for more details). We plan to translate all of this in Sage. But firstable, David Roe and I would like to develop basics for p-adics – including a more powerful control on precision (again ask questions if you are interested).
