Linear algebra over Z p [[ u ]] Xavier Caruso, David Lubicz IRMAR - PowerPoint PPT Presentation

Linear algebra over Z p [[ u ]] Xavier Caruso, David Lubicz IRMAR — University Rennes 1 June 20, 2011

Notations Let W be a discrete valuation ring, that is a ring equipped with a surjective map v W : W → N ∪ { + ∞} such that • for all x ∈ W , v W ( x ) = + ∞ iff x = 0; • for all x , y ∈ W , v W ( xy ) = v W ( x ) + v W ( y ) ; • for all x , y ∈ W , v W ( x + y ) ≥ min ( v W ( x ) , v W ( y )) ; • for all x ∈ W , v W ( x ) = 0 iff x is invertible.

Notations Let W be a discrete valuation ring. Examples: the ring of p -adic integers Z p ; v W , here, is the usual p -adic valuation the ring k [[ X ]] where k is a field; v W , 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 Q p Let π be a uniformizer of W , that is an element such that v W ( π ) = 1.

Notations Let W be a discrete valuation ring. Examples: the ring of p -adic integers Z p ; v W , here, is the usual p -adic valuation the ring k [[ X ]] where k is a field; v W , 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 Q p Let π be a uniformizer of W , and set S = W [[ u ]] . Aim of the talk Describe efficient algorithms to compute with S -modules.

Motivations Well, it is certainly interesting for itself, but concretely we expect applications to : Iwasawa theory Certain abelian Galois groups inherit a structure of Z p [[ u ]] -module (and Iwasawa used this fact to study them). p -adic Hodge theory � � � � lattices in semi-stable modules over Z p [[ u ]] ∼ − → representations of G Q p + additional structures Example: (restrictions to G Q p of) p -adic Galois representations associated to a modular form of level prime to p are semi-stable (and even crystalline).

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 S d for a suitable d . In the sequel, we will always assume that our modules are embedded in some S d and have full rank ( i.e. contain a family of d vectors linearly independant).

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 ( π n , π n − 1 u , . . . , u n ) cannot be gen- erated by less than n + 1 elements.

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 M n × m ( S ) . Example: ( π n π n − 1 u u n ) � = ( ⋆ · · · 0 · · · 0 ) · P with P invertible

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 M d × d ( S ) with non-vanishing determinant. Example: Assume that W = k [[ v ]] ( k is a field) and � � � � u v a b = · P with P invertible. v u 0 d • a , b and d belong to the ideal ( u , v ) • ad = unit · ( u − v ) · ( u + v ) = ⇒ a = unit · ( u ± v ) (since S is factorial)

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 M d × d ( S ) with non-vanishing determinant. Example: Assume that W = k [[ v ]] ( k is a field) and � � � � u v u ± v b = · P with P invertible. v u 0 d = ⇒ ∃ λ, µ ∈ S , λ u + µ v = u ± v and λ v + µ u = 0 By the second equation, u divides λ and v divides µ . Therefore u ± v ∈ u 2 S + v 2 S . Contradiction.

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 M d × d ( S ) with non-vanishing determinant. Problem of precision Find a good notion of precision of course, for elements in S Basic idea: truncate series mod u N and coefficients mod π n . In other words, we describe a serie f by the values of f ( 0 ) mod p n 0 , f ′ ( 0 ) mod p n 1 , . . . , f ( N ) ( 0 ) mod p n N But why not f(x) for another x ? Itself not fully determined? Or even, something more involved?

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 M d × d ( S ) with non-vanishing determinant. Problem of precision Find a good notion of precision of course, for elements in S , but also, for submodules of S d (of full rank). NB: It is definitely not the same than for matrices. � � 1 + O ( p 2 ) O ( p 2 ) Example (over Z p ): The matrix 1 + O ( p 2 ) p + O ( p 2 ) determines a well-defined lattice in Q 2 p .

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 M d × d ( S ) with non-vanishing determinant. Problem of precision Find a good notion of precision of course, for elements in S , but also, for submodules of S d (of full rank). In the sequel, we will give complete solutions to the first two problems, and just hints for the third one.

Solution to the theoretical problem

Iwasawa’s Theorem Let M be a submodule of S d . We define: � � x ∈ S d � ∃ n ∈ N , u n x ∈ M and π n x ∈ M Max ( M ) = � Examples: if M = ( u , π ) ⊂ S , then Max ( M ) = S if M is free, then Max ( M ) = M . Proof: Let ( e 1 , . . . , e h ) be a basis of M . Let x ∈ Max ( M ) . u n · π n x u n α 1 e 1 + · · · + u n α h e h = = π n · u n x π n β 1 e 1 + · · · + π n β h e h = Therefore u n α i = π n β i for all i . Hence u n divides all β i ’s and: x = β 1 u n e 1 + · · · + β h u n e h ∈ M .

Iwasawa’s Theorem Let M be a submodule of S d . We define: � � x ∈ S d � ∃ n ∈ N , u n x ∈ M and π n x ∈ M Max ( 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 ) .

Application to algorithmics � � x ∈ S d ∃ n ∈ N , u n x ∈ M and π n x ∈ M � Max ( 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”: M 1 + free M 2 = Max ( M 1 + M 2 ) Example: uS + free pS = Max ( u , p ) = S NB : We do not need a special “free intersection”

Solution to the problem of representation

More rings � � a i ∈ K , for all i � � a i u i S π = � � v W ( a i ) bounded below i ∈ N � � a i ∈ W , for all i � � a i u i S u = � � lim i →−∞ v W ( a i ) = + ∞ i ∈ Z a i ∈ K , for all i � � � � a i u i E = � v W ( a i ) bounded below � i ∈ Z lim i →−∞ v W ( a i ) = + ∞ S π and S u contain S and are contained in E . Moreover, S π ∩ S u = S .

S π is an euclidean ring � � a i ∈ K , for all i � � a i u i S π = � � v W ( a i ) bounded below i ∈ N � a i u i . Define: Let A = i ∈ N the Gauss valuation : v G ( A ) = min i ∈ N v W ( a i ) • v G ( A ) = + ∞ ⇔ A = 0 • v G ( AB ) = v G ( A ) + v G ( B ) • v G ( A + B ) ≥ min ( v G ( A ) , v G ( B )) • But we do not have v G ( A ) = 0 ⇔ A is invertible

S π is an euclidean ring � � a i ∈ K , for all i � � a i u i S π = � � v W ( a i ) bounded below i ∈ N � a i u i . Define: Let A = i ∈ N the Gauss “valuation” : v G ( A ) = min i ∈ N v W ( a i ) the Weierstrass degree : � � deg W ( A ) = min i ∈ N | v W ( a i ) = v G ( A ) • deg W ( AB ) = deg W ( A ) + deg W ( B )

S π is an euclidean ring � � a i ∈ K , for all i � � a i u i S π = � � v W ( a i ) bounded below i ∈ N � a i u i . Define: Let A = i ∈ N the Gauss “valuation” : v G ( A ) = min i ∈ N v W ( a i ) the Weierstrass degree : � � deg W ( A ) = min i ∈ N | v W ( a i ) = v G ( A ) Proposition (Euclidean division) Let A , B ∈ S π with B � = 0. There exists a unique pair ( Q , R ) with Q ∈ S π , R ∈ K [ u ] such that A = BQ + R and deg ( R ) < deg W ( B ) . Moreover v G ( Q ) = v G ( A ) − v G ( B ) and v G ( R ) ≥ v G ( A ) − v G ( B ) .

Hermite Normal Form for matrices over S π Every matrix A ∈ M d × d ( S π ) of full rank can be decomposed as a product:   T 1 ⋆ T 2     A = · P       T d ⋆ ⋆ P is in GL d ( S π ) T i = u d i + R i where R i ∈ π W [ u ] has degree < d i Lemma: X ∈ S π ⇒ X = UT with U ∈ S × π and T as before. Proof: Assume v G ( X ) = 0. Let d = deg W ( X ) . Write: T = − R + u d = XQ with deg ( R ) < d • d = deg W ( T ) and deg W ( Q ) = 0 • R ∈ π W [ u ] • v G ( Q ) = v G ( u d ) − v G ( X ) = 0 ⇒ Q is invertible