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HAL Id: hal-02501322

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Preprint submitted on 6 Mar 2020

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Presenting isomorphic finitely presented modules by equivalent matrices: a constructive approach

Cyrille Chenavier, Thomas Cluzeau, Alban Quadrat

To cite this version:

Cyrille Chenavier, Thomas Cluzeau, Alban Quadrat. Presenting isomorphic finitely presented modules

by equivalent matrices: a constructive approach. 2020. �hal-02501322�

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Presenting isomorphic finitely presented modules by equivalent matrices: a constructive approach

Cyrille Chenavier a , Thomas Cluzeau b , Alban Quadrat c

a Inria Lille - Nord Europe, Valse team, Villeneuve d’Ascq, France

b Universit´ e de Limoges ; CNRS ; XLIM UMR 7252, MATHIS, Limoges, France

c Inria Paris, Ouragan project, IMJ - PRG, Sorbonne University, France

Abstract

Constructive versions of Fitting and Warfield’s theorems, implementation Keywords:

Linear functional systems, algebraic analysis, isomorphism problem, Fitting’s theorem, Warfield’s theorem, constructive homological algebra, module theory, symbolic computation, mathematical physics

1. Introduction 2. Preliminaries

In the present paper, we consider linear functional systems that are gen- erally written as R η = 0, where R ∈ D q×p is a q × p matrix with coefficients in a non-commutative polynomial ring D of functional operators, and η is a vector of unknown functions. We study such systems through technics from module theory, and the goal of the present section is to recall how module homomorphisms and module isomorphisms are defined in terms of matrix operations.

Email addresses: [email protected] (Cyrille Chenavier),

[email protected] (Thomas Cluzeau), [email protected] (Alban Quadrat)

URL: http://researchers.lille.inria.fr/~cchenavi/ (Cyrille Chenavier),

http://www.unilim.fr/pages perso/thomas.cluzeau/ (Thomas Cluzeau),

http://pages.saclay.inria.fr/alban.quadrat (Alban Quadrat)

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We further assume that D is a general noetherian ring, i.e., every left/right ideal of D is finitely generated as a left/right D-module (see Lam (1999);

Rotman (2009)). The matrix R ∈ D q×p induces the left D-homomorphism .R : D 1×q −→ D 1×p

λ 7−→ λ R, and we consider the left D-module

M := D 1×p /(D 1×q R),

which is defined as the cokernel of the map .R. It is the left D-module finitely presented by R and the following exact sequence (see Rotman (2009)) holds:

D 1×q −→ .R D 1×p −→ π M −→ 0, where the D-linear map π ∈ hom D (D 1×p , M ) is defined by:

π : D 1×p −→ M λ 7−→ π(λ),

and π(λ) denotes the residue class of λ ∈ D 1×p in M . The left D-module M can be defined by generators and relations as follows. If {f j } j=1,...,p is the standard basis of D 1×p (i.e., f j ∈ D 1×p is the vector formed by 1 at the j th position and 0 elsewhere), then one can easily prove that {y j := π(f j )} j=1,...,p

is a family of generators of M which satisfies the left D-linear relations:

∀ i = 1, . . . , q,

p

X

j=1

R ij y j = 0.

For more details, see Chyzak et al. (2005); Cluzeau and Quadrat (2008);

Quadrat (2010).

2.1. Homomorphisms of finitely presented left D-modules

We recall the characterization of left D-homomorphisms of finitely pre- sented left D-modules, see, e.g., Rotman (2009); Cluzeau and Quadrat (2008) and references therein.

Lemma 2.1 (Rotman (2009); Cluzeau and Quadrat (2008)). Let us consider the following finite presentations of the left D-modules M and M 0

D 1×q −→ .R D 1×p −→ π M −→ 0,

D 1×q

0

−→ .R

0

D 1×p

0

−→ π

0

M 0 −→ 0.

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1. The existence of f ∈ hom D (M, M 0 ) is equivalent to the existence of P ∈ D p×p

0

and Q ∈ D q×q

0

satisfying:

R P = Q R 0 .

Then, the following commutative exact diagram (see Rotman (2009)) D 1×q −→ .R D 1×p −→ π M −→ 0

↓ .Q ↓ .P ↓ f

D 1×q

0

−→ .R

0

D 1×p

0

−→ π

0

M 0 −→ 0 holds, where f ∈ hom D (M, M 0 ) is defined by:

∀ λ ∈ D 1×p , f (π(λ)) = π 0 (λ P ).

2. Let R 0 2 ∈ D q

02

×q

0

be such that ker D (.R 0 ) = D 1×q

20

R 2 0 and let P ∈ D p×p

0

and Q ∈ D q×q

0

be two matrices satisfying R P = Q R 0 . Then, the matrices defined by

P = P + Z R 0 , Q = Q + R Z + Z 2 R 0 2 ,

where Z ∈ D p×q

0

and Z 2 ∈ D q×q

02

are two arbitrary matrices, satisfy R P = Q R 0 and:

∀ λ ∈ D 1×p , f (π(λ)) = π 0 (λ P ) = π 0 (λ P ).

Let f : M −→ M 0 be a left D-homomorphism of left D-modules. Then, we can define the kernel, image, coimage and cokernel of f as the following left D-modules:

 

 

 

 

ker f := {m ∈ M | f(m) = 0},

im f := {m 0 ∈ M 0 | ∃ m ∈ M : m 0 = f (m)}, coim f := M/ ker f,

coker f := M 0 /im f.

Let us explicitly characterize the kernel, image, coimage and cokernel of f ∈ hom D (M, M 0 ) when M and M 0 are two finitely presented left D-modules.

Lemma 2.2 (Cluzeau and Quadrat (2008)). Let M := D 1×p /(D 1×q R) (resp.,

M 0 := D 1×p

0

/(D 1×q

0

R 0 )) be a left D-module finitely presented by R ∈ D q×p

(resp., R 0 ∈ D q

0

×p

0

). Moreover, let f ∈ hom D (M, M 0 ) be defined by P ∈ D p×p

0

and Q ∈ D q×q

0

satisfying R P = Q R 0 .

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1. Let S ∈ D r×p and T ∈ D r×q

0

be such that ker D

.

P R 0

= D 1×r (S − T ),

L ∈ D q×r a matrix satisfying R = L S and a matrix S 2 ∈ D r

2

×r such that ker D (.S) = D 1×r

2

S 2 . Then:

ker f = (D 1×r S)/(D 1×q R) ∼ = D 1×r /

D 1×(q+r

2

) L

S 2

.

2. With the above notations, we have:

coim f = D 1×p /(D 1×r S) ∼ = im f =

D 1×(p+q

0

) P

R 0

/(D 1×q

0

R 0 ).

3. coker f = D 1×p

0

/

D 1×(p+q

0

) P T R 0T T

, and thus the left D-module coker f admits the following beginning of a finite free resolution (see Rotman (2009)):

D 1×r . − −−−−− (S −T) D 1×(p+q

0

) .(P

T

R

0T

)

T

−−−−−−→ D 1×p

0

−→ coker f −→ 0. (1) 4. The following commutative exact diagram

0

D 1×r −→ .S D 1×p −→ κ coim f −→ 0

↓ .T ↓ .P ↓ f

]

D 1×q

0

.R

0

−→ D 1×p

0

π

0

−→ M 0 −→ 0

↓ coker f

↓ 0 holds, where f ] : coim f −→ M 0 is defined by:

∀ λ ∈ D 1×p , f ] (κ(λ)) := π 0 (λ P ).

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2.2. Isomorphisms of finitely presented left D-modules

Throughout this paper, we will consider isomorphisms between finitely presented left D-modules. In this subsection, we recall a result of Cluzeau and Quadrat (2008) which allows us to decide when f ∈ hom D (M, M 0 ) is zero, injective, surjective or defines an isomorphism.

Lemma 2.3 (Cluzeau and Quadrat (2008)). With the notations of Lemma 2.2, the left D-homomorphism f : M −→ M 0 is:

1. The zero homomorphism, i.e., f = 0, if and only if one of the following equivalent conditions holds:

(a) There exists Z ∈ D p×q

0

such that P = Z R 0 . Then, there exists a matrix Z 0 ∈ D q×q

02

such that Q = R Z + Z 0 R 0 2 , where R 2 0 ∈ D q

20

×q

0

is any matrix satisfying ker D (.R 0 ) = D 1×q

20

R 0 2 .

(b) The matrix S admits a left inverse, i.e., there exists X ∈ D p×r such that X S = I p .

2. Injective, i.e., ker f = 0, if and only if one of the following equivalent conditions holds:

(a) There exists F ∈ D r×q such that S = F R. Then, if ρ : M → coim f is the canonical projection onto coim f , then we have the following commutative exact diagram:

0 0

↑ ↑

D 1×q −→ .R D 1×p −→ π M −→ 0

↑ .F k ↑ ρ

−1

D 1×r −→ .S D 1×p −→ κ coim f −→ 0

↑ ↑

0 0

(b) The matrix (L T S 2 T ) T admits a left inverse.

3. Surjective, i.e., im f = M 0 , if and only if (P T R 0T ) T admits a left

inverse. Then, the long exact sequence (1) splits (see Rotman (2009)),

i.e., there exist (X Y ) ∈ D p

0

×(p+q

0

) and (U T V T ) T ∈ D (p+q

0

)×r ,

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where X ∈ D p

0

×p , Y ∈ D p

0

×q

0

, U ∈ D p×r and V ∈ D q

0

×r , such that the following identities hold:

 

 

 

 

 

 

X P + Y R 0 = I p

0

, P X + U S = I p , P Y − U T = 0, R 0 X + V S = 0, R 0 Y − V T = I q

0

.

In this case, we have the following commutative exact diagram:

0

D 1×r −→ .S D 1×p −→ κ coim f −→ 0

↑ . − V ↑ .X ↑ f

]−1

D 1×q

0

−→ .R

0

D 1×p

0

−→ π

0

M 0 −→ 0

↑ 0

4. An isomorphism, i.e., M ∼ = M 0 , if and only if the matrices (L T S 2 T ) T and (P T R 0T ) T admit a left inverse. The inverse f −1 of f is then defined by:

∀ λ 0 ∈ D 1×p

0

, f −100 )) := π(λ 0 X),

where X ∈ D p

0

×p is defined in 3. Moreover, we have the following commutative exact diagram:

D 1×q −→ .R D 1×p −→ π M −→ 0

↑ . − V F ↑ .X ↑ f

−1

D 1×q

0

.R

0

−→ D 1×p

0

π

0

−→ M 0 −→ 0

We can now characterize the inverse of an isomorphism which will be useful in the sequel.

Proposition 2.1. Let R ∈ D q×p , R 0 ∈ D q

0

×p

0

and

f : M := D 1×p /(D 1×q R) −→ M 0 := D 1×p

0

/(D 1×q

0

R 0 ) π(λ) 7−→ π 0 (λ P )

be a left D-isomorphism, where P ∈ D p×p

0

is a matrix such that R P = Q R 0

for a certain matrix Q ∈ D q×q

0

.

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1. f admits a right inverse g ∈ hom D (M 0 , M ), namely f ◦ g = id M

0

, i.e., M ∼ = ker f ⊕ M 0 , if and only if there exist three matrices P 0 ∈ D p

0

×p , Q 0 ∈ D q

0

×q and Z 0 ∈ D p

0

×q

0

satisfying the following relations:

R 0 P 0 = Q 0 R, P 0 P + Z 0 R 0 = I p

0

.

Then, there exists Z 2 0 ∈ D q

0

×r

0

satisfying Q 0 Q + R 0 Z 0 + Z 2 0 R 0 2 = I q

0

, where R 0 2 ∈ D r

0

×q

0

is a matrix such that ker D (.R 0 ) = D 1×r

0

R 0 2 .

2. f admits a left inverse g ∈ hom D (M 0 , M ), namely g ◦ f = id M , i.e., M 0 ∼ = M ⊕ coker f, if and only if there exist three matrices P 0 ∈ D p

0

×p , Q 0 ∈ D q

0

×q and Z ∈ D p×q satisfying the following relations:

R 0 P 0 = Q 0 R, P P 0 + Z R = I p .

Then, there exists Z 2 ∈ D q×r satisfying Q Q 0 + R Z + Z 2 R 2 = I q , where R 2 ∈ D r×q is a matrix such that ker D (.R) = D 1×r R 2 .

3. f is a left D-isomorphism, i.e., f ∈ iso D (M, M 0 ), if and only if there exist P 0 ∈ D p

0

×p , Q 0 ∈ D q

0

×q , Z ∈ D p×q and Z 0 ∈ D p

0

×q

0

satisfying the following relations:

R 0 P 0 = Q 0 R, P P 0 + Z R = I p , P 0 P + Z 0 R 0 = I p

0

. (2) Then, there exist Z 2 ∈ D q×r and Z 2 0 ∈ D q

0

×r

0

satisfying the following relations

Q Q 0 + R Z + Z 2 R 2 = I q , Q 0 Q + R 0 Z 0 + Z 2 0 R 0 2 = I q

0

, (3) where R 2 ∈ D r×q (resp., R 0 2 ∈ D r

0

×q

0

) is such that ker D (.R) = D 1×r R 2 (resp., ker D (.R 0 ) = D 1×r

0

R 2 0 ).

Proof. 1. The existence of g ∈ hom D (M 0 , M ) is equivalent to the existence of two matrices P 0 ∈ D p

0

×p and Q 0 ∈ D q

0

×q such that R 0 P 0 = Q 0 R (see 1 of Lemma 2.1). Composing the following two commutative exact diagrams

D 1×q −→ .R D 1×p −→ π M −→ 0

↓ .Q ↓ .P ↓ f

D 1×q

0

−→ .R

0

D 1×p

0

−→ π

0

M 0 −→ 0

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D 1×q −→ .R D 1×p −→ π M −→ 0

↑ .Q

0

↑ .P

0

↑ g

D 1×q

0

.R

0

−→ D 1×p

0

π

0

−→ M 0 −→ 0

and denoting by χ = id M

0

− f ◦ g, we obtain the following commutative exact diagram:

D 1×q

0

−→ .R

0

D 1×p

0

−→ π

0

M 0 −→ 0

↑ .(I

q0

− Q

0

Q) ↑ .(I

p0

− P

0

P ) ↑ χ

D 1×q

0

.R

0

−→ D 1×p

0

π

0

−→ M 0 −→ 0

By 1.a of Lemma 2.3, χ = 0 if and only if there exists Z 0 ∈ D p

0

×q

0

such that I p

0

− P 0 P = Z 0 R 0 , i.e., P 0 P + Z 0 R 0 = I p

0

, which proves the result since the following short exact sequence

0 −→ ker f −→ M −→ f M 0 −→ 0

←− g

then splits (see Rotman (2009)), namely, M ∼ = ker f ⊕ M 0 . According to 1.a of Lemma 2.3, there exists a matrix Z 2 0 ∈ D q

0

×r

0

satisfying the relation I q

0

− Q 0 Q = R 0 Z 0 + Z 2 0 R 0 2 , i.e., Q 0 Q + R 0 Z 0 + Z 2 0 R 2 0 = I q

0

, where R 0 2 ∈ D r

0

×q

0

is such that ker D (.R 0 ) = D 1×r

0

R 0 2 .

2. Repeating the same arguments as in 1 with the left D-homomorphism δ = id M − g ◦ f , we obtain the following commutative exact diagram:

D 1×q −→ .R D 1×p −→ π M −→ 0

↓ .(I

q

− Q Q

0

) ↓ .(I

p

− P P

0

) ↓ δ

D 1×q −→ .R D 1×p −→ π M −→ 0

By 1.a of Lemma 2.3, δ = 0 if and only if there exists Z ∈ D p×q such that I p − P P 0 = Z R, i.e., P P 0 + Z R = I p , which proves the result because the following short exact sequence

0 −→ M −→ f M 0 −→ coker f −→ 0

←− g

then splits, i.e., M 0 ∼ = M ⊕ coker f. Finally, using 1.a of Lemma 2.3, there exists Z 2 ∈ D q×r satisfying the relation I q −Q Q 0 = R Z +Z 2 R 2 , i.e., we have Q Q 0 + R Z + Z 2 R 2 = I q , where R 2 ∈ D r×q is such that ker D (.R) = D 1×r R 2 .

3. This is a direct consequence of 1 and 2.

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Example 2.1. Let D := Q [σ i , σ j ] be the commutative polynomial ring in the two forward shift operators σ i and σ j defined by σ i f(i, j) = f(i + 1, j) and σ j f (i, j) = f(i, j + 1). We consider the linear systems defined by the matrices

R := σ i σ j + 2 σ i − σ j − 7 −3 σ i + 1 −3 σ j − 10

−σ i + σ j + 2 σ i σ j + σ i − σ j − 2 σ j + 4

!

∈ D 2×3 , and

R 0 :=

σ i − 1 0 −1 −1 −3

1 σ i − 1 0 −1 1

−2 −1 σ j + 3 −1 −1

0 −1 −1 σ j 0

∈ D 4×5 .

Note that the system defined by R (resp., R 0 ) is a so-called generalized Fornasini- Marchesini Fornasini and Marchesini (1978) (resp., Roesser Roesser (1975)) model. Using the package OreMorphisms Cluzeau and Quadrat (2009), we can compute the morphisms from the D-module M := D 1×3 /(D 1×2 R) to the D-module M 0 := D 1×5 /(D 1×4 R 0 ). In particular, we find that the matrices

P :=

1 0 0 0 0 0 1 0 0 0 0 0 0 0 1

∈ D 3×5 , Q := σ j + 2 −3 1 1

−1 σ j + 1 0 1

!

∈ D 2×4 ,

satisfy R P = Q R 0 and thus define a morphism f : M → M 0 by f (π(λ)) :=

π 0 (λ P ) for all λ ∈ D 1×3 , where π and π 0 are the canonical projections onto M and M 0 . Let us prove that f is an isomorphism, i.e., M ∼ = M 0 , or in other words, that the linear systems defined by R and R 0 are equivalent. With the notations of Lemmas 2.2 and 2.3, we have:

S = −σ i + σ j + 2 σ i σ j + σ i − σ j − 2 σ j + 4 σ i σ j + 2 σ i − σ j − 7 −3 σ i + 1 −3 σ j − 10

!

∈ D 2×3 , and

L = F = 0 1 1 0

!

∈ D 2×2 ,

with satisfy R = L S and S = F R which proves that f is injective. This

can also be seen by noting that S has full row rank, i.e., S 2 = 0 and L is

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a symmetric and orthogonal matrix, i.e., L L = I 2 . Moreover the matrix (P T R 0T ) T admits the left inverse

1 0 0 0 0 0 0

0 1 0 0 0 0 0

σ i − 2 −σ i + 1 −4 −1 1 0 0 1 σ i − 1 1 0 −1 0 0

0 0 1 0 0 0 0

 ,

which proves that f is surjective and thus an isomorphism from M to M 0 . In particular, for all D-module F, we have ker F (R.) ∼ = ker F (R 0 .) that is, there exists a one-to-one correspondance between the F-solutions of R η = 0 and those of R 0 η 0 = 0. Let us now compute the matrices appearing in 2. of Proposition 2.1. We get:

P 0 =

1 0 0

0 1 0

σ i − 2 −σ i + 1 −4 1 σ i − 1 1

0 0 1

∈ D 5×3 , Q 0 =

 0 0 0 0 1 −1 0 1

∈ D 4×2 ,

which satisfy R 0 P 0 = Q 0 R and define the inverse morphism f −1 : M 0 → M by f −100 )) := π(λ 0 P 0 ) for all λ 0 ∈ D 1×5 . We also find the matrices

Z =

 0 0 0 0 0 0

∈ D 3×2 , Z 0 =

0 0 0 0

0 0 0 0

−1 1 0 0 0 −1 0 0

0 0 0 0

∈ D 5×4 ,

which satisfy the equations (2) and (3) with R 2 = R 0 2 = 0. Note finally that

the one-to-one correspondance between the F -solutions of R η = 0 and those

of R 0 η 0 = 0 is the following: a solution η ∈ F 3 of R η = 0 is sent to the

solution η 0 = P 0 η ∈ F 5 and conversely, a solution η 0 ∈ F 5 of R 0 η 0 = 0 is

sent to the solution η = P η 0 ∈ F 3 .

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3. A constructive version of Fitting’s theorem

In this section, we constructively study the relations between isomor- phisms of finitely presented left D-modules and equivalences of their presen- tation matrices. It is known that the presentation matrices of two isomorphic finitely presented left D-modules are not necessarily equivalent. Note that in general they do not need to have the same row and column dimensions.

However, Fitting’s theorem (see Fitting (1936) and Cox et al. (2005); Kunz (1985) for a modern formulation) asserts that the presentation matrices of isomorphic finitely presented left D-modules can be inflated by blocks of 0 and I so that the resulting matrices are presentation matrices of the same D-modules and are equivalent. We will now provide a constructive version of this theorem of Fitting and give some useful consequences.

We first prove the following theorem which yields a constructive version of Fitting’s theorem in the case where the left D-modules are finitely presented by full row rank matrices.

Theorem 3.1. Let R ∈ D q×p , R 0 ∈ D q

0

×p

0

be two matrices and f : M := D 1×p /(D 1×q R) −→ M 0 := D 1×p

0

/(D 1×q

0

R 0 )

π(λ) 7−→ π 0 (λ P )

be a left D-isomorphism, where P ∈ D p×p

0

is a matrix such that R P = Q R 0 for a certain matrix Q ∈ D q×q

0

. Then, there exist four matrices P 0 ∈ D p

0

×p , Q 0 ∈ D q

0

×q , Z ∈ D p×q and Z 0 ∈ D p

0

×q

0

such that we have R 0 P 0 = Q 0 R, I p = P P 0 + Z R, and I p

0

= P 0 P + Z 0 R 0 . Moreover, the following results hold:

1. We have U :=

I p P

−P 0 I p

0

− P 0 P

∈ GL p+p

0

(D), U −1 =

I p − P P 0 −P P 0 I p

0

. 2. The following commutative exact diagram holds

0 0

↓ ↓

D 1×(q+p

0

) −−−−−−→ .diag(R,I

p0

) D 1×(p+p

0

)π −−− 0

p0

M −→ 0

↓ .V ↓ .U ↓ f

D 1×(p+q

0

) .diag(I

p

,R

0

)

−−−−−−→ D 1×(p+p

0

) 0

p

π

0

− −−− → M 0 −→ 0

↓ ↓

0 0

(4)

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with the following notations V :=

R Q

−P 0 Z 0

∈ D (q+p

0

)×(p+q

0

) ,

diag(R, I p

0

) :=

R 0 0 I p

0

, diag(I p , R 0 ) :=

I p 0 0 R 0

, D 1×(p+p

0

)π −−− 0

p0

M

(λ λ 0 ) 7−→ π(λ),

D 1×(p

0

+p) 0

p

π

0

− −−− → M 0

(λ λ 0 ) 7−→ π 00 ). (5) 3. If the matrices R and R 0 have full row rank, i.e., ker D (.R) = 0 and ker D (.R 0 ) = 0, then q + p 0 = p + q 0 and the matrix V defined above is unimodular, i.e.:

V ∈ GL q+p

0

(D), V −1 =

Z −P Q 0 R 0

. Finally, diag(R, I p

0

) and diag(I p , R 0 ) are equivalent:

diag(I p , R 0 ) = V −1 diag(R, I p

0

) U. (6) Proof. Since f ∈ iso D (M, M 0 ) is defined by two matrices P ∈ D p×p

0

and Q ∈ D q×q

0

such that R P = Q R 0 , using 3 of Proposition 2.1, the existence of the inverse f −1 of f is equivalent to the existence of four matrices P 0 ∈ D p

0

×p , Q 0 ∈ D q

0

×q , Z ∈ D p×q and Z 0 ∈ D p

0

×q

0

such that we have R 0 P 0 = Q 0 R, I p = P P 0 + Z R, and I p

0

= P 0 P + Z 0 R 0 . Moreover, we have

diag(R, I p

0

) U =

R 0 0 I p

0

I p P

−P 0 I p

0

− P 0 P

=

R R P

−P 0 I p

0

− P 0 P

,

V diag(I p , R 0 ) =

R Q

−P 0 Z 0

I p 0 0 R 0

=

R Q R 0

−P 0 Z 0 R 0

, which yields diag(R, I p

0

) U = V diag(I p , R 0 ) by the above relations. More- over, we can check U ∈ GL p+p

0

(D) with the inverse U −1 given by the formula in 1., and

D 1×(p+p

0

) /(D 1×(q+p

0

) diag(R, I p

0

))

∼ = [D 1×p /(D 1×q R)] ⊕ [D 1×p

0

/(D 1×p

0

I p

0

)] = D 1×p /(D 1×q R), D 1×(p+p

0

) /(D 1×(p+q

0

) diag(I p , R 0 ))

∼ = [D 1×p /(D 1×p I p )] ⊕ [D 1×p

0

/(D 1×q

0

R 0 )] = D 1×p

0

/(D 1×q

0

R 0 ),

(14)

which proves 1 and 2. Then, we can easily check that

Z 0 Q 0 R = Z 0 R 0 P 0 = (I p

0

− P 0 P ) P 0 = P 0 (I p − P P 0 ) = P 0 Z R,

i.e., (Z 0 Q 0 − P 0 Z) R = 0, which yields Z 0 Q 0 = P 0 Z when R has full row rank. Using 3 of Proposition 2.1, we get that the identities Q Q 0 + R Z = I q and Q 0 Q + R 0 Z 0 = I q

0

hold, and thus

R Q

−P 0 Z 0

Z −P Q 0 R 0

=

R Z + Q Q 0 −R P + Q R 0

−P 0 Z + Z 0 Q 0 P 0 P + Z 0 R 0

= I q+p

0

. If R 0 has full row-rank, a similar computation shows that

Z −P Q 0 R 0

R Q

−P 0 Z 0

=

Z R + P P 0 Z Q − P Z 0 Q 0 R − R 0 P 0 Q 0 Q + R 0 Z 0

= I p+q

0

, which yields q + p 0 = q 0 + p, V ∈ GL q+p

0

(D) and (6).

Example 3.1. Let us consider again Example 2.1. Using the formula of Theorem 3.1 (which are implemented in the package OreMorphisms Cluzeau and Quadrat (2009)), we obtain:

U =

1 0 0 1 0 0 0 0

0 1 0 0 1 0 0 0

0 0 1 0 0 0 0 1

−1 0 0 0 0 0 0 0

0 −1 0 0 0 0 0 0

−σ i + 2 σ i − 1 4 −σ i + 2 σ i − 1 1 0 4

−1 −σ i + 1 −1 −1 −σ i + 1 0 1 −1

0 0 −1 0 0 0 0 0

∈ GL 8 (D),

(15)

U −1 =

0 0 0 −1 0 0 0 0

0 0 0 0 −1 0 0 0

0 0 0 0 0 0 0 −1

1 0 0 1 0 0 0 0

0 1 0 0 1 0 0 0

σ i − 2 −σ i + 1 −4 0 0 1 0 0

1 σ i − 1 1 0 0 0 1 0

0 0 1 0 0 0 0 1

 ,

and

V =

σ i σ j + 2 σ i − σ j − 7 −3 σ i + 1 −3 σ j − 10 σ j + 2 −3 1 1

−σ i + σ j + 2 σ i σ j + σ i − σ j − 2 σ j + 4 −1 σ j + 1 0 1

−1 0 0 0 0 0 0

0 −1 0 0 0 0 0

−σ i + 2 σ i − 1 4 −1 1 0 0

−1 −σ i + 1 −1 0 −1 0 0

0 0 −1 0 0 0 0

 .

Since R and R 0 are full row rank matrices, the matrix V is unimodular, i.e., V ∈ GL 7 (D) and we have:

V −1 =

0 0 −1 0 0 0 0

0 0 0 −1 0 0 0

0 0 0 0 0 0 −1

0 0 σ i − 1 0 −1 −1 −3

0 0 1 σ i − 1 0 −1 1

1 −1 −2 −1 σ j + 3 −1 −1

0 1 0 −1 −1 σ j 0

.

(16)

Finally, we obtain the equivalence of matrices:

1 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0

0 0 1 0 0 0 0 0

0 0 0 σ i − 1 0 −1 −1 −3

0 0 0 1 σ i − 1 0 −1 1

0 0 0 −2 −1 σ j + 3 −1 −1

0 0 0 0 −1 −1 σ j 0

= V −1

σ i σ j + 2 σ i − σ j − 7 −3 σ i + 1 −3 σ j − 10 0 0 0 0 0

−σ i + σ j + 2 σ i σ j + σ i − σ j − 2 σ j + 4 0 0 0 0 0

0 0 0 1 0 0 0 0

0 0 0 0 1 0 0 0

0 0 0 0 0 1 0 0

0 0 0 0 0 0 1 0

0 0 0 0 0 0 0 1

 U.

Example 3.2. Let us consider two linear PD systems used in the theory of linear elasticity, namely the Lie derivative of the euclidean metric of R 2 and its Spencer operator, see Pommaret (2001):

 

 

1 ξ 1 = 0,

1

2 (∂ 2 ξ 1 + ∂ 1 ξ 2 ) = 0,

2 ξ 2 = 0,

 

 

 

 

 

 

 

 

1 ζ 1 = 0,

2 ζ 1 − ζ 2 = 0,

1 ζ 2 = 0,

1 ζ 3 + ζ 2 = 0,

2 ζ 3 = 0,

2 ζ 2 = 0.

Let D := Q [∂ 1 , ∂ 2 ] be the commutative polynomial ring of PD operators in ∂ 1

(17)

and ∂ 2 with rational constant coefficients,

R :=

∂ 1 0

1 2 ∂ 2 1

2 ∂ 1

0 ∂ 2

 ∈ D 3×2 , R 0 :=

∂ 1 ∂ 2 0 0 0 0 0 −1 ∂ 1 1 0 ∂ 2

0 0 0 ∂ 12 0

T

∈ D 6×3 , (7) M := D 1×2 /(D 1×3 R), and M 0 := D 1×3 /(D 1×6 R 0 ). In Example 3.2 of Cluzeau and Quadrat (2008), it is proven that the matrices

P :=

1 0 0 0 0 1

, Q := 1 2

2 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 2 0

 , define f ∈ iso D (M, M 0 ). Applying Theorem 3.1, we get:

U =

1 0 1 0 0

0 1 0 0 1

−1 0 0 0 0

−∂ 2 0 −∂ 2 1 0

0 −1 0 0 0

∈ GL 5 (D). (8)

Moreover, the matrix V defined in Theorem 3.1 has the form:

V =

1 0 1 0 0 0 0 0

1

2 ∂ 2 1 21 0 1 2 0 1 2 0 0

0 ∂ 2 0 0 0 0 1 0

−1 0 0 0 0 0 0 0

−∂ 2 0 0 −1 0 0 0 0

0 −1 0 0 0 0 0 0

∈ D 6×8 .

Then, we have diag(R, I 3 ) U = V diag(I 2 , R 0 ), where the matrix V is not unimodular because it is not a square matrix. The computations can be done using the package OreMorphisms (Cluzeau and Quadrat (2009)) built upon OreModules (Chyzak et al. (2007)).

Consider two matrices R ∈ D q×p and R 0 ∈ D q

0

×p

0

such that the corre-

sponding presented left D-modules M and M 0 are isomorphic, i.e., M ∼ = M 0 .

(18)

Fitting’s theorem asserts that under these hypotheses, R and R 0 can be en- larged by blocks of 0 and I to get equivalent matrices: letting

n = q + p 0 + p + q 0 , m = p + p 0 the two matrices

L :=

 R 0

0 I p

0

0 0 0 0

∈ D n×m , L 0 :=

 0 0 0 0 I p 0 0 R 0

∈ D n×m

define finite presentations of M and M 0 , i.e., M = D 1×m /(D 1×n L) and M 0 = D 1×m /(D 1×n L 0 ), and are equivalent, i.e., there exist X ∈ GL m (D) and Y ∈ GL n (D) satisfying L 0 = Y −1 L X. We recover this result by giving a completely constructive proof in the sense that the matrices X and Y are explicitly given in terms of the matrices defining the isomorphism M ∼ = M 0 . Theorem 3.2. Let R ∈ D q×p , R 0 ∈ D q

0

×p

0

be two matrices and

f : M := D 1×p /(D 1×q R) −→ M 0 := D 1×p

0

/(D 1×q

0

R 0 ) π(λ) 7−→ π 0 (λ P ),

be a left D-isomorphism, where P ∈ D p×p

0

is a matrix such that R P = Q R 0 for a certain matrix Q ∈ D q×q

0

. Moreover, let R 2 ∈ D r×q (resp., R 0 2 ∈ D r

0

×q

0

) be a matrix such that ker D (.R) = D 1×r R 2 (resp., ker D (.R 0 ) = D 1×r

0

R 2 0 ).

Then, there exist matrices P 0 ∈ D p

0

×p , Q 0 ∈ D q

0

×q , Z ∈ D p×q , Z 0 ∈ D p

0

×q

0

, Z 2 ∈ D q×r and Z 2 0 ∈ D q

0

×r

0

satisfying (2) and (3), and such that the following results hold:

1. With the notations

n := q + p 0 + p + q 0 , m := p + p 0 we have:

X :=

I p P

−P 0 I p

0

− P 0 P

∈ GL m (D), X −1 =

I p − P P 0 −P P 0 I p

0

,

Y :=

I q 0 R Q

0 I p

0

−P 0 Z 0

−Z P 0 P Z 0 − Z Q

−Q 0 −R 0 0 Z 2 0 R 0 2

∈ GL n (D),

(19)

Y −1 =

Z 2 R 2 0 −R −Q P 0 Z − Z 0 Q 0 0 P 0 −Z 0

Z −P I p 0

Q 0 R 0 0 I q

0

. (9)

2. The following commutative exact diagram holds

0 0 0

↓ ↓ ↓

D 1×n −→ .L D 1×mπ −−− 0

p0

M −→ 0

↓ .Y ↓ .X ↓ f

D 1×n .L

0

−→ D 1×m 0

p

π

0

− −−− → M 0 −→ 0

↓ ↓ ↓

0 0 0

(10)

where π ⊕ 0 p

0

and 0 p ⊕ π 0 are defined by (5), and

L :=

 R 0

0 I p

0

0 0

0 0

∈ D n×m , L 0 :=

 0 0 0 0 I p 0 0 R 0

∈ D n×m ,

i.e., we have L X = Y L 0 , and thus:

L 0 = Y −1 L X ⇐⇒ L = Y L 0 X −1 , namely, the matrices L and L 0 are equivalent.

Proof. Since f ∈ iso D (M, M 0 ) is defined by the matrices P ∈ D p×p

0

and Q ∈ D q×q

0

satisfying R P = Q R 0 , then, according to 3 of Proposition 2.1, the existence of f −1 is equivalent to the existence of four matrices P 0 ∈ D p

0

×p , Q 0 ∈ D q

0

×q , Z ∈ D p×q and Z 0 ∈ D p

0

×q

0

satisfying:

R P = Q R 0 , R 0 P 0 = Q 0 R,

P P 0 + Z R = I p , P 0 P + Z 0 R 0 = I p

0

.

The fact that X belongs to GL m (D) can be proved as in Theorem 3.1. More- over, 3 of Proposition 2.1 shows that two matrices Z 2 ∈ D q×r and Z 2 0 ∈ D q

0

×r

0

exist such that:

Q Q 0 + R Z + Z 2 R 2 = I q

Q 0 Q + R 0 Z 0 + Z 2 0 R 2 0 = I q

0

.

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Using the above relations, we can prove the following identities:

Z 2 R 2 +R Z +Q Q 0 = I q , −R P +Q R 0 = 0, P 0 P +Z 0 R 0 = I p

0

, Z R+P P 0 = I p , Z 2 0 R 0 2 R 0 = 0, Q 0 R − R 0 P 0 = 0, Q 0 Q + R 0 Z 0 + Z 2 0 R 0 2 = I q

0

,

(P Z 0 −Z Q) R 0 = P Z 0 R 0 −Z Q R 0 = P Z 0 R 0 −Z R P = P (I p

0

−P 0 P )−(I p −P P 0 ) P = 0,

−Z Z 2 R 2 + P (P 0 Z − Z 0 Q 0 ) + (P Z 0 − Z Q) Q 0 = −Z Z 2 R 2 + (P P 0 ) Z − Z (Q Q 0 )

= −Z Z 2 R 2 + (I p − Z R) Z − Z (I q − R Z − Z 2 R 2 ) = 0,

−Q 0 Z 2 R 2 − R 0 (P 0 Z − Z 0 Q 0 ) + Z 2 0 R 2 0 Q 0 = −Q 0 (Z 2 R 2 ) − (R 0 P 0 ) Z + R 0 Z 0 Q 0 + (Z 2 0 R 0 2 ) Q 0

= −Q 0 (I q − Q Q 0 − R Z ) − Q 0 R Z + R 0 Z 0 Q 0 + (I q

0

− Q 0 Q − R 0 Z 0 ) Q 0 = 0.

We can then check that if we note K the matrix defined in the right-hand side of (9), we have Y K = I n . Similarly, we can easily check that K Y = I n , which proves that K = Y −1 .

Moreover, we can easily check that:

 R 0

0 I p

0

0 0

0 0

I p P

−P 0 I p

0

− P 0 P

=

R R P

−P 0 I p

0

− P 0 P

0 0

0 0

 .

Since (P Z 0 − Z Q) R 0 = 0, we then have

I q 0 R Q

0 I p

0

−P 0 Z 0

−Z P 0 P Z 0 − Z Q

−Q 0 −R 0 0 Z 2 0 R 0 2

0 0 0 0 I p 0 0 R 0

=

R R 0 Q

−P 0 Z 0 R 0 0 (P Z 0 − Z Q) R 0 0 Z 2 0 R 0 2 R 0

=

R R P

−P 0 I p

0

− P 0 P

0 0

0 0

 ,

which yields L X = Y L 0 . Finally, using Theorem 3.1, we get:

D 1×m /(D 1×n L) = D 1×m /(D 1×(q+p

0

) diag(R, I p

0

)) ∼ = D 1×p /(D 1×q R) = M,

D 1×m /(D 1×n L 0 ) = D 1×m /(D 1×(p+q

0

) diag(I p , R 0 )) ∼ = D 1×p

0

/(D 1×q

0

R 0 ) = M 0 .

(21)

The matrices X, X −1 , Y and Y −1 can be computed when the pairs of matrices (P, Q) and (P 0 , Q 0 ) respectively defining f and f −1 are known. The computation of the matrices X and Y and their inverses has been imple- mented in the OreMorphisms package, Cluzeau and Quadrat (2009).

Example 3.3. We consider again Example 3.2. With the notations of The- orem 3.2, the matrix X := U ∈ GL 5 (D) is defined by (8), the matrix Y ∈ GL 14 (D) is defined by

Y :=

1 0 0 0 0 0 ∂ 1 0 1 0 0 0 0 0

0 1 0 0 0 0 1 22 1 21 0 1 2 0 1 2 0 0

0 0 1 0 0 0 0 ∂ 2 0 0 0 0 1 0

0 0 0 1 0 0 −1 0 0 0 0 0 0 0

0 0 0 0 1 0 −∂ 2 0 0 −1 0 0 0 0

0 0 0 0 0 1 0 −1 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0 0 0 0

−1 0 0 −∂ 1 0 0 0 0 0 0 0 0 0 0

0 0 0 −∂ 2 1 0 0 0 0 0 0 0 0 0

−∂ 2 0 0 0 −∂ 1 0 0 0 −∂ 2 ∂ 1 1 0 0 0

0 −2 0 0 −1 −∂ 1 0 0 0 0 0 0 0 0

0 0 −1 0 0 −∂ 2 0 0 0 0 0 0 0 0

0 −2 ∂ 21 0 −∂ 2 0 0 0 0 0 0 −∂ 21 1

,

(22)

and its inverse is defined by:

Y −1 =

0 0 0 0 0 0 −∂ 1 0 −1 0 0 0 0 0

0 0 0 0 0 0 − 1 221 21 0 − 1 2 0 − 1 2 0 0

0 0 0 0 0 0 0 −∂ 2 0 0 0 0 −1 0

0 0 0 0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 −∂ 1 0 0 0 −1 0 0

0 0 0 0 0 0 0 1 0 0 0 0 0 0

0 0 0 −1 0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 −1 0 1 0 0 0 0 0 0

1 0 0 ∂ 1 0 0 0 0 1 0 0 0 0 0

0 2 0 ∂ 2 −1 0 0 0 0 1 0 0 0 0

∂ 2 −2 ∂ 1 0 0 ∂ 1 0 0 0 0 0 1 0 0 0

0 0 0 0 1 ∂ 1 0 0 0 0 0 1 0 0

0 0 1 0 0 ∂ 2 0 0 0 0 0 0 1 0

0 0 −∂ 1 0 ∂ 2 0 0 0 0 0 0 0 0 1

 .

Then the matrix L := (diag(R, I 3 ) T 0 T ) T ∈ D 14×5 is equivalent to the matrix L 0 := (0 T diag(I 2 , R 0 ) T ) ∈ D 14×5 , namely:

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

1 0 0 0 0

0 1 0 0 0

0 0 ∂ 1 0 0 0 0 ∂ 2 −1 0 0 0 0 ∂ 1 0 0 0 0 1 ∂ 1 0 0 0 0 ∂ 2

0 0 0 ∂ 2 0

= Y −1

∂ 1 0 0 0 0

1

2 ∂ 2 1 21 0 0 0 0 ∂ 2 0 0 0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

X. (11)

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Using Theorem 3.2, we find again Schanuel’s lemma (see Rotman (2009)):

ker D (.L) = ker D (.R) ⊕ D 1×(p+q

0

) ∼ = ker D (.L 0 ) = ker D (.R 0 ) ⊕ D 1×(q+p

0

) . 4. A constructive version of Warfield’s theorem

As illustrated by the examples given in the previous section, the size of the equivalent matrices obtained through the constructive versions of Fitting’s theorem given by Theorems 3.1 and 3.2 is generally large. However, a result due to Warfield (see Warfield (1978)) asserts that some zero and identity blocks may be removed from these matrices to get new equivalent matrices presenting the same finitely presented left D-modules. Warfield’s result is based on the properties of an algebraic invariant of the ring D called stable rank. In this section, we aim at providing constructive versions of Warfield’s result (see Theorems 4.1 and 4.2). In fact, we will obtain a slightly more general result than Warfield’s theorem: we show that zero and identity blocks maybe removed as soon as the equation given in (12) admits a solution, which is in particular the case when hypotheses of Warfield’s theorem are fulfilled.

We discuss this in Remark 4.1 and Examples 4.4 and 4.6.

4.1. Stable ranges and stable rank

In Lemma 4.1, we give a sufficient condition in terms of the stable rank of D so that Equation (12) admits a solution. Computing this solution is a hard task in general, but once it is known, other constructions of the section are purely algorithmic. We discuss this problem through the section.

We first recall some definitions and some properties of the stable rank of a ring as it will be useful in what follows. We refer to (McConnell and Robson, 2001, Section 11.3.3) for more details concerning stable ranks.

Definition 4.1. Let D be a ring.

1. A column vector u ∈ D n is said to be unimodular if it admits a left inverse, i.e., there exists a row vector v ∈ D 1×n such that v u = 1.

2. A column vector u = (u 1 , . . . , u n ) T ∈ D n is said to be stable if there ex- ist d 1 , . . . , d n−1 ∈ D such that the column vector (u 1 +d 1 u n , . . . , u n−1 + d n−1 u n ) T ∈ D n−1 is unimodular.

3. An integer r is said to be in the stable range of D if for all integer

n > r, a unimodular vector u ∈ D n is stable.

Références

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