REAL DIFFERENCE GALOIS THEORY.
THOMAS DREYFUS

arXiv:1507.02192v3 [] 13 Feb 2017

Abstract. In this paper, we develop a diﬀerence Galois theory in the setting of real
ﬁelds. After proving the existence and uniqueness of the real Picard-Vessiot extension,
we deﬁne the real diﬀerence Galois group and prove a Galois correspondence.

Contents
Introduction
1. Reminders of difference algebra
2. Existence and uniqueness of Picard-Vessiot extensions over real fields
3. Real difference Galois group
References

1
2
3
10
12

Introduction
Let us consider an equation of the form:
(1)

φY = AY,

where A is an invertible matrix having coefficients in a convenient field k∗ and φ is an
automorphism of k. A typical example is k := C(x) and φy(x) := y(x + 1). The aim of
the difference Galois theory is to study (1) from an algebraic point of view. See [vdPS97]
for details on this theory. See also [BB62, Fra63, HS08, Mor09, MU09]. The classical
framework for difference Galois theory is to assume that C, the subfield of k of elements
invariant under φ, is algebraically closed. The goal of the present paper is to present a
descent result. We explain what happens if we take instead a smaller field k, such that k
is a real field and C is real closed, see §2 for the definitions.
Assume that C is algebraically closed and let us make a brief summary of the difference
Galois theory. An important object attached to (1) is the Picard-Vessiot extension.
Roughly speaking, a Picard-Vessiot extension is a ring extension of k containing a basis
of solutions of (1). The Picard-Vessiot extension always exists, but the uniqueness is
proved in [vdPS97] only in the case where C is algebraically closed. To the Picard-Vessiot
extension, we attach a group, the difference Galois group, that measures the algebraic
relations between solutions belonging to the Picard-Vessiot extension. This group may
be seen as a linear algebraic subgroup of invertible matrices in coefficients in C. We
also have a Galois correspondence. Note that several definitions of the difference Galois
group have been made and the comparison between different Galois groups can be found
in [CHS08].
Date: February 14, 2018.
2010 Mathematics Subject Classification. 12D15,39A05.
Work supported by the labex CIMI. This project has received funding from the European Research
Council (ERC) under the European Union’s Horizon 2020 research and innovation programme under the
Grant Agreement No 648132.
∗In all the paper, all ﬁelds are of characteristic zero.
1

2

THOMAS DREYFUS

From now, we drop the assumption that C is algebraically closed, and we make
the assumptions that k is a real field and C is real closed. Our approach will follow
[CHS13, CHvdP16], which prove similar results in the framework of differential Galois
theory. Let us present a rough statement of our main result, Theorem 7. We prove
that in this setting, a real Picard-Vessiot exists, i.e., a Picard-Vessiot extension that is
additionally a real ring. Then, we also show a uniqueness result: given R1 and R2 two real
Picard-Vessiot extensions, then R1 and R2 are isomorphic over k if and only if R1 ⊗k R2
has no elements x satisfying x2 + 1 = 0. We define a real difference Galois group, which
may be seen as a linear algebraic subgroup of invertible matrices in coefficients in the
algebraic closure of C, and that is defined over C. See Proposition 11. This allows us
to prove a Galois correspondence, see Theorem 12. See also [CH15, Dyc05] for similar
results in the framework of differential Galois theory.
The paper is presented as follows. In §1, we make some reminders of difference algebra.
In §2, we state and prove our main result, Theorem 7, about the existence and uniqueness
of real Picard-Vessiot extensions. In §3, we define the real difference Galois group, and
prove a Galois correspondence.
Acknowledgments. The author would like the thank the anonymous referee for permitting him to correct some mistakes that was originally made in the paper.
1. Reminders of difference algebra
For more details on what follows, we refer to [Coh65]. A difference ring (R, φ) is a ring
R together with a ring automorphism φ : R → R. An ideal of R stabilized by φ is called
a difference ideal of (R, φ). A simple difference ring (R, φ) is a difference ring with only
difference ideals (0) and R. If R is a field then (R, φ) is called a difference field.
Let (R, φ) be a difference ring and m ∈ N∗ . The difference ring R{X1 , . . . , Xm }φ of
difference polynomials in m indeterminacies over R is the usual polynomial ring in the
infinite set of variables
{φν (Xj )}ν∈Z
j≤m ,
and with automorphism extending the one on R defined by:
φ (φν (Xj )) = φν+1 (Xj ).
The ring of constants Rφ of the difference ring (R, φ) is defined by
Rφ := {f ∈ R | φ(f ) = f }.

If Rφ is a field, the ring of constants will be called field of constants.
e
e φ)
A difference ring morphism from the difference ring (R, φ) to the difference ring (R,
e
e
is a ring morphism ϕ : R → R such that ϕ ◦ φ = φ ◦ ϕ.
e is a difference ring extension of a difference ring (R, φ) if R
e φ)
e is a
A difference ring (R,
e
e
ring extension of R and φ|R = φ; in this case, we will often denote φ by φ. Two difference
e ) and (R
e ) of a difference ring (R, φ) are isomorphic over (R, φ)
e1, φ
e2 , φ
ring extensions (R
1
2
e1 ) to (R
e2 ) such that ϕ = Id.
e1, φ
e2, φ
if there exists a difference ring isomorphism ϕ from (R
|R

Let (R, φ) be a difference ring such that X 2 + 1 ∈ R[X] is irreducible, i.e., there is no
x ∈ R such that x2 + 1 = 0. We define, R[i], to be the ring R[i] := R[X]/(X 2 + 1). We
equip R[i] with a structure of difference ring with φ(i) = i. If (R, φ) is a difference ring
with an element x ∈ R satisfying x2 + 1 = 0, we make the convention that R[i] = R.

REAL DIFFERENCE GALOIS THEORY.

3

2. Existence and uniqueness of Picard-Vessiot extensions over real fields
Let (k, φ) be a difference field of characteristic zero. Consider a linear difference system
(2)

φY = AY with A ∈ GLn (k),

where GLn denotes the group of invertible n × n square matrices with entries in k.
Definition 1. A Picard-Vessiot extension for (2) over (k, φ) is a difference ring extension
(R, φ) of (k, φ) such that
(1) there exists U ∈ GLn (R) such that φ(U ) = AU (such a U is called a fundamental
matrix of solutions of (2));
(2) R is generated, as a k-algebra, by the entries of U and det(U )−1 ;
(3) (R, φ) is a simple difference ring.
We may always construct a Picard-Vessiot extension as follows. Take an indeterminate
n × n square matrix X := Xj,k and consider k{X, det(X)−1 }φ which is equipped with a
structure of difference ring with φX = AX. Then, for any I, maximal difference ideal of
k{X, det(X)−1 }φ , the ring k{X, det(X)−1 }φ /I is a simple difference ring and therefore, is
a Picard-Vessiot extension.
According to [vdPS97, §1.1], when the field of constants C := kφ is algebraically closed,
we also have the uniqueness of the Picard-Vessiot extension, up to a difference ring isomorphism. Furthermore, in this case we have C = Rφ and, see [vdPS97, Corollary 1.16],
there exist an idempotent e ∈ R, and t ∈ N∗ , such that φt (e) = e, R =
for all 0 ≤ j ≤ t − 1,

φj (e)R

t−1
M

φj (e)R, and

j=0

is an integral domain.

In [CHS08], it is defined the notion of weak Picard-Vessiot extension we will need in
the next section.
Definition 2. A weak Picard-Vessiot extension for (2) over (k, φ) is a difference ring
extension (R, φ) of (k, φ) such that
(1) there exists U ∈ GLn (R) such that φ(U ) = AU ;
(2) R is generated, as a k-algebra, by the entries of U and det(U )−1 ;
(3) Rφ = kφ = C.
From what is above, we deduce that when the field of constants is algebraically closed,
a Picard-Vessiot extension is a weak Picard-Vessiot extension. Note that the converse is
not true as shows [vdPS97, Example 1.25].
We say that a field k is real when 0 is not a sum of squares in k\{0}. We say that a field
k is real closed when k does not admit an algebraic extension that is real. In particular,
k is real closed if and only if k[i] is algebraically closed and satisfies k[i] 6= k.
Example 3. The field R((x)) of formal Laurent series with real coefficients is real. The
field Q(x) is real. The field of real numbers is real closed.
From now we assume that k is a real field and its field of constants C := kφ is real closed.
Remind that we have seen that we have the existence of (R, φ), Picard-Vessiot extension
for (2) over (k, φ).

4

THOMAS DREYFUS

Lemma 4. Let (R, φ), be a Picard-Vessiot extension for (2) over (k, φ) and assume that
R 6= R[i]. Then, (R[i], φ), is a Picard-Vessiot extension for (2) over (k[i], φ).
Proof. Let (0) 6= I be a difference ideal of (R[i], φ). Note that I ∩ R is a difference ideal of
(R, φ). We claim that I ∩ R 6= (0). Let a, b ∈ R with 0 6= a + ib ∈ I. Then, φ(a) + iφ(b) ∈ I
and for all c ∈ R, ac + ibc ∈ I. Let J be the smallest difference ideal of R that contains
a. From what is above, we may deduce that for all a1 ∈ J, there exists b1 ∈ R such that
a1 + ib1 ∈ I. Since (R, φ) is a simple difference ring, we have two possibilities: J = (0)
and J = R. We are going to treat separately the two cases. Assume that J = (0).
Then a = 0 and ib ∈ I. But ib × (−i) = b ∈ I ∩ R \ {0} which proves our claim when
J = (0). Assume that J = R. Then, there exists b1 ∈ R such that 1 + ib1 ∈ I. But
(1 + ib1 )(1 − ib1 ) = 1 + b21 ∈ I ∩ R. Since R 6= R[i] we find that 1 + b21 6= 0 which proves
our claim when J = R.
Since I ∩ R 6= (0) and (R, φ) is a simple difference ring, I ∩ R = R. We now remark
that I is stable by multiplication by k[i], which shows that I = R[i]. This proves the
lemma.

Proposition 5. Let (R, φ), be a Picard-Vessiot extension for (2) over (k, φ). Then, there
exist an idempotent e ∈ R, and t ∈

N∗ ,

such that

φt (e)

= e, R =

t−1
M

φj (e)R, and for all

j=0

0 ≤ j ≤ t − 1, φj (e)R is an integral domain.
Proof. Let us treat separately two cases. Assume that R 6= R[i]. Due to Lemma 4,
(R[i], φ), is a Picard-Vessiot extension for (2) over (k[i], φ). We remind that by definition,
if R 6= R[i], we extend φ to R[i] by φ(i) = i. Then, the field of constants of R[i] is C[i],
which is algebraically closed. From [vdPS97, Corollary 1.16], we obtain that there exist
a, b ∈ R, with a + ib is idempotent, t ∈ N∗ , such that φt (a + ib) = a + ib,
(3)

R[i] =

t−1
M

φj (a + ib)R[i],

j=0

and for all 0 ≤ j ≤ t − 1, φj (a + ib)R[i] is an integral domain. Let e := a2 + b2 ∈ R. A
straightforward computation shows that a − ib is idempotent. Since e = (a + ib)(a − ib) is
the product of two idempotent elements it is also idempotent. Using φt (a − ib) = a − ib,
we find φt (e) = e.
Let us prove that for all 0 ≤ j ≤ t − 1, φj (a − ib)R[i] is an integral domain. Let
0 ≤ j ≤ t − 1, c + id ∈ R[i] with c, d ∈ R, such that φj (a − ib)(c + id) = 0. It follows
that φj (a + ib)(c − id) = 0 and therefore, c − id = 0 = c + id since for all 0 ≤ j ≤ t − 1,
φj (a+ib)R[i] is an integral domain. We have proved that for all 0 ≤ j ≤ t−1, φj (a−ib)R[i]
is an integral domain. Let us prove that for all 0 ≤ j ≤ t − 1, φj (e)R[i] is an integral
domain. Let 0 ≤ j ≤ t − 1, c ∈ R[i], such that cφj (e) = cφj (a + ib)φj (a − ib) = 0. We
use successively the fact that φj (a + ib)R[i] and φj (a − ib)R[i] are integral domains to
deduce that c = 0, which shows that φj (e)R[i] is an integral domain. Therefore, for all
0 ≤ j ≤ t − 1, φj (e)R is an integral domain.
We claim that {φj (e), 0 ≤ j ≤ t − 1} are linearly independent over R[i]. Let us consider
c0 , . . . , ct−1 ∈ R[i] such that

t−1
X

j=0

cj φj (e) = 0. We have

t−1
X

j=0

cj φj (a − ib)φj (a + ib) = 0. We

use (3) to deduce that for all 0 ≤ j ≤ t − 1, cj φj (a − ib) = 0. We remind that for all
0 ≤ j ≤ t − 1, φj (a − ib)R[i] is an integral domain. This shows that for all 0 ≤ j ≤ t − 1,
cj = 0. This proves our claim.
Using (3), to prove the proposition, it is now sufficient to prove the equality

REAL DIFFERENCE GALOIS THEORY.

t−1
M

(4)

j

φ (a + ib)R[i] =

j=0

The inclusion

t−1
M
j=0

α∈

φj (e)R[i].

j=0

φj (e)R[i] ⊂

t−1
M

φj (a + ib)R[i] is a direct consequence of the fact

j=0

that e = (a − ib)(a + ib) ∈ (a + ib)R[i].
t−1
M

t−1
M

5

φj (a + ib)R[i], and define f :=

j=0

t−1
Y

Let us prove the other inclusion.

Let

φj (e) which is invariant under φ. Therefore,

j=0

f R[i] is a difference ideal of R[i]. We use e = (a + ib)(a − ib) and the fact that (a + ib)R[i]
is an integral domain to obtain that f 6= 0 and f R[i] 6= (0). Since R[i] is a simple difference ring, the difference ideal f R[i] equals to R[i]. This means that the ideal of R[i]
generated by the φj (f ), j ∈ Z, is R[i]. Since φ(f ) = f , there exists β ∈ R[i] such that
f β = α. We again use (3) to find that α = f

t−1
X

j=0

cj φj (a + ib) for some cj ∈ R[i]. Since

e = (a−ib)(a+ib), we may define for all 0 ≤ j ≤ t−1, dj := f /φj (a−ib) ∈ R[i]. A straightforward computation shows that α =

t−1
X

j=0

have proved

t−1
M
j=0

φj (a + ib)R[i] ⊂

t−1
M

cj dj φj (e), which implies α ∈

t−1
M

φj (e)R[i]. We

j=0

φj (e)R[i]. If we combine with the other inclusion, we

j=0

obtain (4). This completes the proof in the case R 6= R[i].
Assume that R = R[i]. Since i2 = −1, we have φ(i) = ±i and then φ2 (i) = i. Hence,
(R, φ2 ) is a ring extension of (k[i], φ2 ), whose field of constants is C[i], which is algebraically closed. Furthermore, by construction, it is also a Picard-Vessiot extension for
φ2 Y = φ(A)AY over (k[i], φ2 ). From [vdPS97, Corollary 1.16], we obtain that there exist
an idempotent e ∈ R, t ∈ N∗ , such that φ2t (e) = e,
R=

t−1
M

φ2j (e)R,

j=0

and for all 0 ≤ j ≤ t − 1, φ2j (e)R is an integral domain. If t = 1, R = eR is an
integral domain, and we may take e = 1 to have the desired decomposition of R. Assume that t > 1. Using the fact that φ is an automorphism we find that for all j ∈ Z,
φj (e)R is an integral domain and φj (e) is idempotent. Let t′ ∈ N∗ maximal such that
′
′
eR, . . . , φt −1 (e)R are in direct sum. This implies that there exists r ∈ R with rφt (e) 6= 0,
′

such that rφt (e) ∈

′ −1
tM

j=0

φj (e)R. We claim that rφt (e) ∈ eR. If t′ = 1 the claim is
′

clear. Assume that t′ > 1. Then, for all 0 < j < t′ , we have eφj (e) = 0 and, since φ is
an automorphism,

′
φt (e)φj (e)

= 0. It follows that

′ −1
tM

j=0

′

φj (e)Rφt (e) ⊂ eR and therefore,

′

′

rφt (e) = r(φt (e))2 ∈ eR, which proves the claim in the t′ > 1 case. In particular, there
′
′
exists r ′ ∈ R such that rφt (e) = r ′ e 6= 0. We use the fact that φt is an automorphism,
′
′
′
′
′
′
to find φt (r)φ2t (e) = φt (r ′ )φt (e) 6= 0. Since φt (e)R is an integral domain and φt (e) is
′
′
′
′
′
idempotent, we have φt (r ′ )rφt (e) = φt (r ′ )φt (e)rφt (e) 6= 0. But the latter inequality im′

′

′

plies φt (r)φ2t (e)r ′ e 6= 0. This shows that φ2t (e)e 6= 0. Since R =

t−1
M
j=0

φ2j (e)R, φ2t (e) = e,

6

THOMAS DREYFUS

and

′ −1
tM

φj (e)R, we find t = t′ . With R =

j=0

t−1
M
j=0

φ2j (e)R, we obtain φt (e) ∈

t−1
M

φ2j (e)R.

j=0

We remind that for all 0 < j < t, we have eφj (e) = 0. Using the fact that φ is an automorphism, we obtain that for all 0 < j < t, we have φt (e)φ2j (e) = 0. It follows that
t−1
M
j=0

φ2j (e)Rφt (e) ⊂ eR and therefore, φt (e) = (φt (e))2 ∈ eR. So there exists r ′ ∈ R such

that φt (e) = er ′ . But an integral domain may have only one non zero idempotent element.
Since e, φt (e) are non zero idempotent and eR is an integral domain we find that e = φt (e).
In particular,

t−1
M

φj (e)R is a difference ideal of the simple difference ring (R, φ). Since

j=0

e 6= 0, we find that the difference ideal is not (0), proving that R =

t−1
M

φj (e)R. This

j=0

completes the proof in the case R = R[i].

Let R be a difference ring that is the direct sum of integral domains R :=

t−1
M

Rj . We

j=0

define K, the total ring of fractions of R, by K :=

t−1
M
j=0

Kj , where for all 0 ≤ j ≤ t − 1, Kj

is the fraction field of Rj .
We say that R is a real ring if for all 0 ≤ j ≤ t − 1, Kj is a real field. Note that by
[Lam84, Theorem 2.8], this is equivalent to the usual definition of a real ring, that is that
0 is not a sum of squares in R \ {0}, see [Lam84, Definition 2.1].
The notion of Picard-Vessiot extension is not well suited in the real case. Following
[CHvdP16], let us define:
Definition 6. A real Picard-Vessiot extension for (2) over (k, φ) is a difference ring extension (R, φ) of (k, φ) such that
(1) (R, φ) is a Picard-Vessiot extension for (2) over (k, φ);
(2) (R, φ) is a real difference ring.
Let us remind that if (R, φ) is a difference ring such that X 2 + 1 ∈ R[X] is irreducible,
then R[i] is the ring R[i] := R[X]/(X 2 + 1). If (R, φ) is a difference ring with x ∈ R
satisfying x2 + 1 = 0, we make the convention that R[i] = R.
We are now able to state our main result:
Theorem 7. Let us consider the equation (2) which has coefficients in (k, φ).
(1) There exists a real Picard-Vessiot extension for (2) over (k, φ).
(2) Let (R, φ) be a real Picard-Vessiot extension for (2) over (k, φ). Then, (R, φ) is a
weak Picard-Vessiot extension for (2) over (k, φ), i.e., the ring of constants of R
is C.
(3) Let (R1 , φ1 ) and (R2 , φ2 ) be two real Picard-Vessiot extensions for (2) over (k, φ).
Let us equip the ring R1 ⊗k R2 with a structure of difference ring as follows:
φ(r1 ⊗k r2 ) = φ1 (r1 ) ⊗k φ2 (r2 ) for rj ∈ Rj . Then, (R1 , φ1 ) is isomorphic to
(R2 , φ2 ) over (k, φ) if and only if R1 ⊗k R2 6= R1 ⊗k R2 [i].

Before proving the theorem, we are going to state and prove a lemma which is inspired
by a lemma of [Sei58].

REAL DIFFERENCE GALOIS THEORY.

7

Lemma 8. Consider a difference field (K, φ) of characteristic zero that is finitely generated
over Q by the elements u1 , . . . , um and let (KR , φ) be a real difference subfield of (K, φ).
Then, there exists h : K → C, injective morphism of fields that induces an injective
morphism from KR to R.
For every 1 ≤ j ≤ m, k ∈ Z, let us write cj,k := h(φk (uj )) ∈ C. Then, the assignment
ej := (cj,k )k∈Z defines (resp. induces) an injective morphism of difference fields
uj 7→ u
between (K, φ) (resp. (KR , φ)) and (CZ , φs ) (resp. (RZ , φs )), where φs denotes the shift.
Proof of Lemma 8. Let us prove that there exists h : K → C, injective morphism of fields.
Let tj be a transcendental basis of K|Q. Since K|Q is generated as a field by a countable
number of elements, the number of elements in the transcendental basis is countable.
Using the fact that R is not countable, we find that there exists h : Q(tj ) → R, injective
morphism of fields. Let us extend h to K. The elements of K|Q(tj ), satisfy a list of
algebraic equations, which have a solution in an extension of C. Since C is algebraically
closed, we find that the equations have a solution in C. In other words, we have the
existence of an embedding of K into C.
Let us prove that KR := h(KR ) ⊂ R. Let tj be a transcendental basis of KR |Q. We
have h(Q(tj )) ⊂ KR ⊂ C. Since h is an injective morphism of fields and KR is a real
field, we find that KR is a real field. Then, we obtain that the real closure of h(Q(tj ))
contains KR . Since by construction h(Q(tj )) ⊂ R we find that the real closure of h(Q(tj ))
is contained in R. Then, we conclude that KR ⊂ R ⊂ C.
e1 , . . . , u
em ) =
Let P ∈ Q{X1 , . . . , Xm }φ .
We have the following equality P (u
e1 , . . . , u
em ) = 0.
P ((c1,k )k∈Z , . . . , (cm,k )k∈Z ). Therefore, P (u1 , . . . , um ) = 0 if and only if P (u
ej := (cj,k )k∈Z defines (resp. induces) an inThis shows that the assignment uj 7→ u
jective morphism of difference fields between (K, φ) (resp. (KR , φ)) and (CZ , φs ) (resp.
(RZ , φs )).

Proof of Theorem 7. (1) Let us prove the existence of a real Picard-Vessiot extension.
We have seen how to construct (R, φ), Picard-Vessiot extension for (2) over (k, φ). Let
U ∈ GLn (R) be a fundamental solution. As we can see in Proposition 5, R is a direct
sum of integral domains and we may define K, the total ring of fractions of R. The ring
K is a direct sum of fields K :=

t−1
M

Kj satisfying φ(Kj ) = Kj+1 , Kt := K0 . Therefore,

j=0

for all 0 ≤ j ≤ t − 1, (Kj , φt ) is a difference field. Let (K, φ) be the difference subring of
(K, φ) generated over Q by the components on the Kj , 0 ≤ j ≤ t − 1, of the entries of
U , det(U )−1 , and the elements in k involved in the algebraic difference relations between
the entries of U and det(U )−1 . In particular, the entries of the matrix A of (2) belong to
K. As we can see from Lemma 8, for all 0 ≤ j ≤ t − 1, there exists e
hj , an embedding of
(K ∩ Kj , φt ) into (CZ , φs ). If t > 1, without loss of generality, we may assume that for
e j+1 (φ(u)) (resp.
all 0 ≤ j ≤ t − 2, (resp. for j = t − 1), for all u ∈ K ∩ Kj , e
hj (u) = h
e (φ(u)) = h
e
e
φs h
0
t−1 (u)). We may define h, an embedding of the difference ring (K, φ)
P

into the difference ring (CZ , φs ) as follows. Let k = t−1
j=0 kj with k ∈ K, kj ∈ Kj and let
Z
e
us define h(k) ∈ C as the sequence which term number c + dt, with 0 ≤ c ≤ t − 1, d ∈ Z,
equals to the term number d of e
hc (kc ). Furthermore, since k is a real field, we find, see
e
Lemma 8, that for all k ∈ (K ∩ k, φ), h(k)
∈ RZ .
Let Cr,1 , . . . , Cr,n , (resp. Ci,1 , . . . , Ci,n ) be the real parts (resp. the imaginary parts) of
the columns of the term number zero of e
h(U ). We remind that U is invertible. Theree
fore, the term number zero of h(U ) is invertible. Then, we may extract n columns
C1 , . . . , Cn ∈ {Cr,1 , . . . , Cr,n , Ci,1 , . . . , Ci,n }, that are linearly independent. Therefore,
e )B
there exists B a matrix with entries in Q[i], such that the term number zero of h(U

8

THOMAS DREYFUS

has columns C1 , . . . , Cn , and is consequently real and invertible. Then, the term number
e )B is invertible. Since the term number zero of e
zero of h(U
h(U ) is also invertible, we
find that B ∈ GLn (Q[i]). Let V := U B be a fundamental solution, which belongs to
e extends to a morphism of difference rings between (K[i], φ) and
GLn (K[i]). The map h
Z
e ) = h(U
e )h(B).
e
(C , φs ). Consequently, we have Ve := h(V
Z
e
e φs )
Let (Q, φs ) be the difference subring of (Q , φs ) of constant sequences. Note that (Q,
e φs ),
e φs ) be the difference subring of (RZ , φs ) generated over (Q,
is a difference field. Let (k,
e
e
by the elements h(k), k ∈ (K∩k, φ). Note that (k, φs ) is a difference field. We remind that
e
since k is a real field, Lemma 8 tells us that h(A)
∈ (GLn (R))Z . Since the term number
e
e
e
e
zero of V belongs to GLn (R), and φs (V ) = h(A)V , we obtain that Ve ∈ (GLn (R))Z . Let
e φs ) by the entries of Ve , and
e φs ) be the difference subring of (RZ , φs ) generated over (k,
(R,
−1
e
det(V ) .
e φs ) is a simple difference ring. To the contrary, assume that there
We claim that (R,
e different from (0) and R.
e It follows that I(R[i])
e
exists I, a difference ideal of R
is different
e
e
e
from (0) and R[i]. We have a natural embedding from (R, φs ) into (R[i], φ). Then, I(R[i])
induces a difference ideal of (R[i], φ), which is different from (0) and R[i]. Let us treat
separately two cases. If R = R[i], then we use the fact that the Picard-Vessiot extension
e φs )
(R[i], φ) is a simple difference ring to conclude that we have a contradiction and (R,
is a simple difference ring. If R 6= R[i], we use Lemma 4, to deduce that (R[i], φ) is a
Picard-Vessiot extension for (2) over (k[i], φ) and therefore, is a simple difference ring.
e φs ) is a simple
We find a contradiction and we have proved our claim, that is that (R,
e is a real ring to prove
difference ring. We additionally use the fact that by construction R
e
e
e
e
e φs ).
that (R, φs ) is a real Picard-Vessiot extension for φY = h(A)Y , over (k,
Let R1 , be the difference ring generated over K ∩ k by the entries of the fundamental
e φs ) is isomorphic to (K ∩ k, φ), and
solution V and det(V )−1 . Using the fact that (k,
e
e
e
e φs ), we obtain that R1
(R, φs ) is a real Picard-Vessiot extension for φY = h(A)
Ye , over (k,
is a real Picard-Vessiot extension for (2) over (K ∩ k, φ).
Let us prove that R2 := R1 ⊗K∩k k is a real Picard-Vessiot extension for (2) over (k, φ).
By construction, R2 is generated over k by the entries of V and det(V )−1 . It is sufficient
to prove that R2 is a simple difference ring which is real.
P
We claim that R2 is a real ring. Let aj ∈ R2 such that j (aj )2 = 0. Let us write
P
P
aj = ℓ rj,ℓ ⊗K∩k kj,ℓ , with rj,ℓ ∈ R1 , kj,ℓ ∈ k. With j (aj )2 = 0, we obtain an algebraic
relation over Q between the rj,ℓ and the kj,ℓ . Note that R1 ⊂ R2 ⊂ K[i]. Then, as a
consequence of the definition of K, we find that K[i] is the difference subring of (K[i], φ)
generated over Q by i, the components on the Kj [i], 0 ≤ j ≤ t − 1, of the entries of V ,
det(V )−1 , and the elements in k involved in the algebraic difference relations between the
entries of V and det(V )−1 . Furthermore, since i in an algebraic number that does not
belong to the real field k, and V = U B, with U ∈ GLn (K), B ∈ GLn (Q[i]), we find that
the elements in k involved in the latter relations are in fact involved in algebraic difference
relations between the entries of U and det(U )−1 , proving that they belong to K. Hence,
we find that for all j, ℓ, rj,ℓ ∈ R1 ∩ K[i] and kj,ℓ ∈ k ∩ K. Therefore, for all j, aj ∈ R1 .
Since R1 is a real ring, we find that for all j, aj = 0, proving that R2 is a real ring.
It is now sufficient to prove that (R2 , φ) is a simple difference ring. Let I 6= (0) be
a difference ideal of (R2 , φ). Since V = U B, with V ∈ GLn (R2 ), U ∈ GLn (R), and
B ∈ GLn (Q[i]), we find that R2 [i] = R[i]. The difference ideal I induces the difference
ideal I[i] of (R[i], φ). Let us treat separately two cases. If R[i] = R, then (R[i], φ) is a
simple difference ring since it is a Picard-Vessiot extension for (2) over (k, φ), proving
that I[i] = R[i] and I = R2 . Therefore, (R2 , φ) is a simple difference ring and a real
Picard-Vessiot extension for (2) over (k, φ). Assume that R[i] 6= R. With Lemma 4,
(R[i], φ) is a Picard-Vessiot extension for (2) over (k[i], φ). Then, (R[i], φ) is a simple

REAL DIFFERENCE GALOIS THEORY.

9

difference ring and I[i] = R2 [i], proving that I = R2 . This shows that (R2 , φ) is a simple
difference ring and a real Picard-Vessiot extension for (2) over (k, φ).
(2) With Lemma 4 we find that (R[i], φ) is a Picard-Vessiot extension for (2) over (k[i], φ).
Remind that by assumption, C[i] is algebraically closed. As we can deduce from [vdPS97,
Lemma 1.8], R[i]φ = C[i]. It follows that Rφ ⊂ C[i]. By assumption, R is a real ring.
This implies that i ∈
/ R. Therefore, C = kφ ⊂ Rφ . Hence, the field of constants of R is C.
(3) Let us assume that, R1 ⊗k R2 6= R1 ⊗k R2 [i] and let us prove that (R1 , φ1 ) is isomorphic
to (R2 , φ2 ) over (k, φ). We remind, see Lemma 4, that for j ∈ {1, 2}, (Rj [i], φj ), is a PicardVessiot extension for (2) over (k[i], φ). We also remind that the field of constants of k[i] is
C[i]. Due to [vdPS97, Proposition 1.9], we find that (R1 [i], φ1 ) is isomorphic to (R2 [i], φ2 )
over (k[i], φ). Let ϕ : R1 → R2 [i] be the restriction of the morphism. Then, we may define
a morphism of difference rings
Ψ : R1 ⊗k R2 → R2 [i]
x⊗y
7→ ϕ(x)y.

The morphism Ψ is a R2 -linear map, and the image of R1 ⊗k R2 under Ψ is a R2 -submodule
of R2 [i], called V .
The assumption R1 ⊗k R2 6= R1 ⊗k R2 [i] implies that there are no f ∈ R1 ⊗k R2 such
that f 2 + 1 = 0. Since Ψ is a morphism of difference ring, there are no g ∈ V such that
g2 + 1 = 0, which proves i ∈
/ V . Combining this fact with the inclusion R2 ⊂ V , we obtain
that V = R2 (we remind that V is a R2 -submodule of R2 [i]). In other words, the image
of R1 under ϕ is included in R2 . This implies that (R1 , φ1 ) is isomorphic to (R2 , φ2 ) over
(k, φ).
Conversely, if (R1 , φ1 ) is isomorphic to (R2 , φ2 ) over (k, φ), then there exists a morphism
of difference rings ϕ : R1 → R2 . As above, let us define Ψ, morphism of difference
rings between R1 ⊗k R2 and R2 defined by Ψ(x ⊗ y) = ϕ(x)y. Since R2 is a real ring,
we find that R2 6= R2 [i]. Since Ψ is a morphism of difference rings, we obtain that
R1 ⊗k R2 6= R1 ⊗k R2 [i].

The following example, who is inspired by [CHS13], illustrates a situation where two
Picard-Vessiot extensions are not isomorphic.
√
Example 9. Let φ := f (z) 7→ f (2z) and consider φY = 2Y √which has √coefficients
in R(x). Let us consider the following fundamental solutions ( h x) and (ii x). Con√ √ −1
sider the corresponding difference ring extensions R1 |R(x) := R x, x
|R(x) and
√
√ −1 
R2 |R(x) := R i x, (i x) . Let us prove that (R1 , φ) is a simple difference ring.
The proof for (R2 , φ) is similar. Let I 6= (0)
√ be a difference ideal of R1 and let
Let k ∈ N be the degree of
P ∈ R[X] with minimal degree such that P ( x) ∈ I.
√ √
√
P . Assume that k 6= 0. We have φ(P ( x)) = P ( 2 x) ∈ I, which shows that
√ k √
√
√
φ(P ( x)) − 2 P ( x) = Q( x) ∈ I where Q ∈ R[X] has degree less than k. This
is in contradiction with the minimality of k, and shows that k = 0. This implies that
I = R1 , which proves that (R1 , φ) is a simple difference ring. Since R1 and R√
2 are real
rings, R1 |R(x) and R2 |R(x) are two real Picard-Vessiot extensions for φY = 2Y over
(R(x), φ). Note that there are no difference ring isomorphism between (R1 , φ) and (R2 , φ)
over R(x) because X 2 = x has a solution in R1 and no solutions in R2 . This is not in
contradiction with Theorem 7 since R1 ⊗R(x) R2 = R1 ⊗R(x) R2 [i], because


√

1
x ⊗R(x) √
i x

2

= −1.

10

THOMAS DREYFUS

3. Real difference Galois group
In this section, we still consider (2). Let (R, φ) be a real Picard-Vessiot extension for (2)
over (k, φ) with fundamental solution U ∈ GLn (R). Consider the difference ring (R[i], φ),
which is different from (R, φ), since R is a real ring. Inspiriting from [CHS13], let us define
the real difference Galois group as follows:
Definition 10. We define GR[i] , as the group of difference ring automorphism of R[i]
letting k[i] invariant. We define G, the real difference Galois group of (2), as the group
{ϕ|R , ϕ ∈ GR[i] }.
Note that elements of G are maps from R to R[i]. Due to Theorem 7, (2), we have an
injective group morphism
ρU : G −→ GLn (C[i])
ϕ 7−→ U −1 ϕ(U ),

which depends on the choice of the fundamental solution U in R. Another choice of a
fundamental solution in R will gives a representation that is conjugated to the first one.
Remind, see Proposition 5, that there exist an idempotent e ∈ R, and t ∈ N∗ , such that
φt (e)

= e, R =

t−1
M
j=0

φj (e)R, and for all 0 ≤ j ≤ t − 1, φj (e)R is an integral domain. Due

to Lemma 4, (R[i], φ) is a Picard-Vessiot extension for (2) over (k[i], φ). Furthermore,
R[i] =

t−1
M

φj (e)R[i] and the total ring of fractions of R[i] equals K[i], where K is the total

j=0

ring of fractions of R. Then, we call GK[i] , the classical difference Galois group of (2), the
group of difference ring automorphism of K[i] letting k[i] invariant. See [vdPS97] for more
details. The difference Galois group of (2) may also be seen as a subgroup of GLn (C[i]).
Furthermore, its image in GLn (C[i]) is a linear algebraic subgroup of GLn (C[i]). We have
the following result in the real case.
Proposition 11. Let (R, φ) be a real Picard-Vessiot extension for (2) over (k, φ) with
fundamental solution U ∈ GLn (R). Let G, be the real difference Galois group of (2) and
GK[i] , be the difference Galois group of (2). We have the following equality
n

o

o

n

Im ρU = U −1 ϕ(U ), ϕ ∈ G = U −1 ϕ(U ), ϕ ∈ GK[i] .
Furthermore, Im ρU is a linear algebraic subgroup of GLn (C[i]) defined over C. We will
identify G with a linear algebraic subgroup of GLn (C[i]) defined over C for a chosen
fundamental solution.
n



o

Proof. Let us prove the equality U −1 ϕ(U ), ϕ ∈ G = U −1 ϕ(U ), ϕ ∈ GK[i] . Remind
that U ∈ GLn (R). Since an element of GK[i] induces an element of G, we obtain the
o

n



inclusion U −1 ϕ(U ), ϕ ∈ GK[i] ⊂ U −1 ϕ(U ), ϕ ∈ G . Let ϕ ∈ G. We may extend ϕ as
an element ϕK[i] ∈ GK[i] by putting ϕK[i] (i) = i and for all 0 ≤ j ≤ t − 1, a, b ∈ φj (e)R[i],
ϕK[i] (a)
ϕK[i] (b) .

ϕK[i] ( ab ) =

Since U ∈ GLn (R), we find that U −1 ϕ(U ) = U −1 ϕK[i] (U ). Therefore,

we obtain the other inclusion
equality



U −1 ϕ(U ), ϕ

∈G =

n



U −1 ϕ(U ), ϕ ∈ G

U −1 ϕ(U ), ϕ

o

⊂

∈ GK[i] .

n

U −1 ϕ(U ), ϕ ∈ GK[i]

With a similar reasoning to what is above, we obtain the equalities:
n

o

n

o

n

o

o

U −1 ϕ(U ), ϕ ∈ G = U −1 ϕ(U ), ϕ ∈ GK[i] = U −1 ϕ(U ), ϕ ∈ GR[i] .

and the

REAL DIFFERENCE GALOIS THEORY.

11

We define GR , as the group of difference ring automorphism of R letting k invariant. Due
to Theorem 7, (2), (R, φ) is a weak Picard-Vessiot
extension for (2) over (k, φ). Applying

[CHS08, Proposition 2.2], we find that U −1 ϕ(U ), ϕ ∈ GR is a linear algebraic subgroup
of GLn (C). Then, we may use [CHS08, Corollary 2.5],nto find that the lattero group, viewed
as a linear algebraic subgroup of GLn (C[i]), equals to U −1 ϕ(U ), ϕ ∈ GR[i] . We conclude
the proof using the equality
n

o

n

o

U −1 ϕ(U ), ϕ ∈ G = U −1 ϕ(U ), ϕ ∈ GR[i] .


We finish this section by giving the Galois correspondence. See [vdPS97, Theorem 1.29]
for the analogous statement in the case where C is algebraically closed.
Theorem 12. Let (R, φ) be a real Picard-Vessiot extension for (2) over (k, φ) with total
ring of fractions K, F be the set of difference rings k ⊂ F ⊂ K, and such that every non
zero divisor is a unit of F . Let G, be the real difference Galois group of (2), G be the set
of linear algebraic subgroups of G.
(1) For any F ∈ F, the group G(K/F ) of elements of G letting F invariant belongs
to G.
(2) For any H ∈ G, the ring K H := {k ∈ K|∀ϕ ∈ H, ϕ(k) = k} belongs to F.
(3) Let α : F → G and β : G → F denote the maps F 7→ G(K/F ) and H 7→ K H .
Then, α and β are each other’s inverses.
Remark 13. If we replace G by GR , see the proof of proposition 11, which is a more natural candidate for the definition of the real difference Galois group, we loose the Galois
correspondence. Take for example φY (x) := Y (x + 1) = exp(1)Y (x), which has solution exp(x). A real Picard-Vessiot extension for Y (x + 1) = exp(1)Y (x) over (R, φ) is
(R[exp(x), exp(−x)], φ). Let K := R(exp(x)) be the total ring of fractions. We have
G ≃ C∗ and GR ≃ R∗ . Note that GR ⊂ GL1 (R), viewed as a linear algebraic subgroup of
GL1 (C), equals to G. On the other hand, we have no bijection with the linear algebraic
subgroups of GR , which are {1}, Z/2Z, R∗ , and the difference subfields of K, which are
R(exp(kx)), k ∈ N.

Proof of Theorem 12. Let F[i] be the set of difference rings k[i] ⊂ F ⊂ K[i], such that
every non zero divisor is a unit of F . Let G[i] be the set of linear algebraic subgroups of
GK[i] . Remind that the field of constants of k[i] is algebraically closed. In virtue of the
Galois correspondence in difference Galois theory, see [vdPS97, Theorem 1.29], we find
that
(a) For any F ∈ F[i] , the group GK[i] (K[i]/F ) of elements of GK[i] letting F invariant
belongs to G[i] .
(b) For any H ∈ G[i] , the ring K[i]H belongs to F[i] .
(c) Let α[i] : F[i] → G[i] and β[i] : G[i] → F[i] denote the maps F[i] 7→ GK[i] (K[i]/F ) and
H 7→ K[i]H . Then, α[i] and β[i] are each other’s inverses.
We use Proposition 5 to find that we have a bijection γ : F → F[i] given by γ(F ) := F [i].
The inverse is γ −1 (F ) = F ∩ K. Now, let us remark that since the fundamental solution
has coefficients in R, for all F ∈ F, G(K/F ) = GK[i] (K[i]/γ(F )). If we combine this fact
with (a) and Proposition 11, we find (1).
Proposition 11 tells us that we may identify the groups in G with the corresponding
groups in G[i] . To prove the point (2), we remark that for all H ∈ G, K[i]H = K H [i].
Combined with (b), this shows the point (2) since K H = γ −1 (K[i]H ) ∈ F .
The point (3) follows from (c) and the fact that for all F ∈ F (resp. H ∈ G) we have
G(K/F ) = GK[i] (K[i]/γ(F )) (resp. K[i]H = γ(K H )).


12

THOMAS DREYFUS

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