<<

. 5
( 33 .)



>>

reproduce it here, with some changes of notation.
First, let L1 = 1, L0 = L. We see that Properties 6(ab) are satis¬ed, so that there exists
a pair (L1 , L0 ) satisfying those properties with L1 of maximal order.

18
Next, note that if L = L1 L0 , then KI includes the full solution set VL1 (resp., VL0 ) of
L1 (y) = 0 (resp., L0 (y) = 0).
Let f ∈ K be a particular solution of L(y) = b. Let σ ∈ GI . Then σ(f ) ’ f ∈ VH . Thus,
we may de¬ne ¦ : GI ’ VL by ¦(σ) = σ(f ) ’ f. Let H = Gal(KI /KH ). Then H is normal
in GI . Since KI /KH is generated by f, we see that ¦ is injective on H.
For any σ ∈ GI , „ ∈ H, we have


¦(σ„ σ ’1 ) = σ„ σ ’1 (f ) ’ f

σ[„ (σ ’1 (f ) ’ f ) + „ (f )] ’ f
=

σ[σ ’1 (f ) ’ f ] + σ„ (f ) ’ f since „ ¬xes VL ⊆ KL elementwise
=

σ[„ (f ) ’ f ]
=

= σ¦(„ ).


This calculation (from the proof of Th´or`m 1 of [Ber92]) implies that ¦ is a GI -module
ee
morphism, where GI acts on H by conjugation. Therefore, ¦(H) is a GI -invariant subspace
W ⊆ VH . Since GI /H is isomorphic to the reductive group GH and H is unipotent, it
follows that GI has Levi decomposition GI = HP (semidirect product of subgroups), where
H = Ru (GI ) and P GH .
˜ ˜
Let L0 ∈ D be the unique monic operator such that VL0 = W and let L1 ∈ D be such that
˜

˜˜ ˜ ˜
L = L1 L0 . Since „ (f ) ’ f ∈ W for all „ ∈ H, we have that L0 („ (f )) = L0 (f ) for all „ ∈ H;
˜ ˜
it follows that L0 (f ) ∈ KH . Let W1 = VL1 ⊆ KH and let WL0 (f ) = W1 + C L0 (f ) ⊆ KH .
˜ ˜

˜ ˜
Given σ ∈ GH , we have that σ(L0 (f )) is another particular solution of that L1 (y) = b, so
˜ ˜
that σ(L0 (f )) = L0 (f ) + w for some w ∈ W1 . It follows that WL0 (f ) is GH -invariant and,
˜

moreover, that WL0 (f ) /W1 is a trivial one-dimensional GH -module. Since GH is a reductive
˜

group, we see that W1 has a GH -invariant complement in WL0 (f ) . This implies that there
˜

˜
exists f0 ∈ KH such that f0 = L0 (f ) + w for some w ∈ W1 and σ(f0 ) = f0 for all σ ∈ GH .
˜ ˜
˜
From these properties, we conclude that f0 ∈ k and L1 (f0 ) = b.
¯
Now let L0 , L1 ∈ D satisfy Properties 6(abc). Let W1 = VL1 ⊆ KH . Since L1 (L0 (f )) = b,
¯
we have L0 (f ) = f1 + w for some w ∈ W1 . It follows that L0 (f ) ∈ KH . Thus, given „ ∈ H,
¯ ¯
we have


L0 („ (f ) ’ f ) = „ (L0 (f ) ’ f ) = „ (f ’ f )
L0 (¦(„ )) =

= 0.

19
˜
This yields φ(H) = VL0 ⊆ VL0 . This implies that L0 divides L0 on the right. We then see
˜

˜ ˜ ˜
that ord(L1 ) ¤ ord(L1 ). In case ord(L1 ) = ord(L1 ) and L1 is monic, then L0 = L0 and
˜
therefore L1 = L1 .




The following examples appeared in [BS99] and were computed by the ¬rst author.

Example 3.2.2 Let k = C(x) and L = D2 ’ 4xD + (4x2 ’ 2) = (D ’ 2x) —¦ (D ’ 2x). A
C—.
2 2
basis for the solution space of the equation L(y) = 0 is {ex , xex }, so that GH
2
For any (c, d) ∈ C 2 , (c, d) = (0, 0), we have that (c + dx)ex is a solution of L(y) = 0,
so that L has a right factor of the form D ’ (2x + d
c+dx ). Furthermore, all right factors of
order one are of this form. Therefore the formula

d d
L = (D ’ (2x ’ )) —¦ (D ’ (2x + ))
c + dx c + dx

with (c, d) = (0, 0) yields a parameterization of all irreducible factorizations of L.
We shall now compute the Galois groups of L(y) = b where b = 4x2 ’ 2, 1 and 1
x.

(i) b = 4x2 ’ 2. In this case the equation L(y) = b has the rational solution y = 1. This
implies that L0 = 1, where L0 is as de¬ned in Proposition 3.2.1. Thus, VL0 is trivial, and
we conclude that the Galois group of L(y) = b is C — .
(ii) b = 1. A partial fraction computation shows that L(y) = 1 has no rational solutions.
Now let us search for ¬rst order left factors L1 of L such that L1 = 1 has a rational solution.
A calculation shows that the equation

d
y ’ (2x ’ )y = 1 (3.10)
c + dx

has a rational solution y = f if and only if z = (c + dx)f is a rational solution of

z ’ 2xz = c + dx (3.11)

(c.f., Lemma 3.2.4). The rational solutions of (3.11) must be polynomials, and we see that
this has a polynomial solution if and only if c = 0. Therefore the operator L0 as de¬ned
2
in Proposition 3.2.1 is equal to D ’ (2x + x ); its solution space is spanned by VL0 = xex .
1


C — , where t.u = tu for t ∈ C — , u ∈ C.
Therefore the Galois group of L(y) = 1 is C
1
(iii) b = x . We shall show that for any (c, d) = (0, 0), the equation

d 1
y ’ (2x ’ )y = (3.12)
c + dx x

20
1
has no rational solution. This implies that L(y) = also has no rational solution and so
x

the W of Proposition 3.2.1 is the full solution space of L(y) = 0. Therefore the Galois group
C — , where t.(u, v) = (tu, tv) for t = inC — , (u, v) ∈ C 2 . Equation (3.12)
is C 2
1
of L(y) = x

has a rational solution y = f if and only if z = (c + dx)f is a rational solution of

c + dx
z ’ 2xz = . (3.13)
x

If c = 0 then any rational solution of (3.13) must have a pole at x = 0. Comparing orders of
the left and right hand side of this equation yields a contradiction. Therefore c = 0. Similar
considerations show that z ’ 2xz = d can never have a rational solution if d = 0.




Proposition 3.2.1 lets us describe KI as follows.


Corollary 3.2.3 Given k, L, b, KI , KH , L1 , L0 as in Proposition 3.2.1. Write

L0 = Dt ’ bt’1 Dt’1 ’ . . . ’ bo , bi ∈ k.

Then KI = KH (z0 , z1 , . . . , zt’1 ), where z0 , z1 , . . . , zt’1 are algebraic indeterminates, zi =
t’1
zi+1 for 0 ¤ i ¤ t ’ 2, and zt’1 = f1 + bi z i .
i=0


ˆ ˆ
Proof. Let L = (D ’ b /b)L. Then KI /k is a Picard-Vessiot extension for L(y) = 0, with full
solution space VL . We see that L0 , viewed as a linear operator on KI , maps VL onto the full
ˆ ˆ

ˆ ˆ
solution space of L1 (y) = 0, where L1 = (D ’ b /b)L1 . Therefore, there exists an element
z ∈ VL with L0 (z) = f1 . We see that L(z) = b and therefore that K = KL < z > . Since
ˆ

Gal(KI /KH ) is a vector group of dimension t, we have that K is a purely transcendental
extension of KL of transcendence degree t. It follows that K = KL (z, z , . . . , z (t’1) ). The
desired result follows easily after setting zi = z (i) . satisfy the conclusion of the Corollary.




To compute GI for a given inhomogeneous equation L(y) = b using Proposition 3.2.1, it
su¬ces to perform the following tasks:

1. Compute GH .

2. Find L1 , L0 satisfying the conditions given in Proposition 3.2.1.

21
The ¬rst of these two tasks is addressed in [CS99] and lies outside the scope of this
dissertation. The second task is dealt with in [BS99], and below we summarize the relevant
results from that article.
Let L ∈ D be completely reducible. Then, by de¬nition, there exist operators T1 , . . . , Ts
such that L = LCLM(T1 , . . . , Ts ). Proposition 3.1.1 then implies that any left or right factor
of L will be equivalent to the least common left multiple of some subset of {T1 , . . . , Ts } .
Suppose k is a ¬nite algebraic extension of C(x), where C is a computable algebraically
closed ¬eld of characteristic zero, then (cf. [CS99] and [Sin96]) one can e¬ectively perform
the following tasks:

1. Factor an arbitrary element L ∈ D = k[D] as a product of irreducible operators.

2. Decide whether L is completely reducible.

3. In case L is completely reducible, compute a set {T1 , . . . , Ts } ⊆ D such that L =
LCLM(T1 , . . . , Ts ).

The article [BS99] approaches the second task above, as follows. First, ¬nd a set T =
{T1 , . . . , Ts } such that L = LCLM(T1 , . . . , Ts ). If L1 is a monic left factor of L, then L1
is equivalent to the least common left multiple of elements from some subset of T . Let
S = Ti1 , . . . , Tiµ ⊆ T be a ¬xed subset and let L2 = LCLM(Ti1 , . . . , Tiµ ). A sequence of
lemmas shows that one can:

A. Parameterize the set ML2 of pairs (L1 , S) with ord S < ord L1 = ord L2 such that
L1 is a left factor of L and, moreover, (3.9) holds for some R; and

B. Determine whether there exists (L1 , S) ∈ ML2 such that L1 (y) = b admits a solution
in k.

If these steps are carried out for all subsets S ⊆ T , then one ¬nds operators L1 , L0 satisfying
conditions 6(abc) of Proposition 3.2.1. [BS99] then shows how to describe the action of GH
on VL1 .
Below, Lemmas 3.2.4, 3.2.5, 3.2.6, 3.2.7 describe how to compute ML2 for a given L2 ,
and Lemma 3.2.8 will be used to decide whether L1 (y) = b admits a k-rational solution for
a given (L1 , S) ∈ ML2 . Lemmas 3.2.5, 3.2.6 and 3.2.7 are proved by technical means that
lie outside the scope of this dissertation; we omit the proofs here.


22
Lemma 3.2.4 Given L1 , L2 , R, S ∈ D such that ord(L1 ) = ord(L2 ) = n, ord(R) <
n, ord(S) < n, and (3.9) holds. Then, the equation L1 (y) = b, b ∈ k has a solution in
k if and only if the equation L2 (y) = S(b) has a solution in k.

˜ ˜ ˜ ˜
Proof. The extended Euclidean algorithm yields R and L1 in D such that RR + L1 L1 = 1
˜
and ord R1 < ord L1 . The map v ’ R(v) is an isomorphism of VL1 onto VL2 , and the map
˜ ˜ ˜
w ’ R(w) is the inverse of this isomorphism. Since L1 R and RR ’ 1 vanish on VL2 , we
˜ ˜
have that L2 divides both of these operators. Therefore there exist S and L2 ∈ D such that
˜ ˜ ˜˜
L1 R = SL2 and RR + L2 L2 = 1.
˜ ˜
We now claim that SS + L1 L1 = 1. We have that

˜ ˜ ˜ ˜
(SS + L1 L1 )L1 = SSL1 + L1 L1 L1
˜ ˜
SL2 R + L1 (1 ’ RR)
=
˜ ˜
SL2 R + L1 ’ L1 RR
=
˜ ˜
SL2 R + L1 ’ SL2 R
=

= L1 ,

and the equation follows after cancelling L1 on the right.

<<

. 5
( 33 .)



>>