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.