Qj and let CQ /C0 be a splitting ¬eld extension for Q. Let σ ∈

Indeed, de¬ne Q = j

Gal(CQ /C0 ). We may apply σ to members of CQ (x); in particular, we have σ(h) = h for all

h ∈ C0 (x). Applying this fact to (4.20) yields

ci

= f

x ’ ±i

j i:Qj (±i )=0

σ(f ) since f ∈ C0 (x)

=

ci

= σ

x ’ ±i

j i:Qj (±i )=0

ci

= .

x ’ σ(±i )

j i:Qj (±i )=0

103

The claim follows easily after observing that Gal(CQ /C0 ) acts transitively on the roots of

Qj for each j.

We now have

nj

f= . (4.21)

x ’ ±i

j i:Qj (±i )=0

We also have that

(x ’ ±i )

=

Qj

i:Qj (±i )=0

1

’ Qj /Qj =

x ’ ±i

i:Qj (±i )=0

nj

’ nj Qj /Qj =

x ’ ±i

j j i:Qj (±i )=0

= f.

Condition 3 now follows from uniqueness of PF decomposition over C0 .

Next we show that Condition 3 implies Condition 4. Assume Condition 3 holds and let

n

Qj j ∈ C0 (x). A straightforward calculation implies that f = g /g, so that Condition

g= j

4 holds.

Finally, it is clear that Condition 4 implies Condition 1.

The following calculations yield an alternative partial fraction or “APF” decomposition

for rational functions de¬ned over C0 . As we shall see, the APF decomposition is more useful

than the PF decomposition for deciding whether a Z-linear combination of two rational

functions is the logarithmic derivative of another rational function; thus, APF is more

useful than PF for making Corollary 4.6.7 e¬ective.

Given f ∈ C0 (x) having PF decomposition (4.18) over C0 . For each j, compute γj ∈

C0 , Bj ∈ C0 [x] such that Aj,1 = γj Qj + Bj and deg Bj < deg Qj . Then compute a decom-

position γj = ±j + βj having the following properties:

1. ±j ∈ Q

2. βj is a Q-linear combination of nontrivial power products of the zi , where C0 =

Q(z1 , . . . , z» ).

104

This decomposition is unique for a given γj . Applying these computations to (4.18) yields

« «

Aj,1 Aj,d

f = P + +

Qd

Qj j

j j,d:d>1

«

«

(±j + βj )Qj + Bj Aj,d

+

= P +

Qd

Qj j

j j,d:d>1

±j Qj βj Qj + Bj Aj,d

= + P+ +

Qd

Qj Qj j

j j j,d:d>1

= frat + firrat . (4.22)

We refer to (4.22) as the APF decomposition of f over C0 . Note that this decomposition is

unique for a given f ∈ C0 (x).

¯

For the remainder of this section, we assume that C = Q.

Before stating the next lemma, we make a new de¬nition which involves a minor abuse

of notation: Given m1 , m2 ∈ Z, GCD(m1 , m2 ) = 1, de¬ne

T(m1 ,m2 ) = diag(tm1 , tm2 , t’m1 ’m2 ) : t ∈ C — C—.

This de¬nition di¬ers from the de¬nition of T(d1 ,d2 ) given in Lemma 4.2.2 only in that we

place fewer restrictions on the subscripts m1 , m2 .

Lemma 4.6.9 Given L a third-order operator whose only singularities in the ¬nite plane

are apparent singularities, such that GL is nontrivial and L = (D +r1 +r2 ’h /h)—¦(D ’r2 )—¦

¯

(D ’ r1 ), r1 , r2 , h ∈ Q(x). Let C0 /Q be a ¬nite algebraic extension such that r1 , r2 , h ∈ C0 (x).

Let ri = rirat +riirrat be the APF decomposition of ri over C0 (x) for i = 1, 2. Let F(r1 ,r2 ) , Lred

and T ⊆ GL be as de¬ned in Lemma 4.6.2.

1. If r1irrat and r2irrat are linearly dependent over Q, then there exist m1 , m2 ∈ Z such

C — . The values of m1 , m2 can be

that GCD(m1 , m2 ) = 1 and [T ]F(r1 ,r2 ) = T(m1 ,m2 )

computed as follows:

(a) If r2irrat = 0, then [T ]F(r1 ,r2 ) = T(1,0) .

(b) If r1irrat /r2irrat = µ1 /µ2 ∈ Q, µ1 , µ2 ∈ Z, GCD(µ1 , µ2 ) = 1, then [T ]F(r1 ,r2 ) =

T(µ1 ,µ2 ) .

2. If r1irrat and r2irrat are linearly independent over Q or, equivalently, if r1irrat /r2irrat ∈ Q,

/

C— — C—.

then [T ]F(r1 ,r2 ) = D3 © SL3

105

Proof. First of all, by Lemma 4.6.2, we have that T GLred . Moreover, after considering

Table 4.1, we see that all nontrivial solvable groups listed in Theorem 4.1.5 have Levi

subgroup isomorphic to either C — or C — — C — ; it follows that T is isomorphic to one of those

groups. Thus, we may apply Corollary 4.6.7 to the problem of computing T in this case.

Suppose r2irrat = 0. This yields

±j Qj

r2 = r2rat = ,

Qj

j

where ±j ∈ Q and Qj ∈ C0 [x] is irreducible for each j. We claim that ±j ∈ Z for each j.

Indeed, if ±j ∈ Q \ Z for some j, then we see that R2 ∈ K2 \ k for some ¬nite algebraic

extension K2 /k, so that T has a ¬nite quotient; this contradicts the fact that T is isomorphic

±

to either C — or C — — C — . We conclude that R2 = Qj j ∈ C0 (x), from which Item 1(a) of

j

the conclusion follows easily.

Next suppose r1irrat /r2irrat = µ1 /µ2 ∈ Q, µ1 , µ2 ∈ Z, GCD(µ1 , µ2 ) = 1. We compute

µ2 r1irrat ’ µ1 r2irrat = 0

’ µ2 r1 ’ µ1 r2 µ2 r1rat ’ µ1 r2rat

=

±j Qj

= ,

Qj

j

where ±j ∈ Q and Qj ∈ C0 [x] is irreducible for each j. As in the previous case, one checks

’µ ±

µ

that ±j ∈ Z for each j. It follows that R1 2 R2 1 = Qj j ∈ C0 (x). This means that, for

j

σ ∈ T with [σ]F(r1 ,r2 ) = diag(tm1 , tm2 , t’m1 ’m2 ), t not a root of unity, we have

’µ ’µ

µ µ

R 1 2 R2 1 σ R1 2 R2 1

=

(tm1 R1 )µ2 (tm2 R2 )’µ1

=

’µ

tm1 µ2 ’m2 µ1 R1 2 R2 1

µ

=

= tm1 µ2 ’m2 µ1

’1

’0 = m1 µ2 ’ m2 µ1

’ m2 µ1 = m1 µ2

µ1 m1

’ = .

µ2 m2

Item 1(b) of the conclusion now follows easily.

Finally, suppose r1irrat /r2irrat ∈ Q, i.e., r1irrat and r2irrat are linearly independent over Q.

/

Then, for all pairs of integers m1 , m2 not both zero, we have that the APF decomposition

of r = m1 r1 + m2 r2 over C0 (x) satis¬es rirrat = 0. On the other hand, let g ∈ C0 (x)

106

n

Qj j , nj ∈ Z, Qj ∈ C0 [x] irreducible; then the APF decomposition of

and write g = j

nj Qj /Qj over C0 (x) satis¬es (g /g)irrat = 0. It follows that there do not exist

g /g = j

(m1 , m2 ) ∈ Z2 \ {(0, 0)} and g ∈ C(x) such that m1 r1 + m2 r2 = g /g. Corollary 4.6.7

C — — C — . Since T is an algebraic subgroup of D3 © SL3 , we conclude that

now implies T

T = D3 © SL3 , and Item 2 of the conclusion is proved.

We are now ready to present our algorithm and prove its correctness.

Algorithm IV

¯ ¯

Input: Two polynomials a, b ∈ Q[x], representing the operator L = D3 + aD + b ∈ Q(x)[D].

Output: An explicit description of GL , the group of L(y) = 0.

Steps: First compute n1 and n2 , where ni is the number of monic ith order right factors

of L in D. This step can be carried out by computing rational solutions of Ricc L and

Ricc(adj L). Next, if (n1 , n2 ) = . . .

• (0, 0) : Compute L PSL2 . Otherwise, test

2 2

. If ord(L ) = 5, then return GL

PSL2 ; otherwise, return GL SL3 .

2

whether L is reducible. If so, then return GL

• (0, 1) (resp., (1, 0)): Here, L has an irreducible (left or right) factor of the form

¯

L2 = D2 + b1 D + b0 , b0 , b1 ∈ Q(x). (4.23)

¯ ¯

for some f ∈ Q(x). If so, then return GL Q2 SL2

f

Compute b1 . Test whether b1 = f

¯

Q2 GL2 with conjugation

with the unique conjugation action; otherwise, return GL

as described in Item 2(a) (resp., Item 2(b)) of Lemma 4.5.8.

¯

• (1, 2) (resp., (2, 1)): Compute r1 , r2 ∈ Q(x) such that

L = (D + r1 + r2 ) —¦ (D ’ r2 ) —¦ (D ’ r1 ). (4.24)

Q— — Q— , then

¯ ¯

Apply Lemma 4.6.9 to compute the maximal torus T of GL . If T

(Q— — Q— ) with conjugation as described in Item 2 (resp., Item

¯ ¯ ¯

Q2

return GL

3) of Lemma 4.5.4. Otherwise, apply Lemma 4.6.9 to compute m1 , m2 such that

Q— , with:

¯ ¯

[T ]F(r1 ,r2 ) = T(m1 ,m2 ) ⊆ D3 © SL3 . Return GL Q2

1. (Int t)(u, v) = (tm1 ’m2 u, t2m1 +m2 v) for t ∈ Q— , u, v ∈ Q, in case (n1 , n2 ) = (1, 2)

¯ ¯

2. (Int t)(u, v) = (t2m1 +m2 u, tm1 +2m2 v) for t ∈ Q— , u, v ∈ Q, in case (n1 , n2 ) =