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’1
= βg 2 y1
L2 (¦(y3 ))
’1
’ βg 2 y1 = L2 (¦(y3 )) = L2 ((hL2 + ±)(y3 ))

(L2 —¦ (hL2 + ±))(y3 ) = L2 (hL2 (y3 )) + ±L2 (y3 )
=
’1 ’1
L2 (hg ’1 y1 ) + ±g ’1 y1 by (4.16)
=
’1
gy1 L2 (hg ’1 y1 ) + ± (after multiplying by gy1 )
’ βg 3 =
’1
(gy1 —¦ L2 —¦ g ’1 y1 )(h) + ±
=

= (shiftr1 +r2 (L2 ))(h) + ±.

The third condition now follows easily after observing that

shiftr1 +r2 (L2 ) = (D ’ r1 ’ 2r2 ) —¦ (D ’ 2r1 ’ r2 ).


Next we show that the third condition implies the second condition. Suppose

(shiftr1 +r2 (L2 ))(h) = g 3 + c

for some c ∈ C \ {0} . De¬ne ¦ = hL2 ’ c. Then ¦(yi ) = ’cyi for i = 1, 2. Also, we
compute

’1
(hL2 ’ c)(y3 ) = hg ’1 y1 ’ cy3 by (4.16)
¦(y3 ) =
’1 ’1
= L2 (hg ’1 y1 ) ’ cg ’1 y1 by (4.16)
’ L2 (¦(y3 ))
’1 ’1 ’1
g ’1 y1 (gy1 L2 (hg ’1 y1 ) ’ c)(after distributing out g ’1 y1 )
=

99
’1
g ’1 y1 ((shiftr1 +r2 (L2 ))(h) ’ c)
=
’1
g ’1 y1 ((g 3 + c) ’ c) by hypothesis
=
’1
g 2 y1 .
=

This result, together with (4.17), implies that ¦(y3 ) = y3 + y0 for some y0 ∈ VL2 . We
ˆ
now see that ¦ maps a basis of VL isomorphically onto a basis of VL , and the second
ˆ

condition follows.

To prove that the third condition implies the fourth condition, simply di¬erentiate
both sides of the equation given in the third condition and divide by 3.

To prove that the fourth condition implies the third condition, suppose

¯
(D —¦ (D ’ r1 ’ 2r2 ) —¦ (D ’ 2r1 ’ r2 ))(h) = r2 g 3

¯ ¯
for some h ∈ C(x). Let h = 3h. It is then easy to check that

((D ’ r1 ’ 2r2 ) —¦ (D ’ 2r1 ’ r2 ))(h) = g 3 + c

for some c ∈ C. We must show that c = 0. Suppose instead that c = 0. Then the
˜ ˜
rational function h = (D ’ 2r1 ’ r2 )(h) satis¬es (D ’ r1 ’ 2r2 )(h) = g 3 . As in the
˜
last step of the proof of Item 3 of the lemma, we see that h ∈ VL © C(x) = {0} and
2

therefore that g = 0, contradicting hypothesis. This completes the proof.




Below, we state the Kolchin-Ostrowski theorem (Theorem 4.6.6) and a corollary (Corol-
¯
lary 4.6.7). In case C = Q, these results lead to an e¬ective criterion (Lemma 4.6.9) to
compute the maximal torus of GL and its representation [T ]F(r1 ,r2 ) , where T and F(r1 ,r2 )
are as de¬ned in Lemma 4.6.2, in case GL is solvable.


Theorem 4.6.6 Given a Picard-Vessiot extension K/k and a set S = {f1 , . . . , fν } ∈ K
fi
∈ k for 1 ¤ i ¤ ν. Then S is algebraically dependent over k if and only if there
such that fi
ν
fimi = g for some g ∈ k.
exist integers mi , not all zero, such that
i=1


Proof. This result is stated and proved in [Kol68].



100
Corollary 4.6.7 Given: L is an operator of the form

L = LCLM(D ’ s1 , D ’ s2 , D + s1 + s2 ’ h /h), s1 , s2 , h ∈ C(x),

such that GL is isomorphic to either C — or C — — C — . Then, the following are equivalent:

C—
1. GL

2. S1 and S2 are algebraically dependent over C(x), where Si /Si = si for i = 1, 2

for some m1 , m2 ∈ Z, not both zero, and g ∈ C(x) \ {0} .
g
3. m1 s1 + m2 s2 = g

’1 ’1
Proof. First, observe that VL is spanned by S1 , S2 and S1 S2 h. It follows that KL /k is
generated by S1 and S2 . Equivalence of the ¬rst two conditions now follows from the well-
known fact (see, for instance, [Mag94]) that the transcendence degree of a Picard-Vessiot
extension is equal to the dimension of the corresponding Galois group. Equivalence of
the second and third conditions follows easily from Theorem 4.6.6 after taking logarithmic
derivatives.




Before proceeding, we provide an example of the criterion for algebraic dependence
¯
given in Corollary 4.6.7. Given: S1 , S2 are members of a Picard-Vessiot extension of Q(x)
satisfying

S1 7+9 2
2
= s1 = 6x + ,
S1 x

S2 1+3 2
2x2 +
= s2 = .
S2 x



Are S1 and S2 algebraically dependent over Q(x)? To answer this question, we write

92
7
+ 6x2 +
s1 = ,
x x

32
1
+ 2x2 +
s2 = .
x x



It is now easy to see that s1 ’ 3s2 = 4/x and therefore that S1 /S2 = x4 . We conclude that
3


S1 and S2 are indeed algebraically dependent over Q(x).
What follows can be viewed as a generalization of the steps taken in the above example.

101
Let C0 be a ¬nite algebraic extension of Q, speci¬ed by a set of generators z1 , . . . , z» and
minimal polynomials

p1 ∈ Q[x], p2 ∈ Q(z1 )[x], . . . , p» ∈ Q(z1 , . . . , z»’1 )[x].

It is a fact (cf. [Bro96]) that every rational function f ∈ C0 (x) admits a unique partial
fraction or “PF” decomposition over C0 , written as follows:
µj
t
Aj,d
f =P + , (4.18)
Qdj
j=1 d=1

where:

1. P ∈ C0 [x]

2. t is a nonnegative integer and µj is a positive integer for all j

3. Qj ∈ C0 [x] is monic and irreducible for all j

4. Aj,d ∈ C0 [x] for all j, d

5. deg Aj,d < deg Qj for all j, d.

There are e¬ective algorithms to carry out this decomposition ([Bro96]). We can also carry
¯
out the PF decomposition of f ∈ C0 (x) over Q :
νi
s
ci,d
f =P + , (4.19)
(x ’ ±i )d
i=1 d=1

where:

1. P is as in (4.18)

2. s is a nonnegative integer and νi is a positive integer for all i

¯
3. ±i ∈ Q for all i

¯
4. ci,d ∈ Q for all i, d.

Lemma 4.6.8 Given f ∈ C0 (x). Suppose the PF decomposition of f over C0 (resp., over
¯
Q) is as given in (4.18) (resp., (4.19)). The following are equivalent:

¯
1. f = g /g for some g ∈ Q(x)

2. The following three conditions hold:

102
(i) P = 0

(ii) ci,d = 0 for all d > 1 and all i

(iii) For all i, we have ci,1 ∈ Z.

3. The following three conditions hold:

(i ) P = 0

(ii ) Aj,d = 0 for all d > 1 and all j

(iii ) For all j, there exists nj ∈ Z such that Aj,1 = nj Qj .

4. f = g /g for some g ∈ C0 (x)

Proof. Equivalence of the conditions numbered 1 and 2 follows after considering Laurent
¯
series expansions of the equation y ’ f y = 0 over Q(x) at its singularities in the ¬nite plane
and at in¬nity.
Next, we show that Condition 2 implies Condition 3. Condition (i ) follows immediately
from Condition (i). Condition (ii) implies that the denominator of f is squarefree; this
yields Condition (ii ). We prove Condition (iii ) as follows: Condition 2 implies that

ci ¯
, ±i ∈ Q, ci ∈ Z for each i.
f=
x ’ ±i
i

We see that for each i, the minimal polynomial of ±i over C0 is Qj for some j. Thus, we
may write
ci
. (4.20)
f=
x ’ ±i
j i:Qj (±i )=0

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