4 – The Riemann Surface of an Algebraic Function
301
The rational functions
s i (q) = P i
q
k + a
/P 0
q
k + a
are meromorphic at the origin; these are elementary symmetric functions
of the roots of (4), up to sign. To show that h(q) has at most one pole at
q = 0, it suffices to show that there is an upper bound of the form h(q) =
O(q
−N ). Now, the coefficients s i (q) are of this form. Therefore, the following
result remains to be proved. It probably dates back to time immemorial and
for example can be found in Dieudonn´ e’s exercise in Calcul infinit´ esimal,
Chap. III :
Lemma 1. Let ζ i (1 ≤ i ≤ n) be the roots of an equation
ζ
n + c 1 ζ
n−1 + . . . + c n = 0
(4.5)
with complex coefficients. Then
sup |ζ i | ≤ max (1, |c 1 | + . . . + |c n |) .
(4.6)
Let M be the left hand side of (6). Each root of (5) satisfies
|ζ i |
n ≤ |c 1 | . |ζ i |
n−1 + . . . + |c n | ≤ |c 1 | .M
n−1 + . . . + |c n | ,
whence
M
n
≤ |c 1 | .M
n−1 + . . . + |c n | .
If M ≤ 1, there is nothing to prove. If M ≥ 1, we then write
M ≤ |c 1 | + . . . + |c n | /M
n−1
≤ |c 1 | + . . . + |c n | ,
qed.
Note that if P 0 (a) = 0, the coefficients s i (q) of (5) are holomorphic at
q = 0, and so are bounded for sufficiently small q, hence so is h(q). As
a result, the function F (z, ζ) = ζ is holomorphic at the point η Y ∈ ˆ
X if
P 0 (a) = 0, in other words if the equation P (a, Y ) = 0 is effectively of degree
n at the point a. Since P (z, ζ) = 0 in Y , the value of ζ at the point η Y is
obtained by passing to the limit as (z, ζ) converges to η Y , hence is a root of
P (a, ζ) = 0.
On the contrary, if R 0 (0) = P 0 (a) = 0, then P 0 (X + a) = X
m Q 0 (X) with
Q 0 (0) = 0, the coefficients
s i (q) = P i
q
k + a
/q
mk Q 0
q
k
can have poles of order ≤ mk at the origin and so are O(q
−mk ). Thus so
is h(q) as well. The exact order of ζ at the point η Y can theoretically be
301
The rational functions
s i (q) = P i
q
k + a
/P 0
q
k + a
are meromorphic at the origin; these are elementary symmetric functions
of the roots of (4), up to sign. To show that h(q) has at most one pole at
q = 0, it suffices to show that there is an upper bound of the form h(q) =
O(q
−N ). Now, the coefficients s i (q) are of this form. Therefore, the following
result remains to be proved. It probably dates back to time immemorial and
for example can be found in Dieudonn´ e’s exercise in Calcul infinit´ esimal,
Chap. III :
Lemma 1. Let ζ i (1 ≤ i ≤ n) be the roots of an equation
ζ
n + c 1 ζ
n−1 + . . . + c n = 0
(4.5)
with complex coefficients. Then
sup |ζ i | ≤ max (1, |c 1 | + . . . + |c n |) .
(4.6)
Let M be the left hand side of (6). Each root of (5) satisfies
|ζ i |
n ≤ |c 1 | . |ζ i |
n−1 + . . . + |c n | ≤ |c 1 | .M
n−1 + . . . + |c n | ,
whence
M
n
≤ |c 1 | .M
n−1 + . . . + |c n | .
If M ≤ 1, there is nothing to prove. If M ≥ 1, we then write
M ≤ |c 1 | + . . . + |c n | /M
n−1
≤ |c 1 | + . . . + |c n | ,
qed.
Note that if P 0 (a) = 0, the coefficients s i (q) of (5) are holomorphic at
q = 0, and so are bounded for sufficiently small q, hence so is h(q). As
a result, the function F (z, ζ) = ζ is holomorphic at the point η Y ∈ ˆ
X if
P 0 (a) = 0, in other words if the equation P (a, Y ) = 0 is effectively of degree
n at the point a. Since P (z, ζ) = 0 in Y , the value of ζ at the point η Y is
obtained by passing to the limit as (z, ζ) converges to η Y , hence is a root of
P (a, ζ) = 0.
On the contrary, if R 0 (0) = P 0 (a) = 0, then P 0 (X + a) = X
m Q 0 (X) with
Q 0 (0) = 0, the coefficients
s i (q) = P i
q
k + a
/q
mk Q 0
q
k
can have poles of order ≤ mk at the origin and so are O(q
−mk ). Thus so
is h(q) as well. The exact order of ζ at the point η Y can theoretically be