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X – The Riemann Surface of an Algebraic Function
where T is an indeterminate, shows that the f i (z) are the roots of the polynomial
P
(T ) = T
k + c 1 (z)T
k−1 + . . . + c k (z) = 0
(4.11)
with coefficients in the ring of holomorphic functions on D (nothing to do
with the derivative of P ). But for all z ∈ D, the c i (z) are the elementary
symmetric functions of the f i (z), i.e.the roots of the equation P (z, ζ) = 0
such that (z, ζ) ∈ X
. It follows that the c i (z) are the same for all discs
D ⊂ B and hence are the restrictions to D of holomorphic functions on B.
We show that these are rational and to do this, that they only have polar
singularities at every a ∈ ˆ
C − B.
As in (ii), consider the disc D(a) = D ; over D
∗ , X is the union of open
connected sets Y ; hence the same holds for X
. If Y ⊂ X
and if q = ϕ Y (z, ζ)
is the corresponding local uniformizer, ζ is a meromorphic function of q on ˆ
Y
with at most one pole at q = 0. For z ∈ D
∗ the roots ζ i of P (z, ζ) = 0 such
that (z, ζ i ) ∈ Y give an upper bound ζ i = O(q
−N ) as q tends to 0. But if the
order of the covering space Y of D
∗ is equal to r, then q
r = z − a or 1/z as
the case may be, whence ζ = O((z − a)
−N ) or O(z
N ) for another integer N .
Hence elementary symmetric functions of ζ i = f i (z) such that (z, ζ i ) ∈ X
satisfy similar upper bounds properties. Thus, being holomorphic on B and
at most of polynomial growth in the neighbourhood of ˆ
C − B, the coefficients
of (11) are indeed rational functions of z.
As a result, for each connected component X
of X, (11) is an algebraic
equation with coefficients in the field K of rational functions of z. Denoting by
X j the connected components of X and by P j the corresponding polynomials,
it becomes clear that
P j (T ) =
(T − ζ) = T
n + s 1 (z)T
n−1 + . . . + s n (z) = P (z, T ) ,
(4.12)
where the product is extended to all roots ζ of P (z, ζ) = 0 and where, in
consequence, the coefficients are the rational functions s i (z) = P i (z)/P 0 (z)
already encountered in (4). Multiplied by P 0 (z), this identity between polynomials in T with coefficients in the field K = C(z) of one variable rational
fractions shows that in the ring K[T ], the P j (T ) divide P (T ). The coefficients
of the P j are not necessarily polynomials in z, but we know (exercise below)
that if P is irreducible, i.e. does not have any non-trivial divisors with coefficients in C[z], then neither does it have any with coefficients in C(z). Hence
there is a unique index j, and X is connected, qed.
Exercise 2. Let f (Y ) =
a k Y
k be a polynomial with coefficients a k ∈ Z.
The gcd c(f ) of these coefficients is said to be the content of f . (a) Let
f, g ∈ Z[Y ] and let p be a prime number dividing c(fg), i.e. all the coefficients
of fg. Show that p divides c(f ) or c(g). [As p divides a 0 b 0 , it divides a 0
or b 0 ; if p divides a 0 but not all the coefficients of f , let r be the largest
integer such that p divides a 0 , . . . , a r−1 . By calculating the coefficients of
Y
r , Y
r+1 , . . . in fg, show that p divides b 0 , then b 1 , etc.] (b) f is said to
X – The Riemann Surface of an Algebraic Function
where T is an indeterminate, shows that the f i (z) are the roots of the polynomial
P
(T ) = T
k + c 1 (z)T
k−1 + . . . + c k (z) = 0
(4.11)
with coefficients in the ring of holomorphic functions on D (nothing to do
with the derivative of P ). But for all z ∈ D, the c i (z) are the elementary
symmetric functions of the f i (z), i.e.the roots of the equation P (z, ζ) = 0
such that (z, ζ) ∈ X
. It follows that the c i (z) are the same for all discs
D ⊂ B and hence are the restrictions to D of holomorphic functions on B.
We show that these are rational and to do this, that they only have polar
singularities at every a ∈ ˆ
C − B.
As in (ii), consider the disc D(a) = D ; over D
∗ , X is the union of open
connected sets Y ; hence the same holds for X
. If Y ⊂ X
and if q = ϕ Y (z, ζ)
is the corresponding local uniformizer, ζ is a meromorphic function of q on ˆ
Y
with at most one pole at q = 0. For z ∈ D
∗ the roots ζ i of P (z, ζ) = 0 such
that (z, ζ i ) ∈ Y give an upper bound ζ i = O(q
−N ) as q tends to 0. But if the
order of the covering space Y of D
∗ is equal to r, then q
r = z − a or 1/z as
the case may be, whence ζ = O((z − a)
−N ) or O(z
N ) for another integer N .
Hence elementary symmetric functions of ζ i = f i (z) such that (z, ζ i ) ∈ X
satisfy similar upper bounds properties. Thus, being holomorphic on B and
at most of polynomial growth in the neighbourhood of ˆ
C − B, the coefficients
of (11) are indeed rational functions of z.
As a result, for each connected component X
of X, (11) is an algebraic
equation with coefficients in the field K of rational functions of z. Denoting by
X j the connected components of X and by P j the corresponding polynomials,
it becomes clear that
P j (T ) =
(T − ζ) = T
n + s 1 (z)T
n−1 + . . . + s n (z) = P (z, T ) ,
(4.12)
where the product is extended to all roots ζ of P (z, ζ) = 0 and where, in
consequence, the coefficients are the rational functions s i (z) = P i (z)/P 0 (z)
already encountered in (4). Multiplied by P 0 (z), this identity between polynomials in T with coefficients in the field K = C(z) of one variable rational
fractions shows that in the ring K[T ], the P j (T ) divide P (T ). The coefficients
of the P j are not necessarily polynomials in z, but we know (exercise below)
that if P is irreducible, i.e. does not have any non-trivial divisors with coefficients in C[z], then neither does it have any with coefficients in C(z). Hence
there is a unique index j, and X is connected, qed.
Exercise 2. Let f (Y ) =
a k Y
k be a polynomial with coefficients a k ∈ Z.
The gcd c(f ) of these coefficients is said to be the content of f . (a) Let
f, g ∈ Z[Y ] and let p be a prime number dividing c(fg), i.e. all the coefficients
of fg. Show that p divides c(f ) or c(g). [As p divides a 0 b 0 , it divides a 0
or b 0 ; if p divides a 0 but not all the coefficients of f , let r be the largest
integer such that p divides a 0 , . . . , a r−1 . By calculating the coefficients of
Y
r , Y
r+1 , . . . in fg, show that p divides b 0 , then b 1 , etc.] (b) f is said to