Abstract. Let K be a subfield of the meromorphic function field on C

A remark on algebraic addition theorems

Yukitaka Abe

Abstract. Let K be a subfield of the meromorphic function field on C

which has the transcendence degree n over C. There are two ways of stating an algebraic addition theorem for K. The author wrote the relation of these two ways in a book [Toroidal Groups, Yokohama Publishers, Inc., Yokohama, 2018]. However, the explanation was too rough. In this paper, the detailed complete proof is given.

1. Introduction

Let M(C

) be the field of meromorphic functions on C

. We studied and determined an algebraic function field K(⊂ M(C

)) of n variables over C ([1, 2]). This study makes Weierstrass’ statement established and clear. Weierstrass’ statement is stated in [6] as follows:

Tout syst` eme de n fonctions (ind´ ependantes) ` a n variables qui admet un th´ eor` eme d’addition est une combinaison alg´ ebrique de n fonctions ab´ eliennes (ou d´ eg´ en´ erescences) ` a n arguments et aux mˆ emes p´ eriodes.

We consider the following condition (T) for a subfield K of M(C

).

(T) K is finitely generated over C and Trans

K = n.

If K satisfies condition (T), then there exist f

, f

, . . . , f

∈ K such that K = C(f

, f

, . . . , f

) and f

, . . . , f

are algebraically independent

2010 Mathematics Subject Classification. 32A20 .

Key words and phrases. algebraic addition theorem, algebraic function fields of n variables .

37

over C. To simplify the description, we write C(F ) = C(f

, f

, . . . , f

) and C(F

) = C(f

, . . . , f

) setting F := {f

, f

, . . . , f

} and F

:= {f

, . . . , f

}.

Furthermore we write

C(F(x), F (y)) = C(f

(x), f

(x), . . . , f

(x), f

(y), f

(y), . . . , f

(y)), where x = (x

, . . . , x

) and y = (y

, . . . , y

) are two independent tuples of n complex variables. We note that there are two ways of stating an algebraic addition theorem. The first one is the following.

Definition 1. Let K = C(F ) be as above. We say that K admits (AAT) if f

(x + y) ∈ C(F(x), F (y)) for any j = 0, 1, . . . , n.

The above definition is independent of the choice of generators f

, f

, . . . , f

(cf. Lemma 6.2.2 in [3]).

An algebraic addition theorem stated in Weierstrass’ original statement is slightly different. That is the following type.

Definition 2. Let f

, . . . , f

∈ M(C

) be algebraically independent over C. We say that f

, . . . , f

admit (AAT

) if f

(x + y) is algebraic over C(F

(x), F

(y)) for all j = 1, . . . , n.

We assume that algebraically independent functions f

, . . . , f

∈ M(C

) admit (AAT

). Let K/C(F

) be a finite algebraic extension. If g

, . . . , g

∈ K are algebraically independent, then g

, . . . , g

admit (AAT

) (Lemma 6.2.6 in [3]). Therefore, (AAT

) for subfields K is defined as follows.

Definition 3. Let K be a subfield of M(C

) satisfying condition (T). We say that K admits (AAT

) if there exist f

, . . . , f

∈ K algebraically inde- pendent over C such that f

, . . . , f

admit (AAT

).

The following proposition is immediate from the definitions.

Proposition 1 (Proposition 6.2.8 in [3]) Let K be a subfield of M(C

) satisfying condition (T). If K admits (AAT), then it admits (AAT

).

The converse is the following theorem.

Theorem 1 (Proposition 6.2.9 in [3]) Let K be a subfield of M(C

) satisfying condition (T). If K admits (AAT

), then there exists an algebraic extension K ^f of K such that K ^f admits (AAT).

The statement of the above theorem is correct. However, the proof in [3]

was written too roughly. Recently, Baro, de Vicente and Otero [4] published an extension result for maps admitting an algebraic addition theorem. We note that a similar argument is needed to complete the proof of the above theorem. In [4] they considered germs of meromorphic maps. Therefore, they had to investigate carefully domains of convergence. Our objects are meromorphic functions on C

. Then, some parts of their argument are not needed in our case. Although Theorem 1 is considered as a corollary of the main theorem of [4], we will give the detailed proof of Theorem 1 modifying their argument.

2. Proof of Theorem 1

We assume that a subfield K of M(C

) satisfies condition (T) and admits (AAT

) through this section. We set K = C(F ), where F = {f

, f

, . . . , f

} and f

, . . . , f

are algebraically independent over C.

Lemma 1. There exists an open dense subset Ω of C

such that for any f ∈ K, f (x + a) is algebraic over K for all a ∈ Ω.

Proof. Let P

be the polar set of f

for i = 0, 1, . . . , n. We set P

:=

S

i=0

P

. Since K admits (AAT

), f

(x+y) is algebraic over C(F

(x), F

(y)) for any j = 1, . . . , n. Then there exists a non-zero polynomial of the mini- mal degree

P

(X) =

mj

X

k=0

A

(x, y)X

with A

(x, y) ∈ C[f

(x), . . . , f

(x), f

(y), . . . , f

(y)] such that A

(x, y), A

(x, y), . . . , A

(x, y) have no common divisor except constants and P

(f

(x+

y)) = 0. We define

N

:= {a ∈ C

\ P

; A

(x, a) = 0, k = 0, 1, . . . , m

}

and N := ^S

N

. If we set Ω := C

\ (P

∪ N ), then Ω is an open dense subset of C

. Take any a ∈ Ω. By the definition of Ω, f

(x + a) is algebraic over K for any j = 1, . . . , n. Since f

(x + a) is algebraic over C(F

(x + a)), it is algebraic over K.

Lemma 2. For any f ∈ K, f(−x) is algebraic over K.

Proof. It suffices to show that f

(−x) is algebraic over K for any j = 0, 1, . . . , n.

By the assumption, f

(x + y) is algebraic over C(F (x), F (y)). There- fore we have Trans

C(F (x), F (y), F (x + y)) = 2n. On the other hand we have Trans

C(F (x), F (x + y)) = 2n. Hence f

(y) is algebraic over C(F (x), F (x + y)). Take a ∈ Ω. Let y = −x + a. Then f

(−x + a) is algebraic over C(F (x), F (a)) = K. Substituting x + a for x, we obtain that f

(−x) is algebraic over C(F(x + a)). By Lemma 1, f

(x + a) is algebraic over K for all i = 0, 1, . . . , n. Thus we conclude that f

(−x) is algebraic over K.

Lemma 3. There exist a finite subset C of Ω ∪ {0} with 0 ∈ C and C = −C, and a finite number of functions A

, A

, . . . , A

∈ C({F(x + a), F (y + a); a ∈ C}) satisfying the following conditions.

(a) For any f ∈ K, f (x + y) is algebraic over C(A

, A

, . . . , A

).

(b) For any j = 0, 1, . . . , N we have

A

(x, y) = A

(x + a, y − a) (1) for any a ∈ C

.

Proof. Take any i = 0, 1, . . . , n. We set S

:= {0} and K

:= C(F (x), F (y)).

Let

P

(X) = X

+

`0

X

j=0

A

(x, y)X

, A

(x, y) ∈ K

,

be the minimal polynomial of f

(x + y) over K

. If all of A

satisfy (1), then we denote C

:= S

and A

(x, y) := A

(x, y) for j = 0, 1, . . . , `

. Otherwise, there exists a

∈ C

such that

Q

(X) := P

(X) −



 X

+

`0

X

j=0

A

(x + a

, y − a

)X



 6= 0.

Since Ω is dense in C

, we may assume a

∈ Ω. We set S

:= S

∪ {a

, −a

} and K

:= C({F (x + a), F (y + a); a ∈ S

}) in this case. Then K

⊂ K

and f

(x + y) is algebraic over K

. We take the minimal polynomial

P

(X) = X

+

`1

X

j=0

A

(x, y)X

of f

(x + y) over K

. Since deg Q

< `

+ 1, we have deg P

< deg P

. If all of A

satisfy (1), then we set C

:= S

and A

:= A

for j = 0, 1, . . . , `

.

If it is not the case, we repeat the above procedure. Then we obtain sequences {S

}, {K

} and {Q

} such that S

= S

∪{a+a

, a−a

; a ∈ S

} for some a

∈ Ω with Q

(X) 6= 0 and K

= C({F (x + a), F (y + a); a ∈ S

}). If Q

(X) 6= 0, then deg Q

< deg Q

. Therefore, this procedure stops by a finite number of steps. Let s be the first number with Q

(X) = 0. Then for the minimal polynomial

P

(X) = X

+

`s

X

j=0

A

(x, y)X

of f

(x + y) over K

, the coefficients A

, A

, . . . , A

satisfy (1). We set C

:= S

and A

:= A

for j = 0, 1, . . . , N

, where we write N

=

`

. Let C := ^S

C

and {A

, A

, . . . , A

} := ^S

{A

, A

, . . . , A

i

}.

Then we obtain the desired conclusion.

Proof of Theorem 1. Let C and A

, A

, . . . , A

∈ C({F (x + a), F (y + a); a ∈ C}) be as in Lemma 3. Take b ∈ C

such as A

(x, b) ∈ M(C

) for all j = 0, 1, . . . , N. We define Ω

:= ^T

(Ω − b − c). Then Ω

is also an open dense subset of C

. Let B

(x) := A

(x, b) for j = 0, 1, . . . , N. We define two fields K ^f and L by

K f := C({B

(x + a), B

(−x + a); a ∈ Ω

, j = 0, 1, . . . , N }) and

L := C({F(x + c), F (−x + c); c ∈ C}).

We show that K ^f ⊂ L and any f ∈ K f is algebraic over K. From (1) it follows that for any a ∈ C

A

(x + a, b) = A

(x, a + b) (2)

for j = 0, 1, . . . , N . Let a ∈ Ω

. Since a + b + c ∈ Ω for any c ∈ C, we have B

(x + a) = A

(x, a + b) ∈ C({F (x + c); c ∈ C}). By Lemma 1, any element of C({F (x + c); c ∈ C}) is algebraic over K. Because of C = −C, we have that B

(−x + a) ∈ C({F (−x + c); c ∈ C}) and any element of C({F(−x+c); c ∈ C}) is algebraic over C(F (−x)). Since f

(−x) is algebraic over K for j = 0, 1, . . . , n (Lemma 2), any element of C({F (−x +c); c ∈ C}) is algebraic over K.

Next we show that K ^f admits (AAT). Since Trans

^f K = n, we can take g

, g

, . . . , g

∈ K f such that K ^f = C(G) and g

, . . . , g

are algebraically independent over C. Let f ∈ K. It is obvious by the definition of f K ^f that f(x + a) ∈ K f for any a ∈ Ω

. We define g(x, y) := f (x + y). Then, g(x, a) ∈ C(G(x)) for any a ∈ Ω

. Similarly, we have g(a, y) ∈ C(G(y)) for any a ∈ Ω

. It follows from Theorem 3 in [5] (the proof is the same as that of Theorem 6.6.5 in [3]) that f (x + y) = g(x, y) ∈ C(G(x), G(y)) on Ω

× Ω

. Since Ω

× Ω

is an open dense subset of C

× C

, it holds on

C

× C

by the uniqueness theorem. 2

References

[1] Y. Abe, A statement of Weierstrass on meromorphic functions which admit an algebraic addition theorem, J. Math. Soc. Japan, 57(2005), 709-723.

[2] Y. Abe, Explicit representation of degenerate abelian functions and related topics, Far East J. Math. Sci. (FJMS), 70(2012), 321-336.

[3] Y. Abe, Toroidal Groups, Yokohama Publishers, Inc., Yokohama, 2018.

[4] E. Baro, J. de Vicente and M. Otero, An extension result for maps

admitting an algebraic addition theorem, J. Geom. Anal., 29(2019),

316-327.

[5] S. Bochner, On the addition theorem for multiply periodic functions, Proc. Am. Math. Soc., 3(1952), 99-106.

[6] P. Painlev´ e, Sur les fonctions qui admettent un th´ eor` eme d’addition, Acta Math., 27(1903), 1-54.

Yukitaka Abe

Department of Mathematics Faculty of Science

University of Toyama

Gofuku, Toyama 930-8555, JAPAN e-mail: [email protected]

(Received December 3, 2019)