An analogue of Hilbert’s tenth problem for fields of

meromorphic functions over non-Archimedean valued

fields

Xavier VIDAUX

Journal of Number Theory, Volume 101, Issue 1 (2003) pp 48-73.

Abstract

Let K be a complete and algebraically closed valued field of characteristic 0.We prove that the set of rational integers is positive existentially definable in thefield M of meromorphic functions on K in the language L∗

z of rings augmentedby a constant symbol for the independent variable z and by a symbol for theunary relation “the function x takes the value 0 at 0”. Consequently, we provethat the positive existential theory of M in the language L∗

z is undecidable. Inorder to obtain these results, we obtain a complete characterization of all analyticprojective maps (over K) from an elliptic curve E minus a point to E , for anyelliptic curve defined over the field of constants.

This research was supported in part by the greek foundation of state scholarship(IKY) and was done at the university of Crete whose hospitality I acknowledge.

1

Contents

1 Introduction 3

1.1 History - Setting of the problems . . . . . . . . . . . . . . . . . . . . . 3

1.2 Main results - Related problems . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Sketch of the proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Rationality of analytic projective maps from an elliptic curve E minusa point to E 6

2.1 Relation between the speeds of convergence of the general term and ofthe partial sums of an ultrametric power series. . . . . . . . . . . . . . 6

2.2 Meromorphic functions on K − {0} . . . . . . . . . . . . . . . . . . . . 7

2.3 Extensions of Berkovich’s Theorem for elliptic curves . . . . . . . . . . 13

2.4 Rationality of global meromorphic solutions of Equation (MD) . . . . . 14

2.5 Proof of Theorem A . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 Proof of Theorem B 21

2

Notation : The letter K will denote a complete and algebraically closed valuedfield of characteristic 0 and the letter M will denote the field of meromorphic functionson K, of the variable z. By Lz we will denote the language of rings augmented by aconstant symbol for the variable z, and by L∗

z we will denote the augmentation of Lz

by a symbol for the unary relation “ord0(x) > 0” (i.e. : the function x takes the value0 at 0).

1 Introduction

1.1 History - Setting of the problems

Hilbert’s tenth problem asked for an algorithm which decides, for any given Diophantineequation (polynomial equation in several variables with integer coefficients), whetherthe equation has or does not have integer solutions. Y. Matiyasevich proved in 1970that Hilbert’s tenth problem has a negative answer (see [5, 17]). In the terminology ofmathematical logic, the positive existential theory of the ring Z of rational integers isundecidable (see [4]).

The analogue of Hilbert’s tenth problem for the field Q of rational numbers is stillan open problem. The first analogues of Hilbert’s tenth problem for rings other than thering of rational integers were obtained by J. Denef (see, for instance, [7]). He obtainedundecidability results for various rings of algebraic integers (e.g. : Z[i]).

Extensions to rings or fields of functions :The language Lz is a natural language in which one may consider problems analo-gous to Hilbert’s tenth problem for rings of functions of the variable z. Let R be such aring. The analogue of Hilbert’s tenth problem for R in the language Lz is the following :

Is there an algorithm which, given any finite system of polynomial equations

(fi = 0)i=1,...,k

where fi ∈ Z[z, T1, . . . , Tn], determines whether the system has or does not have a solu-tion in R?

If the answer is yes, then we say that the positive existential theory of R is decidable,else it is undecidable. The similar problem for systems (fi = 0)i=1,...,k ∧ g 6= 0 whichcontain also inequalities is the question whether the existential theory of R in Lz isdecidable or undecidable. The decidability question for the existential theory of R inL∗

z is the similar problem where one allows the addition to the system of conditions ofrelations of the form ord0(x) > 0, which are interpreted as “the function x takes thevalue 0 at z = 0”.

A natural language in which one may consider questions of decidability is Lz,C whichresults from the augmentation of Lz by the predicate C(x) for “x is a constant”. In thefollowing tables, we list some results about some rings of interest (if not specified, inthe language Lz,C). We denote by A(X) the ring of functions, analytic on an open setcontaining the set X and by M(X) the field of fractions of A(X). By U we denote theunit disc (with or without boundary), and by a an element of X.

3

Complex case :

C[z] C(z) A(C) M(C) A(U) A({a})pos. ex. th. und. [8] ? ? ? ? dec.

ex. th. und. ? ? ? ? dec.full th. und. ? und. ? und. [21] dec.

ex. th. in L∗z und. und. ? ? ? ?

The existential decidability of the ring A({a}) is a consequence of [14, §7, Thm. 3].The same theorem implies the analogous result in the p-adic case.

L. Rubel, in [22], proves that the positive existential theory of A(U), in the languageof rings augmented by a symbol for the relation “x is not a constant”, is decidable. Noanalogue of this result is known for the p-adic case. Some other relevant results can befound in [13, 25, 26].

Non-Archimedean case :

The following table lists the analogous state of affairs in the p-adic case. We denoteby Cp the completion of an algebraic closure of the field of p-adic numbers. By Ap(X)and Mp(X) we denote respectively the ring of analytic functions and the field of mero-morphic functions on the disc X ⊂ Cp.

Cp[z] Cp(z) Ap(Cp) Mp(Cp) Ap(U) Ap({a})pos. ex. th. und. [8] ? und. [16] ? ? dec.

ex. th. und. ? und. ? ? dec.full th. und. ? und. ? ? dec.

ex. th. in L∗z und. und. und. und. ? ?

The decidability result for the positive existential theory of A({a}) and Ap({a}) followsfrom [10, 11] and Artin’s approximation (see [1] and [18, Sect. 7]). No relevant resultsare known for the fields of fractions M(U), M({a}), Mp(U) and Mp({a}) in any ofthe languages Lz, L

∗z or Lz,C.

We also note that the power series rings C[[z]] and Cp[[z]], as well as their fieldof fractions C((z)) and Cp((z)), have decidable theories in Lz,C (see [14]). The ringC[[z, w]] of power series in two variables has an undecidable full theory in Lz,C ∪ {w}(see [6]).

Finally, observe that the existential theories of A({a}) and Ap({a}) are undecidablein the language of rings augmented by a symbol d

dzfor the derivation, since one can

define the integers by an existential formula :

c ∈ Z ⇐⇒ C(c) ∧ ∃x(x 6= 0 ∧ cdx

dz= x).

In this paper we prove the undecidability of the positive existential theory of thefield Mp(Cp), in the language L∗

z. This result follows from Theorem B (see below).

1.2 Main results - Related problems

Let us give a few comments on our choice of the language L∗z. Here we deal with solving

systems of the form

Pi(f1, · · · , fk) = 0, ord0(fj) > 0, i = 1, · · · , n and j = 1, · · · , k

4

where Pi, for i = 1, · · · , n, are polynomials with coefficients in Z(z), and the variablesf1,. . . , fk range in M. One may consider that we deal with systems of differential equa-tions “of order zero”, together with initial conditions (ord0(fj) > 0). From this pointof view, the language L∗

z is of wide use in everyday mathematical practice. Of course itwould be preferable to obtain the analogue of our theorem (see below) in the languageLz. However, the problem of defining existentially the relation ord0(f) > 0 in the lan-guage Lz over a field of functions is in general not trivial. In particular the problemis open for C(z) as well as for the field of complex meromorphic functions (see [18, §2.5]).

We prove the following theorems :

Theorem A Let K be a complete algebraically closed valued field of characteristic 0.Let E be an elliptic curve defined over K and let P ∈ E . Any analytic projective map(over K) on E − {P} and into E is rational.

Theorem A contrasts sharply the complex case in which there are many non-rationalanalytic projective maps from E − {P} into E .

Theorem B Let K be a complete algebraically closed valued field of characteristic 0and M the field of meromorphic functions on K. Let L∗

z be the language of rings aug-mented by a constant symbol for the variable z and by a symbol for the unary relationord0(x) > 0 (i.e. : the function x takes the value 0 at 0).(a) The set Z of rational integers is positive existentially definable in M, in L∗

z.(b) The positive existential theory of M in L∗

z is undecidable.

Fact (a) of Theorem B implies Fact (b) by Matiyasevich’s Theorem. A consequenceof our result is the similar result for global analytic functions, in the language Lz, whichwas proved before (see [16]). The analogue of Theorem B in the language Lz, as wellas the analogue for the field M(D) of meromorphic functions on a disc D, are openproblems. The analogue of Theorem B for complex meromorphic functions is also open.

1.3 Sketch of the proof

We obtain Theorems A and B by proving that equations of the form

(MD) (z3 + δz2 + z)y2 = x3 + δx2 + x

where z is the independent variable, have only rational solutions over M (Theorem2.22), for δ ∈ K − {±2}. This result together with the fact that elliptic curves over Kcannot be parametrized by global meromorphic functions (see Theorem 2.11) impliesTheorem A. Theorem B follows from Theorem A (actually from the weaker Theorem2.22) by techniques developed by J. Denef in [8].

In order to prove Theorem 2.22, we do as follows : let E be the elliptic curve

(1) Y 2 = X3 + δX2 + X

5

and let s be such that s2 = z3 + δz2 + z. A pair (x, y) ∈ M2 is a solution of Equation(MD) if and only if (x, sy) is a solution of (1). The rational solutions of Equation (MD)are of the form (xn, yn) where (z, s) 7→ (xn(z), syn(z)) is the n-th endomorphism of E ,for n 6= 0. For any endomorphism n 6= 0 of E , we have the property

x2n(z−1) = x2n(z) and x2n+1(z−1) = x−1

2n+1(z).

Now let (x, y) be a solution of Equation (MD) over M. We consider the map τ definedby

τ(x, sy) = (x(z−1),−s

z2y(z−1))

and we look at the quantity (x, sy) = (x, sy) ⊖ τ(x, sy) in order to understand howclose a solution over M is to having the mentioned property of rational solutions. Itwill turn out that (x, sy) is a point of order 1 or 2 on E . From this result, we will haveenough information on x and y to conclude that they have to be rational (see Sect. 2.4).

In order to see that the quantity (x, sy) has to be a point of order 1 or 2, weinvestigate the solutions of Equation (MD) over the field M∗ of functions which aremeromorphic on K∗ = K − {0}. For this purpose, crucial are the results of Sect. 2.1and 2.2.

2 Rationality of analytic projective maps from an

elliptic curve E minus a point to E

2.1 Relation between the speeds of convergence of the gen-eral term and of the partial sums of an ultrametric power

series.

In what follows, we denote by | · | the ultrametric absolute value of the field K. Sincethe field K is complete, a power series in K[[T ]] converges if and only if its generalterm converges to zero. The following proposition gives a relation between the speed ofconvergence of the general term and the speed of convergence of the partial sums of agiven numerical series. First observe that if H =

∑

n≥0 hn is a convergent series, thenthe set {|hn| | n ∈ N} has a maximum.

Proposition 2.1 Let (hn)n≥0 be a sequence of elements of K such that n√

|hn| convergesto 0 as n goes to infinity. Let Sn denote the sequence

∑n

i=0 hi. Then S =∑

i≥0 hi exists

and the sequence n√

|Sn − S| converges to 0 as n goes to infinity.

6

Proof : Clearly the limit S of Sn exists. For any integer n, we have :

n√

|Sn − S| = n

√

√

√

√

∣

∣

∣

∣

∣

∑

i≥n+1

hi

∣

∣

∣

∣

∣

≤ n

√

maxi≥n+1

|hi|

= n

√

maxj≥1

|hn+j|

= n

√

|hn+j0|

=

(

n+j0

√

|hn+j0|

)

n+j0n

−−−→n→∞

0

for some integer j0 ≥ 1. Note that the integer j0 depends a priori on n, but only thefact that it is positive matters. 3

Corollary 2.2 Let∑

hnTn be a power series in K[[T ]] with infinite radius of conver-

gence, such that∑

hn = S, and let Sn denote the partial sum∑n

i=0 hi. Then the powerseries

∑

(Sn − S)T n has infinite radius of convergence.

Actually we can obtain something better (this is what we will need later on) :

Lemma 2.3 For any integers n and k, let cn,k be an element of the valuation ring ofK (that is |cn,k| ≤ 1). Assume that (hn)n≥0 is a sequence of elements of K such thatn√

|hn| converges to 0 as n goes to infinity. For each n, let Rn denote the sequence

∑

k≥n

cn,khk.

Then the sequence n√

|Rn| converges to 0.

Proof : Note that if the cn,k are all 1, then we obtain Proposition 2.1, setting Rn =S − Sn−1. The proof of the generic case is similar to that of Proposition 2.1 and is leftto the reader. 3

2.2 Meromorphic functions on K − {0}

We refer to [12], [15], or [20] for the following definitions and basic facts. A function his analytic on K if there exists a formal power series

∑

n∈NhnT

n ∈ K[[T ]] convergingeverywhere in K such that, for all z ∈ K, we have h(z) =

∑

n∈Nhnzn. We will say that

a function is meromorphic on K if it can be written as the quotient of two analyticfunctions on K. Let K∗ denote the set K−{0}. A function h is analytic on K∗ if thereexists a formal Laurent series

∑

n∈ZhnT

n ∈ K[[T, T−1]] converging everywhere in K∗

such that, for all z ∈ K∗, we have h(z) =∑

n∈Zhnzn. We will say that a function is

meromorphic on K∗ if it can be written as the quotient of two analytic functions onK∗. Analytic (resp. meromorphic) functions on K will be called also global analyticfunctions (over K) (resp. global meromorphic functions (over K)).

7

The integral domain of analytic functions on K (resp. on K∗) will be denoted by A(resp. by A∗) and the field of fractions of A (resp. of A∗) will be denoted by M (resp.by M∗).

We will gather what we need in the following lemma. The order of a zero or a poleof a meromorphic function is well defined. See [20, chap.6, p. 305], for a proof of thisfact and of the following.

Lemma 2.4 1. A meromorphic function f in M or in M∗ has no accumulation ofpoles and, if f 6= 0, then f has no accumulation of zeros.

2. A global analytic function either is a polynomial or has infinitely many zeros.

3. A global meromorphic function avoids at most one value, in which case it can bewritten as C + 1

hwhere C is the avoided value and h is a global analytic function.

4. A meromorphic function without any pole is an analytic function.

5. A global meromorphic function having finitely many zeros and finitely many polesis a rational function.

6. If a global meromorphic function has infinitely many zeros (resp. poles), then theycan be arranged as a sequence of zeros (resp. poles) whose absolute value goes toinfinity.

7. Meromorphic functions in M or in M∗ can be written as the quotient of twoanalytic functions with no common zeros.

We will say that a function f : K∗ 7→ K is invariant under the map z 7→ z−1 if, forany z in K∗, we have f(z) = f(1

z).

Lemma 2.5 Let h and f be two meromorphic functions on K. If for every non-zeroz in K, we have h(z) = f(z−1), then the functions h and f are rational functions. Ifthe functions h and f are analytic functions on K, then they are constant.

Proof : Observe that if 6= 0 is a zero (resp. a pole) of h, then 1

is a zero (resp. a pole)

of f . Assume that h has infinitely many zeros (resp. poles). Then it has a sequencen of zeros (resp. poles) whose absolute value converges to infinity. Then the sequence1n

is a sequence of zeros (resp. poles) of f whose absolute value converges to 0, andthis is not possible. Consequently the function h, therefore f as well, has finitely manyzeros and finitely many poles. So they must be rational functions. This finishes theproof of the first assertion. The second assertion is then easy and is left to the reader. 3

The next lemma is a consequence of the uniqueness of the expansion of analyticfunctions on K∗ (see [15]).

Lemma 2.6 Let h =∑

n∈Zanzn be a non-zero analytic function on K∗, invariant

under z 7→ z−1. Then for all n ∈ Z we have an = a−n.

8

Lemma 2.7 Let h 6= 0 be a function in A∗. Then h can be written as

h(z) = Czm∏

|ρ|≥1

(

1 −z

ρ

)νρ∏

|ρ|<1

(

1 −ρ

z

)νρ

,

the products being taken over all non-zero zeros ρ of h, the integer νρ being the multi-plicity of h at ρ.

Proof : See [20, chap. 6, p. 320]. 3

Lemma 2.8 Let h be an analytic function on K∗, invariant under z 7→ z−1. Thenthere exists a unique function g, analytic on K, such that for all z in K∗, g satisfies

h(z) = g(z + z−1)

Proof : Let us write w = z + z−1. By lemma 2.6, we have :

h(z) = a0 +∑

n≥1

anz−n +

∑

n≥1

anzn

Fix an integer N ≥ 1. Let us write

hN(z) = a0 +N∑

n=1

anz−n +

N∑

n=1

anzn

Clearly, the function hN(z) is a polynomial in w of degree at most N . Let us write thispolynomial as

GN(w) =N∑

n=0

bn,Nwn

If there exists an integer N0 such that, for any integer n > N0, we have an = 0, thenhN0

= h, and in this case, the polynomial GN0is the function g we are looking for.

From now on, we will suppose that infinitely many an are not 0.

The proof is done in three steps. First, for N fixed, we will express bn,N as alinear combination of the an’s with integer coefficients, and this will imply that, forany fixed integer n, the sequence bn,N converges as N goes to infinity. We will writebn = limN→+∞ bn,N . Secondly, we will prove that the function g defined by the powerseries

∑

n≥0 bnT n is an analytic function on K. Finally, we will prove that this functionsatisfies g(z + z−1) = h(z). In other words, we will find the function g by successiveapproximation of the coefficients of its power series expansion.

Fix the integer N . Let k be any non-negative integer such that n+2k ≤ N . Write :

(z + z−1)n+2k =n+2k∑

j=0

(

n + 2k

j

)

zn+2k−jz−j

=n+2k∑

j=0

(

n + 2k

j

)

zn+2k−2j

9

We observe that, on the right hand side of the equation, the term zn corresponds toj = k and the term z−n corresponds to j = n + k. Then the coefficient of zn is

(

n+2k

k

)

and the coefficient of z−n is(

n+2k

n+k

)

. Also observe that

(

n + 2k

k

)

=

(

n + 2k

n + k

)

Since the coefficients of zn + z−n in hN(z) and GN(z + z−1) must be equal, we see thatthe unknowns bn,N satisfy the following system of N + 1 equations :

(SN) an =

[ N−n2

]∑

k=0

(

n + 2k

k

)

bn+2k,N , for n = 0, . . . , N,

where [x] denotes the integral part of x. This is a triangular system of N +1 equationsand N + 1 unknowns (the bn+2k,N). We obtain bN,N from the equation aN = bN,N , weobtain bN−1,N from the equation aN−1 =

(

N−10

)

bN−1,N , and we obtain bn,N from

an =

(

n

0

)

bn,N +

(

n + 2

1

)

bn+2,N + · · ·

Observe that since(

n

0

)

= 1, we have

bn,N =

N∑

k=n

cn,N,kak

for some rational integers cn,N,k. These cn,N,k depend a priori on N . In order to seethat they are actually independent of N , we treat an and bn as variables, we considerthe expression of an in the systems SN and SN+2, and we subtract them :

0 = an − an (1)

=

[ N+2−n2

]∑

k=0

(

n + 2k

k

)

bn+2k,N+2 −

[ N−n2

]∑

k=0

(

n + 2k

k

)

bn+2k,N

=

[ N−n2

]∑

k=0

((

n + 2k

k

)

(bn+2k,N+2 − bn+2k,N)

)

+

(

N + 2

[N+2−n2

]

)

bN+2,N+2

For n = N , we obtain k = 0 and

0 = bN,N+2 − bN,N +

(

N + 2

1

)

bN+2,N+2

that is

bN,N+2 − bN,N = −

(

N + 2

1

)

aN+2

By downwards induction (n = N −2, . . .), we have, for any integer n such that 0 ≤ n ≤N ,

bn,N+2 − bn,N = c · aN+2

10

for some integer c depending on n and N . This proves that ak, for k = n, . . . , N , appearswith the same coefficient in bn,N and in bn,N+2 and we conclude that the integers cn,N,k

do not depend on N . So we can write cn,N,k = cn,k, and then

bn,N =N∑

k=n

cn,kak

Let us now fix the integer n. Since h converges on K∗, the sequence |ak| convergesto 0 as k goes to infinity. Since the coefficients cn,k are integers, the sequence |cn,kak|converges also to zero as k goes to infinity and then the sequence bn,N converges as Ngoes to infinity; let bn denote the limit of bn,N as N goes to infinity, that is

bn =∑

k≥n

cn,kak

Let g(T ) denote the formal power series :

g(T ) =∑

n≥0

bnT n ∈ K[[T ]]

By Lemma 2.3, the power series g(T ) has an infinite radius of convergence (applyLemma 2.3 for hk = ak and Rn = bn).

It remains to show that g(z + z−1) = h(z) for any non-zero z in K. Let us write :

gN(w) =

N∑

n=0

bnwn

On the one hand, for any integer N and any z ∈ K∗, we have

GN(z + z−1) = hN(z)

and then, as N goes to infinity, GN(z + z−1) converges to h(z). On the other hand,gN(z + z−1) converges to g(z + z−1) as N goes to infinity. So we have to prove thatgN(w) − GN(w) converges to 0 as N goes to infinity. Fix an arbitrary N . We have :

|gN(w) − GN(w)| =

∣

∣

∣

∣

∣

N∑

n=0

bnwn −

N∑

n=0

bn,Nwn

∣

∣

∣

∣

∣

≤N

maxn=0

{|bn − bn,N ||w|n}

=N

maxn=0

{∣

∣

∣

∣

∣

∑

k≥N+1

cn,kak

∣

∣

∣

∣

∣

.|w|n

}

Let u(N) denote the integer n for which the maximum is reached. Observe that wehave 0 ≤ u(N) ≤ N . The inequality becomes :

|gN(w) − GN (w)| ≤

∣

∣

∣

∣

∣

∑

k≥N+1

cu(N),k · ak

∣

∣

∣

∣

∣

|w|u(N) = |RN+1| · |w|u(N)

11

whereRN+1 =

∑

k≥N+1

cu(N),k · ak

Apply Lemma 2.3 with n replaced by N + 1 and hk replaced by ak, to obtain :

• If |w| < 1 then we have : |RN+1| · |w|u(N) −−−→N→∞

0

• If |w| ≥ 1 then we have |w|u(N) ≤ |w|N , and so

|RN+1| · |w|u(N) ≤ |RN+1| · |w|N

By Lemma 2.3, we have N√

|RN+1| −−−→N→∞

0, thus

N√

|RN+1| · |w|N −−−→N→∞

0

which implies that|RN+1| · |w|N −−−→

N→∞0

The uniqueness of g is easily seen (by the fact that two distinct meromorphic func-tions cannot take the same value at each point of a set containing an accumulationpoint). 3

Lemma 2.9 Let h be a meromorphic function on K∗ invariant under the map z 7→ z−1.There exist functions h1 and h2 analytic on K∗ and invariant under z 7→ z−1 such thath = h1

h2.

Proof : We use Lemmas 2.7 and 2.4. Let us write h as the quotient h1

h2of two functions

in A∗ with no common zeros. We can suppose without loss of generality that the factorzm does not appear in the product expansion of h2 (see Lemma 2.7, if m 6= 0 , wemultiply both h1 and h2 by z−m). If is any element of K∗ and n a positive integer, is a zero of order n of h1 if and only if is a zero of order n of h, which happens if andonly if 1

is a zero of order n of h, and this happens if and only if 1

is a zero of order n

of h1.Let us write

π(z) =∏

|ρ|=1h1(ρ)=0

(

1 −z

ρ

)νρ

π+(z) =∏

|ρ|>1h1(ρ)=0

(

1 −z

ρ

)νρ

and π−(z) =∏

|ρ|<1h1(ρ)=0

(

1 −ρ

z

)νρ

By Lemma 2.7 we have, for some constant C and integer m

(⋆) h1(z) = Czmπ(z)π+(z)π−(z)

For an arbitrary ρ 6= 0, writing ρ = 1µ, we have seen that h1(ρ) = 0 if and only if

h1(µ) = 0 and νρ = νµ. Then we have

π(z) =∏

|ρ|=1h1(ρ)=0

(

1 −z

ρ

)νρ

=∏

|µ|=1h1(µ)=0

(1 − zµ)νµ = π(z−1)

12

π+(z) =∏

|ρ|>1h1(ρ)=0

(

1 −z

ρ

)νρ

=∏

|µ|<1h1(µ)=0

(1 − zµ)νµ = π−(z−1)

and

π−(z) =∏

|ρ|<1h1(ρ)=0

(

1 −ρ

z

)νρ

=∏

|µ|>1h1(µ)=0

(1 −1

zµ)νµ = π+(z−1)

Therefore we have

h1(z)

Czm= π(z)π+(z)π−(z) = π(z−1)π−(z−1)π+(z−1) =

h1(z−1)

Cz−m

which implies thath1(z

−1) = z−2mh1(z)

We prove in the same way that h2(z−1) = h2(z) (we do not have for h2 the factor

zm because we have supposed that z does not divide h2). Thus we have

h(z) = h(z−1) =h1(z

−1)

h2(z−1)=

z−2mh1(z)

h2(z)= z−2mh(z),

which implies that m = 0. 3

Corollary 2.10 Let h be a meromorphic function on K∗ invariant under z 7→ z−1.Then there exists a function g in M such that, for all z in K∗, g satisfies h(z) =g(z + z−1).

Proof : By Lemma 2.9, h can be written as h1

h2, for some functions h1 and h2 in A∗

invariant under z 7→ z−1. By lemma 2.8, there exist functions g1 and g2 in A such that,for all z ∈ K∗, h1(z) = g1(z + z−1) and h2(z) = g2(z + z−1). Writing g = g1

g2∈ M, we

have h(z) = g(z + z−1). 3

2.3 Extensions of Berkovich’s Theorem for elliptic curves

From now on the letter F denotes the polynomial T 3 + δT 2 + T .

Theorem 2.11 (Berkovich) Let x and y be two meromorphic functions on K whichsatisfy Equation (1), that is,

y2 = x3 + δx2 + x

Then x and y are constant.

Proof : See [2, Chap. 4, Thm. 4.5.1] for a proof of a more general result; [3, Cor. A,p. 753] for a proof involving p-adic Nevanlinna’s theory; or [24, Sect. 8] for a proof usinga characterization of non-global solutions of (1). 3

Theorem 2.12 Let x and y be two meromorphic functions on K∗, invariant underz 7→ z−1, which satisfy Equation (1), that is,

y2 = x3 + δx2 + x

Then x and y are constant.

13

Proof : By Corollary 2.10, there exist functions g1 and g2 in M such that, for allz ∈ K∗, x(z) = g1(z + z−1) and y(z) = g2(z + z−1). It is obvious that (g1, g2) satisfiesthe same equation as (x, y). Thus by Berkovich’s Theorem, g1 and g2 must be constant.Therefore, x and y must be constant. 3

Theorem 2.13 Let x and y be two meromorphic functions on K∗, invariant underz 7→ z−1, which satisfy

(z + δ + z−1)y2 = x3 + δx2 + x

Then (x, y) is a point of order 2 of E .

Proof : Write w = z + z−1. We know by Corollary 2.10 that there exist global mero-morphic functions g and h such that x(z) = g(w) and y(z) = h(w). Then, in terms ofthe variable w, our equation becomes

(w + δ)h2 = F (g)

Set w = t2 − δ, then the equation becomes

(t · h ◦ (t2 − δ))2 = F (g ◦ (t2 − δ))

and we can conclude by Berkovich’s Theorem that both functions t · h ◦ (t2 − δ) andg◦(t2−δ) are constant. Since h◦(t2−δ) = a constant

tis, as a function of t, both even and

odd, the constant must be zero. Since t2 − δ is surjective (K is algebraically closed),we conclude that h = 0, and then y = 0. Then F (x) = 0, and so x can take onlythree values, those corresponding to the roots of the polynomial F . Thus x is constant.Finally, we obtain that (x, y) is one of the three points of order 2 of E : (0, 0), (ξ, 0) or(ξ−1, 0), where ξ is one of the non-zero roots of the polynomial F . 3

2.4 Rationality of global meromorphic solutions of Equation(MD)

We will study now equations of the form

(MD) (z3 + δz2 + z)y2 = x3 + δx2 + x

where z is the independent variable, x and y are functions of z. We fix a δ ∈ K−{±2},so that Equation (MD) defines an elliptic curve E∗. These curves have been introducedby Y. Manin and J. Denef. We show that Equation (MD) has only rational solutionsover M (see Theorem 2.22).

Remark 2.14 Let us consider, over any field, a solution (x, y) of Equation (MD) :

F (z)y2 = F (x)

Let s be an element of an algebraic closure of K(z) such that s2 = F (z). Then (x, sy)is a point on the elliptic curve E (it is a solution of Equation (1)). From now on, (x, sy)is both a point on E , considered as an elliptic curve over M, and a map from E − {∞}to E :

(z, s) 7→ (x(z), sy(z))

14

Let us fix a solution (x, y) of Equation (MD) over the field M. Observe that thecomposition of elements of M is not, in general, in M. But considering a function hin M as a meromorphic function on K∗, we can compose it with the function of M∗

which sends z to z−1; the function h ◦ (z−1) still lies in M∗.The map ι : z 7→ z−1 is, obviously, an automorphism over K of the field K(z)

of rational functions. This automorphism ι extends to an automorphism of the fieldextension K(z, s) of K(z) in two ways. Since

ι(s2) = ι(F (z)) = F (z−1) =1

z4F (z) =

s2

z4,

s may be mapped to any of ± sz2 . Let ι denote the automorphism of K(z, s) which sends

s to − sz2 . Observe that, under composition, ι is nilpotent of order 2, that is, ι ◦ ι is the

identity function.If (z, s) ∈ E(K∗), then the pair (ι(z), ι(s)) lies in E(K∗). Therefore, to ι corresponds

naturally a map τ0 : E(K∗) → E(K∗), which sends the point (z, s) to

(ι(z), ι(s)) = (z−1,−s

z2).

Obviously, the map τ0 is of order 2, that is τ0 ◦ τ0 is the identity of E(K∗). Letτ : E(M) → E(M∗) denote the map which sends the point (x, sy) to

(x ◦ z−1,−s

z2y ◦ z−1) = (x, sy) ◦ τ0

Observe that, since τ(x, sy) is a solution of Equation (1) over M∗, the image of themap τ is included in E(M∗). Since the map τ0 is of order 2, the map τ , also, is of order2.

Remark 2.15 Consider (z, s) as a point on the elliptic curve E . Note that, by theaddition law on E , we have

(z, s) ⊕ (0, 0) = (z−1,−s

z2)

Then, considering (x, sy) as a function of the variable (z, s), we could also define themap τ by :

τ(x, sy) = (x, sy) ◦ [(z, s) ⊕ (0, 0)]

If P is any point on the elliptic curve E (over any field), we will write ±P to meanthat we consider either the point P or its opposite ⊖P .

Lemma 2.16 For any solution (x, y) of Equation (MD), with x and y in M∗, we have :

1. τ(2(x, sy)) = 2τ(x, sy)

2. τ(2(x, sy) ⊕ (z, s)) = 2τ(x, sy) ⊕ τ(z, s)

3. If (x, sy) 6= ±τ(x, sy) then τ((x, sy) ⊖ τ(x, sy)) = τ(x, sy) ⊖ (x, sy)

Proof : In order to simplify the formulae, we will write x = x ◦ z−1 and y = y ◦ z−1.See [23, Chap. III, p. 58] for the general algorithm giving the addition formula of anelliptic curve.

15

1. Write (a, sb) = 2(x, sy) and (u, v) = τ(2(x, sy)). By the addition law on E , wehave

a =

(

3x2 + 2δx + 1

2sy

)2

− δ − 2x

and

sb = −(3x2 + 2δx + 1)(x2 − 1)2 − 4s2y2(x3 − x)

8y3s3

So we have :

u =

(

3x2 + 2δx + 1

−2 sz2 y

)2

− δ − 2x

and

v = −(3x2 + 2δx + 1)(x2 − 1)2 − 4 s2

z4 y2(x3 − x)

−8y3 s3

z6

Therefore we have

τ(2(x, sy)) = 2(x,−s

z2y) = 2τ(x, sy)

2. Write

(a, sb) = 2(x, sy)

(u, v) = (a, sb) ⊕ (z, s)

(t, w) = τ((a, sb) ⊕ (z, s))

a = a ◦ z−1 and b = b ◦ z−1. Note that a cannot be equal to z, because (z, s)cannot be written as the double of a solution (see Definition 2.18 and Lemma2.19). We have

u =

(

s − sb

z − a

)2

− δ − z − a

and

v = −

(

s − sb

z − a

)

u(z, s) −sbz − as

z − a

Therefore we have

t =

(

− sz2 + s

z2 b

z−1 − a

)2

− δ − z−1 − a

and

w = −

(

− sz2 + s

z2 b

z−1 − a

)

u(z−1,−s

z2) −

− sz2 bz

−1 + a sz2

z−1 − a

We conclude that

(t, w) = (a,−s

z2b) ⊕ (z−1,−

s

z2) = τ(a, sb) ⊕ τ(z, s)

Use 1, above, to obtain the result.

16

3. Write (u, v) = (x, sy) ⊖ τ(x, sy). We have

u =

( sz2 y − sy

x − x

)2

− δ − x − x

and

v = −sz2 y − sy

x − xu −

syx − sz2 xy

x − x

Therefore the quantity (u, v) becomes (u,−v) = ⊖(u, v) if we change z to z−1 ands to −s

z2 , that is, if we apply the function τ .

3

Lemma 2.17 Assume that x and y are functions in M and (x, y) is a solution ofEquation (MD). Then, either (x, sy) = ±τ(x, sy), or the point (x, sy) ∈ E(M∗) definedby

(x, sy) = (x, sy) ⊖ τ(x, sy)

is a point of order 2 of E .

Proof : Assume that (x, sy) 6= ±τ(x, sy). Then the pair (x, y) satisfies Equation (MD),which we can write as

(z + δ + z−1)(zy)2 = F (x)

By Lemma 2.16 (3), we have

τ(x, sy) = ⊖(x, sy)

Then, by the definition of τ , for any (z0, s0) ∈ E(K), with z0 6= 0, we have x(z−10 ) = x(z0)

and

−s0

z20

y(z−10 ) = −s0y(z0)

that is y(z−10 ) = z2

0 y(z0). The latter implies that the function zy is invariant underz 7→ z−1. We then apply Theorem 2.13 to the pair (x, zy) to obtain the conclusion. 3

Definition 2.18 We will call a global meromorphic solution (x, y) of Equation (MD)even (resp. odd) if there exists a global meromorphic solution (a, b) of Equation (MD)such that (x, sy) = 2(a, sb) (resp. (x, sy) = 2(a, sb) ⊕ (z, s)). We will say also that(x, sy) is even (resp. odd) if (x, y) is even (resp. odd). We will say that a solution(x, y) of Equation (MD) has the even property, if (x, y) satisfies :

x(z−1) = x(z) and y(z−1) = ±z2y(z).

We will say that a solution (x, y) of Equation (MD) has the odd property, if (x, y)satisfies :

x(z−1) = x−1(z) and y(z−1) = z2 y

x2.

Lemma 2.19 A point of order 2 on E is not an even solution. A non-trivial solutioncannot be both even and odd.

17

Proof : The proof is easy and is left to the reader. 3

The next Corollary is not necessary for proving Theorem 2.22. We present it nev-ertheless because it gives a nice correspondence between two kinds of properties of thesolutions of Equation (MD). Note that Corollary 2.20 was known for rational solutions.Our proof does not use rationality. This amazing property of rational solutions was thestarting point of our investigation.

Corollary 2.20 If (x, y) is an even (resp. odd) global meromorphic solution of Equa-tion (MD), then (x, y) has the even (resp. odd) property.

Proof : If (x, sy) is even, then by Lemma 2.16 (1), τ(x, sy) is also even. If (x, sy) werenot equal to ±τ(x, sy), then the pair (x, sy) defined in Lemma 2.17 would be even,which is impossible by Lemma 2.19. Then we are in the case (x, sy) = ±τ(x, sy). Nowfrom this equality, we obtain that the pair (x, sy) has the even property just by thedefinition of τ .

If (x, sy) = 2(a, sb) ⊕ (z, s), then by Lemma 2.16 (2), τ(x, sy) = 2τ(a, sb) ⊕ τ(z, s),and it follows that :

(x, sy) = 2(a, sb) ⊖ τ(2(a, sb)) ⊕ (z, s) ⊖ τ(z, s)

= (z, s) ⊖ [(z, s) ⊕ (0, 0)]

= (0, 0)

Then we have τ(x, sy) = (x, sy) ⊕ (0, 0) = (x−1,− sy

x2 ), and we obtain that the pair(x, sy) has the odd property just by the definition of τ . 3

Lemma 2.21 A function in M which is algebraic over K(z) is a rational function.

Proof : The proof is easy and is left to the reader. 3

We can now prove the key theorem of this article.

Theorem 2.22 The solutions (x, y) of Equation (MD) over the field M of global mero-morphic functions are rational.

Proof : Assume that (x, y) is a solution of Equation (MD), x and y being functions inM. By Lemma 2.17, the pair

2(x ◦ z−1,−s

z2y ◦ z−1) ∈ E(M∗)

must be equal to ±2(x, sy). We get

x2 ◦ x(z−1) = x2 ◦ x,

which implies by Lemma 2.5 that x2 ◦ x is a rational function. We conclude by Lemma2.21 that x must be rational. 3

Note that Lemma 2.21 is not necessary for proving Theorem 2.22, because x2 hasconstant coefficients. Actually we will need Lemma 2.21 in order to finish the proof ofTheorem A.

18

2.5 Proof of Theorem A

We will prove that Theorems 2.22 and Berkovich’s Theorem, together, give a completecharacterization of all the analytic projective maps from E minus a point to E .

Proposition 2.23 Any elliptic curve defined over a field F of characteristic 6= 2, 3 isisomorphic (over an algebraic closure F of F ) to the curve with affine equation

(1) y2 = x3 + δx2 + x

for some δ ∈ F .

Proof : Let us write Eδ the elliptic curve defined by Equation (1). Replacing x by x− δ3

in Equation (1), we obtain

y2 = x3 + (−δ2

3+ 1)x + (2

δ3

27−

δ

3)

It is then easy to compute the j-invariant of the curve Eδ (see for example [23] for adefinition and some basic properties of the j-invariant of an elliptic curve). We obtain

j(Eδ) = 17284(−δ2 + 3)3

27(4 − δ2)

It is clear that the mapj : K − {±2} −→ K

δ 7−→ j(Eδ)

is surjective. 3

We denote by A(z, w) the ring of analytic functions on K2 and by M(z, w) the fieldof meromorphic functions on K2, in the variable (z, w). In this section, by A(z) andM(z) we denote respectively the ring of global analytic functions and the field of globalmeromorphic functions of the variable z.

Let E be the elliptic curve defined by the affine equation

w2 = z3 + δz2 + z.

Let ∼ denote the equivalence relation on A(z, w) and M(z, w) defined by : f(z, w) ∼g(z, w) if and only if, for all (z, w) ∈ K2, whenever w2 = z3 + δz2 + z, we havef(z, w) = g(z, w). Let A(E) = A(z, w)/ ∼ and M(E) = M(z, w)/ ∼. An element ofA(E) is called an analytic function on E and an element of M(E) is called a meromorphicfunction on E . Let s be an element in an algebraic closure of M(z) satisfying

s2 = z3 + δz2 + z.

It is trivial to see that

• The polynomial w2 = z3 +δz2 +z is irreducible over M(z). Hence s is an elementof degree 2 over M(z) and integral over A(z).

• We may identify A(E) with the ring A(z)[s] and M(E) with the field M(z)[s].

19

Definition 2.24 A map G from E −∞ into the projective curve E is called analyticprojective if there exist functions g1, g2, g3 in A(E), not all identically zero, such that

(1’) g22g3 = g3

1 + δg21g3 + g1g

23.

Note that the relation (1’) is obtained by homogenizing Equation (1).Let P ∈ E and (zP , sP ) = (z, s) ⊖ P . We will say that G is an analytic projective

map on E − P and into E if G, as a function of (zP , sP ), is an analytic projective mapfrom E −∞ into E . If G is not the constant (0, 0, 1), we will represent it by the pair(g1

g3, g2

g3).

Let G = (u, v) be an analytic projective map on E −∞ into E . It is obvious fromthe remark above that u and v can be written as u = u0 + su1 and v = v0 + sv1 with ui

and vi in M(z). We will say that G is rational if the functions ui and vi, for i = 0, 1,are rational functions (they lie in K(z)).

Proof of Theorem A :

Without loss of generality, we will work with P = ∞. Let G be an analytic projectivemap on E − {∞} into E . Define

G+(z, s) = G(z, s) ⊕ G(z,−s)

G−(z, s) = G(z, s) ⊖ G(z,−s)

Assume that G+ and G− are not ∞. Write G+ = (a+, b+) andG− = (a−, b−). It is clear from the addition formula that a+, b+, a− and b− liein M(E), and that the maps G+ = (a+, b+) and G− = (a−, b−) can be written asG+ = (a+

0 + sa+1 , b+

0 + sb+1 ) and G− = (a−

0 + sa−1 , b−0 + sb−1 ). Moreover we have :

G+(z,−s) = G+(z, s)

G−(z,−s) = ⊖G−(z, s)

which implies that :• If G+ is not ∞, then a+

1 = b+1 = 0, that is, we have G+ = (a+

0 , b+0 ) and then G+

depends only on z.• If G− is not ∞, then a−

1 = b−0 = 0, that is, G− = (a−0 , sb−1 ).

By Berkovich’s Theorem, G+ is a constant. Moreover, the coordinates of G− satisfy

(sb−1 )2 = (a−0 )3 + δ(a−

0 )2 + a−0

and then we have(z3 + δz2 + z)(b−1 )2 = (a−

0 )3 + δ(a−0 )2 + a−

0

Therefore, (a−0 , b−1 ) is a solution of Equation (MD) over M. By Theorem 2.22, G− is

rational. Observe thatG+(z, s) ⊕ G−(z, s) = 2G(z, s)

It follows that 2G is rational. Write 2G = (a, b) and G = (u, v) so that

(u2 − 1)2

4F (u)= a ∈ K(z, s)

20

Observe that K(z, s) is an algebraic extension of degree 2 of K(z) and then u is algebraicover K(z). Write u = u0 + su1 and u = u0 − su1, where u0 and u1 are functions in M.It is clear that u is algebraic over K(z), therefore u + u = 2u0 and u − u = 2su1 arealgebraic over K(z). So u0 and u1 are both algebraic over K(z). By Lemma 2.21, u0

and u1 are rational functions. So u is a rational function. By similar arguments, it iseasy to see that v is also rational. Therefore G is a rational map. The cases in whichany of G+ or G− is the point ∞ are similar.

3 Proof of Theorem B

What remains to be proved in order to obtain Theorem B follows by techniques by J.Denef in [8].

Equation (MD)(z3 + δz2 + z)y2 = x3 + δx2 + x

defines an elliptic curve E∗ over the field of rational functions K(z), and the point (z, 1)lies on E∗. For any n in the ring End (E) of endomorphisms of E , let us write

(xn, yn) = n(z, 1)

Note that the addition on E∗ is induced by the addition on E :

(x, sy) ∈ E ⇐⇒ (x, y) ∈ E∗.

We could define (xn, yn) by : (xn, syn) = n(z, s), where the addition is now meant onE . We know by the addition formula that xn and yn lie in K(z).

Lemma 3.1 For n ∈ End (E) − {0}, we have

x′n

yn

= n

Proof : See [19, lemma 1.2] or [24, Sect. 8.6]. 3

Corollary 3.2 For n ∈ End (E) − {0}, the order of xn at 0 is equal either to 1 or to−1 and ord0(yn) = ord0(xn) − 1. Moreover, for n ∈ End (E) − {0}, we have

xn

zyn

|z=0 =

{

n if ord0(xn) > 0

−n if ord0(xn) < 0

Proof : From Equation (MD), we see that the order at 0 of xn cannot be zero (since bythe definition, xn is not constant). By Lemma 3.1 we have

ord0(yn) = ord0(x′n) = ord0(xn) − 1.

From Equation (MD), equating the order at 0 of both sides, we get

1+2ord0(yn) = ord0(x3n+δx2

n+xn) =

{

ord0(xn) = ord0(yn) + 1 if ord0(xn) > 0

3ord0(xn) = 3(ord0(yn) + 1) if ord0(xn) < 0.

If ord0(xn) > 0, we find ord0(yn) = 0 and ord0(xn) = 1. While if ord0(xn) < 0, we findord0(yn) = −2 and ord0(xn) = −1. We conclude by Lemma 3.1. If ord0(xn) = 1, weobtain xn

zyn|z=0 = n, and if ord0(xn) = −1, we obtain xn

zyn|z=0 = −n. 3

21

Corollary 3.3 For n ∈ End (E) − {0}, the order at 0 of x2n is −1, and the order at 0of x2n+1 is 1.

Proof : It is clear by the definition that ord0(x1) = 1. From the duplication formula,we get

x2(z) =(z2 − 1)2

4(z3 + δz2 + z),

therefore we have ord0(x2) = −1. On the one hand, we have

x2n = x2 ◦ xn =(x2

n − 1)2

4xn(x2n + δxn + 1)

.

By Corollary 3.2, we have only two cases : either ord0(xn) = 1 or ord0(xn) = −1. Inboth cases, one can see that ord0(x2n) = −1.

Let us prove by induction that ord0(x2n+1) = 1. The addition formula gives

x2n+1 = (z3 + δz2 + z)(zy2n − x2n)2

zx2n(z − x2n)2.

We writezy2n − x2n

z − x2n

=1 − x2n

zy2n

1y2n

− x2n

zy2n

Since ord0(x2n) = −1, we have, by Corollary 3.2, ord0(yn) = −2. We know also byCorollary 3.2 that x2n

zy2n|z=0 = −n. Therefore we obtain

ord0

(

zyn − xn

z − xn

)

= 0

which implies that

ord0(xn+1) = ord0(z3 + δz2 + z) − ord0(zxn).

It is then clear that ord0(xn+1) = −ord0(xn). 3

For the next theorem, one can refer to [8, part 3], or to [18].

Theorem 3.4 [Denef] Suppose that the elliptic curve E has no complex multiplication.

1. All the rational solutions of Equation (MD) are of the form

(xn, yn) ⊕ (a, b)

where (a, b) is either the neutral or a point of order 2 of E∗.

2. The set {(xn, yn) | n ∈ End (E)} is existentially definable in the field of rationalfunctions K(z).

22

Proof of Theorem B :

Choose an elliptic curve E such that End (E) = Z. Let µ be a function in M. Thenµ ∈ Z if and only if the following formula ϕ(µ) (which depends on δ) holds :

µ = 0 ∨ ∃x, y, a, b, v, w [F (z)b2 = F (a) ∧ (x, y) = 2(a, b) ∧ y 6= 0

∧ vzy = x ∧ ord0(v + 2µ) > 0 ∧ w2 = F (2µ)]

Note that the relation (x, y) = 2(a, b) can be expressed by an existential (actually,quantifier free) formula, using the addition formula on E .

Let µ ∈ M − {0} be such that the formula ϕ(µ) is true in M. Then there exista, b, x, y ∈ M such that (a, b) and (x, y) = 2(a, b) satisfy Equation (MD). By Theorem2.22, they must be rational. By Lemma 3.4 and since it is an even solution, (x, y) is ofthe form

(xm, ym) = m(z, 1) = 2(a, b)

for some non-zero integer m. Since (z, 1) is not an even solution (see Lemma 2.19), it isclear that the integer m must be even; say m = 2n. Set v ∈ M such that v = x

zy. From

Corollaries 3.2 and 3.3, we deduce that v(0) = −2n. Since we have ord0(v + 2µ) > 0,the function −2µ must take the same value as the function v at 0. Therefore we haveµ(0) = n. Since (2µ, w) satisfies Equation (1), we know by Berkovich’s Theorem thatthe function µ must be constant. So we have µ = n.

Let us prove the converse. Suppose µ = n is a non-zero integer. Choose x = x2n,y = y2n, a = xn, b = yn, v = x2n

zy2nand w such that w2 = F (2n). Using the properties of

(xn, yn), it is easy to see that the formula ϕ(µ) is satisfied.

The following corollary was proved in [16].

Corollary 3.5 (Lipshitz-Pheidas) The positive existential theory of the ring A, inthe language Lz, is undecidable.

Proof : We represent each meromorphic function x of M as the quotient of two analyticfunctions, x = x1

x2, with x1, x2 ∈ A and x2 6= 0; note that, by Berkovich’s Theorem, if

c ∈ A then c ∈ K if and only if there exists a d ∈ A such that c2 = d3 + d2 + d; andc ∈ K is non-zero if and only if there exists d ∈ K such that cd = 1. Also ord0(x) isgreater than 0 if and only if the following holds (division is understood in A) :

“There exists c ∈ K∗ such that z divides x2 − c and z divides x1.”It is then obvious that every existential formula of L∗

z satisfied in M is equivalent toan existential formula satisfied in A. Finally note that if x2 ∈ A, then x2 6= 0 if and onlyif there exist c, e, f ∈ K such that ef = 1 and z−c divides x2−e in A. Therefore everyexistential formula of Lz satisfied in A is equivalent to a positive existential formula ofLz satisfied in A. Hence the result follows. 3

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