Decomposition of p-adic meromorphic functions

Given a meromorphic function, we are interested in how many ways it can be expressed as a composite of other meromorphic functions. Is this always possible? And, when it is the case, what about unicity of such an expression? Etc... In the case of polynomials, the discussion is relatively easy and has been studied in details by Ritt [13]. As soon as one passes to rational functions the problem becomes much more difficult . Various aspects of the problem concerning the complex meromorphic functions have been studied by many authors [3], [7], [8], [9], [10], [11], [12] and [14]. In this work, we deal with p-adic meromorphic functions. We will see that one neither can always use the same methods that those of the complex case nor obtain the same results. This is due to the fact that the distribution of the singularities is not the same in the two cases. For instance, in C one can consider entire functions whose all zeros lie on a single straight line (real numbers for example). Then many results concerning such functions and based upon a theorem of Edrei [6] are obtained (11J, [12]. This result of Edrei plays an important role in the complex decomposition theory and there is no possible p-adic version of it. 1991 Mathematics subject classification: : Primary 12H25; Secondary 46S10. I. p-adic meromorphic functions We note A( Cp) the ring of entire functions in Cp and M(Cp) the field of meromorphic functions in all Cp. Definition 1.1. Let f ; f l, ... , f n e M(Cp) such that : f ‘f10~2~...p fn. We say that this is a decomposition of f and that f i, ... fn are factors of f .

DECOMPOSITION OF p-ADIC MEROMORPHIC FUNCTIONS J.P. Bézivin and A. Boutabaa Ann. Math. Blaise Pascal, Vol. 2, N° 1, 1995, pp.51-60 Abstract. Given a meromorphic function, we are interested in how many ways it can be expressed as a composite of other meromorphic functions. Is this always possible? And, when it is the case, what about unicity of such an expression? Etc...
In the case of polynomials, the discussion is relatively easy and has been studied in details by Ritt [13]. As soon as one passes to rational functions the problem becomes much more difficult .
In this work, we deal with p-adic meromorphic functions. We will see that one neither can always use the same methods that those of the complex case nor obtain the same results. This is due to the fact that the distribution of the singularities is not the same in the two cases. For instance, in C one can consider entire functions whose all zeros lie on a single straight line (real numbers for example). Then many results concerning such functions and based upon a theorem of Edrei [6] are obtained (11J, [12]. This result of Edrei plays an important role in the complex decomposition theory and there is no possible p-adic version of it.  of zeros of f in the closed (resp. open) disc of radius p and center 0 is given by the right-derivative 03B8+f (log p) (resp le f t derivative 03B8-f(log 03C1)) of 8 f(u) at the point log p.
For the proof see ~1~ for example : : Proposition 1.7. Let F, G, H be three meromorphic functions . Suppose that H is not a rational function and that F = H o G. Then G is entire.

Proof. We have H( x) = A(x) B(x)
where A( x) and B( x) are entire functions without common zeros. The fact that H ( x) is not rational implies that there exist at most one value 03C90 E p such that A( x) -is a polynomial. Let w be a value different of this eventual one.
Then the entire function A(x) -03C9B(x) is transcendental and hence has infinitely many zeros, which we note Suppose that G(x) has a pole x0. We set G(x) = (.r 2014 xo) where W(x) is an analytic function that has no zeros in an open disc ~ xxa ( R and f. an entire number ~ 1. For any k set Tk(x) = W(x) -ck(x -For one of these k let us choice p ~]0, R[ such that 1 Ck | pl =| W(xo) (. Hence we have : On the other hand, the fact that W has no zero in |xxo | R implies that : Therefore the function Tk(x) has at least one zero xk in the circle I xxo 1= p. Thus we have G(Xk) = ck ; and so F(xk) = w. consequentely, since the xk are infinitely many, the function F(x) -w has infinitely many zeros in the disc j xxo ~ R, which is a contradiction. Hence G(x) has no poles and so is entire.
Remark 1.8. The above proposition justifies the fact that, subsequently, in any decomposition of a meromorphic function f in the forme f = h o g, we suppose h meromorphic and g entire or h rational and g meromorphic. The following result will enable us to show the existence of indecomposable or pseudoindecomposable transcendent meromorphic functions . For an entire fonction f and a positive number p, we note m(f, p) the number of zeros of f in the open disc ~ x ~ p, M( f , p) the number of zeros of f in the closed disc I x ~ p and A( f, p) the number of zeros of f in the circle ' x ~= p; each one of these zeros being computed with its multiplicity. Proof. From F = H o G; we deduce that, for all r, we have : ( F (r) = H i (~ G ~ (r)). Since the function G | (r) is strictly increasing for r > 03C10, we deduce that, for u > log pa, we have : (1) 03B8-F(u) ' 03B8-H(03B8G(u)).03B8-G(u), and (2) = 03B8+H(03B8G(u)).03B8+G(u).
By proposition 1.6, we have relations i) and ii). Substracting i) from ii), we obtain relation iii).
and a a positive entire number. We suppose that on infinitely many circles of center 0 of Cp, F has a number of zeros included between 1 and a. Then all decomposition of F of the form F = H o G with H, G E A(p) implies that H or G is a polynomial of degree between 1 and a.
Proof. If any of the functions Hand G is a polynomial of degree between 1 and ~, we have for p enough large : M(G, p) > ~ + 1 and m(H, ( G ( p)) > a + 1.
On the other hand, the hypothesis of the Theorem and the formula iii) of the proposition 1.9 show that for an infinity of p's arbitrarily large, we have A(H, [ G ~ ( p)) # 0 or 0. Then for such a p we have A(F, p) > a + 1. This is a contradiction with the hypothesis. Hence H or G is necessarily a polynomial of degree between 1 and a. Corollary 1.11. Let F E and .1 an entire > l. We suppose that on an infinity of circles of center 0, F has a number of zeros included between 1 and a and that on an infinity of circles of center 0, F has a number of poles included between 1 and a. Then any decomposition of F in the form F = H o G with H E and G E A(p) implies that : G is a polynomial of degree between 1 and a, or H is a rational function of degree between 1 and . Proof. We apply the relation ii) of the proposition 1.9. Corollary 1.14. Let F E We suppose that in an infinity of discs of center 0 and radius arbitrarily large, F has a number of zeros equal to the product of two prime numbers (not necessarily distinct). Then, either F is indecomposable, or it is a composite of two indecomposable factors.
Proof. We use the previous proposition. Corollary 1.15. Let F E Cp[z] and 03BB ~ IN *. We suppose that in an infinity of discs of center 0 and radius arbitrarily large, F has a number of zeros equal to the product of a prime numbers (not necessarily distinct). Then F is a composite of at most A indecomposable factors.
Proof. We proceed gradually using the corollary 1.14.
Given a meromorphic function, it is not easy to know if it is decomposable or not. The first of the two next examples shows that a decomposable function may have non equivalent decompositions. The second example shows that a meromorphic function can have an infinity of prime factors. n~+~ Let fn(x) = x(1an ). We define the sequence (cpn(x)) by : = x and ~pn o fn(x). Let R > 0. We will show that the sequence of CPn( x) is of Cauchy in the space of functions analytic in the closed disc aj R. We have : 0 3 C 6 n + 1 ( x ) = 0 3 C 6 n ( x -x 2 a n ) = 0 3 A 3 ( -x 2 a n ) k 0 3 C 6 ( k ) n ( x ) k l .
On the other hand we have : (R) ~| k! | 03C6n | (R). Let N be the first index such that n > N implies that I an |> R. Hence we have : 03C6n '(R) |0 3 C 6 n (R) a .
Consequently we have, (R) ==) (R) for all n > N. It follows that the sequence (cpn(x)) is of Cauchy in the space of functions analytic in the closed disc x ( R. Hence the sequence (03C6n(x)) converges to an entire function 03C6(x).
We can write for all k and n > ~ : : = f0 p fi o ... p fn 0 Bn(x), where 8n(x) is a function of the same type as and hence will converge to an entire function ~(:c). Hence we will have : (x) = fo o and this relation shows that the f k( x), which are indecomposable are factors of 03C6(x).
Hence has infinitely many indecomposable factors.
II. Comparison with polynomials and complex case.
We know that P(x) being a given polynomial, we have : (1) P(x) has at least one decomposition in indecomposable factors.
(2) The number of indecomposable factors is finite and, is the same in any decomposition of P(x).
(3) P(x) has only a finite number of non equivalent decompositions.
Do these properties remain true for entire or meromorphic functions? Example 1.16 shows that (2) is no longer so. In fact, even if we assume that all decomposition of f has a finite number of factors, we can not show if this number is bounded and if it is the same in any decomposition. This is not obvious at all because only for rational functions the following result stated by Ritt [13] in 1922 has only been proved recently [2]. Proposition 2.1. (Ritt). There exists a rational function which has two decompositions into indecomposable functions each having a different number of factors.
On the other hand Ritt [13] had shown that for any decomposition of a polynomial into rational functions, there exists an equivalent decomposition into polynomials. The following result shows that this is extended to p-adic entire functions. This result is false in Indeed, Ozawa [11] has shown that the function F(z) = (~ -l)e~"~ is indecomposable in ~(C), but that F = fog where f = and g = This means that in M(Gj), the function jF is not even pseudo-decomposable.

III. Common right factors of F and F(n).
Suppose that a p-adic meromorphic function F(z) and its derivative F'(z) have an entire function g as their common right factor, it is easily shown from F = f o g and F' = h o g that g must be a polynomial of degree one. It is no longer a simple problem of searching the possible forms of any common right factor of F and F~"~. However, we have the following general result : Let g E A( Cp) such that T(r, ho) + T(r, h1) +... + T(r, hm) = 0(Log | 9 | (r)). . We suppose that : + ... + F,n(9)hm(x) = 0.