EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 11, No. 3, 2018, 803-814 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global On the existence of roots of some p-adic exponential-polynomials Amran Dalloul Department of Mathematics, Beirut Arab University, Beirut, Lebanon Abstract. In this paper, we apply Newton polygon method in order to derive sufficient conditions for the existence of zeros of some p-adic exponential-polynomials. 1. Introduction Let K be an algebraically closed field of characteristic zero, and let exp be a (partial) exponential map exp : E → K×, E ⊂ K being the domain of the exponential map. An exponential-polynomial is any expression of the form a(X) = P1(X) exp(w1X) + ...+ Pd(X) exp(wdX), (1) where the Pi’s (i = 1, . . . , d) are polynomials in K[X], and wi ∈ K for i = 1, 2, . . . , d. The theory of exponential polynomials is an important topic in transcendental number theory. A remarkable work has been made by P. D’A quino, A. Macintyre and G. Terzo [1], where they proved that Shapiro’s Conjecture (over an algebraically closed exponential field of characteristic zero having an infinite cyclic group of periods and the exponential is surjective onto the multiplicative group) is true with an extra assumption, Schanuel’s Conjecture. In 2017, they proved the following [2]: Assume Schanuel’s Conjecture. Let Z(f) be the zero set of the exponential polynomial f(X) = α1 exp(w1X) + ...+ αd exp(wdX), where αi, wi are constants in K. If Z(f) is infinite, then each infinite subset X ⊆ Z(f) has an infinite transcendence degree over Q. In this paper, we work in the non-archimedean fields, namely the p−adic fields. Here, the situation is different to some extend. In fact, Poorten and Rumely [PR] proved that each exponential polynomial of the form (1.1) has at most finitely many roots in the domain of convergence. Their proof relies on a geometric approach, namely The Newton polygon of power series, where they proved that The Newton polygon of the exponential DOI: https://doi.org/10.29020/nybg.ejpam.v11i3.3245 Email address: amrandalloul@hotmail.com (A. Dalloul) http://www.ejpam.com 803 c© 2018 EJPAM All rights reserved. A. Dalloul / Eur. J. Pure Appl. Math, 11 (3) (2018), 803-814 804 polynomial (1.1) ends with a straight line. That guarantees the existence of a bound on the number of zeros. Furthermore, the p-adic exponential function and the p-adic trigonometric functions are not periodic. Many results, however, have been made in the p−adic exponential polynomials. For example, Poorten, [5] , used Strassmann Theorem to prove that the exponential polynomial b(z) = P1[z] exp(w1z) + ...+ Pd[z] exp(wdz), (2) with Pi[z] ∈ Cp[z] and ord(wi) > 1 p−1+ε, i = 1, 2, .., d, has at most (d−1+ ∑d i=1 degPi[z])(1+ 1 ε(p−1)) roots in the unit disk. Poorten and Rumely, [6], proved that the exponential poly- nomial (1.2) over Qp (for large enough p) has at most (d − 2 + ∑d i=1 degPi[z])p roots in its domain of convergence using the Newton polygon method and concepts of recurrence sequences and generalized sums. In this work, we consider some p−adic exponential polynomials, where we put a least bound on the number of zeros (counting multiplicity) of the p−adic exponential polyno- mials. We mainly use here the Newton Polygon of power series method which assures that the roots of the power series yield from the finite segments of the polygon. In other words, if the Newton polygon of the power series f has a finite segment with projection length on the x-axis equals to m, then there exist m roots (counting multiplicity) of f of the same order. Similarly, we use The Newton polygon method to put sufficient conditions on some polynomials P [X,Y ] ∈ Q[X,Y ] to have roots of the form (x, exp(x)). 2. Notation and preliminaries Let p be a prime number, Qp the completion of Q with respect to the p−adic absolute value |.| and Cp the completion of an algebraic closure of Qp. The absolute value | · | on Cp is the extension of the p−adic absolute value |.|. Starting from | · |, one can define a map ord : Cp → Q∪{∞}, as follows: ord(0) =∞, and ord(x) = − log(|x|). This map satisfies the properties: ord(x± y) ≥ min {ord(x), ord(y)}, ord(xy±1) = ord(x)± ord(y), if ord(x) 6= ord(y), then ord(x± y) = min{ord(x), ord(y)}. The set O := {x ∈ Cp : ord(x) ≥ 0} forms a local ring called the ring of integers in Cp. Let n ∈ N with n ≥ 1. It is well known that ord(n!) = n− Sn p− 1 , where Sn is the sum of digits of n when it is written in the base p. In particular, if n = pm,m ≥ 1, then Sn = 1. It is clear that Sn ≥ 1,∀n ≥ 1. This implies that ord( 1 n! ) ≥ −n− 1 p− 1 . We also recall some basic concepts and results concerning The Newton polygon method. For more details, see [4] and [3]. A. Dalloul / Eur. J. Pure Appl. Math, 11 (3) (2018), 803-814 805 2.1. The Newton polygon for polynomials Let f(X) = 1 + a1X + · · ·+ anX n ∈ 1 +XCp[X] be a polynomial with degree n and the constant term is 1. We plot the following points in the Euclidean space R2: (0, 0), (1, ord(a1)), (2, ord(a2)), . . . , (n, ord(an)). If ai = 0 for some i, we omit this point (considering it as a point at infinity). The Newton polygon of the polynomial f is defined as the convex hull of the points (0, 0), (1, ord(a1)), (2, ord(a2)), . . . , (n, ord(an)). That is the highest convex polygonal line joining (0, 0) with (n, ord(an)) and passing through or below all the points (i, ord(ai)), i = 1, 2, .., n− 1. Practically, the Newton polygon of polynomials is obtained by the following steps: 1) Start with the vertical half-line which is the negative part of the y-axis. 2) Rotate the line counter-clockwise until it hits one of the points we have plotted. 3) Break the line at that point, and continue rotating the remaining part until another point is hit. 4) Continue until all the points have either been hit or lie strictly above a portion of the polygon. A vertex of the Newton polygon is a point (i, ord(ai)) where the slopes change. If a segment joins the point (i,m) to the point (i′,m′), then the slope is the quantity m−m′ i−i′ . By the length of the slope we mean the quantity i− i′. If the polygon has a segment ends by a point (i, ord(ai)) and continues by another segment of different slope, we say that the Newton polygon has a ”break” at the point (i, ord(ai)). Theorem 1. Let f(X) = 1 + a1X + · · · + anX n ∈ 1 + XCp[X]. If λ is a slope of the Newton polygon associated to the polynomial f with the length m, then there exist the numbers α1, α2, .., αm ∈ Cp (counting multiplicity) such that f(αi) = 0 and ord(αi) = −λ, ∀i = 1, 2, ..,m. 2.2. The Newton polygon for power series The definition is formally identical to that given for polynomials: Consider the power series f(X) = 1 + a1X + a2X 2 + · · ·+ anX n + . . . We plot the points (i, ord(ai)), i = 1, 2, ..., ignoring as before any points where ai = 0. The Newton polygon of f(X) is again obtained by the rotating line procedure. In this case, the things become more complicated than the case of polynomials. For example, the Newton polygon of the power series f(X) = 1 + pX + pX2 + .. + pXn + ... is just the horizontal line OX which does not hit any of the points (i, ord(ai)), i = 1, 2, .... For this case and other cases, we must modify the rules to obtain the Newton polygon of power series as the following steps: Start with the half-line which is the negative part of the y-axis. Rotate that line counter- clockwise until one of the following happens: A. Dalloul / Eur. J. Pure Appl. Math, 11 (3) (2018), 803-814 806 i) The line simultaneously hits infinitely many of the points we have plotted. In this case, stop and the polygon is complete. For example the Newton polygon of the power series f(X) = 1 + ∑∞ i=1 p iXi is just the line Y = X. ii) The line reaches a position where it contains only one of our points that serves as a center of rotation, but can be rotated no further without leaving behind some points. In this case, stop and the polygon is complete. We will counter this case in Section 4. iii) The line hits a finite number of points. In this case, break the line at the last point it was hit, and repeat the whole procedure again. Using the above procedure, it can be seen that the Newton polygon of power series either ends by a ray (see the Appendix) or has an infinite number of finite segments (for example, the Newton polygon of the power series 1 + ∑∞ i=1 p i2Xi). Furthermore, it is well known that if the Newton polygon of a power series f ends by a ray, then f has at most finitely many zeros in its disk of convergence. The following lemma is a connection between the domain of convergence of a power series and the slopes of its polygon. Lemma 1. Let m be the sup of all slopes appearing in the Newton polygon of a power series f(X) = 1 + ∑∞ i=1 aiX i. Then, the domain of f is the set {x ∈ Cp : ord(x) > −m}. In the case m is infinite. Then f converges on all of Cp. In particular, if the Newton polygon of f ends by a ray of slope m, then the domain is {x ∈ Cp : ord(x) > −m}. Finally, we need the following: Corollary 1. ([4], p.106) If a segment of the Newton polygon of f(X) ∈ 1 + Cp[[X]] has finite length N and slope λ, then there are precisely N values of x counting multiplicity for which f(x) = 0 and ord(x) = −λ. We summarize what we need as the following: Fact 1. The points (i,ord( 1 i!)); i > 1 are on or above the line Y = −1 p−1(X − 1). It algebraically means that ord( 1 i!) > −1 p−1(i− 1), ∀i > 1. Fact 2. A finite segment of the length m of the Newton polygon of the power series f determines at least m roots (counting multiplicity) of f of the same order. 3. The Main Results Keep the notation as above. We prove the following: Theorem 2. Consider the polynomials over O: P1(z) = n1∑ j=0 a (1) j zj , P2(z) = n2∑ j=0 a (2) j zj , . . . ., Pd(z) = nd∑ j=0 a (d) j zj , A. Dalloul / Eur. J. Pure Appl. Math, 11 (3) (2018), 803-814 807 where nd > max1≤i≤d−1{degPi, 1}. Let w1, . . . , wd ∈ Cp with ord(wi) > 1 p−1 , i = 1, 2, .., d. Then, the exponential polynomial b(z) = P1(z) exp(w1z) + P2(z) exp(w2z) + . . . .+ Pd(z) exp(wdz), with ord(a (1) 0 + · · ·+ a (d) 0 ) = ord(a (d) nd ) = 0, has at least nd roots (counting multiplicity) in the unit disk. Proof. As ord(wj) > 1 p−1 for j = 1, 2, .., d, the domain of convergence is the unit disk. Let c = a (1) 0 + ....+ a (d) 0 , then, by expanding a(z) as a power series in z, one finds: a(z) c = 1 + c−1 ∞∑ i=1 ( d∑ j=1 a (j) 0 wij (i)! + ....+ a(j)nj w i−nj j (i− nj)! ) zi =: 1 + ∞∑ i=1 miz i. We have ord(wj) > 1 p− 1 = i i(p− 1) > i− Si i(p− 1) , ∀i ≥ 1,∀j = 1, 2, .., d. Hence, i ord(wj) > i− Si p− 1 = ord(i!)⇒ i ord(wj)− ord(i!) > 0. Therefore, ord( wij i! ) > 0. (3) Using (3), the assumptions of the theorem that ord(c) = 0 and the coefficients of Pj , j = 1, 2, ..., d are in O, we find that ord(mi) ≥ 0, i = 1, 2, ... That means the points (i, ord(mi)), i = 1, 2, .. are on or above the x-axis. Consider the coefficient mnd in the series a(z) c . The assumption that ord(and ) = ord(c) = 0 and (3) guarantee that ord(mnd ) = 0. Assume that ord(w1) = min{ord(wj), j = 1, 2, .., d} (the other cases can be done similarly). Therefore, we have for all i > nd min { ord( 1 (i− j)! ) : 0 ≤ j ≤ nd } = ord( 1 i! ). Hence, ord(mi) ≥ min { ord(ord( wi−kj (i− k)! )) : j = 1, . . . , d, 0 ≤ k ≤ nj } ≥ (i− nd)ord(w1) + ord( 1 i! ) A. Dalloul / Eur. J. Pure Appl. Math, 11 (3) (2018), 803-814 808 Figure 1: The Newton polygon of a(z) c ≥ (i− nd)ord(w1)− i− 1 p− 1 . Therefore, for all i > nd the points (i, ord(mi)) are on or above the line L : Y + nd.ord(w1)− 1 p− 1 = (ord(w1)− 1 p− 1 )X. This line has a positive slope λ = ord(w1)− 1 p−1 > 0 and intersects with the x-axis in the point ( nd.ord(w1)− 1 p−1 ord(w1)− 1 p−1 , 0) which lies on the right of the point (nd, 0) since nd > 1. The points (i, ord(mi)), i = 1, 2, 3, . . . are thus distributed as follows: 1) The points (1, ord(m1)), . . . , (nd − 1, ord(mnd−1)) are on or above the x-axis. 2) The point (nd, ord(mnd )) is on the x-axis. 3) The points (i, ord(mi)), i > nd are on or above the line L and above x-axis. It follows that there exists a finite number of points (i, ord(mi)) lying on a horizontal line above the x-axis. Now, we apply the previous steps to obtain The Newton polygon of a(z)c as follows: Rotate the vertical half-line of the negative part of the y-axis until it hits the point (nd, 0). Break the line at this point (the existence of the break is because there is at most finitely many points (i, ord(mi)) lying on a horizontal line above the x-axis). Rotate it around this point until it hits another point or continues until it reaches a position parallel to a line with a positive slope. In all cases, the Newton polygon of a(z)c starts with a segment of the length nd and has a break at the point (nd, 0) (see figure 1). Therefore, using Fact 2, we find that a(z) c (and hence a(z)) has at least nd roots. A. Dalloul / Eur. J. Pure Appl. Math, 11 (3) (2018), 803-814 809 Remark 1. Theorem 2 covers only certain cases of exponential polynomials. The as- sumption ord(wj) = 0,∀j = 1, 2, .., d is crucial. In fact, there exists a very big class of exponential polynomials that have no roots. For example, the exponential polynomial over O: a(z) = a exp(w1z) + (a1z + · · ·+ anz n) exp(w2z) + (b1z + · · ·+ bmz m) exp(w3z), where ord(wj) = 0,∀j = 1, 2, 3, . . . , n < m, has no roots it its domain even if ord(a) = ord(bm) = 0. The above follows since, if z0 ∈ {z ∈ Cp : ord(z) > 1 p−1} is a root of a(z), then a exp(w1z0) = − ( (a1z0 + · · ·+ anz n 0 ) exp(w2z0) + (b1z0 + · · ·+ bmz m 0 ) exp(w3z0) ) . Therefore, |a exp(w1z0)| = ∣∣∣(a1z0 + · · ·+ anz n 0 ) exp(w2z0) + (b1z0 + · · ·+ bmz m 0 ) exp(w3z0) ∣∣∣ ≤ max{|(a1z0 + · · ·+ anz n 0 ) exp(w2z0)|, |(b1z0 + · · ·+ bmz m 0 ) exp(w3z0)|} < p −1 p−1 < 1, since | exp(wjz0)| = 1, j = 2, 3 and |z0| < p −1 p−1 < 1. On the other hand, we have |a exp(w1z0)| = | exp(w1z0)| = 1. This contradiction shows that a(z) has no roots in its domain. Corollary 2. Consider the polynomial P [X,Y ] = a+ bY m + a(i1,j1)X i1Y j1 + · · ·+ a(id,jd)X idY jd ∈ Z[X,Y ], with p| gcd(m, j1, .., jd), 0 < i1 < · · · < id, and (a+ b, p) = (a(id,jd), p) = 1. Then, P has at least id roots of the form (x, exp(x)). Proof. Consider the exponential polynomial b(z) = (a · z0) + (b · z0) exp(mz) + a(i1,j1)z i1 exp (j1z) + · · ·+ a(id,jd)z id exp (jdz). Then, we have ord(m) ≥ 1 > 1 p− 1 , ord(jk) ≥ 1 > 1 p− 1 , k = 1, 2, . . . , d. Also, the polynomial Pd(z) = a(id,jd)z id has the largest degree with ord(a(id,jd)) = ord(a+ b) = 0. This implies, by Theorem 2, that b(z) has at least id roots. This proves the Corollary. Corollary 3. Let P [X,Y ] ∈ Q[X,Y ] be polynomial defined by the conditions of the pre- vious Corollary. Then there exists a tuple (x, exp(x)), x ∈ E (domain of the exponential function) such that P (x, exp(x)) = 0. In other words, the elements x, exp(x) are Q−algebraically dependent. Hence, tdQQ(x, exp(x)) ≤ 1, where td stands for the transcendence degree. A. Dalloul / Eur. J. Pure Appl. Math, 11 (3) (2018), 803-814 810 4. Further Applications of the Newton polygon method One can also use the Newton polygon of power series in order to obtain sufficient conditions on a polynomial P [X,Y ] ∈ Q[X,Y ] to have roots of the form (x, exp(x)), as done in what follows. Theorem 3. For any polynomial P [X,Y ] having the form P [X,Y ] = dY n + cXm + c1X m1Y n1 + · · ·+ crX mrY nr ∈ Q[X,Y ], with mi,m, n, ni ∈ Z≥1, (ni, p) = (n, p) = 1, ord(d) = ord(ci) = 0, for i = 1, 2, . . . r, and such that ord(c) < −1 p−1m, has a root of the form (x, exp(x)), x ∈ Cp with ord(x) > 1 p−1 . Proof. Let f(X) := P [X, exp(X)] be the corresponding power series associated with the original polynomial P [X,Y ]. Then, f(X) can be written as f(X) = d(1 + b1X + ..+ bm−1X m−1 + (bm + c.d−1)Xm + bm+1X m+1 + . . . ), where bi = ni i! + e1d −1 n i−m1 1 (i−m1)! + · · ·+ erd −1 ni−mr r (i−mr)! ; ek = 0 or ck, for k = 0, .., r. According to the assumption of theorem (the coefficients d, ci, and the degrees nj have the same order which is zero) and fact (1), we find that the numbers bi satisfy the inequality ord(bi) ≥ −1 p− 1 (i− 1). Since ord(c) < −1 p−1m < −1 p−1(m− 1), it follows that ord(bm + cd−1) = min{ord(bm), ord(cd−1)} = ord(c) < −1 p− 1 m. Also, if i is sufficiently large index of the form pj , then ord(n i i! ) = ord( 1 i!) = − i−1 p−1 . Clearly, mk + Si−mk > 1,∀k = 1, 2, .., r (since Sn ≥ 1,∀n ≥ 1). This is equivalent to − i−1 p−1 < − i−mk−Si−mk p−1 . In other words, ord( 1 i! ) < ord( 1 (i−mk)! ), k = 1, 2, . . . , r. Hence, ord(bi) = min { ord( ni i! ), ord( ni−m1 1 (i−m1)! + ...+ ni−mr r (i−mr)! ) } = ord( 1 i! ). Therefore, the points of the power series f(X) d are distributed as follows (for more details, see the Appendix): The points (i, ord(bi)) are on or above the line Y = −1 p−1(X−1), the subsequence (pi, ord(bpi)), for large enough i lies on the previous line and the point (m, ord(bm + d−1c)) is below the line Y = − 1 p−1X. A. Dalloul / Eur. J. Pure Appl. Math, 11 (3) (2018), 803-814 811 Figure 2: The Newton polygon of f(X) d Now, we apply the previous steps to obtain The Newton polygon of the power series f(X) d as follows: Rotate the vertical half-line of the negative part of the y-axis until it hits the point (m, ord(bm+d−1c)). Then rotate it around this point until it reaches a position parallel to the line Y = − 1 p−1(X − 1). Stop here and the polygon is complete. Any further rotation would leave behind some points (i, ord(bi)) (see figure 2 and the Appendix). Therefore, the Newton polygon of f(X) d has a break at the point (m, ord(bm + d−1c)). This implies, that the Newton polygon of f(X) d has a finite segment of the length m. Using Fact 2 , f(X) d (and hence f) has at least m roots. So, the original polynomial P [X,Y ] has m roots in Cp × C∗p of the form (x,exp(x)). Remark 2. Newton polygon method does not only guarantee the existence of polynomials that admit roots of the form (x, exp(x)), but it also gives us information about the order of x. Example 1. Consider the polynomial P [X,Y ] = p−1X2 + Y 2; p > 5. Let f(X) := P [X, exp(X)] = p−1X2 + (exp(X))2 = p−1X2 + exp(2X). Then we have, f(X) = 1 + 2X 1! + ( 22 2! + p−1)X2 + 23 3! X3 + · · ·+ 2i i! Xi + . . . . A. Dalloul / Eur. J. Pure Appl. Math, 11 (3) (2018), 803-814 812 Therefore, ord( 2 1! ) = 0, ord( 2i i! ) = ord( 1 i! ),∀i > 3. This is because, ord(2) = 0 for p > 5. Furthermore, we find ord( 22 2! + p−1) = −1. The assumption p > 5 guarantees that the point (2,−1) is below the line Y = −1 p−1X. This means that the Newton polygon of f(X) starts with a segment of the slope (−12 ) and ends with the point (2,-1)while the second segment is a half line which starts with the point (2,-1) and has the slope −1 p−1 . Therefore, the Newton polygon of f(X) has a break at the point (2,-1). Using fact (2), we find that the power series f(X) has at least two roots of the order 1 2 . So, the original polynomial P [X,Y ] has at least two roots of the form (x,exp(x)). 5. Appendix The following known Lemma, (see, e.g., [4], p. 143), determines the distribution of vertices (i, ord( 1 i!)) that appear in the Newton polygon of the exponential map. Lemma 2. The Newton polygon of the exponential function is a straight line from (0, 0) with the slope −1 p−1 . Proof. We know that the exponential function exp(X) is defined as exp(X) = 1 + X 1 + X2 2! + · · ·+ Xi i! + . . . We first show that, for all i > 0, the points (pi,ord(api)) belong to the line Y = −1 p−1(X−1) and the other points are on or above this line. That is, for all j > 1, the points (j,ord(aj)) are on or above the line Y = −1 p−1(X − 1). To do that, we prove that the slope of the line which passes through any two points (pi,ord(api)), (pj ,ord(apj ));(i > j) has a slope independent of i and j and indeed has the value −1 p−1 . Let m be the slope of the line through any points (pi,ord(api)), (pj ,ord(apj )). Then m = ord(api)− ord(apj ) pi − pj = −ord(pi!) + ord(pj !) pi − pj = ord(pj !)− ord(pi!) pi − pj = pj−S pj p−1 − pi−Spi p−1 pi − pj . Therefore, m = pj−1 p−1 − pi−1 p−1 pi − pj = pj−pi p−1 pi − pj = −1 p− 1 . REFERENCES 813 So, all the points (pi, ord(api)); i > 0 belong to the line with the slope −1 p−1 and passes through the point (p0,ord(ap0))(which is the point (1,0)). This line has the equation Y = −1 p−1(X − 1). Also, for all i > 1, we have ord(ai) = ord( 1 i! ) ≥ − 1 p− 1 (i− 1). This implies that all the points (i,ord(ai)); (i > 1) are on or above the line Y = −1 p−1(X−1). From that argument, we deduce that the points (i,ord(ai)); (i ≥ 1) are distributed as follows: 1) The subsequence (pj , ord(apj )), j ≥ 0 lies on the line Y = −1 p−1(X − 1) which has the slope − 1 p−1 . 2) The other points are on or above this line. Now apply the previous steps to obtain The Newton polygon of exp(X) as follows: Rotate the vertical half-line of the negative part of the y-axis until it reaches to a position parallel to the line Y = − 1 p−1(X − 1). We stop here without any further rotation. This is because, for any ε > − 1 p−1 , the line Y = εX would leave behind it some points of the form (pi, ord( 1 pi! )). Since ε > −1 p−1 , it follows that there exists some positive real number δ > 0 such that ε = −1 p−1 + δ. Therefore, ord(ai) < εi⇔ − i− Si p− 1 < εi⇔ − i− Si p− 1 < ( −1 p− 1 + δ)i⇔ i− Si > i− (p− 1)δi⇔ (p− 1)δi > Si ⇔ i > Si (p− 1)δ . We can choose i to be sufficiently large and has the form pj . In this case, we find that Si = 1, so the relation i > 1 δ(p−1) holds true for the index i := pj . This implies that there is no further rotation of the line Y = −1 p−1X. Hence, the Newton polygon of the exponential function is the straight line Y = −1 p−1X from (0, 0). Acknowledgements I would like to thank the referees for their constructive comments. Also, I would like to thank Ali Bleybel for proposing this subject, as well as his constant help and support throughout the preparation of this paper. References [1] P. D’A quino, A. Macintyre and G. Terzo. From Schanuels Conjecture to Shapiros Conjecture, available at: http://arxiv.org/abs/1206.6747v1. REFERENCES 814 [2] P. D’A quino, A. Macintyre and G. Terzo. Comparing C and Zilbers Exponential Fields: Zero Sets of Exponential Polynomials, available at: http://arxiv.org/abs/1310.6891v1. [3] F. Q. Gouvea, p-adic Numbers: An Introduction, Springer-Verlag, Berlin, Heidelberg, New York, Second Edition, Universitext, 2000. [4] N. Koblitz. p-adic Numbers, p-adic Analysis, and Zeta-functions. Springer-Verlag, Berlin, Heidelberg, New York, second edition, 1984. [5] A.J.Van Der Poorten, Zeros of p-adic Exponential Polynomials I, School of Mathe- matics the university of NSW Kensington, NSW 2033, Australia, 1975. [6] A.J.Van Der Poorten and Roberts Rumely, Zeros of p-adic Exponential Polynomials II, J. London Math. Soc. (2) 36 (1987) 1-15.