On Hensel's Roots and a Factorization Formula in Z [[X]]

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2. Series solutions of algebraic equations The main results of this paper rely on the following consequence of the inversion formula of Lagrange for formal power series. For a detailed proof and other applications, we refer the reader to [8, Section 3.8] or [7, Section 11.6]. In the sequel, Bn,j(x1,x2,... ) denotes the (n, j)-th partial Bell polynomial, see the appendix.

Lemma 2.1 (cf. [7, Corollary 11.3]). If φ(t) is a power series of the form

∞ tr φ(t)= t 1+ α , r r! Xr=1 then its formal inverse is given by

∞ un φ−1(u)= u 1+ β , n n! nX=1 where n (n + j)! β = (−1)j B (α , α ,... ). n (n + 1)! n,j 1 2 Xj=1 Inversion formulas of this type have been studied by many authors in the search for solutions of algebraic equations. For instance, a series solution for the equation xm +px = q was already given by Lambert in 1758, cf. [12]. The most general formulas we found in the literature were obtained by Birkeland around 1927. In [3], the author studied arbitrary polynomial equations and obtained explicit solutions in terms of hypergeometric functions, see also [13]. It turns out that, if f(x) is a power series over a commutative ring R with an invertible linear coeﬃcient, then formal series solutions for the equation f(x) = 0 can be obtained from Lemma 2.1 as follows.

2 Proposition 2.2. Given a power series f(x) = a0 + a1x + a2x + · · · ∈ R[[x]] with a1 invertible in R, the equation f(x) = 0 has the formal root

∞ n n−k+1 (−1) 2n + 1 a0 n+1 x = k Bn+k,k(1!a1, 2!a2,... ) . (2.3) a (n + 1)! n − k a1 nX=0 Xk=0 1 Proof. Let ∞ xℓ φ(x)= x 1+ α with α = ℓ! a /a . ℓ ℓ! ℓ ℓ+1 1 Xℓ=1 ON HENSEL’S ROOTS AND A FACTORIZATION FORMULA IN Z[[x]] 3

Thus f(x)= a1 a0/a1 +φ(x) and f(x)=0if φ(x)= −a0/a1. By Lemma 2.1, this equation has the formal root ∞ a (−1)nβ a n x = φ−1(−a /a )= − 0 1+ n 0 0 1 a n! a 1 nX=1 1 ∞ n a (−1)n (n + j)! a n = − 0 1+ (−1)j B (α , α ,... ) 0 a n! (n + 1)! n,j 1 2 a 1 nX=1 Xj=1 1 ∞ n (−1)n+j+1 (n + j)! a n+1 = B (1!a , 2!a ,... ) 0 . (2.4) j (n + 1)! n,j 2 3 a Xn=0 Xj=0 a1 n! 1

Now, if we set xj = j!aj , then n! B (1!a , 2!a ,... )= B ( x2 x3 ,... )= B (0,x ,x ,... ) n,j 2 3 n,j 2 3 (n + j)! n+j,j 2 3 by (A.1). Moreover, by means of (A.2) and (A.3), we have n + j B (0,x ,x ,... )= B (x ,x ,... )B − − (−x , 0,... ) n+j,j 2 3 ν ν,k 1 2 n+j ν,j k 1 Xk≤j ν≤n+j n + j = (−x )j−kB (x ,x ,... ). n + k 1 n+k,k 1 2 Xk≤j Therefore, n (−1)n+j+1 (n + j)! B (1!a , 2!a ,... ) j (n + 1)! n,j 2 3 Xj=0 a1 n! n (−1)n+j+1 = j Bn+j,j(0, 2!a2, 3!a3,... ) Xj=0 a1 (n + 1)! n (−1)n−k+1 n + j = Bn+k,k(1!a1, 2!a2,... ) ak (n + 1)! n + k Xj=0 Xk≤j 1

n n (−1)n−k+1 n + j = Bn+k,k(1!a1, 2!a2,... ) ak (n + 1)! n + k Xk=0 1 Xj=k n (−1)n−k+1 2n + 1 = Bn+k,k(1!a1, 2!a2,... ). ak (n + 1)! n − k Xk=0 1 Inserting this expression into (2.4), we arrive at (2.3).

The simplicity (or complexity) of formula (2.3) clearly depends on the structure of the partial Bell polynomials. 4 DANIELBIRMAJER,JUANB.GIL,ANDMICHAELD.WEINER

For example, for xm + px − q = 0 with p,q ∈ R, p 6= 0, m > 1, the root (2.3) takes the form ∞ n (−1)n−k+1 2n + 1 −q n+1 x = B (p, 0,..., 0,m!, 0,... ) , pk (n + 1)! n − k n+k,k p Xn=0 Xk=0 which by means of (A.2) and (A.3) reduces to ∞ (−1)k mk 1 q (m−1)k+1 x = . (2.5) pk k (m − 1)k + 1 p Xk=0 This formula includes Eisenstein’s series solution for x5 + x = q, cf. [15], ∞ 5k 1 x = (−1)k q4k+1. k 4k + 1 Xk=0 Of course, this series does not converge for all values of q, so further analysis is required to understand and possibly make sense of (2.3). While convergence in general is not the focus of this paper, we want to brieﬂy discuss f(x) = x3 + x − q in order to illustrate a possible analytic approach. For this polynomial, the sum (2.5) becomes ∞ ∞ 3k 1 3k 1 x = (−1)k q2k+1 = q (−q2)k, k 2k + 1 k 2k + 1 Xk=0 Xk=0 which converges only when q2 ≤ 4/27. However, ∞ 3k 1 (−q2)k = F 1 , 2 ; 3 ; − 27 q2 , k 2k + 1 2 1 3 3 2 4 Xk=0 1 2 3 and since the hypergeometric function 2 F1 3 , 3 ; 2 ; z extends analytically to C \ (1, ∞), we can actually evaluate the formal root for larger values of q. For example, if q = 2, then the 3 1 2 3 root of x + x − 2 = 0 provided by (2.3) is precisely x = 2 · 2 F1 3 , 3 ; 2 ; −27 = 1. In the next section we will fully discuss the use of (2.3) to ﬁnd roots of polynomials over the p-adic integers.

3. Hensel’s roots In this section, we use Proposition 2.2 to give a version of Hensel’s lemma that provides an explicit formula for the p-adic root of a polynomial in Zp[x]. We start by recalling some basic facts about the p-adic numbers. For a comprehensive treatment of this subject, the reader is referred to [10, 11, 14]. Let p be a prime integer. For any nonzero integer a, let vp(a) (the p-adic valuation of a) be the highest power of p which divides a, i.e., the greatest m such that a ≡ 0 (mod pm); we agree to write vp(0) = ∞. Note that vp (a1 a2) = vp(a1)+ vp(a2) for all a1, a2 ∈ Z. For any rational number x = a/b, deﬁne vp(x) = vp(a) − vp(b). Note that this expression depends only on x and not on its representation as a ratio of integers. ON HENSEL’S ROOTS AND A FACTORIZATION FORMULA IN Z[[x]] 5

−vp(x) The p-adic norm in Q is deﬁned as kxkp = p if x 6= 0, and k0kp = 0. This norm is non-Archimedean, that is kx + ykp ≤ max(kxkp, kykp). The p-adic completion of Q with respect to k·kp is denoted by Qp. Every a ∈ Qp admits a unique p-adic expansion, a a a − a = 0 + 1 + · · · + m 1 + a + a p + a p2 + · · · , pm pm−1 p m m+1 m+2 with 0 ≤ ai

2 be prime and let f(x)= a0 + a1x + · · · + amx be a polynomial of degree m in Zp[x]. If r0 ∈ Z is such that ′ f(r0) ≡ 0 (mod p) and vp(f (r0)) = 0, then r0 lifts to a p-adic root r of f given by ∞ n n−k+1 (−1) 2n + 1 c0 n+1 r = r0 + k Bn+k,k(1!c1, 2!c2,... ) , (3.2) c (n + 1)! n − k c1 nX=0 Xk=0 1 (j) f (r0) where cj = j! for j = 0, 1,...,m. Note that vp(c1) = 0 implies 1/c1 ∈ Zp.

Proof. Given f(x) and r0 ∈ Z as above, consider the function g(x)= f(r0 + x). The Taylor expansion of g(x) at x = 0 gives

′′ (m) ′ f (r0) 2 f (r0) m 2 m g(x)= f(r0)+ f (r0)x + 2! x + · · · + m! x = c0 + c1x + c2x + · · · + cmx , ′ with the property that vp(c0) = vp(f(r0)) ≥ 1 and vp(c1) = vp(f (r0)) = 0. Thus c1 6= 0, and by Proposition 2.2, g(x) has a formal root ∞ cn+1 ̺ = γ 0 , n (n + 1)! nX=0 where n (−1)n−k+1 2n + 1 γ = B (1!c , 2!c ,... ). n k+n+1 n − k n+k,k 1 2 Xk=0 c1 Since vp(c1) = 0 and each j!cj is a p-adic integer, we have γn ∈ Zp for every n. Moreover, 2 if n + 1 has the p-adic expansion n +1= n0 + n1p + n2p + · · · , we have n + 1 − s (n + 1) n + 1 v ((n + 1)!) = p < , p p − 1 p − 1 where sp(n +1) = n0 + n1 + n2 + · · · . Therefore, since vp(c0) ≥ 1, we get

cn+1 n + 1 v 0 = v (cn+1) − v ((n + 1)!) >n + 1 − = p−2 (n + 1) →∞ as n →∞, p (n+1)! p 0 p p − 1 p−1 6 DANIELBIRMAJER,JUANB.GIL,ANDMICHAELD.WEINER

cn+1 0 Z which implies that γn (n+1)! converges in p p. In conclusion, the formal root ̺ is indeed a p-adic root of g(xP) and r0 + ̺ ∈ Zp is a root of f(x). More generally, we have:

m Theorem 3.3. Let p> 0 be prime and let f(x)= a0 + a1x + · · · + amx be a polynomial in Zp[x]. Let ν, κ ∈ Z such that 0 ≤ 2κ<ν. If r0 ∈ Z is such that ν ′ f(r0) ≡ 0 (mod p ) and vp(f (r0)) = κ, then r0 lifts to a p-adic root r of f given by ∞ n n−k+1 n+1 κ (−1) 2n + 1 c0 r = r0 + p k Bn+k,k(1!c1, 2!c2,... ) , c (n + 1)! n − k c1 nX=0 Xk=0 1

(j) (j−2)κ f (r0) where cj = p j! for j = 0, 1,...,m.

Proof. The proof is similar to the one for the previous theorem. Let r0 be a root of f ν ′ −2κ κ modulo p and let κ be the p-adic valuation of f (r0). Consider g(x)= p f(r0 + p x). A Taylor expansion of g(x) at 0 gives

′′ (m) −2κ −κ ′ f (r0) 2 (m−2)κ f (r0) m g(x)= p f(r0)+ p f (r0)x + 2! x + · · · + p m! x 2 m = c0 + c1x + c2x + · · · + cmx . ′ If 0 ≤ 2κ<ν, then vp(c0) ≥ 1 and vp(c1) = 0 since vp(f(r0)) ≥ ν > 2κ and vp(f (r0)) = κ. At this point, we can proceed as in the proof of Theorem 3.1 and conclude that the formal root of g(x) provided by (2.3) is indeed a p-adic root of g(x). If we denote that root by ̺, κ then r = r0 + p ̺ is a p-adic root of the polynomial f(x). In the case of quadratic and cubic polynomials, one can use known properties of Bell polynomials to give a simpler representation of the corresponding Hensel’s roots.

2 Quadratic Polynomials. Let p> 2 and f(x)= a0 + a1x + a2x ∈ Zp[x], a2 6= 0. If there ′ is an r0 ∈ Z such that f(r0) ≡ 0 (mod p) and vp(f (r0)) = 0, then by Theorem 3.1 and elementary Bell polynomial identities, the p-adic lift of r0 may be written as ∞ c 2n 1 c c n r = r − 0 0 2 ∈ Z , (3.4) 0 c n n + 1 c2 p 1 nX=0 1 ′ 2n 1 where c0 = f(r0), c1 = f (r0), and c2 = a2. Note that n n+1 ∈ Z are the well-known Catalan numbers. 2 Example 3.5. Let us consider f(x) = 1 + 11x − 5x over Z7. This polynomial has two ′ simple roots mod 7, r0 = 1, 4. Since c0 = f(1) = 7, c1 = f (1) = 1, and c2 = −5, the lift of r0 =1 in Z7 is given by ∞ 2n (−5)n r = 1 − 7n+1. n n + 1 nX=0 ON HENSEL’S ROOTS AND A FACTORIZATION FORMULA IN Z[[x]] 7

′ On the other hand, since f(4) = −35 and f (4) = −29, the lift of r0 =4 in Z7 is given by ∞ 2n 1 5 2n+1 r = 4 − 7n+1. n n + 1 29 nX=0 2 3 4 5 6 Note that 1/29 = 1 + 3 · 7 + 7 + 7 + 2 · 7 + 5 · 7 + O(7 ) is an element of Z7. 2 Example 3.6. We now consider f(x) = 17 + 6x + 2x over Z5. Modulo 5 this polynomial 2 ′ has a double root, r0 = 1. Since f(1) = 5 and f (1) = 2 · 5, we cannot apply any of the above theorems directly. However, the polynomial 1 2 g(x)= 25 f(1 + 5x)=1+2x + 2x has 1 and 3 as simple roots modulo 5, so using (3.4), we get the lifts ∞ ∞ 5 2n 1 5 n 25 2n 1 25 n 1 − and 3 − in Z . 6 n n + 1 18 14 n n + 1 98 5 Xn=0 nX=0 Therefore, the 5-adic roots of f(x) are given by ∞ ∞ 25 2n 1 5 n 125 2n 1 25 n 1 + 5 − and 1 + 3 · 5 − . 6 n n + 1 18 14 n n + 1 98 nX=0 Xn=0 2 3 Cubic Polynomials. Let p > 2 and f(x) = a0 + a1x + a2x + a3x ∈ Zp[x], a3 6= 0. ′ Once again, if there is an r0 ∈ Z such that f(r0) ≡ 0 (mod p) and vp(f (r0)) = 0, then Theorem 3.1 gives a formula for the p-adic lift of r0. However, for cubic polynomials, it is more convenient to use the equation (2.4) and write the root as ∞ n n+k+1 (−1) (n + k)! c0 n+1 r = r0 + k Bn,k(c2, 2c3, 0,... ) , c n! (n + 1)! c1 nX=0 Xk=0 1 (j) where cj = f (r0)/j! for j = 0, 1, 2, 3. Note that Bn,k(c2, 2c3, 0,... )=0 for k

n κ = c B − − (0, 2c , 0,... ) κ 2 n κ,k κ 3 κX≤k n [2(n − k)]! = c2k−n cn−k 2k − n 2 (n − k)! 3 n! = c2k−ncn−k. (2k − n)!(n − k)! 2 3 Therefore, ∞ n n+k+1 n+1 (−1) (n + k)! n! 2k−n n−k c0 r = r0 + k c2 c3 , c n! (n + 1)! (2k − n)!(n − k)! c1 nX=0 kX≥n/2 1 8 DANIELBIRMAJER,JUANB.GIL,ANDMICHAELD.WEINER and with the change n = 2k − j,

∞ k k−j+1 2k−j c0 (−1) k 3k − j j k−j c0 = r0 + k c2c3 . c1 c (2k − j + 1)j k c1 Xk=0 Xj=0 1 2 3 In summary, if r0 is a simple root mod p of f(x) = a0 + a1x + a2x + a3x ∈ Zp[x], then the p-adic lift of r0 is given by ∞ k k−j j k−j k c0 (−1) c2 k 3k − j c0c3 c0 r = r0 − 2 , (3.7) c1 2k − j + 1j k c1 c Xk=0 Xj=0 1 ′ ′′ where c0 = f(r0), c1 = f (r0), c2 = f (r0)/2, and c3 = a3.

Following the same steps as for cubic polynomials, we can obtain the following result.

ℓ m Proposition 3.8. Let p> 2 and 1 <ℓ 2 and f(x)= x − 1. Assume that r0 is a single root mod p of m m m−j f. Then r0 lifts to a p-adic root of the form (3.2) with c0 = r0 − 1 and cj = j r0 for j = 1,...,m. Now, since m! j!c = rm−j(m) with (m) = , j 0 j j (m − j)! homogeneity properties of the Bell polynomials together with identity (A.4) give m−1 m−2 Bn+k,k(1!c2, 2!c3,... )= Bn+k,k(r0 (m)1, r0 (m)2,... ) mk−(n+k) = r0 Bn+k,k((m)1, (m)2,... ) k − 1 k = rmk (n+k) (−1)k−j (jm) . 0 k! j n+k Xj=0 Therefore, n (−1)n−k+1 2n + 1 Bn+k,k(1!c1, 2!c2,... ) ck (n + 1)! n − k Xk=0 1 n n−k+1 k (−1) 2n + 1 mk−(n+k) 1 k−j k = r (−1) (jm)n+k (rm−1m)k (n + 1)! n − k 0 k! j Xk=0 0 Xj=0 n k (−1)n−j+1 2n + 1 1 k = (jm) . rn mk (n + 1)! n − k k!j n+k Xk=0 Xj=0 0 ON HENSEL’S ROOTS AND A FACTORIZATION FORMULA IN Z[[x]] 9

Finally, the p-adic lift of r0 is given by ∞ n k n−j c0 (−1) 2n + 1 1 k c0 n r = r0 − k (jm)n+k , c1 m (n + 1)! n − k k!j r0c1 nX=0 Xk=0 Xj=0 ′ where c0 = f(r0) and c1 = f (r0).

For m = p − 1, the above expression gives an explicit formula for the Teichm¨uller lifts. Proposition 3.9. Let p> 2. Every integer q ∈ {1,...,p − 1} is a (p − 1)-st root of unity mod p and lifts to a p-adic root of unity ξq given by

∞ n k n−j c0 (−1) 2n + 1 1 k c0 n ξq = q − k (j(p − 1))n+k , c1 (p − 1) (n + 1)! n − k k!j qc1 nX=0 Xk=0 Xj=0 p−1 p−2 where c0 = q − 1 and c1 = (p − 1)q .

4. Factorization of polynomials over Z[[x]] 2 Let f(x) = f0 + f1x + f2x + . . . be a formal power series in Z[[x]]. It is easy to prove that f(x) is invertible in Z[[x]] if and only if |f0| = 1. A natural question, initially discussed in [4], is whether or not a non-invertible element of Z[[x]] can be factored over Z[[x]]. In recent years, this question has been investigated by several authors, leading to suﬃcient and in some cases necessary reducibility criteria, see e.g. [2, 5, 9]. In particular, [9] deals with the factorization of formal power series over principal ideal domains. For the case at hand, the following elementary results are known. The formal power 2 w series f(x) = f0 + f1x + f2x + . . . is irreducible in Z[[x]] if |f0| is prime, or if |f0| = p with p prime, w ∈ N, and gcd(p,f1) = 1. On the other hand, if f0 is neither a unit nor a prime power, then f(x) is reducible. In this case, the factorization algorithm is simple and relies on a recursion and a single diophantine equation, see [4, Prop. 3.4]. Finally, in the remaining case when f0 is a prime power and f1 is divisible by p, the reducibility of f(x) in Z[[x]] is linked to the existence of a p-adic root of positive valuation. The goal of this section is to give an explicit factorization over Z[[x]] for reducible polynomials of the form w m 2 d f(x)= p + p γ1x + γ2x + · · · + γdx , m ≥ 1, w ≥ 2, d ≥ 2, (4.1) where γ1, . . . , γd ∈ Z and gcd(p, γ1) = 1. This is the only type of polynomial for which the reducibility and factorization over Z[[x]] is not straightforward. Theorem 4.2. Let p be an odd prime and let f(x) be a polynomial of the form (4.1). ℓ ∞ ℓj Assume that f has a simple root r ∈ pZp with vp(r) = ℓ ≤ m and r = p (1 + j=1 ejp ) with ej ∈ Z. Then f(x) admits the factorization P ∞ ∞ f(x)= pℓ − x − x a xn pw−ℓ + (pw−2ℓ + pm−ℓγ )x + x b xn , n 1 n nX=1 nX=1 ℓn where the coeﬃcients an are given by (4.4), and bn = ˆbn/p with ˆbn as in (4.11). 10 DANIELBIRMAJER,JUANB.GIL,ANDMICHAELD.WEINER

ℓn Remark. (a) As shown in Lemma 4.12, ˆbn is divisible by p , so bn ∈ Z for every n. (b) If r ∈ pZp is a root of f with vp(r)= ℓ ≤ m, then 2ℓ ≤ w. (c) If w ≤ 2m and f has a root r ∈ pZp, then vp(r) = ℓ ≤ m holds. If w > 2m, then 0 lifts to a p-adic root of f, but it is not necessarily true that f has a root of valuation less than or equal to m. This property depends on the coeﬃcients γ2, γ3,... . However, even if that condition fails, f(x) is still reducible and a factorization can be obtained through the algorithm given in [5, Prop. 2.4]. ℓ ∞ ℓj (d) A p-adic integer r with vp(r)= ℓ can always be written as r = p (e0 + j=1 ejp ) with ∗ e0 ∈ Zp. For factorization purposes, we can assume without loss of gPenerality e0 = 1. ∗ ∗ ∗ ℓ Otherwise, consider g(x)= f(x/e0), where e0 is such that e0e0 = 1 (mod p ). (e) As discussed in [5], the existence of a root in pZp is in many cases (e.g. when d ≤ 3) a necessary condition for the polynomial (4.1) to factor over Z[[x]].

Remark. If f has a multiple root in pZp, then f(x) admits the simpler factorization

f(x)= G(x)fred(x), ′ where G(x) = gcd(f(x),f (x)) ∈ Z[x] and fred(x)= f(x)/G(x). ℓ ∞ ℓj Proof of Theorem 4.2. Let r = p 1+ j=1 ejp be the p-adic root of f and define P ∞ j φ(x)= xE(x) with E(x)=1+ ejx . (4.3) Xj=1 ℓ ℓ −1 ℓ −1 Thus r = φ(p ) in Zp and therefore p = φ (r). Define A(x) = p − φ (x). So A(r) = 0 in Zp, and by Lemma 2.1, we have ∞ ℓ −1 ℓ n A(x)= p − φ (x)= p − x 1+ anx , nX=1 where n 1 (n + k)! a = (−1)k B (1!e , 2!e ,... ) ∈ Z. (4.4) n n! (n + 1)! n,k 1 2 Xk=1 Our goal is to find B(x) ∈ Z[[x]] such that f(x)= A(x)B(x). For convenience, consider fˆ(x)= p−2ℓf(pℓx) and Aˆ(x)= p−ℓA(pℓx). Thus ∞ ℓn n Aˆ(x) = 1 − x − x p anx . nX=1 Proposition 4.5. The reciprocal of Aˆ(x) is a power series in Z[[x]] of the form ∞ 1 Aˆ(x)−1 = =1+ x + x t xn ˆ n A(x) nX=1 with n k n + 1 − k (n + j)! t =1+ pℓk (−1)j B (1!e , 2!e ,... ) ∈ Z. (4.6) n k! (n + 1)! k,j 1 2 Xk=1 Xj=1 ON HENSEL’S ROOTS AND A FACTORIZATION FORMULA IN Z[[x]] 11

Proof. By Theorem B in [8, Section 3.5], and using basic properties of partial Bell polynomials, we have ∞ n k! Aˆ(x)−1 =1+ x + B (1, 2!a pℓ, 3!a p2ℓ,... ) xn n! n,k 1 2 nX=2 Xk=1 ∞ n k n n! ℓ 2ℓ x =1+ x + k! B − − (1!p a , 2!p a ,... ) (n − k)!j! n k,k j 1 2 n! nX=2 Xk=1 Xj=0 ∞ n n+1−k (n + 1 − k)! ℓk n+1 =1+ x + 1+ p Bk,j(1!a1, 2!a2,... )  x k!(n + 1 − k − j)! nX=1 Xk=1 Xj=1   ∞ n ℓk k p (n + 1 − k)! n =1+ x + x 1+ B (1!a1, 2!a2,... )  x . k! (n + 1 − k − j)! k,j nX=1 Xk=1 Xj=1   In the last step, we declare the interior sum to be zero if j > min(k,n + 1 − k). Thus

n k pℓk (n + 1 − k)! t =1+ B (1!a , 2!a ,... ) n k! (n + 1 − k − j)! k,j 1 2 Xk=1 Xj=1 n k n + 1 − k n − k =1+ pℓk (j − 1)!B (1!a , 2!a ,... ) k! j − 1 k,j 1 2 Xk=1 Xj=1

Now, if we write k!ak as k k + j k!a = (j − 1)!B (−1!e , −2!e ,... ), k j − 1 k,j 1 2 Xj=1 then by means of Theorem 15 in [6] we get

k n − k (j − 1)!B (1!a , 2!a ,... ) j − 1 k,j 1 2 Xj=1 k n + j = (j − 1)!B (−1!e , −2!e ,... ) j − 1 k,j 1 2 Xj=1 k (n + j)! = (−1)j B (1!e , 2!e ,... ). (n + 1)! k,j 1 2 Xj=1

In other words, tn has the form claimed in (4.6). Now, motivated by (4.6), for n ≥ 1 we consider

∞ k n + 1 − k (n + j)! T (x)=1+ (−1)j B (1!e , 2!e ,... ) xk. n k! (n + 1)! k,j 1 2 Xk=1 Xj=1 12 DANIELBIRMAJER,JUANB.GIL,ANDMICHAELD.WEINER

Lemma 4.7. With E(x) as in (4.3), we have −n−2 ′ Tn(x)= E(x) E(x)+ xE (x) . Proof. Fix n ≥ 1 and denote k 1 (n + j)! τ = (−1)j B (1!e , 2!e ,... ). k k! n! k,j 1 2 Xj=1 Then ∞ ∞ ∞ k 1 T (x)=1+ 1 − τ xk =1+ τ xk − kτ xk. n n + 1 k k n + 1 k Xk=1 Xk=1 Xk=1 ∞ k −n−1 It is easy to check that 1 + k=1 τkx = E(x) . Therefore, P −n−1 1 d −n−1 Tn(x)= E(x) − x E(x) n + 1 dx = E(x)−n−1 + xE(x)−n−2E′(x)= E(x)−n−2 E(x)+ xE′(x) .

As a direct consequence of this lemma we get the recurrence relation

Tn−1(x)= E(x)Tn(x), which can be used to deﬁne T0(x) and T−n(x) for n ≥ 1. More precisely, we let n+1 T0(x)= E(x)T1(x) and T−n(x)= E(x) T1(x) for n ≥ 1. Given that d −2ℓ ℓ w−2ℓ m−ℓ ℓ(n−2) n fˆ(x)= p f(p x)= p + p γ1x + p γnx , (4.8) Xn=2 j the relation Tn−j(x)= E(x) Tn(x) gives d w−2ℓ m−ℓ ℓ(j−2) p Tn(x)+ p γ1Tn−1(x)+ p γjTn−j(x)= fˆ(E(x))Tn(x). (4.9) Xj=2

Moreover, since E(x) is a unit in Z[[x]], for every ν ∈ Z the function Tν(x) is in Z[[x]] and ℓ so Tν(p ) ∈ Zp. ℓ Lemma 4.10. For ν ≥−1, the p-adic numbers Tν(p ) satisfy ℓ ℓ(ν+2) Tν (p ) − tν ≡ 0 (mod p ) with tν as in (4.6) for ν > 0 and t0 = t−1 = 1.

Proof. For ν = n ≥ 1 the statement is a consequence of the fact that tn is the n-th partial ℓ n+1 sum of Tn(p ) and the coeﬃcient of x in Tn(x) is zero. Further, given that ℓ ℓ 2ℓ ℓ ℓ 2ℓ E(p )=1+ p e1 + O(p ) and T1(p ) = 1 − p e1 + O(p ), we have ℓ ℓ 2ℓ ℓ 2ℓ 2ℓ T0(p )= 1+ p e1 + O(p ) 1 − p e1 + O(p ) ≡ 1 (mod p ). ON HENSEL’S ROOTS AND A FACTORIZATION FORMULA IN Z[[x]] 13

ℓ 2ℓ This implies T0(p ) − t0 ≡ 0 (mod p ). ℓ ℓ 2 ℓ ℓ 2 ℓ Finally, since T−1(p ) = E(p ) T1(p ), and because E(p ) and T1(p ) are both of the ℓ ℓ ℓ ℓ ℓ form 1+ O(p ), we get T−1(p ) ≡ 1 (mod p ), hence T−1(p ) − t−1 ≡ 0 (mod p ).

Using fˆ(x) as in (4.8), we now deﬁne

d ∞ −1 w−2ℓ m−ℓ ℓ(n−2) n n Bˆ(x)= fˆ(x)Aˆ(x) = p + p γ1x + p γnx 1+ x + x tnx , nX=2 nX=1 and write it as ∞ w−2ℓ w−2ℓ m−ℓ n Bˆ(x)= p + (p + p γ1)x + x ˆbnx nX=1 with d w−2ℓ m−ℓ ℓ(j−2) ˆbn = p tn + p γ1tn−1 + p γjtn−j ∈ Z, (4.11) Xj=2 where tn is given by (4.6), t0 = t−1 = 1, and t−n = 0 for n> 1.

ℓn Lemma 4.12. The coeﬃcients ˆbn are divisible by p . Proof. First of all, since p−ℓr = E(pℓ) is a p-adic root of fˆ, identity (4.9) implies

d w−2ℓ ℓ m−ℓ ℓ ℓ(j−2) ℓ p Tn(p )+ p γ1Tn−1(p )+ p γjTn−j(p )=0 in Zp. Xj=2 Therefore, for n ≥ d − 1,

d w−2ℓ m−ℓ ℓ(j−2) ˆbn = p tn + p γ1tn−1 + p γjtn−j Xj=2 d w−2ℓ ℓ m−ℓ ℓ ℓ(j−2) ℓ = p tn − Tn(p ) + p γ1 tn−1 − Tn−1(p ) + p γj tn−j − Tn−j(p ) , Xj=2 which by Lemma 4.10 is congruent to 0 mod pℓn. Similarly, for 1 ≤ n < d − 1,

n+1 w−2ℓ m−ℓ ℓ(j−2) ˆbn = p tn + p γ1tn−1 + p γjtn−j Xj=2 d ℓ(j−2) ℓ ℓn ≡− p γjTn−j(p ) ≡ 0 (mod p ). j=Xn+2

Finally, deﬁning B(x)= pℓBˆ(x/pℓ), we arrive at the factorization f(x)= A(x)B(x). 14 DANIELBIRMAJER,JUANB.GIL,ANDMICHAELD.WEINER

Remark. It is worth mentioning that our method for factorization in Z[[x]] is not restricted to polynomials and can be applied to power series. As an example, consider ∞ 8x3 f(x) = 9 + 12x + 7x2 + 8x3 xk = 9 + 12x + 7x2 + , 1 − x Xk=0 discussed by B´ezivin in [2]. This series is reducible in Z[[x]] and factors as (3 − x)2(1 + x) f(x)= . 1 − x The reader is invited to conﬁrm that the power series version of Theorem 4.2 gives the (3−x)(1+x) factorization f(x)= A(x)B(x) with A(x) = 3 − x and B(x)= 1−x . 2 An interesting feature of this example is that the partial sums fd(x) = 9+12x+7x +· · · of f(x) of degree d ≥ 2 are all irreducible in Z[[x]]. This fact was proved in [2, Prop. 8.1], but it can also be derived from Proposition 3.4 of [6] together with the observation that for d ≥ 2, the polynomial fd(x) has no roots in pZp.

Appendix: Some properties of Bell polynomials Throughout this paper, we make extensive use of the well-known partial Bell polynomials. For any sequence x1,x2,... , the (n, k)-th partial Bell polynomial is deﬁned by

n! x1 i1 x2 i2 Bn,k(x)= · · · , i1!i2! · · · 1! 2! i∈Xπ(n,k) where π(n, k) is the set of all sequences i = (i1, i2,... ) of nonnegative integers such that

i1 + i2 + · · · = k and i1 + 2i2 + 3i3 + · · · = n. Clearly, these polynomials satisfy the homogeneity relation 2 3 k n Bn,k(abx1, ab x2, ab x3,... )= a b Bn,k(x1,x2,x3,... ). Here are other elementary identities (cf. [8, Section 3.3]) needed in this paper: n! B ( x2 , x3 ,... )= B (0,x ,x ,... ), (A.1) n,k 2 3 (n + k)! n+k,k 2 3

′ ′ n ′ ′ B (x + x ,x + x ,... )= B (x ,x ,... )B − − (x ,x ,... ), (A.2) n,k 1 1 2 2 ν ν,κ 1 2 n ν,k κ 1 2 κX≤k ν≤n (jk)! B (0,..., 0,x , 0,... ) = 0, except B = xk. (A.3) n,k j jk,k k!(j!)k j Also of special interest is the identity k 1 k B ((a) , (a) ,... )= (−1)k−j (ja) , (A.4) n,k 1 2 k! j n Xj=0 where (a)n = a(a − 1) · · · (a − n + 1). This is a special case of [16, Example 3.2]. For more on Bell polynomials and their applications, see e.g. [1, 7, 8, 16]. ON HENSEL’S ROOTS AND A FACTORIZATION FORMULA IN Z[[x]] 15

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