FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization properties in Z[[x]], the ring of formal power series over Z We treat polynomials of arbitrary degree and give sufficient conditions for their reducibility as power series Moreover, if a polynomial is reducible over Z[[x]], we provide an explicit factorization algorithm For polynomials whose constant term is a prime power, our study leads to the discussion of p-adic integers 1 Introduction In this paper we consider polynomials with integer coefficients and discuss their factorization as elements of Z[[x]], the ring of formal power series over Z We treat polynomials of arbitrary degree and give sufficient conditions for their reducibility in Z[[x]] For polynomials of degree two or three these conditions are also necessary If the constant term of the polynomial is not a prime power, the reducibility discussion is straightforward We briefly address these cases in Section 2 On the other hand, if the constant term of the polynomial is a nontrivial prime power, say p n with p prime and n 2, the question of reducibility in Z[[x]] leads in some cases (when n is no greater than twice the p-adic valuation of the linear coefficient) to the discussion of p-adic integers In this context, our main result is that, if such a polynomial has a root in Z p whose p-adic valuation is positive, then it is reducible in Z[[x]] This particular case is presented in Section 3 Our proofs are constructive and provide explicit factorization algorithms All the needed material regarding the ring Z p of p-adic integers can be found in [5, 6] It is important to note that irreducible elements in Z[x] and in Z[[x]] are, in general, unrelated For instance, 6+x+x 2 is irreducible in Z[x] but can be factored in Z[[x]], while 2 + 7x + 3x 2 is irreducible in Z[[x]] but equals (2 + x)(1 + 3x) as a polynomial Observe that the latter is not a proper factorization in Z[[x]] since 1 + 3x is an invertible element More examples and a few remarks concerning the hypotheses of our main theorem are given in Section 4 A first study of the factorization theory of quadratic polynomials in Z[[x]] was presented in [3] In that paper we established necessary and sufficient conditions for a quadratic polynomial to be irreducible in Z[[x]] In particular, for polynomials whose constant term is a power of a prime p, we showed in [3] that reducibility in Z[[x]] is related to reducibility in Z p [x] Some of the basic ideas from [2] and [3] are summarized in Section 2 2010 Mathematics Subject Classification 13P05;13F25,11Y05,11D88 Key words and phrases Factorization, integral polynomials, power series, p-adic numbers D Birmajer would like to acknowledge the support and hospitality of the Universidad Nacional de San Luis, San Luis, Argentina 1

2 DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER The results of this paper contain and expand those obtained in [3] to polynomials in Z[x] of arbitrary degree The special case of cubic polynomials is explicitly and fully discussed in Section 4 Recently, J Elliott posted a preprint [4] that discusses the irreducibility and factoring of polynomials in the ring of formal power series over a principal ideal domain His methods are different from ours and his results rely on a generalization of the p-adic Weierstrass preparation theorem Finally, we would like to acknowledge our correspondence with J-P Bézivin We appreciate his initiative to share with us an early draft of his unpublished work We also thank the IJNT referee for all the useful suggestions 2 Preliminaries We start by reviewing some of the basic properties of the factorization theory of power series over Z For a more extensive discussion, we refer the reader to [2] Let f(x) = f 0 + f 1 x + + f d x d be a polynomial with integer coefficients It is easy to check that f(x) is invertible in Z[[x]] if and only if f 0 = ±1 In particular, the reducibility of f(x) is equivalent to the reducibility of f(x) Also observe that if f 0 = 0, then f(x) is either an associate of x or f(x) has x as a proper factor For this reason, it suffices to look at the cases when f 0 > 0 We say that f(x) is reducible in Z[[x]] if there exist power series, A(x) = a k x k and B(x) = b k x k with a k, b k Z, k=0 k=0 such that a 0 ±1, b 0 ±1, and f(x) = A(x)B(x) A proof of the following basic proposition can be found in [2] Proposition 21 Let f(x) be a non-invertible polynomial in Z[[x]] with f 0 > 0 (a) If f 0 is prime, then f(x) is irreducible in Z[[x]] (b) If f 0 is not a prime power, then f(x) is reducible in Z[[x]] (c) If f 0 = p n with p prime, n 2, and p f 1, then f(x) is irreducible in Z[[x]] As an immediate consequence, we have: Proposition 22 (Linear polynomials) A polynomial f(x) = p n + f 1 x with n 2 is reducible in Z[[x]] if and only if p f 1 It remains to examine the reducibility of nonlinear polynomials of the form f(x) = p n + p m γ 1 x + γ 2 x 2 + + γ d x d, γ d 0, (23) with p prime, n 2, m 1, gcd(p, γ 1 ) = 1 or γ 1 = 0, and gcd(p, γ 2,, γ d ) = 1 As it turns out, the reducibility of such polynomials depends on the relation between the parameters n and m in (23) and on their factorization properties as elements of Z p [x] Accordingly, we divide our study into two cases: The case when n > 2m, see Proposition 25, and the case when n 2m, which is more involved and will be investigated in the next section Remark 24 If a polynomial of the form (23) has no linear term, ie if γ 1 = 0, we can assume m as large as needed and can therefore discuss its reducibility as for the case when n 2m

FACTORING POLYNOMIALS IN Z[[x]] 3 Proposition 25 If n > 2m and gcd(p, γ 1 ) = 1, then the polynomial f(x) in (23) is reducible in both Z[[x]] and Z p [x] Proof First of all, observe that f(p n ) 0 (mod p n ), and f (p n ) 0 (mod p m ) because n > m Moreover, since n > 2m m + 1 and gcd(p, γ 1 ) = 1, we have f (p n ) 0 (mod p m+1 ) Then, by Hensel s lemma, p n lifts to a root of f(x) in Z p Thus f(x) is reducible in Z p [x] To show the reducibility of f(x) in Z[[x]], we give an inductive procedure to find a k, b k Z such that f(x) = ( p m + a 1 x + a 2 x 2 + )( p n m + b 1 x + b 2 x 2 + ) First of all, observe that n > 2m implies n m > m As a first step, we need p m γ 1 = p m (b 1 + p n 2m a 1 ) γ 2 = p m s 2 + a 1 (γ 1 p n 2m a 1 ) (26) with s 2 = b 2 + p n 2m a 2 If we let h(x) = γ 2 γ 1 x + p n 2m x 2, then solving the last equation is equivalent to finding a 1 Z such that h(a 1 ) 0 (mod p m ) Since the discriminant of h(x) is a square mod p, there exists r Z p such that h(r) = 0 We let a 1 Z be the reduction of r mod p m With this choice of a 1 we can now use (26) to determine that b 1 = γ 1 p n 2m a 1 and s 2 = h(a 1 )/p m With the notation the rest of the equations become s j = b j + p n 2m a j, j = 2, 3, (27) γ 3 = p m s 3 + a 2 (γ 1 2p n 2m a 1 ) + a 1 s 2 γ 4 = p m s 4 + a 3 (γ 1 2p n 2m a 1 ) + a 1 s 3 + a 2 b 2 γ d = p m s d + a d 1 (γ 1 2p n 2m a 1 ) + a 1 s d 1 + a 2 b d 2 + + a d 2 b 2 0 = p m s d+1 + a d (γ 1 2p n 2m a 1 ) + a 1 s d + a 2 b d 1 + + a d 1 b 2 0 = p m s k+1 + a k (γ 1 2p n 2m a 1 ) + a 1 s k + a 2 b k 1 + + a k 1 b 2 Since γ 1 2p n 2m a 1 is not divisible by p, there are integers a 2 and s 3 such that the equation for γ 3 is satisfied Now that a 2 has been chosen, we combine it with s 2 in (27) to determine b 2 For each of the above equations, the same argument can be used to solve inductively for the pair a k, s k+1 Z At each iteration, equation (27) then gives b k In the next section, we will make use of the following elementary result: Lemma 28 (Theorem 142 in [5]) A polynomial with integer coefficients has a root in Z p if and only if it has an integer root modulo p k for any k 1

4 DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER 3 Factorization in the presence of a p-adic root In this section, we finish our discussion of reducibility in Z[[x]] for polynomials with integer coefficients Our results rely on the existence of a root ϱ pz p, and the nature of the factorization depends on the multiplicity of ϱ and on its p-adic valuation, denoted by v p (ϱ) As we will show later, the existence of a root in pz p is a necessary condition for quadratic and cubic polynomials If ϱ pz p is a multiple root of f(x), then f(x) is reducible in Z[x]: Proposition 31 Let f(x) be a polynomial of degree d 2 If f(x) has a multiple root ϱ Z p with v p (ϱ) = l 1, then f(x) admits a proper factorization f(x) = G(x) f red (x) where G(x) = gcd(f, f ) in Z[x], and f red (x) = f(x)/g(x) Proof Let ϱ Z p with v p (ϱ) = l be the multiple root of f(x) Let G(x) = gcd(f, f ) Z[x] Then f red (x) = f(x)/g(x) is an element of Z[x] and f(x) = G(x)f red (x) This factorization is also valid in Q p [x], and every root of f(x) is a simple root of f red (x) Thus, together with f red (ϱ) = 0, we must also have G(ϱ) = 0 Hence p l divides both f red (0) and G(0) giving that the factorization f(x) = G(x)f red (x) is a proper factorization of f(x) in Z[[x]] We now deal with the case when the polynomial f(x) = p n + p m γ 1 x + γ 2 x 2 + + γ d x d, γ d 0, d 2, (32) has a simple root ϱ pz p with v p (ϱ) = l 1 As before, we assume p prime, n 2, m 1, gcd(p, γ 1 ) = 1 or γ 1 = 0, and gcd(p, γ 2,, γ d ) = 1 We also assume n 2m; the case when n > 2m is covered by Proposition 25 Our goal is to prove that f(x) admits a factorization f(x) = ( p l + a 1 x + a 2 x 2 + )( p n l + b 1 x + b 2 x 2 + ) in Z[[x]] (33) Clearly, this factorization is equivalent to the solvability of the infinite system of equations obtained by comparing coefficients We set a 2 = a 3 = = a d 1 = 0 and start by solving the first d equations: p m γ 1 = p l (b 1 + p n 2l a 1 ) γ 2 = p l b 2 + a 1 b 1 γ d 1 = p l b d 1 + a 1 b d 2 γ d = p l (b d + p n 2l a d ) + a 1 b d 1 (34) We set s d = b d + p n 2l a d and replace the last equation by γ d = p l s d + a 1 b d 1 In our next lemma, we solve this finite system of equations and take a first step towards the factorization of f(x) Lemma 35 Under the above assumptions, there exist b 1, b 2,, b d 1 Z, s d Z, and a 1 Z with gcd(p, a 1 ) = 1, such that f(x) = (p l + a 1 x) ( p n l + b 1 x + b 2 x 2 + + b d 1 x d 1) + p l s d x d

FACTORING POLYNOMIALS IN Z[[x]] 5 Proof Let ϱ Z p, with v p (ϱ) = l, be the root of f(x) Since n 2m and p n + p m γ 1 ϱ + γ 2 ϱ 2 + + γ d ϱ d = 0, we must have l m and 2l n Consider the polynomial in Z[x] g d (x) = p (d 2)l γ d p (d 3)l γ d 1 x + p (d 4)l γ d 2 x 2 + + ( 1) d 2 γ 2 x d 2 + ( 1) d 1 p m l γ 1 x d 1 + ( 1) d p n 2l x d obtained from f(x) via g d ( x) = xd p 2l f(p l /x) We define recursively g k (x) = p(k 1)l γ k+1 g k+1 (x) for k = (d 1), (d 2),, 2 x In particular, g 2 (x) = γ 2 p m l γ 1 x + p n 2l x 2 Solving for g k+1 (x) and taking its derivative, we obtain g k+1(x) = ( g k (x) + xg k(x) ) for every 2 k d 1 (36) Let r = p l /ϱ Z p Then v p (r) = 0 and g d (r) = 0 Moreover, since f (ϱ) 0, we have g d (r) 0 Let θ = v p(g d (r)) Let a 1 Z be a reduction of r mod p (2d 3)l+2θ Thus gcd(p, a 1 ) = 1, g d (a 1 ) 0 (mod p (2d 3)l+2θ ), and v p (g d(a 1 )) = θ (37) Note that the particular choice of p (2d 3)l+2θ is not necessary for this lemma In fact, choosing a reduction of r mod p (d 1)l would suffice here However, the exponent (2d 3)l + 2θ will play a crucial role in the proof of Theorem 39 Since g d (a 1 ) = p (d 2)l γ d a 1 g d 1 (a 1 ), we have that p (d 2)l g d 1 (a 1 ) If we proceed iteratively, we obtain that p (k 1)l g k (a 1 ) for every 2 k d 1 Now that we have a 1, the first equation in (34) gives and the second equation there becomes b 1 = p m l γ 1 p n 2l a 1, γ 2 = p l b 2 + a 1 (p m l γ 1 p n 2l a 1 ) This equation can then be solved with b 2 = p l g 2 (a 1 ) In fact, if we set b k = p (k 1)l g k (a 1 ) for 2 k d 1, and s d = p (d 1)l g d (a 1 ), (38) the integers a 1, b 1, b 2,, b d 1, s d, solve the system (34) and we get the claimed identity f(x) = (p l + a 1 x) ( p n l + b 1 x + b 2 x 2 + + b d 1 x d 1) + p l s d x d Theorem 39 Let f(x) be a polynomial of the form (32) with n 2m If f(x) has a simple root ϱ Z p with v p (ϱ) = l 1, then f(x) admits a factorization in Z[[x]] of the form (33) Proof We let a 2 = a 3 = = a d 1 = 0 and choose a 1 as in (37) Following the proof of the previous lemma, we can then find integers b 1, b 2,, b d 1, s d, that solve the system (34) If we define s k = b k + p n 2l a k for every k = 2, 3, 4,,

6 DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER the remaining equations needed for the factorization of f(x) (obtained by comparing coefficients) can be written as 0 = p l s d+1 + a d b 1 + a 1 b d 0 = p l s d+2 + a d+1 b 1 + a d b 2 + a 1 b d+1 (310) Since b 1 = p m l γ 1 p n 2l a 1, b k = s k p n 2l a k for k 2, and since the polynomial g 2 (x) = γ 2 p m l γ 1 x + p n 2l x 2 satisfies g 2(a 1 ) = (p m l γ 1 2p n 2l a 1 ), we have a k b 1 + a 1 b k = a k g 2(a 1 ) + a 1 s k for every k d This identity allows us to rewrite each equation in (310) as k 1 0 = p l s k+1 a k g 2(a 1 ) + a 1 s k + a j b k+1 j We proceed with an inductive algorithm to find a d+j and s d+1+j for j 0 This will then determine b d+j Together these two sequences of integers provide a solution for (34) and (310), and consequently a factorization of f(x) Base step Choosing a d and s d+1 We consider the next block of d 1 consecutive equations in (310) As mentioned above, these equations can be written as 1 0 = p l s d+1 a d g 2(a 1 ) + a 1 s d j=d 0 = p l s d+2 a d+1 g 2(a 1 ) + a d s 2 + a 1 s d+1 0 = p l s d+3 a d+2 g 2(a 1 ) + a d+1 s 2 + a d s 3 + a 1 s d+2 (311) 0 = p l s 2d 1 a 2d 2 g 2(a 1 ) + a 2d 3 s 2 + + a d s d 1 + a 1 s 2d 2 The first two equations combined give 0 = p 2l s d+2 p l a d+1 g 2(a 1 ) + p l a d s 2 + a 1 (a d g 2(a 1 ) a 1 s d ) = p 2l s d+2 p l a d+1 g 2(a 1 ) + a d ( g2 (a 1 ) + a 1 g 2(a 1 ) ) a 2 1s d = p 2l s d+2 p l a d+1 g 2(a 1 ) a d g 3(a 1 ) a 2 1s d, (312) because p l s 2 = g 2 (a 1 ) and g 2 (a 1 ) + a 1 g 2(a 1 ) = g 3(a 1 ), cf (36) By combining (312) with the third equation in (311), and using the identities p 2l s 3 = g 3 (a 1 ) and g 3 (a 1 ) + a 1 g 3(a 1 ) = g 4(a 1 ), we then arrive at 0 = p 3l s d+3 p 2l a d+2 g 2(a 1 ) a d+1 p l g 3(a 1 ) a d g 4(a 1 ) + a 3 1s d Continuing this process, using the relations (36) and (38) in each iteration, the system (311) becomes 0 = p l s d+1 a d g 2(a 1 ) + a 1 s d 0 = p 2l s d+2 p l a d+1 g 2(a 1 ) a d g 3(a 1 ) a 2 1s d 0 = p 3l s d+3 p 2l a d+2 g 2(a 1 ) a d+1 p l g 3(a 1 ) a d g 4(a 1 ) + a 3 1s d (313) 0 = p (d 1)l s 2d 1 p (d 2)l a 2d 2 g 2(a 1 ) a d g d(a 1 ) + ( 1) d a d 1 1 s d 1 Observe that bk = s k for k = 2,, d 1

FACTORING POLYNOMIALS IN Z[[x]] 7 We now set t d = p (d 2)l s 2d 1 p (d 3)l a 2d 2 g 2(a 1 ) a d+1 g d 1(a 1 ) (314) and rewrite the last equation in (313) as 0 = p l t d a d g d(a 1 ) + ( 1) d a d 1 1 s d (315) 0 = p l t d p θ[ a d p θ g d(a 1 ) + ( 1) d 1 a d 1 1 p θ s d ] (316) Since p (2d 3)l+2θ g d (a 1 ) and s d = p (d 1)l g d (a 1 ), cf (37) and (38), we have that p (d 2)l+2θ s d In particular, p θ s d Z Now, since v p (p θ g d (a 1)) = 0, we can choose a d Z such that the expression inside the brackets in (316) is divisible by p (d 1)l+θ We then solve for t d and obtain that t d is divisible by p (d 2)l+2θ Furthermore, from equation (315) we deduce that a d must be divisible by p (d 2)l+θ With a d having been chosen, we go back to the first equation of (313) and solve for s d+1 It follows that, at the very least, p θ s d+1 Note that if d = 2, then s d+1 = s 2d 1 = t d, so the choice of t d in equation (316) already determines s d+1 Inductive step Choosing a d+j, t d+j and s d+1+j Consistent with the above definition of t d, we now let t d+j = p (d 2)l s 2d 1+j p (d 3)l a 2d 2+j g 2(a 1 ) a d+1+j g d 1(a 1 ) Following the pattern in (311), the equation for s 2d can then be written as 0 = p l s 2d a 2d 1 g 2(a 1 ) + a 2d 2 s 2 + + a d+1 s d 1 + a d b d + a 1 s 2d 1 If we multiply both sides of this equation by p (d 2)l and use the relations (36), (38), and (314), we obtain the equivalent equation 0 = p l t d+1 a d+1 g d(a 1 ) + a 1 t d + p (d 2)l a d b d Repeating this process iteratively, for every j 1, the equation for s 2d+j 1 in the system (310) can be reduced to 0 = p l t d+j a d+j g d(a 1 ) + a 1 t d+j 1 + p (d 2)l R d+j 1, (317) where R d+j 1 = a d b d+j 1 + + a d+j 1 b d Assume that the numbers a d+i, t d+i and s d+1+i have been chosen for all i with 0 i j 1, in such a way that the corresponding equations are all satisfied, and such that p (d 2)l+2θ t d+i, p (d 2)l+θ a d+i, p θ s d+1+i Note that we then have b d+i as well and, in particular, p θ b d+i for all 0 i j 1 Thus each R d+i is determined and p 2θ R d+i We now proceed to choose a d+j and t d+j Since p θ divides g d (a 1), t d+j 1, and R d+j 1, we can rewrite (317) as 0 = p l t d+j p θ[ a d+j p θ g d(a 1 ) a 1 p θ t d+j 1 p (d 2)l θ R d+j 1 ] without leaving Z Now, as for the base step, we can choose a d+j Z such that the bracket is divisible by p (d 1)l+θ (since v p (p θ g d (a 1)) = 0) We then solve for t d+j and get a number which is divisible by p (d 2)l+2θ From equation (317), it follows that a d+j must be divisible by p (d 2)l+θ Once again, if d = 2 we can stop here since s d+1+j = s 2d 1+j = t d+j To choose s d+1+j we consider two cases If j d 2, the definition of t j+2 gives p (d 2)l s d+1+j = t j+2 + p (d 3)l a d+j g 2(a 1 ) + + a j+3 g d 1(a 1 )

8 DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Since the numbers a j+3,, a d+j are all divisible by p (d 2)l+θ, we can solve for s d+1+j in Z, and get that it is divisible by p θ If j < d 2, then d + 1 + j < 2d 1 and we can use the equations in (311) to solve for s d+1+j More precisely, we consider the equation 0 = p l s d+j+1 a d+j g 2(a 1 ) + a d+j 1 b 2 + + a d b j+1 + a 1 s d+j Again, the numbers a d,, a d+j are all divisible by p (d 2)l+θ, in particular they are divisible by p l+θ On the other hand, since p (d 2)l+2θ divides s d, an iteration of the equations in (311) gives that each s d+j is divisible by p (d 2 j)l+θ for every 1 j d 2, and therefore p l+θ s d+j for j < d 2 That means, also in this case, we can solve for s d+1+j in Z and obtain p θ s d+1+j Altogether, we have found a d+j, t d+j, s d+1+j Z satisfying the equations under consideration, and such that p (d 2)l+2θ t d+j, p (d 2)l+θ a d+j, p θ s d+1+j This completes the induction and proves the assertion of the theorem In general, having a root in Z p is certainly not necessary for a polynomial to factor in Z[[x]] For instance, f(x) = 49 + 98x + 63x 2 + 14x 3 + x 4 is reducible in Z[[x]], namely f(x) = (7 + 7x + x 2 ) 2, but it has no roots in Z 7 However, as we discuss below, if the polynomial f(x) = p n + p m γ 1 x + γ 2 x 2 + + γ d x d, γ d 0, is as in Theorem 39, with the additional condition that gcd(p, γ 2, γ 3 ) = 1, then its reducibility in Z[[x]] gives the existence of a root in pz p Note that if n 2m and f(x) is reducible in Z[[x]], then there exist l, a k, b k Z, with 1 l n/2, such that f(x) = ( p l + a 1 x + a 2 x 2 + )( p n l + b 1 x + b 2 x 2 + ) (318) Proposition 319 Let f(x) be a polynomial of the form (32) with n 2m, and such that gcd(p, γ 2, γ 3 ) = 1 If f(x) factors as above, then f(x) has a root ϱ Z p with v p (ϱ) = l In order to prove this proposition, we need the following lemma Lemma 320 If f(x) as above is reducible in Z[[x]], then the factorization (318) can be arranged such that gcd(p, a 1 ) = 1 and a 2 = = a k = 0 for any k 2 Proof From equation (318) we deduce the following equalities: p m γ 1 = p l (b 1 + p n 2l a 1 ), γ 2 = p l (b 2 + p n 2l a 2 ) + a 1 b 1, γ 3 = p l (b 3 + p n 2l a 3 ) + a 1 b 2 + a 2 b 1, which imply that either gcd(p, a 1 ) = 1 or gcd(p, b 1 ) = 1 (since gcd(p, γ 2, γ 3 ) = 1) If 2l < n, then n 2m implies l < m, and so p b 1 Hence gcd(p, a 1 ) = 1 If 2l = n, then the two factors in (318) are exchangeable and we can assume that a 1 is the one relatively prime to p We will proceed by induction in k 2, assuming that gcd(p, a 1 ) = 1

FACTORING POLYNOMIALS IN Z[[x]] 9 Base step (k = 2): Since gcd(p, a 1 ) = 1, there exist u 1, u 2 Z such that u 1 a 1 + u 2 p l = a 2 If we write f(x) = A(x)B(x) with A(x), B(x) Z[[x]] as in (318), then f(x) = ( (1 + u 1 x + u 2 x 2 )A(x) )( (1 + u 1 x + u 2 x 2 ) 1 B(x) ) Since (1 + u 1 x + u 2 x 2 ) 1 B(x) Z[[x]] and (1 + u 1 x + u 2 x 2 )A(x) = (1 + u 1 x + u 2 x 2 ) ( p l + a 1 x + a 2 x 2 + ) = p l + (a 1 + u 1 p l )x + (a 2 + u 1 a 1 + u 2 p l )x 2 + = p l + c 1 x + c 3 x 3 +, we now have a factorization of f(x) with the desired properties Inductive step: Suppose that (318) can be written as f(x) = A k (x)b(x) with A k (x) = p l + a 1 x + a k+1 x k+1 + a k+2 x k+2 +, B(x) = p n l + b 1 x + b 2 x 2 + Since gcd(p, a k 1) = 1, there exist u 1, u k+1 Z such that ( 1) k 1 u 1 a k 1 + u k+1 p kl = a k+1 p (k 1)l (321) In particular, p (k 1)l u 1 If we let u 2 = p l u 1 a 1, then u 1 a 1 + u 2 p l = 0 and p (k 2)l u 2 For j = 3,, k, we recursively define u j = p l u j 1 a 1 and obtain u j 1 a 1 + u j p l = 0 with p (k j)l u j Finally, we let U(x) = 1 + u 1 x + u 2 x 2 + + u k+1 x k+1 and write f(x) = ( U(x)A k (x) )( U(x) 1 B(x) ) Observe that U(x) 1 B(x) Z[[x]] Moreover, U(x)A k (x) = (1 + u 1 x + + u k+1 x k+1 ) ( p l + a 1 x + a k+1 x k+1 + ) = p l + (a 1 + u 1 p l )x + (u 1 a 1 + u 2 p l )x 2 + (u 2 a 1 + u 3 p l )x 3 + + (u k 1 a 1 + u k p l )x k + (a k+1 + u k a 1 + u k+1 p l )x k+1 + By construction, the coefficients of x 2,, x k are all zero, and u k a 1 = ( 1) k 1 p (k 1)l u 1 a k 1 Thus, by (321), the coefficient of x k+1 is also zero and we get U(x)A k (x) = p l + c 1 x + c k+2 x k+2 + as desired Proof of Proposition 319 Let k N with k d By Lemma 320 we can assume that f(x) factors as in (318) with gcd(p, a 1 ) = 1 and a 2 = = a d = = a k = 0

10 DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Note that the choice of a 1 depends on k Using the notation s j = b j + p n 2l a j, we then get the equations p m γ 1 = p l s 1, γ 2 = p l s 2 + a 1 (p m l γ 1 p n 2l a 1 ), γ j = p l s j + a 1 s j 1 for 2 j d Solving for s d in terms of a 1 and the coefficients of f(x), these equations give the identity g d (a 1 ) = p (d 1)l s d, where g d (x) is the polynomial g d (x) = ( 1)d x d p 2l f( p l /x) (cf proof of Lemma 35) We also consider the next block of equations 0 = p l s d+1 + a 1 s d 0 = p l s k+1 + a 1 s k Using that gcd(p, a 1 ) = 1, the last equation then gives p l s k This implies p 2l s k 1 = p 3l s k 2 = = p (k d+1)l s d, and therefore g d (a 1 ) 0 (mod p kl ) In conclusion, for every k d we have an integer a 1,k such that g d (a 1,k ) 0 (mod p kl ) By Lemma 28, there exists r Z p such that g d (r) = 0 Moreover, v p (a 1,k ) = 0 implies v p (r) = 0 In particular, 1/r Z p Finally, because of the relation between f(x) and g d (x), it follows that ϱ = p l /r Z p is a root of f(x) with v p (ϱ) = l 4 Examples and further remarks In this section, we intend to illustrate our results for the special cases of quadratic and cubic polynomials We also discuss some particular aspects of the factorization of integer polynomials over Z[[x]] and Z p [x] Once again, the only difficulty is to understand the factorization properties of polynomials of the form (32) Every other case can be treated with Proposition 21 Thus we consider f(x) = p n + p m γ 1 x + γ 2 x 2 + γ 3 x 3, γ 2 + γ 3 0, (41) with p prime, n 2, m 1, gcd(p, γ 1 ) = 1 or γ 1 = 0, and gcd(p, γ 2, γ 3 ) = 1 If n > 2m, everything is said in Proposition 25 If n 2m, we combine Proposition 31, Theorem 39, and Proposition 319 to formulate the following result Theorem 42 Let f(x) be a polynomial of the form (41) with n 2m or γ 1 = 0 Then f(x) admits a factorization f(x) = ( p l + a 1 x + a 2 x 2 + )( p n l + b 1 x + b 2 x 2 + ) in Z[[x]] with 1 l n/2 if and only if it has a root ϱ Z p with v p (ϱ) = l 1

FACTORING POLYNOMIALS IN Z[[x]] 11 As the polynomials (7+7x+x 2 ) 2 and (7+7x+x 2 )(7+7x+x 3 ) show, Theorem 42 is not necessarily valid for polynomials of degree higher than 3 Observe that neither 7 + 7x + x 2 nor 7 + 7x + x 3 have roots in Z 7 On the other hand, under the assumption that f(x) has a root in Z p, it is indeed necessary to assume that its p-adic valuation is positive For instance, the polynomial 7 + 21x + 15x 2 + x 3 = (1 + x)(7 + 14x + x 2 ) has 1 as root in Z 7, but it does not admit a proper factorization in Z[[x]] since 1 + x is a unit and 7 + 14x + x 2 is irreducible in Z[[x]] Note that v 7 ( 1) = 0 As discussed in Section 2, the conditions n 2 and m 1 are necessary for Theorem 42 to hold In fact, any power series f with f(0) = 1 is a unit in Z[[x]], and any power series f for which f(0) is a prime number is irreducible in Z[[x]] Moreover, if gcd(p, γ 1 ) = 1, then any power series of the form p n + γ 1 x + is irreducible in Z[[x]] However, the polynomial p n + γ 1 x + γ 2 x 2 is reducible in Z p [x] for every γ 1, γ 2 Z with gcd(p, γ 1 ) = 1 Thus, in general, the reducibility of polynomials in Z[[x]] does not follow from their reducibility in Z[x] or Z p [x] However, in certain cases, the reducibility in Z[[x]] of polynomials of the form p n + d k=1 α kx k is indeed equivalent to their reducibility in Z p [x] Such cases are discussed by J-P Bézivin [1] and J Elliott [4] We finish this section with a brief discussion about the factorization of quadratic and cubic polynomials in the presence of a multiple p-adic root Multiple roots As mentioned at the beginning of Section 3, if f(x) has a multiple root in Z p, then a proper factorization of f(x) in Z[[x]] can be achieved with polynomial factors We now proceed to illustrate what this factorization looks like for the cases at hand Recall that if f(x) has a multiple root then its discriminant must be zero First, if f(x) in (41) is quadratic and has a double root, then p 2m γ1 2 4p n γ 2 = 0 Hence n = 2m, γ 1 is an even number, and f(x) = p n + p m γ 1 x + γ 2 x 2 = (p m + γ1 2 x)2 Consider now f(x) = p n + p m γ 1 x + γ 2 x 2 + γ 3 x 3 as in (41) with γ 3 0, n 2m, and suppose that f(x) has a double root, say ϱ Z p Thus p n + p m γ 1 ϱ + γ 2 ϱ 2 + γ 3 ϱ 3 = 0, p m γ 1 + 2γ 2 ϱ + 3γ 3 ϱ 2 = 0, and so p must divide ϱ 2 (γ 2 + γ 3 ϱ) and ϱ(2γ 2 + 3γ 3 ϱ) Since gcd(p, γ 2, γ 3 ) = 1, we then conclude that p divides ϱ Let l = v p (ϱ) 1 Note that n 2m implies l m and 2l n Moreover, using that 9γ 3 f(ϱ) (3γ 3 ϱ + γ 2 )f (ϱ) = 0, we obtain ϱ = pm γ 1 γ 2 9p n γ 3 2(3p m γ 1 γ 3 γ 2 2 ) Q Write ϱ = b/a with a, b Z such that gcd(a, b) = 1 Then a γ 3 and b p n, hence b = p l and ϱ = p l /a We conclude that f(x) admits the factorization f(x) = (p l + ax) ( p n l + (p m l γ 1 p n 2l a)x + γ3 a x2), which is a proper factorization of f(x) in Z[[x]] References [1] Bézivin, Jean-Paul, Unpublished communication, 2010 [2] Birmajer, Daniel, and Gil, Juan B, Arithmetic in the ring of formal power series with integer coefficients, Amer Math Monthly 115 (2008), no 6, 541 549

12 DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER [3] Birmajer, Daniel; Gil, Juan B, and Weiner, Michael D, Factorization of quadratic polynomials in the ring of formal power series over Z, J Algebra Appl 6 (2007), no 6, 1027 1037 [4] Elliott, Jesse, Factoring formal power series over principal ideal domains, preprint, arxiv:11074860v3 [mathac], 2011 [5] Katok, Svetlana, p-adic analysis compared with real, Mathematics Advanced Study Semesters (Student Mathematical Library, 37) American Mathematical Society, Providence, RI, 2007 [6] Serre, Jean-Pierre, A course in arithmetic, Graduate Texts in Mathematics, No 7, Springer Verlag, New York-Heidelberg, 1973 Department of Mathematics, Nazareth College, 4245 East Ave, Rochester, NY 14618 E-mail address: abirmaj6@nazedu Penn State Altoona, 3000 Ivyside Park, Altoona, PA 16601 E-mail address: jgil@psuedu Penn State Altoona, 3000 Ivyside Park, Altoona, PA 16601 E-mail address: mdw8@psuedu