Factorization of quadratic polynomials in the ring of formal power series over $\Z$
aa r X i v : . [ m a t h . A C ] J un FACTORIZATION OF QUADRATIC POLYNOMIALS IN THERING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B. GIL, AND MICHAEL D. WEINER
Abstract.
We establish necessary and sufficient conditions for a quadraticpolynomial to be irreducible in the ring Z [[ x ]] of formal power series withinteger coefficients. For n, m ≥ p prime, we show that p n + p m βx + αx isreducible in Z [[ x ]] if and only if it is reducible in Z p [ x ], the ring of polynomialsover the p -adic integers. Introduction If K is a field, the question of whether or not a quadratic polynomial is reduciblein the polynomial ring K [ x ] is well understood: A polynomial f ( x ) = c + bx + ax ,with a = 0, can be written as a product of two linear factors in K [ x ] if and onlyif its discriminant b − ac is a square in K . More generally, if D is a uniquefactorization domain, a primitive quadratic polynomial in D [ x ] is reducible if andonly its discriminant is a square in D .If we consider the polynomials in Z [ x ] as elements of Z [[ x ]], the ring of formalpower series over Z , the factorization theory has a different flavor. A power seriesover an integral domain D is a unit in D [[ x ]] if and only if its constant term is a unitin D , so irreducible elements in Z [ x ], such as 1 + x , are invertible as power series.On the other hand, any power series whose constant term is not a unit or a primepower, is reducible in Z [[ x ]], hence we can produce many examples of polynomialsthat are reducible as power series, yet irreducible in Z [ x ].Similarly, when considering polynomials with integer coefficients as elements of Z [[ x ]] and as polynomials over Z p , the ring of p -adic integers, we also observedifferent behaviors in their arithmetic properties. For instance, the polynomial p + x + x , which is irreducible as a power series, is reducible in Z p [ x ] for anyprime p . On the other hand, 6 + 2 x + x is reducible in Z [[ x ]] and in Z [ x ], but itis irreducible in Z [ x ] and Z [ x ].In this paper, we discuss the factorization theory of quadratic polynomials inthe ring of formal power series over Z . Based on the above examples, it is naturalto ask whether the question of reducibility of polynomials in Z [[ x ]] can be reduced,at least in some cases, to the reducibility in D [ x ] for some integral domain D . Theimplication of such a reduction in the quadratic case is clear: a reducibility criterionthat relies on the discriminant being a square in D . In this direction, we found thefollowing connection between Z [[ x ]] and Z p [ x ]: Main Theorem.
Let p be prime. Let α, β ∈ Z be such that gcd( p, α ) = 1 and gcd( p, β ) = 1 . Let f ( x ) = p n + p m βx + αx with n, m ≥ . Then f ( x ) is reduciblein Z [[ x ]] if and only if it is reducible as a polynomial over Z p . Mathematics Subject Classification.
We present the proof of this theorem in Sections 3 and 4. Since the conditions forbeing a square in Z p are well known, our approach provides an effective procedurefor deciding whether or not a quadratic polynomial is reducible as a power seriesand, in the affirmative case, our proofs give an algorithmic method (whose founda-tions are based on the Euclidean Algorithm) for finding a proper factorization.In addition to our main theorem, we give a complete picture of the factorizationtheory for quadratic polynomials in Z [[ x ]]. Some basic cases are treated in Section 2where we discuss the necessary background and develop some preliminary results.In Section 5 we finish with some simple reducibility criteria that apply to powerseries whose quadratic part is of the form discussed in the other sections.A standard reference for an introduction to divisibility over integral domainsis [3]. For an extensive treatment of the arithmetic on the ring of formal powerseries over an integral domain the reader is referred to [4] and [5]. All the necessarymaterial about the ring Z p of p -adic numbers, can be found for instance in [2, 4, 6].2. Factorization in the ring of power series
In order to place our main result in the appropriate context, and for the reader’sconvenience, we review some elementary facts about the factorization theory in Z [[ x ]]. First, recall that Z [[ x ]] is a unique factorization domain. Moreover, if f ( x )is a formal power series in Z [[ x ]] and f ∈ Z is its constant term, then:i. f ( x ) is invertible if and only if f = ± f is prime then f ( x ) is irreducible.iii. If f is not a unit or a prime power then f ( x ) is reducible.iv. If f ( x ) = f is a constant then it is irreducible if and only if f is prime.v. If f ( x ) = p m + f x , with p prime and m ≥
1, then f ( x ) is irreducible if andonly if gcd( p, f ) = 1.For an accessible and more detailed treatment of the divisibility theory in Z [[ x ]]the reader is referred to [1].At this point, we have definitive criteria for deciding irreducibility in Z [[ x ]] forconstant and linear polynomials. The next natural step is to examine quadraticpolynomials, say f ( x ) = f + f x + f x . Unless f is a prime power, we knowthat f ( x ) is either a unit or it is reducible in Z [[ x ]]. On the other hand, if f = p n , n > p prime, and if f ( x ) = a ( x ) b ( x ) is a proper factorization, then we must have a = p s , b = p t with s, t ≥ s + t = n . This implies f = p s b + p t a , so weconclude that f ( x ) is irreducible unless p | f . Finally, if p divides all coefficients,then f ( x ) is either reducible or associate to p . Thus the interesting case is when f ( x ) is primitive.Therefore, in the next sections we will focus on polynomials of the form f ( x ) = p n + p m βx + αx , with n, m ≥
1, gcd( p, α ) = 1, and gcd( p, β ) = 1 or β = 0.As stated in the introduction, we will analyze the factorization of such polyno-mials by considering them as elements in Z p [ x ] and Z [[ x ]], and the main tool forestablishing this link will be the discriminant. Our strategy will be to produceexplicit factorizations in Z [[ x ]], when appropriate. To this end, it will be helpfulto assume that one of the factors in f ( x ) = a ( x ) b ( x ) has a certain simplified form.The basis for this assumption is the following lemma. ACTORIZATION OF QUADRATIC POLYNOMIALS IN Z [[ x ]] 3 Lemma 2.1.
Let a ( x ) ∈ Z [[ x ]] such that a = p and gcd( p, a ) = 1 . For every t ≥ there exists an associate q ( x ) to a ( x ) such that q = a , q ≡ a (mod p ) ,and q = q = · · · = q t = 0 . More precisely, we will show that there exists a polynomial u ( x ) = 1 + u x + u x + · · · + u t x t , invertible in Z [[ x ]], such that u ( x ) a ( x ) = p + λx + q t +1 x t +1 + q t +2 x t +2 + · · · , with λ ≡ a (mod p ). In order to find u ( x ), we set up the t × t system of equations: λ = a + pu , a + a u + pu , a + a u + a u + pu , ...0 = a t + t − X j =1 a j u t − j + pu t , (2.2)in the unknowns u , . . . , u t . Since the determinant of the matrix associated withthis system of equations is p t , it is clear that (2.2) admits a unique solution overthe rationals for any integer λ . Our goal is to prove that for any t ≥
2, there exista suitable λ ∈ Z such that (2.2) admits a solution over the integers. This followsfrom the following two propositions. Proposition 2.3.
If for some λ ∈ Z , the solution ( u , . . . , u t ) of the system (2.2) is such that u i ∈ Z for all ≤ i ≤ t , then for every k ∈ Z the system λ + kp t = a + pu , a + a u + pu , a + a u + a u + pu , ... a t + t − X j =1 a j u t − j + pu t , (2.3.a) also has a ( unique ) solution over the integers. Moreover, the solution ( u k , . . . , u kt ) of (2.3.a) and the solution ( u , . . . , u t ) of (2.2) are related as follows: u ki ≡ u i (mod p ) for ≤ i ≤ t − ,u kt ≡ u t + ( − t +1 ka t − (mod p ) . Proposition 2.4.
If the t × t system of equations λ = a + pu , a + a u + pu , a + a u + a u + pu , ... a t + t − X j =1 a j u t − j + pu t , DANIEL BIRMAJER, JUAN B. GIL, AND MICHAEL D. WEINER has a solution ( u , . . . , u t ) over the integers, then there exists k ∈ Z such that the ( t + 1) × ( t + 1) system λ + kp t = a + pu , a + a u + pu , a + a u + a u + pu , ... a t +1 + t X j =1 a j u t +1 − j + pu t +1 , also has its solution ( u k , . . . , u kt +1 ) over the integers. Proof of Lemma 2.1.
Choose λ ∈ Z such that λ ≡ a (mod p ). Then the equa-tion λ = a + pu can be solved for u ∈ Z . Then proceed to the desired value of t by applying repeatedly Proposition 2.3 and Proposition 2.4. (cid:3) Proof of Proposition 2.3.
Let A be a ( t × t )-matrix and let B k be a t -dimensionalcolumn vector defined as A = p · · · a p · · · a a p · · · a t − a t − a t − · · · a p and B k = kp t . If A , A , . . . , A t are the columns of A then, by Cramer’s rule, the unique solution(over the rationals) of the system (2.3.a) is given by u ki = u i + 1 p t det (cid:0) A , . . . , A i − , B k , A i +1 , . . . , A t (cid:1) = u i + ( − i +1 k det A i , where A i denotes the ( t − × ( t −
1) matrix obtained from A by deleting its firstrow and i th column. Thus u ki ∈ Z for every 1 ≤ i ≤ t .The entries q irr , r = 1 , . . . , t −
1, in the principal diagonal of A i , are as given by q irr = ( a if 1 ≤ r ≤ i − ,p if i ≤ r ≤ t − , and the entries in the super-diagonal of A i are q ir,r +1 = ( p if 3 ≤ i ≤ t, ≤ r ≤ i − , . Of course, all the entries above the super-diagonal in A i are 0. Thus, when ex-panding det A i as a sum over all permutations in the symmetric group of t − p t − i a i − , 1 ≤ i ≤ t , and theterm corresponding to any other permutation is a multiple of p . Then, u ki = u i + ( − i +1 k det A i ≡ u i (mod p ) for 1 ≤ i ≤ t − ,u kt = u t + ( − t +1 k det A t ≡ u t + ( − t +1 ka t − (mod p ) . (cid:3) ACTORIZATION OF QUADRATIC POLYNOMIALS IN Z [[ x ]] 5 Proof of Proposition 2.4.
Since gcd( p, a ) = 1, we can choose k such that a t +1 + a u t + t X j =2 a j u t +1 − j + ( − t +1 ka t ≡ p ) . By Proposition 2.3, the system λ + kp t = a + pu a + a u + pu , a + a u + a u + pu , ...0 = a t + t − X j =1 a j u t − j + pu t , (2.5)has its solution ( u k , . . . , u kt ) over the integers. Moreover u ki = u i (mod p ) , ≤ i ≤ t − ,u kt = u t + ( − t +1 ka t − (mod p ) . Therefore, a t +1 + t X j =1 a j u kt +1 − j = a t +1 + a u kt + t X j =2 a j u kt +1 − j = a t +1 + a (cid:0) u t + ( − t +1 ka t − (cid:1) + t X j =2 a j u t +1 − j = a t +1 + a u t + t X j =2 a j u t +1 − j + ( − t +1 ka t ≡ p ) . Hence, we can solve the equation 0 = a t +1 + P t − j =1 a j u t +1 − j + pu t +1 for u t +1 . (cid:3) The case when the constant term is an odd prime power
Let p be an odd prime, let α, β ∈ Z be such that gcd( p, α ) = 1 and gcd( p, β ) = 1. Proposition 3.1.
Let f ( x ) = p n + p m βx + αx with n, m ≥ . ( i ) If m < n , then f ( x ) is reducible in both Z p [ x ] and Z [[ x ]] . ( ii ) If m > n and n is odd, then f ( x ) is irreducible in both Z p [ x ] and Z [[ x ]] .Proof. ( i ) Observe first that the discriminant of f ( x ) is p m β − αp n = p m ( β − αp n − m ) , a nonzero square in Z p , and so f ( x ) is reducible in Z p [ x ]. To show that f ( x ) isreducible as a power series, we will find sequences { a k } and { b k } such that f ( x ) = ( p m + a x + a x + · · · )( p n − m + b x + b x + · · · ) . DANIEL BIRMAJER, JUAN B. GIL, AND MICHAEL D. WEINER
For k ≥ t k = b k + p n − m a k . For the above factorization to hold, we need p m β = p n − m a + p m b , so we have t = β. Since β − αp n − m is a square in Z p , the polynomial g ( x ) = p n − m x − βx + α isreducible in Z p [ x ]. In particular, g ( x ) has a root in Z /p m Z , hence there are integers a and t such that p n − m a − βa + α = p m t . Suppose that we have defined a k , t k +1 for k = 1 , . . . , N − N ≥
2, and let v N = a t N + N − X k =2 a k ( t N +1 − k − p n − m a N +1 − k ) . We want to define a N and t N +1 in such a way that P N +1 k =0 a k b N +1 − k = 0 for N ≥ N +1 X k =0 a k b N +1 − k = a b N +1 + a N +1 b + a N b + a b N + N − X k =2 a k b N +1 − k = p m t N +1 + ( β − p n − m a ) a N + a ( t N − p n − m a N ) + N − X k =2 a k b N +1 − k = p m t N +1 + ( β − p n − m a ) a N + v N . At last, since gcd( p, β ) = 1, this equation can be solved for t N +1 , a N ∈ Z . Thisshows that f ( x ) is reducible in Z [[ x ]].( ii ) In this case, the discriminant of f ( x ), p m β − αp n = p n ( p µ − n β − α ) , is not a square in Z p . Thus f ( x ) is irreducible as a polynomial over Z p . To showthat f ( x ) is irreducible as a power series, assume f ( x ) = ( p s + a x + a x + · · · )( p t + b x + b x + · · · )with t > s ≥ s + t = n . Note that t = s because n is odd. Then we must have p m β = p t a + p s b ,α = p t a + a b + p s b . Since p and α are coprime, it follows that gcd( p, a ) = 1 = gcd( p, b ). Therefore, itmust be s = m , and so 2 m = 2 s < s + t = n . (cid:3) It remains to analyze the cases when n is even, say n = 2 ν , and m ≥ ν ≥ Proposition 3.2.
Let m ≥ ν . The polynomial f ( x ) = p ν + p m βx + αx is reduciblein Z [[ x ]] if and only if b f ( x ) = p + p m − ν +1 βx + αx is reducible in Z p [ x ] . This follows from the following three lemmas.
Lemma 3.3. If f ( x ) is reducible in Z [[ x ]] , then b f ( x ) is reducible in Z [[ x ]] . Lemma 3.4.
Let ℓ ≥ . If the polynomial p + p ℓ βx + αx is reducible in Z [[ x ]] ,then it is reducible in Z p [ x ] . ACTORIZATION OF QUADRATIC POLYNOMIALS IN Z [[ x ]] 7 Lemma 3.5. If b f ( x ) is reducible in Z p [ x ] , then f ( x ) is reducible in Z [[ x ]] . Proof of Lemma 3.3.
We first observe that if f ( x ) = a ( x ) b ( x ) is a proper fac-torization in Z [[ x ]], then a = b = p ν . To see this, assume that a = p s , b = p t with s, t ≥ s + t = 2 ν . Then we have that α = p s b + a b + p t a . Since gcd( p, α ) = 1, we conclude that gcd( p, a ) = gcd( p, b ) = 1. We also have p m β = p s b + p t a . If s < t , then we would have s < ν ≤ m , implying from the above equation that p | b , a contradiction. Similarly, we can rule out the case t < s , hence s = t = ν .We now write f ( x ) = a ( x ) b ( x ) with a = b = p ν . Since p ν − b f ( x ) = f ( p ν − x ) = a ( p ν − x ) b ( p ν − x ) , it follows that b f ( x ) = (cid:16) a ( p ν − x ) p ν − (cid:17)(cid:16) b ( p ν − x ) p ν − (cid:17) is a proper factorization of b f ( x ) in Z [[ x ]]. (cid:3) Proof of Lemma 3.4.
To prove that g ( x ) = p + p ℓ βx + αx is reducible in Z p [ x ],we must show that its discriminant p ℓ β − αp is a square in Z p .Write p ℓ − β − α = p t u with gcd( p, u ) = 1. Suppose that g ( x ) is reducible in Z [[ x ]]. By Lemma 2.1 we can assume that g ( x ) admits a factorization of the form p + p ℓ βx + αx = a ( x ) b ( x ) with a = b = p, a = a = · · · = a t +2 = 0 . With the notation s j = a j + b j for j ≥
1, we must have p ℓ β = ps ,α = ps + a s − a . Then s = p ℓ − β and a is a root of y − s y + α ≡ p ). Note that p a .For n = 3 we have 0 = ps + a s . (3.6)Then p | s and a − a s + α ≡ p ). For n = 4 we have0 = ps + a s . Then p | s , which by (3.6) implies that p | s , and so a − a s + α ≡ p ).Working inductively, the equation0 = ps t +3 + a s t +2 implies that p t +1 | s , and so a − a s + α = p t +2 v for some v .Now, since(2 a − s ) = ( p ℓ − β − α ) + 4 p t +2 v = p t u + 4 p t +2 v = p t ( u + 4 p v ) , and since gcd( p, u ) = 1, we have that t is even and that u is a square mod p . Hence p ℓ − β − α is a square in Z p , and so is p ( p ℓ − β − α ). Therefore, g ( x ) isreducible in Z p [ x ]. (cid:3) DANIEL BIRMAJER, JUAN B. GIL, AND MICHAEL D. WEINER
Proof of Lemma 3.5.
We will consider the cases m = ν and m > ν separately.In both cases we will prove the reducibility of f ( x ) in Z [[ x ]] by providing an ex-plicit factorization algorithm. More precisely, we will give inductive algorithms(depending on m and ν ) to find sequences { a k } and { b k } in Z such that f ( x ) = (cid:16) ∞ X k =0 a k x k (cid:17)(cid:16) ∞ X k =0 b k x k (cid:17) . (3.7)For k ≥ s k = a k + b k . Case 1:
Let m > ν . Since b f ( x ) = p + p m − ν +1 βx + αx is reducible in Z p [ x ], thepolynomial b g ( x ) = x − p m − ν βx + α is reducible in Z p [ x ], too. Observe that thediscriminant of b f ( x ) is p times the discriminant of b g ( x ).Let a = p ν = b and s = p m − ν β. Since b g ( x ) is reducible in Z p [ x ], it has a root in Z /p ν Z . Let a , s ∈ Z be such that a − p m − ν βa + α = p ν s . Now, m > ν and gcd( p, α ) = 1 imply gcd( p, a ) = 1 and gcd( p ν , p m − ν β − a ) = 1.We let a and s be integer numbers such that0 = p ν s + ( p m − ν β − a ) a + a s . Suppose we have defined a k and s k +1 for k = 1 , . . . , N − N ≥
3, and let v N = a s N + N − X k =2 a k ( s N +1 − k − a N +1 − k ) . We know that gcd( p ν , p m − ν β − a ) = 1, so the equation0 = p ν s N +1 + ( p m − ν β − a ) a N + v N can be solved for a N , s N +1 ∈ Z . For k = 1 , . . . , N we now have a k and b k , and itcan be easily checked that the sequences { a k } and { b k } give (3.7). Case 2: If m = ν , then b f ( x ) = p + pβx + αx and f ( x ) = p ν + p ν βx + αx . Since b f ( x ) is reducible in Z p [ x ], so is b g ( x ) = x − βx + α . Thus there are numbers ℓ ∈ N and q ∈ Z such that β − α = p ℓ q with gcd( p, q ) = 1 . Moreover, b g ( x ) has a root in Z /p n Z for every n ∈ N . In particular, for n =3 max( ℓ, ν ), there are integers a and r such that a − βa + α = p µ r with gcd( p, r ) = 1 , (3.8)for some µ ≥ ℓ, ν ). Since ( β − a ) − ( β − α ) = 4 b g ( a ), we get( β − a ) = 4 p µ r + p ℓ q = p ℓ (4 p µ − ℓ r + q ) , hence we can write β − a = p ℓ t with gcd( p, t ) = 1 . (3.9) ACTORIZATION OF QUADRATIC POLYNOMIALS IN Z [[ x ]] 9 Again, our goal is to construct sequences { a k } and { b k } such that (3.7) holds.This will be done with slightly different algorithms for ν > ℓ and ν ≤ ℓ . In bothcases we let a = p ν = b , s = β,a = a, s = p µ − ν r, where a and r are the integers from (3.8). With these choices, the first three termsin the expansion of (3.7) coincide with f ( x ).Assume ν > ℓ . Let˜ a = 0 , u = s = p µ − ν r, and u = − p µ − ν ra . Let t be as in (3.9). For k ≥ a k and u k +1 such that the sequencesdefined by a k = p ν − ℓ ˜ a k and b k = u k − − t ˜ a k − − a k (3.10)give the factorization (3.7). Note that s k +1 = a k +1 + b k +1 = u k − t ˜ a k .Let ˜ a = p µ − ν and u = − p µ − ν (cid:2) p ν − ℓ ( s − a ) − ta (cid:3) . Thus p ν u + (cid:2) p ν − ℓ ( s − a ) − ta (cid:3) ˜ a = 0 . Suppose we have defined ˜ a k and u k +1 for k = 1 , . . . , N − N ≥
3, and let v N = a u N + a s N + N − X k =3 a k ( s N +2 − k − a N +2 − k ) . Since gcd( p, β ) = 1, the relation (3.9) impliesgcd( p, a ) = 1 and gcd( p ν − ℓ , p ν − ℓ ( s − a ) − ta ) = 1 . Therefore, there are ˜ a N , u N +1 ∈ Z such that p ν u N +1 + (cid:2) p ν − ℓ ( s − a ) − ta (cid:3) ˜ a N + ˜ v N = 0 . The sequences { a k } and { b k } defined by (3.10) give (3.7) when ν > ℓ .Assume now ν ≤ ℓ . In this case, for k ≥ a k and ˜ s k +1 such thatthe sequences defined by a k = p ℓ ˜ a k and b k = p ℓ − ν ˜ s k − a k give a factorization of f ( x ). Let r and t be as in (3.8) and (3.9), respectively. Sincegcd( p ν , t ) = 1, there are y, z ∈ Z such that p ν y + tz + r = 0 . Let ˜ a = p µ − ℓ − ν za , ˜ s = p µ − ℓ r , and ˜ s = p µ − ℓ ya . Note that p ℓ ˜ s + t ˜ a + a p ℓ − ν ˜ s = 0 . Suppose we have defined ˜ a k and ˜ s k +1 for k = 1 , . . . , N − N ≥
3, and let˜ v N = a p ℓ − ν ˜ s N + N − X k =2 ˜ a k ( p ℓ − ν ˜ s N +1 − k − ˜ a N +1 − k ) . Finally, since gcd( p ℓ , t ) = 1, the equation0 = p ℓ ˜ s N +1 + t ˜ a N + ˜ v N can be solved for ˜ a N , ˜ s N +1 ∈ Z . This implies0 = p ℓ ˜ s N +1 + p ℓ t ˜ a N + p ℓ ˜ v N = p ν s N +1 + p ℓ ta N + a s N + N − X k =2 a k ( s N +1 − k − a N +1 − k )= p ν s N +1 + ( β − a ) a N + a s N + N − X k =2 a k b N +1 − k = N +1 X k =0 a k b N +1 − k , as desired. This completes the proof. (cid:3) The main result of this section is the following.
Theorem 3.11.
Let p be an odd prime and let n, m ≥ . Let α, β ∈ Z be suchthat gcd( p, α ) = 1 and gcd( p, β ) = 1 . The polynomial f ( x ) = p n + p m βx + αx isreducible in Z [[ x ]] if and only if it is reducible in Z p [ x ] .Proof. Using the fact that f ( x ) is reducible in Z p [ x ] iff b f ( x ) is reducible in Z p [ x ],the statement of the theorem follows from Proposition 3.1 and Proposition 3.2. (cid:3) Remark 3.12.
The previous theorem is not valid when m = 0 . In fact, if p β ,any power series of the form p n + βx + · · · is irreducible in Z [[ x ]] . However, anypolynomial p n + βx + αx with gcd( p, β ) = 1 is reducible in Z p [ x ] . We finish this section with the remaining case: β = 0. Proposition 3.13.
Let p be an odd prime and let n ≥ . Let α ∈ Z be such that gcd( p, α ) = 1 . The polynomial f ( x ) = p n + αx is reducible in Z [[ x ]] if and only ifit is reducible in Z p [ x ] .Proof. Recall that f ( x ) = p n + αx is reducible in Z p [ x ] if and only if its discriminant − αp n is a nonzero square in Z p . This in turn is the case if and only if n is evenand − α is a square in Z /p Z . We will show that these conditions on n and α areequivalent to f ( x ) being reducible in Z [[ x ]].For f ( x ) to admit a factorization of the form p n + αx = ( a + a x + a x + · · · )( b + b x + b x + · · · )it is necessary to solve the equations a = p t and b = p s with t + s = n, p t b + p s a ,α = p t b + a b + p s a . Since gcd( p, α ) = 1, these three equations can be solved in Z only when s = t , thatis, when n is even. Now, if n = 2 ν , we must have a = b = p ν , s = a + b = 0,and α = p ν ( a + b ) − a . Thus, if f ( x ) is reducible in Z [[ x ]], then − α is a squarein Z /p Z . On the other hand, if − αp ν is a nonzero square in Z p , so is − α , i.e., y + α has a root in Z p . Let a and s be integers such that a + α = p ν s . Note that gcd( p ν , a ) = 1. Therefore, there are integers a and s such that0 = p ν s − a a + a s . ACTORIZATION OF QUADRATIC POLYNOMIALS IN Z [[ x ]] 11 Finally, a factorization of f ( x ) in Z [[ x ]] can be obtained with the sequences { a k } and { s k +1 } defined inductively for N ≥ p ν s N +1 − a a N + v N , where v N = a s N + P N − k =2 a k ( s N +1 − k − a N +1 − k ). (cid:3) The case when the constant term is a power of 2
In this section we consider polynomials of the form f ( x ) = 2 n + 2 m βx + αx with n, m ≥ , where α, β are assumed to be odd integers. Observe that β ≡ Proposition 4.1.
Let f ( x ) = 2 n + 2 m βx + αx with n, m ≥ . ( i ) If m < n , then f ( x ) is reducible in both Z [ x ] and Z [[ x ]] . ( ii ) If m > n and n is odd, then f ( x ) is irreducible in both Z [ x ] and Z [[ x ]] .Proof. ( i ) If 2 m < n , then 4 α n − m ≡ β − α n − m ≡ m ( β − α n − m ) of f ( x ) is a square in Z and f ( x ) isreducible in Z [ x ].Similarly, the polynomial g ( x ) = 2 n − m x − βx + α is also reducible in Z [ x ].Therefore, the factorization algorithm given in the proof of Proposition 3.1( i ) workshere as well and we can conclude that f ( x ) is reducible in Z [[ x ]].( ii ) In this case, the discriminant of f ( x ) can be written as∆ = 2 n (2 m − n β − α ) . Recall that n is odd. If n = 2 m −
1, then ∆ = 2 n +1 ( β − α ). Since α is odd and β ≡ β − α Z . If n ≤ m −
3, then ∆ = 2 n +2 (2 m − n − β − α ) and we get, once again, that∆ is not a square in Z since 2 m − n − β − α and n + 2 are both odd numbers. Inconclusion, f ( x ) is irreducible in Z [ x ].That f ( x ) is irreducible in Z [[ x ]] follows verbatim from the arguments in theproof of Proposition 3.1( ii ). (cid:3) Proposition 4.2. If n = 2 m , then f ( x ) is irreducible in both Z [ x ] and Z [[ x ]] .Proof. Since β ≡ α is odd, we have β − α m ( β − α ), the discriminant of f ( x ), is not a square in Z which impliesthat f ( x ) is irreducible in Z [ x ].Suppose now that f ( x ) is reducible in Z [[ x ]]. Then there are power series a ( x ), b ( x ) ∈ Z [[ x ]] such that f ( x ) = (2 m + a x + a x + · · · )(2 m + b x + b x + · · · )with β = a + b , α = 2 m ( a + b ) + a ( β − a ) . Since β is odd, the number a ( β − a ) is always even, a contradiction. (cid:3) It remains to analyze the case when 2 m > n and n = 2 ν for some ν ∈ N . Tothis end, we will consider the cases m > ν + 1 and m = ν + 1, separately. Proposition 4.3.
Let m > ν + 1 . The polynomial f ( x ) = 4 ν + 2 m βx + αx isreducible in Z [[ x ]] if and only if it is reducible in Z [ x ] . Proof.
First of all, observe that the discriminant of f ( x ) can be written as∆ = 4 ν +1 (4 m − ( ν +1) β − α ) . For ∆ to be a square in Z , we need 4 m − ( ν +1) β − α ≡ m − ( ν +1) ≥ − α ≡ m − ( ν +1) = 1, the discriminant ∆ is a squarein Z iff 4 − α ≡ • if m − ( ν + 1) ≥ f ( x ) is reducible in Z [ x ] ⇐⇒ α ≡ • if m − ( ν + 1) = 1, f ( x ) is reducible in Z [ x ] ⇐⇒ α ≡ Z [[ x ]].Assume first that f ( x ) is reducible in Z [[ x ]] and can be factored as f ( x ) = (cid:16) ∞ X k =0 a k x k (cid:17)(cid:16) ∞ X k =0 b k x k (cid:17) . (4.3.a)Then we must have a = 2 ν = b , and using the notation s k = a k + b k ,2 m β = 2 ν s , which implies s = 2 m − ν β,α = 2 ν s + (2 m − ν β − a ) a , ν s + (2 m − ν β − a ) a + a s . (4.3.b)The second equation implies that a is odd, and the third one gives that s is even.Thus we get a − m − ν βa + α = 2 ν s = 2 ν +1 t for some t ∈ Z . (4.3.c)Since ν ≥ m − ν ≥
2, this equation implies a + α ≡ α ≡ α ≡ . If ν > α ≡ ≡ ν +1 t = a − m − ν βa + α ≡ − m − ν βa (mod 8) . This implies 2 m − ν βa ≡ m − ν = 2, that is, m − ( ν + 1) = 1.If ν > α ≡ − m − ν βa ≡ m − ( ν + 1) ≥ ν = 1, then f ( x ) = 4 + 2 m βx + αx and (4.3.c) becomes a − m − βa + α = 2 s . By Lemma 2.1 we can choose a = a = 0, so the third equation in (4.3.b), and thenext one, take the form 0 = 2 s + a s , s + a s . Thus s is even, hence s ≡ a − m − βa + α ≡ α ≡ m − α ≡ m − ≥ f ( x ) in Z [[ x ]] under the conditions on m , ν , and α specified above for the reducibility in Z [ x ]. To this end, consider the polynomial g ( x ) = x − m − ν βx + α whose discriminant is 2 (4 m − ( ν +1) β − α ). If m − ( ν +1) = 1and α ≡ m − ( ν +1) β − α = 4 β − α ≡ m − ( ν +1) ≥ α ≡ m − ( ν +1) β − α ≡ ACTORIZATION OF QUADRATIC POLYNOMIALS IN Z [[ x ]] 13 discriminant of g ( x ) is a square in Z . Hence g ( x ) is reducible and thus has a rootin Z . Let ˜ a , t ∈ Z be such that˜ a − m − ν β ˜ a + α = 2 ν +1 t . Let a = 2 ν = b , a = ˜ a , s = 2 m − ν β , and s = 2 ν +1 t . With these choices, f ( x )coincides with the first three terms of the product in (4.3.a). Since m > ν + 1, wehave that 2 m − ν − β − a is odd, hence there are integers ˜ a and t such that0 = 2 ν t + (2 m − ν − β − a )˜ a + a t . If we let a = 2 ν ˜ a and s = 2 ν +1 t , then0 = 2 ν +1 (cid:0) ν t + (2 m − ν − β − a )˜ a + a t (cid:1) = 2 ν (2 ν +1 t ) + (2 m − ν β − a )(2 ν ˜ a ) + a (2 ν +1 t )= 2 ν s + (2 m − ν β − a ) a + a s . Suppose we have defined ˜ a k and t k +1 for k = 1 , . . . , N − N ≥
3, and let w N = a t N + 2 ν − N − X k =2 ˜ a k (2 t N +1 − k − ˜ a N +1 − k ) . As before, there are ˜ a N , t N +1 ∈ Z such that0 = 2 ν t N +1 + (2 m − ν − β − a )˜ a N + w N . If we let a k = 2 ν ˜ a k and s k = 2 ν +1 t k for every k ≥
3, then the sequences { a k } and { b k } with b k = s k − a k give a factorization of f ( x ). (cid:3) Proposition 4.4.
The polynomial f ( x ) = 4 ν + 2 ν +1 βx + αx is reducible in Z [[ x ]] if and only if it is reducible in Z [ x ] .Proof. Assume first that f ( x ) is reducible in Z [ x ]. Then 4 ν +1 ( β − α ) must be asquare in Z , which implies β − α = 2 ℓ q with ℓ ∈ N and q ≡ . Thus the polynomial g ( x ) = x − βx + α is also reducible in Z [ x ], and so it hasa root in Z . Let ˜ a , t ∈ Z be such that˜ a − β ˜ a + α = 2 ℓ + ν +2 t . Let u be the odd integer such that β − ˜ a = 2 ℓ u . As in the proof of Proposition 4.3,a factorization of f ( x ) in Z [[ x ]] can be obtained as follows. We let a = 2 ν = b , a = ˜ a , s = 2 β , s = 2 ℓ +2 t , and for N ≥
2, we define ˜ a N and t N +1 inductivelyby means of the equation 0 = 2 ν t N +1 + u ˜ a N + w N , where w N = a t N + N − X k =2 ˜ a k (2 ℓ +1 t N +1 − k − ˜ a N +1 − k ) . For k ≥
2, we then let a k = 2 ℓ +1 ˜ a k and b k = s k − a k = 2 ℓ +2 t k − a k .Assume now that f ( x ) is reducible in Z [[ x ]] and consider b f ( x ) = 4 + 4 βx + αx .By Lemma 3.3 and Lemma 3.4 with p = 2, m = ν + 1, and ℓ = 2, we get that b f ( x ) is reducible in Z [ x ]. Finally, since the discriminant of f ( x ) is 4 ν − times thediscriminant of b f ( x ), we conclude that f ( x ) is reducible in Z [ x ], too. (cid:3) Theorem 4.5.
Let α, β ∈ Z be odd. The polynomial f ( x ) = 2 n + 2 m βx + αx with n, m ≥ is reducible in Z [[ x ]] if and only if it is reducible in Z [ x ] . Remark 4.6.
The previous theorem is not valid when m = 0 . In fact, n + βx + αx is always reducible in Z [ x ] , but irreducible in Z [[ x ]] . Proposition 4.7.
Let α ∈ Z be odd. The polynomial f ( x ) = 2 n + αx is reduciblein Z [[ x ]] if and only if it is reducible in Z [ x ] .Proof. As in Proposition 3.13, it can be easily checked that if n is odd, then f ( x )is irreducible in both Z [ x ] and Z [[ x ]]. If n = 2 ν , then the statement follows fromthe arguments in the proof of Proposition 4.3 for the case when m > ν + 2. (cid:3) Further reducibility criteria
In this last section we briefly discuss the factorization in Z [[ x ]] of power serieswhose quadratic part is a polynomial like the ones studied in the previous sections.More precisely, we consider power series of the form f ( x ) = p n + p m βx + αx + ∞ X k =3 c k x k , (5.1)where α and β are integers such that gcd( p, α ) = 1 = gcd( p, β ).For simplicity, we only discuss the case when p is an odd prime. We will focuson the situations for which the arguments in Section 3 extend with little or noadditional effort. For instance, if m = n , the reducibility of f ( x ) in Z [[ x ]] followsthe same pattern as the reducibility of its quadratic part. In fact, we can use theexact same arguments from Section 3 to prove the following two propositions. Proposition 5.2. If m < n , then (5.1) is reducible in Z [[ x ]] . If m > n and n isodd, then (5.1) is irreducible. Proposition 5.3. If m > n and n is even, then (5.1) is reducible in Z [[ x ]] if andonly if − α is a quadratic residue mod p . If 2 m = n , the situation is in general more involved and the reducibility of f ( x )depends on the roots of x − βx + α . The following proposition is easy to prove. Proposition 5.4. If m = n and the polynomial x − βx + α has a simple root in Z /p m Z , then (5.1) is reducible in Z [[ x ]] . If x − βx + α has a double root in Z /p m Z , it is not enough to look at the quadraticpart of f ( x ) and its reducibility depends on the coefficients c k . To illustrate thisfact, consider for example the power series f ( x ) = p + pβx + αx + c x + c x + · · · , with α, β ∈ Z such that β − α = p q , where q is a quadratic residue mod p withgcd( p, q ) = 1. In order to get a proper factorization f ( x ) = a ( x ) b ( x ) in Z [[ x ]], wemust have a = p = b , β = s , as well as α = ps + a ( β − a ) ,c = ps + ( β − a ) a + a s , where s k = a k + b k . Then ( β − a ) − ( β − α ) = 4 ps , which implies p | s .Therefore, f ( x ) is irreducible in Z [[ x ]] unless p | c . ACTORIZATION OF QUADRATIC POLYNOMIALS IN Z [[ x ]] 15 On the other hand, if p | c k for every k ≥
3, then with the same assumptions on α and β as above, we can find a k , b k ∈ Z such that a ( x ) = P a k x k and b ( x ) = P b k x k give a proper factorization f ( x ) = a ( x ) b ( x ). Note that β − α is a square in Z p ,so the polynomial g ( x ) = x − βx + α is reducible in Z p [ x ]. In particular, g ( x ) hasa root in Z /p Z , so there are a, r ∈ Z such that a − βa + α = p r. Moreover, since ( β − a ) − ( β − α ) = 4 p r and p | ( β − α ), we have p | ( β − a ).In fact, there is an integer t with gcd( p, t ) = 1 such that β − a = pt. Choose a = a , ˜ s = r , and write c k +1 = p ˜ c k +1 . Since gcd( p, t ) = 1, for k ≥ a k and ˜ s k +1 inductively as integer solutions of the equation˜ c k +1 = p ˜ s k +1 + t ˜ a k + a ˜ s k + k − X j =2 ˜ a j ( p ˜ s k +1 − j − ˜ a k +1 − j ) . If we let a k = p ˜ a k and s k = p ˜ s k , then multiplication by p gives c k +1 = ps k +1 + pta k + a s k + k − X j =2 a j ( s k +1 − j − a k +1 − j )= ps k +1 + ( β − a ) a k + a s k + k − X j =2 a j b k +1 − j = k +1 X j =0 a j b k +1 − j . In other words, a ( x ) and b ( x ) provide a factorization of f ( x ) in Z [[ x ]], as claimed. References [1] Birmajer, Daniel, and Gil, Juan B.,
Arithmetic in the ring of formal power series with integercoefficients , to appear in American Mathematical Monthly.[2] Borevich, Z. I., and Shafarevich, I. R.,
Number Theory (trans. from Russian by N. Greenleaf),Academic Press, New York, 1966.[3] Dummit, David, and Foote, Richard,
Abstract Algebra , Prentice-Hall, 1991.[4] Eisenbud, David,
Commutative algebra with a view toward algebraic geometry , GraduateTexts in Mathematics, Springer, 1995.[5] Kaplansky, Irving,
Commutative rings , Allyn and Bacon, Inc., Boston, 1970.[6] Serre, Jean-Pierre,
A course in arithmetic , Graduate Texts in Mathematics, Springer, 1973.
Department of Mathematics, Nazareth College, 4245 East Avenue, Rochester, NY14618
E-mail address : [email protected] Penn State Altoona, 3000 Ivyside Park, Altoona, PA 16601.
E-mail address : [email protected] Penn State Altoona, 3000 Ivyside Park, Altoona, PA 16601.
E-mail address ::