aa r X i v : . [ m a t h . N T ] A ug A NEW CLASS OF IRREDUCIBLE POLYNOMIALS
JITENDER SINGH † AND SANJEEV KUMAR ‡ , ∗ Abstract.
In this article, we propose a few sufficient conditions on polynomials havinginteger coefficients all of whose zeros lie outside a closed disc centered at the origin in thecomplex plane and deduce the irreducibility over the ring of integers. Introduction
Testing polynomials for irreducibility over a given domain is an arduous task. Of particularinterest are the polynomials having integer coefficients for which some well–known classicalirreducibility criteria due to Sch¨onemann, Eisenstein, and Dumas exist (see [1, 2, 4] and foran insightful historical account of Sch¨onemann and Eisenstein criteria, see [3]). Recently, theelegant criteria established in [5, 6] turn out to be extremely significant keeping in view theirintimate connection with prime numbers. Moreover, the notion of locating the zeros of thegiven polynomial being tested for irreducibility is quite informative (see [7]). In this regard,one can infer that if for each zero ζ of g ∈ Z [ x ], | ζ | ≤ r holds for some r >
0, then each zero θ of f = g ( x − c ) is given by θ = ζ + c which on applying the triangle inequality yields | θ | > c whose absolute value exceeds r + 2. Also, the translational invariance ofirreducibility of polynomials in the ring Z [ x ] ensures the irreducibility of g vis–´a–vis fromthat of f . Proceeding in this manner, one can frame the following irreducibility criterionfrom that of the one given in [6, Theorem 1]. Theorem A.
Let f ∈ Z [ x ] be such that each zero θ of f satisfies | θ | > d . If f (0) = ± pd forsome positive integer d and prime p ∤ d , then f is irreducible in Z [ x ] .Proof. If possible, let f ( x ) = f ( x ) f ( x ), where f and f are non–constant polynomials in Z [ x ]. By hypothesis on f , f (0) = f (0) f (0) = ± pd which shows that p divides exactlyone of the factors f (0) or f (0). Assume without loss of generality that p | f (0). Then | f (0) | ≤ d . On the other hand if c = 0 is the leading coefficient of f , then we may write(1) f ( x ) = c Y θ ( x − θ ) , where the product runs over all zeros of f . By the hypothesis on zeros of f we must havefrom (1) that | f (0) | = | c | Q θ | θ | > | c | d deg f ≥ d , a contradiction. (cid:3) In Theorem A, the primality of | f (0) | /d is necessary to deduce the irreducibility. In anattempt to weaken the hypothesis, we confront the following natural question: Given | θ | > d for each zero θ of f , is it still possible to recover the irreducibility of f if instead | f (0) | /d is Mathematics Subject Classification.
Primary 12E05; 11C08. † [email protected] ‡ , ∗ Corresponding author: sanjeev kumar [email protected] prime power ? Nevertheless, under certain mild conditions on the coefficients of f , we showthat the answer to the above question is in the affirmative.Recall that a polynomial f having integer coefficients is primitive if the greatest commondivisor of all its coefficients is 1. Our main results are the following: Theorem 1.
Let f = a + a x + · · · + a n x n ∈ Z [ x ] be a primitive polynomial such that eachzero θ of f satisfies | θ | > d , where a = ± p k d for some positive integers k and d , and aprime p ∤ d . If j ∈ { , . . . , n } is such that gcd( k, j ) = 1 , p k | gcd( a , a , . . . , a j − ) and for k > , p ∤ a j , then f is irreducible in Z [ x ] . Theorem 2.
Let f = a + a x + · · · + a n x n ∈ Z [ x ] be a primitive polynomial such that eachzero θ of f satisfies | θ | > d , where a n = ± p k d for some positive integer k and d , and a prime p ∤ d . Let j ∈ { , . . . , n } be such that gcd( k, j ) = 1 , p k | gcd( a n − j +1 , a n − j +2 , . . . , a n ) andfor k > , p ∤ a n − j . If | a /q | ≤ | a n | where q is the smallest prime divisor of a , then f isirreducible in Z [ x ] . To prove Theorems 1-2, elementary divisibility theory for integers is devised. The cogenttechniques involved in the proofs are of independent interest as well. Further, the notationsspecified below are imperative and shall be used in the sequel.
Notations. If f ( x ) = f ( x ) f ( x ), unless otherwise specified, we write f = a + a x + · · · + a n x n ∈ Z [ x ]; f = b + b x + · · · + b m x m and f = c + c x + · · · + c n − m x n − m are non–constantpolynomials in Z [ x ]. Define further that b m +1 = b m +2 = · · · = b n = 0; c n − m +1 = c n − m +2 = · · · = c n = 0 , so that we may write(2) a t = b c t + b c t − + · · · + b t c , for each t = 0 , , . . . , n. Proofs of Theorems 1-2
To prove Theorems 1-2, we first prove the following crucial result.
Lemma 3.
Let f = a + a x + · · · + a n x n , f = b + b x + · · · + b m x m , and f = c + c x + · · · + c n − m x n − m be non–constant polynomials in Z [ x ] such that f ( x ) = f ( x ) f ( x ) .Suppose that there is a prime number p and positive integers k ≥ and j ≤ n such that p k | gcd( a , a , . . . , a j − ) , p k +1 ∤ a , and gcd( k, j ) = 1 . If p | b and p | c , then p | a j . Proof of Lemma 3.
In view of the hypothesis that p | b and p | c , there exists a positiveinteger ℓ ≤ k − ℓ such that p ℓ | b and p k − ℓ | c , where ℓ and k − ℓ are highest powers of p dividing b and c respectively. To proceed we define the nonnegative integer κ such that κ = ( j − / j is even and κ = ( j − / j is odd. We now arrive at the following cases: Case I: ℓ < k − ℓ . In this case we have the following subcases: Subcase I: p | b i for all i = 0 , . . . , κ . Using the expressions for a i and a i successively foreach i = 0 , . . . , κ , we find that p divides c , c , . . . , c κ . If α i and β i are the highest powersof p dividing b i and c i respectively, then α = ℓ and β = k − ℓ . We claim that α i ≥ ℓ and β i ≥ k − ℓ for all i ≤ κ . For proof, we consider a = b c + b c which tells us that ℓ + β ≥ k, β ≤ k − ℓ + α , which further give α ≥ ℓ and β ≥ k − ℓ with α < β since ℓ < k − ℓ . Then p k | ( a − b c ) = b c + b c which for the similar reasons shows that α ≥ ℓ and β ≥ k − ℓ with α < β . ontinuing in this manner, suppose for some positive integer i ∗ < κ that the following havebeen proved successively(3) α i ≥ ℓ, β i ≥ k − ℓ, α i < β i , for each i = 0 , , . . . , i ∗ . Then consider a i ∗ +1 = b c i ∗ +1 + ( b c i ∗ + · · · + b i ∗ c ) + b i ∗ +1 c , where from (3) we get p ℓ | b i and p k − ℓ | c i ∗ +1 − i for each i = 1 , . . . , i ∗ so that p k | b i c i ∗ +1 − i . Consequently, p k | ( b c i ∗ + · · · + b i ∗ c ).Also, by the hypothesis, p k | a i ∗ +1 . So we get p k | ( a i ∗ +1 − b c i ∗ −· · ·− b i ∗ c ) = b c i ∗ +1 + b i ∗ +1 c .This proves that α i ∗ +1 ≥ ℓ and β i ∗ +1 ≥ k − ℓ with α i ∗ +1 < β i ∗ +1 since ℓ < k − ℓ . With this,we conclude that(4) α i ≥ ℓ, β i ≥ k − ℓ, α i < β i for all i = 0 , . . . , κ. To proceed further, we first assume that κ = ( j − /
2. Using (4) in the expression for a j − in (2), we have p k | ( a j − − b c j − − · · · − b ( j − / c ( j +2) / − b ( j +2) / c ( j − / − · · · − b j − c )= b ( j − / c j/ + b j/ c ( j − / , which shows that p k − ℓ | c j/ . Consequently p | { b c j + · · · + b ( j − / c ( j +2) / + b j/ c j/ + b ( j +2) / c ( j − / + · · · + b j − c } = a j , where the equality follows from (2).For κ = ( j − / p ℓ | ( b c j + b c j − + · · · + b ( j − / c ( j +1) / + b ( j +1) / c ( j − / + · · · + b j c ) = a j . Subcase II:
There is a smallest positive integer i ≤ κ for which p ∤ b i . From the SubcaseI, p ℓ divides each of b , . . . , b i − and p k − ℓ divides each of c , . . . , c i − . Let q j be the positiveinteger, such that iq j ≤ j − < (1 + q j ) i . Let β s denote the highest power of p dividing c s for i ≤ s ≤ j −
1. We will show that β ti + r = k − ( t + 1) ℓ , for each t = 1 , . . . q j and r = 0 , . . . , i − b c t = a t − C ( c , c , . . . , c t − ) , where C ( c , . . . , c t − ) is the integer combination of c , . . . , c t − which we define as follows:(6) C ( c ) = 0; C ( c , c , . . . , c t − ) = b t c + b t − c + · · · + b c t − for t > . Since p k − ℓ | c t for each t = 0 , . . . , i −
1, it follows from (6) that p k − ℓ | C ( c , . . . , c i − ), whichin view of (5) and the fact that p k | a i gives β i = k − ℓ since p ∤ b i . Suppose we haveproved successively that β i + r = k − ℓ for 0 ≤ r < i −
1. Then p k − ℓ | ( b i + r c + · · · + b i c r ) and p k − ℓ | ( b i +1 c r + · · · + b c i + r ) so that from (6), we get p k − ℓ | C ( c , . . . , c i + r ), which in view of(5) gives p k − ℓ | c i + r +1 or β i + r +1 ≥ k − ℓ . Since p ∤ b i , we must also have β i + r +1 ≤ k − ℓ .So, β i + r +1 = k − ℓ . This proves the claim for t = 1 and all r = 0 , . . . , i − β ti + r = k − ( t + 1) ℓ for each t = 0 , . . . , t ∗ and r = 0 , . . . , i − t ∗ ≤ q j . Then we have α s = α ; β ti + s = k − ( t + 1) ℓ for s = 0 , . . . , i − t = 0 , . . . , t ∗ . (7)For convenience, we define(8) h ( s ) = b s c i (1+ t ∗ )+ r − s , s = 0 , . . . , i (1 + t ∗ ) + r. rom (7)–(8), we have for r = 0 and each s = 0 , . . . , i − p ℓ + k − (1+ t ∗ ) ℓ | h ( s ); p k − (1+ t ∗ ) ℓ | h ( i + s ); p k − t ∗ ℓ | h (2 i + s ); . . . ; p k − ℓ | h ( i (1 + t ∗ ) + s ) , Also, from (6) and (8) we have C ( c , . . . , c i (1+ t ∗ )+ r − ) = i − X s =1 h ( s ) + i − X s = i h ( s ) + · · · + i (1+ t ∗ ) − X s = it ∗ h ( s ) + i (1+ t ∗ )+ r X s = i (1+ t ∗ ) h ( s )= i − X s =1 h ( s ) + i − X s =0 { h ( i + s ) + · · · + h ( it ∗ + s ) } + r X s =0 h ( i (1 + t ∗ ) + s )= i − X s =1 h ( s ) + t ∗ X s ′ =1 i − X s =0 h ( is ′ + s ) + r X s =0 h ( i (1 + t ∗ ) + s ) . (10)Using (9) in (10) for r = 0, we get p k − (1+ t ∗ ) ℓ | C ( c , . . . , c i (1+ t ∗ ) − ). Consequently, from (5),we have p k − (1+ t ∗ ) ℓ | ( a i (1+ t ∗ ) − C ( c , . . . , c i (1+ t ∗ ) − )) = b c i (1+ t ∗ ) . This further gives p k − (2+ t ∗ ) ℓ | c i (1+ t ∗ ) . Thus,(11) β i (1+ t ∗ )+ r = k − (2 + t ∗ ) ℓ > r = 0. In view of (11), the assertion in (9) holds for r = 1, using which furtherin (10) proves (11) for r = 1. Suppose then that (11) holds for each r = 0 , . . . , r ∗ for somepositive integer r ∗ < i −
1. Then in view of (11) we have that (9) holds for r = r ∗ . Using thisfurther in (10) proves that (11) holds for r = r ∗ + 1. This proves the claim. So, p k − (1+ q j ) ℓ | c s ,where k > (1 + q j ) ℓ for all s = 0 , . . . , j − p k − (1+ q j ) ℓ | ( b c j + b c j − + . . . + b i c j − i + · · · + b j c ) = a j . Case II: ℓ = k − ℓ . Here k is even. Then j is odd since gcd( k, j ) = 1. In this case, we usethe fact that for any two integers a and b , and prime p , if p | ( a + b ) and p | ab , then p | a and p | b .In view of the above fact, we have from the expressions for a and a in (2) that p | b and p | c . Similarly from the expressions for a and a in (2) we get p | b and p | c . Continuingthis way, having proved that p divides each of the integers b , c , b , c , . . . , b ( j − / , c ( j − / ,it follows from the expressions for a ( j − / and a j − in (2) that p | b ( j − / and p | c ( j − / . Soin view of (2), we get the following: p | ( b c j + · · · + b ( j − / c ( j +1) / + b ( j +1) / c ( j − / + · · · + b j c ) = a j . This completes the proof of Lemma 3. (cid:3)
Remark.
Proof of Lemma 3 becomes considerably short if one assumes gcd( k, j !) = 1. Inthat case, the condition gcd( k, j !) = 1 implies k > j and k − tℓ > t = 1 , . . . , j .Consequently in view of (2), one immediately finds recursively that(12) p k − ( t − ℓ | ( a t − − b c t − − b c t − − · · · − b t − c ) = b c t − , t = 1 , . . . , j. So from (12) it follows that p | c t for each t = 0 , . . . , j − p | b yields the desired conclusion p | ( b c j + b c j − + · · · + b j c ) = a j . roof of Theorem 1. If possible, assume that f ( x ) = f ( x ) f ( x ) where f and f are asin the notation. Then in view of (4), we have(13) a = b c = ± p k d ; a m = b m c n − m . Since each zero θ of f satisfies | θ | > d , we must have | b /b m | > d and | c /c n − m | > d whichfurther give | b | > d and | c | > d .If p ∤ c , then p k | b and consequently the second equality in (13) yields | c | < d , acontradiction. On the other hand if p | b and p | c then k > p | a j . (cid:3) Proof of Theorem 2.
Suppose to the contrary that f ( x ) = f ( x ) f ( x ) where f and f are as in the notation. Then b c = a and b m c n − m = a n = ± p k d . Since each zero θ of f satisfies | θ | > d , we must have | b /b m | > d and | c /c n − m | > d . If p ∤ b m then p k | c n − m sothat | b m | ≤ d and we have (cid:12)(cid:12) a /a n (cid:12)(cid:12) = | b /b m | × (cid:12)(cid:12) c /c n − m (cid:12)(cid:12) > | b /d | d = | b | ≥ q, which contradicts the hypothesis.On the other hand if p | b m and p | c n − m , then k ≥ p | a n − j . (cid:3) Remark.
In view of Theorems 1-2, the hypothesis on zeros of f is not required in the casewhen j = n , wherein the hypothesis on a is also not required in Theorem 2 and we thenhave: Theorem B.
Let f = a + a x + · · · + a n x n ∈ Z [ x ] be a primitive polynomial. For a prime p and positive integers k and n , if gcd( k, n ) = 1 , p k | gcd( a , a , . . . , a n − ) , p ∤ a n , and p k +1 ∤ a , then f is irreducible in Z [ x ] . Theorem B is well known and is generally proved using Newton polygons (see [4]). Howeverhere, we provide an alternative proof based on Lemma 3.
Proof of Theorem B.
To the contrary assume that f ( x ) = f ( x ) f ( x ) where f and f are as in the notation. In view of Lemma 3, it is enough to show that p | b and p | c inorder to get the desired contradiction. Since p | a = b c , we may assume without loss ofgenerality that p | b . Since p ∤ a n = b m c n − m , we have p ∤ b m and p ∤ c n − m . So, there exists aleast positive integer t ≤ m such that p ∤ b t . This in view of (2) yields the following: p | ( a t − b c t − b c t − − · · · − b t − c ) = b t c , so that p | b t c , which further gives p | c . (cid:3) Examples For a prime p , positive integers n and k with gcd( k, j ) = 1, consider the polynomial(14) X j,k = p k +1 (1 + x + x + · · · + x j − ) + ( p k − x j + p k − x j +1 (1 + x + · · · + x n − j − ) . We will show that each zero ζ of X j,k satisfies | ζ | >
1. Observe that(15) ( x − X j,k = − p k +1 + ( p k +1 − p k + 1) x j + ( p k − p k − − x j +1 + p k − x n +1 . o that the coefficients of x j , x j +1 , and x n +1 in ( x − X j,k are all positive. If | ζ | < p k +1 = ( p k +1 − p k + 1) ζ j + ( p k − p k − − ζ j +1 + p k − ζ n +1 ≤ ( p k +1 − p k + 1) | ζ | j + ( p k − p k − − | ζ | j +1 + p k − | ζ | n +1 < ( p k +1 − p k + 1) + ( p k − p k − −
1) + p k − = p k +1 , (16)which is absurd. So we must have | ζ | ≥ | ζ | = 1 for some zero ζ of X j,k , then ζ = e ιt for some real number t . Now from (16),( p k +1 − p k + 1)(1 − e jt ) + ( p k − p k − − − e ( j +1) t ) + p k − (1 − e ( n +1) t ) = 0 , which oncomparing real parts gives( p k +1 − p k + 1) sin { ( jt/ } + ( p k − p k − −
1) sin { ( j + 1) t/ } + p k − sin { ( n + 1) t/ } = 0which is possible only if jt, ( j + 1) t, ( n + 1) t ∈ π Z . Thus we have ζ j = ζ j +1 = ζ n +1 = 1,which give ζ = 1. But from (14), X j,k (1) > ζ of X j,k satisfies | ζ | > X j,k satisfies rest of the hypotheses of Theorem 1. So X j,k is irreducible in Z [ x ]. For a prime p , positive integers k , n , m < p , and j ≤ n with gcd( k, j ) = 1, the polynomial Y j,k,m = p k ( n + x + x + · · · + x n − j − ) + mx n − j + p k x n − j +1 (1 + · · · + x j − )satisfies the hypotheses of Theorem 2. So Y j,k,m is irreducible in Z [ x ]. Let d be a positive integer and f = a + a x + · · · + a n x n ∈ Z [ x ] such that | a | > | a | d + | a | d + · · · + | a n | d n . Then for | x | ≤ d , we have | f ( x ) | ≥ | a | − | a || x | − · · · − | a n || x | n > | a | − | a | d − | a | d − · · · − | a n | d n > , which shows that each zero θ of f satisfies | θ | > d . Now imposing the conditions of Theorem1 or Theorem 2 on f , the irreducibility of f in Z [ x ] is immediate. References [1] T. Sch¨onemann, Von denjenigen Moduln, welche Potenzen von Primzahlen sind,
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