Download On the irreducibility of polynomials that take a prime power

Bull. Math. Soc. Sci. Math. Roumanie
Tome 54(102) No. 1, 2011, 41–54
On the irreducibility of polynomials that take
a prime power value
by
A.I.Bonciocat, N.C. Bonciocat, and A. Zaharescu
In memory of Laurenţiu Panaitopol
Abstract
We provide several irreducibility criteria for polynomials with integer
coefficients that take a prime power value and either have one large coefficient, or have all the coefficients of modulus 1. We also obtain upper
bounds for the total number of irreducible factors of such polynomials, by
studying their higher derivatives.
Key Words: Estimates for polynomial roots, prime power, irreducible polynomials.
2010 Mathematics Subject Classification: Primary 11C08; Secondary 11R09.
1
Introduction
Some classical irreducibility criteria rely on the existence of a suitable prime
divisor of the value that a given polynomial takes at a specified integral argument.
One of the most elegant results of this type, given by Pólya and Szegö in [12], is
due to A. Cohn:
Theorem
A. If a prime number p is expressedP
in the decimal system as
Pn
n
p = i=0 ai 10i with 0 ≤ ai ≤ 9, then the polynomial i=0 ai X i is irreducible in
Z[X].
This irreducibility criterion was generalized to an arbitrary base b by Brillhart,
Filaseta and Odlyzko [4]:
Theorem B. P
If a prime number p is expressed in the numberPsystem with
n
n
base b ≥ 2 as p = i=0 ai bi , 0 ≤ ai ≤ b − 1, then the polynomial i=0 ai X i is
irreducible in Z[X].
42
A.I.Bonciocat, N.C. Bonciocat, and A. Zaharescu
For several nice connections between prime numbers and irreducible polynomials, the reader is referred to Ram Murty [10] and Girstmair [8]. Murty [10]
obtained an elementary proof of these irreducibility criteria, and also established
an analogous result for polynomials with coefficients in Fq [t], with Fq a finite
field. Inspired by these results, in [2] several irreducibility criteria are proved for
integer polynomials that take a prime value and either have one large coefficient,
or have all the coefficients of modulus 1. We mention here the following two:
Theorem C. If weP
write a prime number as a sum ofPintegers a0 , . . . , an with
n
n
a0 an 6= 0 and |a0 | > i=1 |ai |2i , then the polynomial i=0 ai X i is irreducible
over Q.
Theorem D. If all the coefficients of a polynomial f are ±1 and f (m) is a
prime number for an integer m with |m| ≥ 3, then f is irreducible over Q.
Similar irreducibility criteria for multivariate polynomials over an arbitrary
field have been obtained in [3]. One may naturally ask whether Cohn’s result will
still hold true if we replace the prime number p with ps , s ≥ 2. This is by no means
necessarily true, as one can see by taking p = 11 and considering the polynomial
f (X) obtained by replacing the powers of 10 by the corresponding powers of X in
the decimal representation of 112 . In this case f (10) = 121, and the polynomial
f (X) = X 2 + 2X + 1 is obviously reducible. For another example one may
consider the decimal representation of 117 . Here f (10) = 117 = 19 487 171, and
the polynomial f (X) is also reducible, being divisible by X + 1.
The first aim of this paper is to find some additional conditions that guarantee
us the irreducibility of a polynomial that takes a prime power value, and thus to
complement the results in [2], by extending them to a larger class of polynomials.
This will be achieved by adding a natural condition on the derivative of our
polynomials. Our second goal is to derive upper bounds for the total number
of irreducible factors of such polynomials, instead of irreducibility criteria, by
considering their higher derivatives. Our first result extends Theorem B to prime
powers, as follows.
Theorem 1.1. If a prime
ps , s ≥ 2 is expressed in P
the number system
Pn power
n
s
i
with base b ≥ 2P
as p = i=0 ai b , 0 ≤ ai ≤ b − 1 and p - i=1 iai bi−1 , then
n
i
the polynomial i=0 ai X is irreducible over Q.
The following three results give irreducibility conditions for polynomials that
have one coefficient of sufficiently large modulus and take a value divisible by a
prime power ps , s ≥ 2.
Pn
i
Theorem 1.2. Let f (X) =
i=0 ai X ∈ Z[X], a0 an 6= 0. Suppose that
s
f (m) = p · q for some integers m, s, q and a prime number p, with s ≥ 2,
p - qf 0 (m) and
n
X
|a0 | >
|ai | · (|m| + |q|)i .
i=1
Then f is irreducible over Q.
Irreducibility criteria
43
Pn
Theorem 1.3. Let f (X) = i=0 ai X di ∈ Z[X], with 0 = d0 < d1 < · · · < dn
and a0 a1 · · · an 6= 0. Suppose that f (m) = ps · q for some integers m, s, q and
a prime number p, with s ≥ 2, |m| > |q| and p - qf 0 (m). If for an index
j ∈ {1, . . . , n − 1} we have
X
|ai |,
|aj | > (|m| + |q|)dn −dj ·
i6=j
then f is irreducible over Q.
Pn
i
Theorem 1.4. Let f (X) =
i=0 ai X ∈ Z[X], a0 an 6= 0. Suppose that
s
f (m) = p · q for some integers m, s, q and a prime number p, with s ≥ 2,
|m| > |q|, p - qf 0 (m) and
|an | >
n−1
X
|ai | · (|m| − |q|)i−n .
i=0
Then f is irreducible over Q.
Roughly speaking, the above three results show in particular that if f (m) is a
prime power for an integer m with |m| ≥ 2, f (m) and f 0 (m) are relatively prime,
and f has one coefficient of sufficiently large modulus, then f must be irreducible
over Q.
Moreover, from Theorems 1.2 and 1.4 one obtains the following irreducibility
criteria, that extend Theorems C and D to polynomials that take a prime power
value.
Corollary 1.5. If we write P
a prime power ps , s ≥ 2, as a sum of integers
n
a0 , . . . , an with a0 an 6= 0, |a
|ai |2i , and a1 + 2a2 + · · · + nan not divisible
P0 n| > i=1
i
by p, then the polynomial i=0 ai X is irreducible over Q.
Corollary 1.6. Let f be a Littlewood polynomial. If f (m) is a prime power
ps , s ≥ 2, for an integer m with |m| ≥ 3, and p - f 0 (m), then f is irreducible over
Q.
As it was done in [2], we will also prove some related results, that depend on
some suitable sets of parameters.
Pn
Proposition 1.7. Let f (X) = i=0 ai X di ∈ Z[X], with 0 = d0 < d1 < · · · <
dn and a0 a1 · · · an 6= 0. Suppose that f (m) = ps · q for some integers m, s, q and a
prime number p, with s ≥ 2, p - qf 0 (m), and let µ0 = 0, µn = 1 and µ1 , . . . , µn−1
be arbitrary positive constants. If
|m| − |q| > max
1≤j≤n
then f is irreducible over Q.
(1 + µj−1 )|aj−1 |
µj |aj |
1
j −dj−1
d
,
44
A.I.Bonciocat, N.C. Bonciocat, and A. Zaharescu
Pn
Proposition 1.8. Let f (X) = i=0 ai X di ∈ Z[X], with 0 = d0 < d1 < · · · <
dn and a0 a1 · · · an 6= 0. Suppose that f (m) = ps · q for some integers m, s, q and a
prime number p, with s ≥ 2, p - qf 0 (m), and let µ0 = 1, µn = 0 and µ1 , . . . , µn−1
be arbitrary positive constants. If
|m| + |q| < min
1≤j≤n
µj−1 |aj−1 |
(1 + µj )|aj |
1
j −dj−1
d
,
then f is irreducible over Q.
Pn
Proposition 1.9. Let f (X) = i=0 ai X i ∈ Z[X], with a0 an 6= 0. Suppose
that f (m) = ps · q for some integers m, s, q and a prime number p, with s ≥ 2,
p - qf 0 (m), and let µ1 , . . . , µn be arbitrary positive constants. If


n
µ µ

X
µ
µ
|a
|
2
3
n
j
j−1
|m| − |q| > max
,
, ,...,
,
·
 µ1 µ2
µn−1 j=1 µn |an | 
then f is irreducible over Q.
Pn
Proposition 1.10. Let f (X) = i=0 ai X i ∈ Z[X], with a0 an 6= 0. Suppose
that f (m) = ps · q for some integers m, s, q and a prime number p, with s ≥ 2
and p - qf 0 (m). Let µ0 = 0 and µ1 , . . . , µn be arbitrary positive constants. If
µj
µn
|aj |
|m| − |q| > max
+
·
,
0≤j≤n−1 µj+1
µj+1 |an |
then f is irreducible over Q.
In particular, for µ1 = µ2 = . . . = µn = 1 we obtain the following irreducibility
criterion.
Pn
Corollary 1.11. Let f (X) = i=0 ai X i ∈ Z[X], with a0 an 6= 0. Suppose
that f (m) = ps · q for some integers m, s, q and a prime number p, with s ≥ 2
and p - qf 0 (m). If
|a1 |
|a2 |
|an−1 |
|a0 |
,1 +
,1 +
,...,1 +
,
|m| − |q| > max
|an |
|an |
|an |
|an |
then f is irreducible over Q.
The above results are flexible and may be useful in certain applications where
other irreducibility criteria fail. In order to keep this paper self-contained, we
will give in Section 2 below detailed proofs, that employ the methods used in [2],
[10] and [12], and we will also include the proofs given in [2] for the estimates on
polynomial roots needed in our results. At the end of Section 2, in Lemma 2.3
we will give a method to obtain upper bounds for the total number of irreducible
factors of a given polynomial, by studying its higher derivatives. We will also
present a series of examples in the last section of the paper.
Irreducibility criteria
2
45
Proof of the main results
Our proofs rely on the following lemma, which complements Lemma 1.1. in [2]:
Lemma 2.1. Let f ∈ Z[X] and suppose that f (m) = ps · q for some integers
m, s, q and a prime number p, with s ≥ 2 and p - qf 0 (m). If for two positive real
numbers A and B we have A < |m| − |q| < |m| + |q| < B, and f has no roots in
the annular region A < |z| < B, then f is irreducible over Q.
The desired results will be obtained by combining Lemma 2.1 or parts of its
proof with some suitable estimates for polynomial roots.
Pn
Proof of Lemma 2.1. Let f (X) = i=0 ai X i and assume that f factors as
f (X) = f1 (X) · f2 (X), with f1 , f2 ∈ Z[X], deg f1 ≥ 1 and deg f2 ≥ 1. Then,
since f (m) = ps · q = f1 (m) · f2 (m), f 0 (m) = f10 (m)f2 (m) + f1 (m)f20 (m) and
p - qf 0 (m), one of the integers f1 (m), f2 (m) must be divisible by ps , so the other
must divide q, say f1 (m) | q. In particular, we have |f1 (m)| ≤ |q|.
Assume now that f factors as f (X) = an (X−θ1 ) . . . (X−θn ), with θ1 , . . . , θn ∈
C. Since f1 is a factor of f , it will factor over C as f1 (X) = bt (X −θ1 ) . . . (X −θt ),
say, with t ≥ 1 and |bt | ≥ 1. Then one has
|f1 (m)| = |bt | ·
t
Y
|m − θi | ≥
i=1
t
Y
|m − θi |.
(1)
i=1
The fact that the roots of f lie outside the annulus A < |z| < B shows that for
each index i ∈ {1, . . . , t} we either have
|m − θi | ≥ |m| − |θi | ≥ |m| − A,
if |θi | ≤ A,
|m − θi | ≥ |θi | − |m| ≥ B − |m|,
if |θi | ≥ B.
or
Since by hypothesis we have A < |m| − |q| < |m| + |q| < B, we conclude that
|m − θi | > |q| for each i = 1, . . . , t, so by (1) we obtain |f1 (m)| > |q|, which is a
contradiction. This completes the proof of the lemma. Proof of Theorem 1.1. Note first that since f (b) = ps and p - f 0 (b), f must be
a primitive polynomial. Let us assume that f factors as f (X) = f1 (X) · f2 (X),
with f1 , f2 ∈ Z[X], deg f1 ≥ 1 and deg f2 ≥ 1. Reasoning as in the proof of
Lemma 2.1 with q = 1 and m = b, we deduce that |f1 (b)| = 1.
We will employ now the elegant method used in [10] and [12] for the proof of
Theorem B. First, we show that any complex zero θ of f either has nonpositive
real part or satisfies
√
1 + 4b − 3
.
(2)
|θ| <
2
46
A.I.Bonciocat, N.C. Bonciocat, and A. Zaharescu
To see this, we observe that if |θ| > 1 and R(θ) > 0 we obtain
1
an−1 1
+ ··· + n
0 ≥ an +
− (b − 1)
θ
|θ|2
|θ|
an−1
b−1
> R an +
− 2
θ
|θ| − |θ|
b−1
|θ|2 − |θ| − (b − 1)
> 1− 2
=
,
|θ| − |θ|
|θ|2 − |θ|
√
which gives a contradiction for |θ| ≥ (1 + 4b − 3)/2. Note that if |θ| ≤ 1, (2)
holds trivially, so we may assume |θ| > 1. Then we either have R(θ) ≤ 0, or θ
must satisfy (2).
In the former case we have |b − θ| ≥ b > 1, while in the latter case we have
|θ| <
1+
√
4b − 3
≤b−1
2
if b ≥ 3, which also yields |b − θ| > 1. In view of (1) we then obtain |f1 (b)| > 1,
a contradiction. We have therefore proved the irreducibility of f for b ≥ 3.
Let us assume now that b = 2. By the
√ argument above, if θ is a root of f ,
we either have R(θ) ≤ 0, or |θ| < (1 + 5)/2. We will prove that in fact, all
the roots of f have real part less than 3/2. To see this, we√first observe
that
√
a root θ of f with
|
arg
θ
|
>
π/4
must
have
R(θ)
<
(1
+
5)/(2
2)
<
3/2,
√
since |θ| < (1 + 5)/2. We may therefore assume that | arg θ | ≤ π/4, and hence
R(1/θ) ≥ 0 and R(1/θ2 ) ≥ 0. Moreover, we have to consider only polynomials f
with deg f ≥ 3, since no polynomial of degree 2 with coefficients 0 or 1 satisfies
all the hypotheses in the statement of the theorem. Let us now assume to the
contrary that R(θ) ≥ 3/2. Then |θ| ≥ 3/2, which yields
f (θ) an−2 1
an−1
1
0 = n ≥ 1 +
+ 2 −
+
·
·
·
+
θ
θ
θ
θ3
θn
an−1
an−2
1
> R 1+
+ 2
− 2
θ
θ
|θ| (|θ| − 1)
1
|θ|3 − |θ|2 − 1
≥ 1− 2
=
> 0,
|θ| (|θ| − 1)
|θ|2 (|θ| − 1)
a contradiction.
The fact that all the roots of f have real part less than 3/2 shows now that
the polynomial f1 (X + 3/2) = bt (X + 3/2 − θ1 ) · · · (X + 3/2 − θt ) has positive
coefficients. Indeed, if θi is a real root of f1 , then the linear factor X +3/2−θi has
positive coefficients, while if θi is not real, we pair it with its complex conjugate
θi and notice that
(X + 3/2 − θi )(X + 3/2 − θi ) = X 2 + 2R(3/2 − θi )X + |3/2 − θi |2
Irreducibility criteria
47
has too positive coefficients. Therefore f1 (−X + 3/2) has alternating coefficients,
and hence for any X > 0 we have |f1 (−X + 3/2)| < f1 (X + 3/2). Letting now
X = 1/2 we deduce that f1 (2) > |f1 (1)|. Since 1 is not a root of f1 and f1 (1) is
an integer, we obtain |f1 (2)| > 1, which is a contradiction. This completes the
proof of the theorem. Proof of Theorem 1.2. Here we only need to observe that our assumption on
the size of |a0 | forces the absolute values of the θi ’s to be greater than
+ |q|.
P|m|
n
Indeed, if |θj | ≤ |m|+|q| for an index j ∈ {1, . . . , n}, then since a0 = − i=1 ai ·θji ,
Pn
Pn
we would obtain |a0 | ≤ i=1 |ai | · |θj |i ≤ i=1 |ai | · (|m| + |q|)i , a contradiction.
The rest of the proof follows now in a manner similar to that given for Lemma
2.1. Proof of Theorem 1.3. We will actually prove a more general result, that
depends on a suitable set of parameters:
Pn
Lemma 2.2. Let f (X) = i=0 ai X di ∈ Z[X], with 0 = d0 < d1 < · · · < dn
and a0 a1 · · · an 6= 0. Suppose that f (m) = ps · q for some integers m, s, q and
a prime number p, with s ≥ 2 and p - qf 0 (m). Suppose also that there exist a
sequence
P of positive real numbers µ0 , µ1 , . . . , µn and an index j ∈ {0, . . . , n} such
that k6=j µk ≤ 1 and
max
k<j
1 |ak |
·
µk |aj |
1
j −dk
d
|aj |
< |m| − |q| < |m| + |q| < min µk ·
k>j
|ak |
1
k −dj
d
.
Then f is irreducible over Q.
Here we obviously have to ignore the left-most inequality if j = 0, and the
right-most one if j = n. Note that the inequalities in the statement of Lemma
2.2 are satisfied if
|m| > |q| and |aj | > max
k6=j
|ak | · (|m| + |q| · sign(k − j))dk −dj
,
µk
so if f (m) = ps , s ≥ 2 for an integer m with |m| ≥ 2 and a prime number p
such that p - f 0 (m), and f has one sufficiently large coefficient, then it must be
irreducible over Q.
By choosing in Lemma
2.2 different sequences of positive real numbers µ0 , µ1 ,
P
. . . , µn that satisfy k6=j µk ≤ 1, we may obtain various irreducibility conditions.
For instance, we may simply choose µk = 1/n for k 6= j, or µk = 2−n nk for
k 6= j. ForPan example when the µk ’s depend on the coefficients of f , we take
µk = |ak |/ i6=j |ai | for k 6= j, and obtain Theorem 1.3.
We also note that a direct use of the triangle inequality as in Theorems 1.2
and 1.4 gives sharper conditions than those exhibited by Lemma 2.2 in the cases
j = 0 and j = n.
48
A.I.Bonciocat, N.C. Bonciocat, and A. Zaharescu
Proof of Lemma 2.2. Assume that f factors as f (X) = an (X − θ1 ) . . . (X −
θdn ), with θ1 , . . . , θdn ∈ C, let
A = max
k<j
1 |ak |
·
µk |aj |
d
1
j −dk
1
|aj | dk −dj
and B = min µk ·
,
k>j
|ak |
and note that according to our hypotheses A must be strictly smaller than B.
M. Fujiwara proved in [7] the following flexible result on the location of the
roots of a complex polynomial:
Pn
Let P (z) = i=0 ai z di ∈ C[z], with 0 = d0 < d1 < · · · < dn and a0 a1 . . . an 6=
1
≤ 1. Then all the
0. Let also µ0 , . . . , µn−1 ∈ (0, ∞) such that µ10 + · · · + µn−1
roots of P are contained in the disk |z| ≤ R, where
R=
max
0≤j≤n−1
|aj |
µj
|an |
1
n −dj
d
.
We will adapt here the classical method of Fujiwara to find information on the
location of the roots of f . More precisely, we will prove that f has no roots in
the annular region A < |z| < B, as required in Lemma 2.1. To see this, let us
assume that A < |θi | < B for some index i ∈ {1, . . . , dn }. Then from A < |θi | we
deduce that µk |aj | · |θi |dj > |ak | · |θi |dk for each k < j, while from |θi | < B we
for each k > j. Adding term by term these
find that µk |aj | · |θi |dj > |ak | · |θi |dk P
inequalities and using the fact that k6=j µk ≤ 1, we obtain
|aj | · |θi |dj >
P
|ak | · |θi |dk .
(3)
k6=j
On the other hand, since f (θi ) = 0 we must have
P
P
0 ≥ |aj | · |θi |dj − | ak θidk | ≥ |aj | · |θi |dj −
|ak | · |θi |dk ,
k6=j
k6=j
which contradicts (3). The conclusion follows now by Lemma 2.1. Proof of Theorem 1.4. In this case our assumption on the size of |an | forces
all the θi ’s to have absolute value smaller than |m| − |q|, for otherwise, if |θj | ≥
|m| − |q| for an index j ∈ {1, . . . , n}, we would have
n
n−1
n−1
X
X
X
0=
ai θji−n ≥ |an | −
|ai | · |θj |i−n ≥ |an | −
|ai | · (|m| − |q|)i−n ,
i=0
i=0
i=0
a contradiction. Pn−1
Proof of Corollary 1.6. Here we use the fact that 1 > i=0 (|m| − |q|)i−n
for |q| = 1 and |m| ≥ 3. Notice that in the statement of Corollary 1.6 one may
consider integer polynomials with coefficients of modulus at most 1 instead of
Littlewood polynomials. Irreducibility criteria
49
Proof of Proposition 1.7. In order to find information on the location of the
roots of f , we use now a classical result of Cowling and Thron (see [5], [6]):
Let P (z) = a0 z d0 + a1 z d1 + · · · + an z dn ∈ C[z] with all aj 6= 0, 0 = d0 <
d1 < · · · < dn , and mj = (dj − dj−1 )−1 , j = 1, 2, . . . , n. Let µ0 = 0, µn = 1 and
µ1 , . . . , µn−1 be arbitrary positive constants. Then all the zeros of P lie in the
disc
mj
(1 + µj−1 ) |aj−1 |
|z| ≤ A = max
·
.
1≤j≤n
µj
|aj |
Indeed, if P would have one root z0 with |z0 | > A, then we would obtain
µ1 |a1 | · |z0 |d1 > (1 + µ0 )|a0 | · |z0 |d0
µ2 |a2 | · |z0 |d2 > (1 + µ1 )|a1 | · |z0 |d1
µ3 |a3 | · |z0 |d3 > (1 + µ2 )|a2 | · |z0 |d2
······
µn |an | · |z0 |
dn
> (1 + µn−1 )|an−1 | · |z0 |dn−1 ,
which after summation
Pn−1and cancellation of equal terms on each side would imply
that |an | · |z0 |dn > i=0 |ai | · |z0 |di . On the other hand, since P (z0 ) = 0, we must
Pn−1
have |an | · |z0 |dn ≤ i=0 |ai | · |z0 |di , which is a contradiction.
This result shows that the roots of our polynomial f satisfy |θi | ≤ A for
i = 1, . . . , dn , and the conclusion follows by Lemma 2.1. Proof of Proposition 1.8. We will prove here that the roots of f satisfy
|θi | ≥ B = min
1≤j≤n
µj−1 |aj−1 |
(1 + µj )|aj |
1
j −dj−1
d
,
uniformly for i = 1, . . . , dn . To see this, let us assume that |θi | < B for some
index i. Then we obtain successively
(1 + µ1 )|a1 | · |θi |d1 < µ0 |a0 | · |θi |d0
(1 + µ2 )|a2 | · |θi |d2 < µ1 |a1 | · |θi |d1
(1 + µ3 )|a3 | · |θi |d3 < µ2 |a2 | · |θi |d2
······
(1 + µn )|an | · |θi |dn < µn−1 |an−1 | · |θi |dn−1 .
Recalling that µ0 = 1 and µn = 0, adding term by term these P
inequalities and
n
canceling the equal terms on both sides, we find that |a0 |·|θi |d0 > j=1 |aj |·|θi |dj .
P
n
On the other hand, since f (θi ) = 0 we must have |a0 | · |θi |d0 ≤ j=1 |aj | · |θi |dj ,
which is a contradiction.
Let us assume now as in the proof of Lemma 2.1 that f decomposes as f =
f1 f2 , with deg f1 ≥ 1 and deg f2 ≥ 1. Then we obtain |f1 (m)| ≤ |q|, while the
50
A.I.Bonciocat, N.C. Bonciocat, and A. Zaharescu
roots of f1 satisfy
|m − θi | ≥ |θi | − |m| ≥ B − |m| > |q|,
i = 1, . . . , t,
which by (1) gives the contradiction |f1 (m)| > |q| and completes the proof. Proof of Proposition 1.9. For the proof we use the following classical result
given in [11]:
If µ = (µ1 , µ2 , . . . , µn ) is an arbitrary set of positive numbers, then all the
characteristic roots of the n×n complex matrix M = (aij ) lie in the disk |z| ≤ Aµ
where
n
X
µj
|aij |.
(4)
Aµ = max
1≤i≤n
µ
j=1 i
Indeed, for any characteristic root λ of M the system of equations
n
X
aij xj = λxi ,
i = 1, 2, . . . , n
(5)
j=1
has a non-trivial solution (x1 , x2 , . . . , xn ). Let us set xj = µj yj and denote by
ym the yj of maximum modulus. By the mth equation of (5) we then infer that


n
n
X
X
|λµm ym | ≤
|amj |µj |yj | ≤ 
|amj |µj  |ym |.
j=1
j=1
Hence, |λ| ≤ Aµ .
If we apply this result to the companion matrix of the polynomial f¯(X) =
1
an f (X):



Mf¯ = 


0
0
·
0
− aan0
1
0
·
0
− aan1
0
1
·
0
− aan2
···
0
···
0
···
·
···
0
· · · − an−2
an
0
0
·
1
− an−1
an



,


we find that all the roots of f lie in the disk


n
µ µ
X
µn
µj |aj−1 | 
2
3
, ,...,
,
·
,
|z| ≤ A = max
 µ1 µ2
µn−1 j=1 µn |an | 
so the roots of f1 satisfy
|m − θi | ≥ |m| − |θi | ≥ |m| − A > |q|,
i = 1, . . . , t,
which by (1) gives the contradiction and completes the proof. Irreducibility criteria
51
Proof of Proposition 1.10. For the proof we use in this case a classical result
of Ballieu (see [1], [9]) on the location of the roots of a complex polynomial:
Let P (z) = a0 + a1 z + · · · + an z n ∈ C[z] with a0 an 6= 0 and let µ0 = 0 and
µ1 , . . . , µn be arbitrary positive constants. Then all the roots of P lie in the disc
µj
µn
|aj |
|z| ≤ A = max
+
·
.
0≤j≤n−1 µj+1
µj+1 |an |
This result follows immediately by using (4) for the transposed of Mf¯.
Using again the same notations as in the proof of Lemma 2.1, we have
|f1 (m)| ≤ |q|, while the roots of f1 satisfy
|m − θi | ≥ |m| − |θi | ≥ |m| − A > |q|,
i = 1, . . . , t,
which by (1) gives the desired contradiction. The last result we will prove is the following variant of Lemma 2.1, which
allows one to consider higher derivatives in the study of the total number of
irreducible factors of a given polynomial. Here for a polynomial f ∈ Z[X] the
notation Ω(f ) ≤ n means that f is a product of at most n irreducible factors over
Q, counting multiplicities.
Lemma 2.3. Let f ∈ Z[X] and suppose that f (m) = ps · q for some integers
m, s, q and a prime number p, with s ≥ 1 and p - q. Suppose also that p - f (i) (m)
for some i ≥ 1. If for two positive real numbers A and B we have A < |m| − |q| <
|m| + |q| < B, and f has no roots in the annular region A < |z| < B, then
Ω(f ) ≤ min(i, s).
Proof: We will first prove that for each factor f1 of f we must have |f1 (m)| > q.
To see this, let us assume that f factors as f (X) = an (X − θ1 ) . . . (X − θn ), with
θ1 , . . . , θn ∈ C. This implies that f1 , as a factor of f , must factor over C as
f1 (X) = bt (X − θ1 ) . . . (X − θt ), say, with t ≥ 1 and |bt | ≥ 1. Then, like in the
proof of Lemma 2.1 one has
|f1 (m)| = |bt | ·
t
Y
|m − θi | ≥
i=1
t
Y
|m − θi |.
(6)
i=1
The fact that the roots of f lie outside the annulus A < |z| < B shows that for
each index i ∈ {1, . . . , t} we either have
|m − θi | ≥ |m| − |θi | ≥ |m| − A,
if |θi | ≤ A,
|m − θi | ≥ |θi | − |m| ≥ B − |m|,
if |θi | ≥ B.
or
Since by hypothesis we have A < |m| − |q| < |m| + |q| < B, we conclude that
|m − θi | > |q| for each i = 1, . . . , t, so by (6) we obtain |f1 (m)| > |q|, as desired.
Moreover, this shows that in fact f1 (m) must be divisible by p, so Ω(f ) ≤ s.
52
A.I.Bonciocat, N.C. Bonciocat, and A. Zaharescu
We will use now the condition on the ith derivative of f to prove that f cannot
be written as a product of more than i irreducible factors. Assume by contrary
that f decomposes as f = f1 · · · fk with f1 , . . . , fk irreducible over Q and k > i,
(i )
(i )
and notice that in the expansion of f (i) the terms f1 1 · · · fk k may be indexed by
partitions of i: i1 + · · · + ik = i, i1 ≥ 0, . . . , ik ≥ 0. Since k > i, each term in this
(i )
expansion must contain at least one factor fl l with il = 0, so by the argument
(i1 )
(ik )
above, each term f1 (m) · · · fk (m) must be divisible by p, which contradicts
our hypothesis that p - f (i) (m) and completes the proof of the lemma.
Using Lemma 2.3, one may easily rephrase the irreducibility criteria stated so
far, to derive bounds for the total number of irreducible factors for polynomials
that take a value divisible by a prime power. We end by noting that one may also
easily formulate separability criteria, by applying the irreducibility conditions in
the results above to the derivative of a given polynomial.
3
Examples.
i) Let f (X) = X 7 +4X 6 +3X 5 +4X 4 +8X 3 +9X 2 +7 and note that f is obtained
by replacing the powers of 10 with the powers of X in the decimal representation
of 315 . Since 3 - f 0 (10) = 9568580, f is irreducible by Theorem 1.1.
ii) Let f (X) = ps q+a1 X+a2 X 2 +. . .+an X n ∈ Z[X],
Pn with p a prime number, q
a nonzero integer, s ≥ 2, an 6= 0 and p - qa1 . If ps > i=1 |ai |·|q|i−1 , then f must
be irreducible over Q. This follows immediately by taking m = 0 in Theorem 1.2.
One such polynomial is for instance f (X) = 250 + 2X + 3X 2 − 3X 3 + 5X 4 + 4X 5 .
P5
Here we have p = 5, s = 3, q = 2 and 125 > i=1 |ai |2i−1 = 124, so f is an
irreducible polynomial.
iii) Let f (X) = 1 − 2X − X 2 + X 3 + 424X 4 − X 5 − X 6 − X 7 and note
that f (2) = 38 , a prime power. Here m = 2, q = 1, f 0 (2) = 12 854 is not
divisible by 3, and
P we may take j = 4 in Theorem 1.3, since 424 = |a4 | >
(|m| + |q|)d7 −d4 · i6=4 |ai | = 33 · 8 = 216. We therefore conclude that f is an
irreducible polynomial.
iv) Let us take f (X) = 3967 − 401X − 251X 2 − 171X 3 − 14X 4 − 5X 5 . Here
P5
P5
P5
5
i
i=1 iai = −1497, which is not
i=0 ai = 3125 = 5 , |a0 | >
i=1 |ai |2 and
divisible by 5, so f is irreducible by Corollary 1.5.
v) Let f (X) = −1 + X − X 2 − X 3 − X 4 + X 5 . Here we have f (3) = 27 and
2 - f 0 (3) = 265, so f is irreducible by Corollary 1.6.
vi) If we take µj P
= 1 for j = 1, . . . , n in Proposition 1.7, we see that a
n
i
polynomial f (X) =
i=0 ai X ∈ Z[X] with a0 a1 . . . an 6= 0, |a0 | < |a1 | and
2|aj−1 | < |aj | for j = 2, 3, . . . , n is irreducible over Q if f (m) is a prime power ps ,
s ≥ 2 for an integer m with |m| ≥ 2 and p - f 0 (m). One such polynomial is for
instance f (X) = 1 − 2X − 5X 2 + 11X 3 + 24X 4 + 61X 5 , since f (2) = 2401 = 74 ,
and 7 - f 0 (2) = 5758.
Irreducibility criteria
53
vii) From Proposition
µ1 = µ2 = . . . = µn = 1 it follows that
Pn 1.9 with
i
a polynomial f (X) =
a
X
∈
Z[X] with a0 an 6= 0, |an | < |a0 | + |a1 | +
i=0 i
· · · + |an−1 | and such that f (m) = ps q for integers m, p, s, q with p a prime
number, s ≥ 2, |m| > |q| + (|a0 | + |a1 | + · · · + |an−1 |)/|an |, and p - qf 0 (m)
must be irreducible over Q. For such an example, take for instance f (X) =
−7 − 7X + 6X 2 + 9X 3 + 3X 4 + 8X 5 and m = 10. Here f (10) = 839 523 = 3 · 234 ,
so if we let p = 23 and q = 3 we see that p - qf 0 (10) = 32 · 7 · 19 753, hence f must
be irreducible.
viii) For an example related to Corollary 1.11, let us consider now the polynomial f (X) = 6−5X +5X 2 +X 3 +9X 4 +2X 5 +6X 6 . Here f (10) = 6 291 456 = 221 3
and f 0 (10) = 3 736 395, so we may take m = 10, p = 2 and q = 3. Since p - qf 0 (m),
and |m| − |q| = 7 > max (1 + |ai |/|a6 |) = 5/2, f is an irreducible polynomial.
0≤i≤5
Acknowledgements This work was partially supported by CNCSISUEFISCSU, project PNII-IDEI 51, code 304/2007.
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A.I.Bonciocat, N.C. Bonciocat, and A. Zaharescu
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Received: 04.09.2010
Accepted: 10.10.2010.
A.I.Bonciocat, N.C. Bonciocat:
Institute of Mathematics of the Romanian Academy,
P.O. Box 1-764,
Bucharest 014700 Romania
E-mail: [email protected]
[email protected]
A. Zaharescu:
Department of Mathematics,
University of Illinois at Urbana-Champaign,
Altgeld Hall, 1409 W. Green Street,
Urbana, IL, 61801, USA.
E-mail: [email protected]