Journal de Théorie des Nombres de Bordeaux

Stanley J. Gurak

Kloosterman sums for prime powers in <i>P</i>-adic fields

Kevin G. Hare

David McKinnon

Christopher D. Sinclair

Patterns and periodicity in a family of resultants

Aristides Kontogeorgis

Field of moduli versus field of definition for cyclic covers of the projective line

Aleksandar Ivić

Aleksandar IVIĆ
On the mean square of the divisor function in short intervals
Tome 21, no 2 (2009), p. 251-261.
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Journal de Théorie des Nombres
de Bordeaux 21 (2009), 251-261
On the mean square of the divisor function in
short intervals
par Aleksandar IVIĆ
Résumé. On donne des estimations pour la moyenne quadratique
de ∫ 2X
X
(∆k(x + h)−∆k(x))2 dx,
où h = h(X)  1, h = o(x) quand X →∞ et h se trouve dans un
intervalle convenable. Pour k > 2 un entier fixé, ∆k(x) et le terme
d’erreur pour la fonction sommatoire de la fonction des diviseurs
dk(n), generée par ζk(s).
Abstract. We provide upper bounds for the mean square inte-
gral ∫ 2X
X
(∆k(x + h)−∆k(x))2 dx,
where h = h(X)  1, h = o(x) as X → ∞ and h lies in a
suitable range. For k > 2 a fixed integer, ∆k(x) is the error term in
the asymptotic formula for the summatory function of the divisor
function dk(n), generated by ζk(s).
1. Introduction
Let, for a fixed integer k > 2, dk(n) denote the (generalized) divisor func-
tion which denotes the number of ways n can be written as a product of k
factors. This is a well-known multiplicative function (dk(mn) = dk(m)dk(n)
for coprime m,n ∈ N, and d2(n) ≡ d(n) is the classical number of divisors
function). Besides this definition one has the property that dk(n) is gener-
ated by ζk(s), where ζ(s) is the Riemann zeta-function, defined as
ζ(s) =
∞∑
n=1
n−s (σ = <e s > 1),
and otherwise by analytic continuation. Namely
ζk(s) =
( ∞∑
n=1
n−s
)k
=
∞∑
n=1
dk(n)n−s (σ = <e s > 1),
252 Aleksandar Ivić
and this connects the problems related to dk(n) to zeta-function theory.
Let as usual
(1.1)
∑
n6x
dk(n) = Res
s=1
xsζk(s)
s
+ ∆k(x) = xPk−1(log x) + ∆k(x),
where Pk−1(t) is a polynomial of degree k − 1 in t, all of whose coef-
ficients can be calculated explicitly (e.g., P1(t) = log t + 2γ − 1, where
γ = −Γ′(1) = 0.577215 . . . is Euler’s constant). Thus ∆k(x) represents the
error term in the asymptotic formula for the summatory function of dk(n),
and a vast literature on this subject exists (see e.g., [2] or [6]). Here we
shall be concerned with the “short difference”
(1.2) ∆k(x + h)−∆k(x) (1  h  x, h = o(x) as x →∞).
The meaning of “short” comes from the condition h = o(x) as x → ∞, so
that h is indeed much smaller in comparison with x.
As in analytic number theory one is usually interested in the averages
of error terms, where the averaging usually smoothens the irregularities
of distribution of the function in question, we shall be interested in mean
square estimates of (1.2), both discrete and continuous. To this end we
therefore define the following means (to stress the analogy between the
discrete and the continuous, x is being kept as both the continuous and the
integer variable):
(1.3)
∑
k
(X, h) :=
∑
X6x62X
(
∆k(x + h)−∆k(x)
)2
,
(1.4) Ik(X, h) :=
∫ 2X
X
(
∆k(x + h)−∆k(x)
)2
dx.
The problem is then to find non-trivial upper bounds for (1.3)–(1.4), and to
show for which ranges of h = h(k, X) (= o(X) ) they are valid. Theorem 1
(see Section 2) provides some results in this direction. Namely the “trivial”
bound in all cases is the function X1+εh2, where here and later ε > 0
denotes arbitrarily small positive constants, not necessarily the same ones
at each occurrence. This comes from the elementary bound dk(n) ε nε
and (1  h  x)
∆k(x + h)−∆k(x) =
∑
x<n6x+h
dk(n) + xPk−1(log x)(1.5)
− (x + h)Pk−1(log(x + h))
ε xε
∑
x<n6x+h
1 + h logk−1 x ε xεh,
where we used the mean value theorem. Here and later ε (> 0) denotes ar-
bitrarily small constants, not necessarily the same ones at each occurrence,
Mean square of the divisor function in short intervals 253
while a ε b (same as a = Oε(b)) means that the implied constant depends
on ε. Note that, from the work of P. Shiu [5] on multiplicative functions,
one has the bound∑
x<n6x+h
dk(n)  h logk−1 x (xε 6 h 6 x),
hence in this range (1.5) can be improved a bit, and thus we can also
consider
(1.6) ∆k(x + h)−∆k(x)  h logk−1 x (xε 6 h 6 x)
as the “trivial” bound. It should be mentioned that the cases k = 2 of
(1.3) and (1.4) have been treated by Coppola–Salerno [1] and M. Jutila [4],
respectively, so that we shall concentrate here on the case when k > 2. In
these papers it had been shown that, for Xε 6 h 6 12
√
X, L = log X,
(1.7)
∑
2
(X, h) =
8
π2
Xh log3
(√X
h
)
+ O(XhL5/2
√
L),
I2(X, h) =
1
4π2
∑
n6 X
2h
d2(n)
n3/2
2X∫
X
x1/2
∣∣∣∣exp(2πih√nx
)
− 1
∣∣∣∣2 dx(1.8)
+ Oε(X1+εh1/2).
From (1.8) Jutila deduces (a  b means that a  b  a) that
(1.9) I2(X, h)  Xh log3
(√X
h
)
(Xε 6 h 6 X1/2−ε),
but it is not obvious that I2(X, h) is asymptotic to the main term on the
right-hand side of (1.7). This, however, is certainly true, and will follow
from our Theorem 2 (see Section 2) and from (1.7). Theorem 2 says that,
essentially, the sums
∑
k(X, h) and Ik(X, h) are of the same order of mag-
nitude. It is also true that, for Xε 6 h 6 12
√
X, L = log X,
(1.10)
∫ 2X
X
(
E(x + h)−E(x)
)2
dx =
8
π2
Xh log3
(√X
h
)
+ O(XhL5/2
√
L),
implying in particular that
(1.11) E(x + h)− E(x) = Ω
{
√
h log3/2
(√x
h
)}
(xε 6 h 6 x1/2−ε).
Here, as usual,
E(T ) :=
∫ T
0
|ζ(12 + it)|
2 dt− T
(
log
T
2π
+ 2γ − 1
)
254 Aleksandar Ivić
represents the error term in the mean square formula for |ζ(12 + it)| (see
e.g., Chapter 15 of [2] for a comprehensive account), while f(x) = Ω(g(x))
means that limx→∞ f(x)/g(x) 6= 0. Namely Jutila (op. cit.) has shown that
the integral in (1.10) equals the expression on the right-hand side of (1.8),
hence the conclusion follows from Theorem 2 and the above discussion. The
omega-result (1.11) shows that the difference E(x + h) − E(x) cannot be
too small in a fairly wide range for h. An omega-result analogous to (1.11)
holds for ∆(x + h)−∆(x) as well, namely
(1.12) ∆(x + h)−∆(x) = Ω
{
√
h log3/2
(√x
h
)}
(xε 6 h 6 x1/2−ε).
Concerning the true order of ∆k(x + h)−∆k(x), we remark that on the
basis of (1.9) M. Jutila [4] conjectured that
(1.13) ∆(x + h)−∆(x) ε xε
√
h (xε 6 h 6 x1/2−ε),
which would be close to best possible, in view of (1.12). The range xε 6
h 6 x1/2−ε is essential here, since for h much larger than x1/2, one expects
∆(x + h) and ∆(x) to behave like independent random variables, and in
that case the quantities in question may not be “close” to one another.
Perhaps one has (1.13) for ∆(x) replaced by ∆k(x) in a suitable range of
h as well, but this is a difficult question.
Further sharpenings of (1.7), (1.9) and (1.10) were recently obtained by
the author in [3]. Namely, for 1  U = U(T ) 6 12
√
T we have (c3 = 8π−2)
∫ 2T
T
(
∆(x + U)−∆(x)
)2
dx = TU
3∑
j=0
cj logj
(√T
U
)(1.14)
+ Oε(T 1/2+εU2) + Oε(T 1+εU1/2),
and the result remains true if ∆(x+U)−∆(x) is replaced by E(x+U)−E(x)
(with different cj ’s). From (1.14) and Theorem 2 one obtains that a formula
analogous to (1.14) holds also for
∑
2(X, h) in (1.7).
2. Statement of results
First we define σ(k) as a number satisfying 12 6 σ(k) < 1, for which
(2.1)
∫ T
0
|ζ(σ(k) + it)|2k dt ε T 1+ε
holds for a fixed integer k > 2. From zeta-function theory (see [2], and in
particular Section 7.9 of [6]) it is known that such a number exists for any
Mean square of the divisor function in short intervals 255
given k ∈ N, but it is not uniquely defined, as one has
(2.2)
∫ T
0
|ζ(σ + it)|2k dt ε T 1+ε (σ(k) 6 σ < 1).
From Chapter 8 of [2] it follows that one has σ(2) = 12 , σ(3) =
7
12 , σ(4) =
5
8 , σ(5) =
41
60 etc., but it is not easy to write down (the best known value of)
σ(k) explicitly as a function of k. Note that the famous, hitherto unproved
Lindelöf hypothesis that ζ(12 + it) ε |t|
ε is equivalent to the fact that
σ(k) = 12 (∀k ∈ N). Our aim is to find an upper bound for Ik(X, h) in (1.4)
which is better than the the trivial bound Oε(X1+εh2) by the use of (1.5),
or O(Xh2 log2k−2 X), by the use of (1.6). Now we can formulate
Theorem 1. Let k > 3 be a fixed integer. If σ(k) = 12 then, for X
ε 6 h =
h(X) 6 X1−ε,
(2.3)
∫ 2X
X
(
∆k(x + h)−∆k(x)
)2
dx ε X1+εh4/3.
If 12 < σ(k) < 1, and θ(k) is any constant satisfying 2σ(k)− 1 < θ(k) < 1,
then there exists ε1 = ε1(k) > 0 such that, for Xθ(k) 6 h = h(X) 6 X1−ε,
(2.4)
∫ 2X
X
(
∆k(x + h)−∆k(x)
)2
dx ε1 X1−ε1h2.
Corollary 1. If the Lindelöf hypothesis is true, then the bound (2.3) holds
for all k.
Corollary 2. From the known values of σ(k) mentioned above it transpires
that one may unconditionally take θ(3) = 16 + ε, θ(4) =
1
4 + ε, θ(5) =
11
30 + ε,
etc.
Remark 1. The bound in (2.3) is fairly sharp, while the bound in (2.4) is
a little better than the trivial bound Xh2(log X)2k−2 (cf. (1.6)).
Remark 2. Theorem 1 holds also for k = 2, but in this case a sharper
result follows from (1.9) in the range Xε 6 h 6 X1/2−ε. It could be true
that, for k > 2 fixed, the weak analogue of (1.7), namely the bound X1+εh
holds in a suitable range for h (depending on k), but this seems unattainable
at present.
Remark 3. Note that the integrals in (2.3)–(2.4) are trivially bounded by
X1+2βk+ε, where as usual
βk := inf
{
bk :
∫ X
1
∆2k(x) dx  X1+2bk
}
for fixed k > 2. It is known (see Chapter 13 of [2]) that βk > (k − 1)/(2k)
for every k, βk = (k − 1)/(2k) for k = 2, 3, 4, β5 6 9/20 (see W. Zhang
[7]), β6 6 12 , etc. This gives an insight when Theorem 1 gives a non-trivial
result.
256 Aleksandar Ivić
Our second result is primarily a technical one. It establishes the connec-
tion between the discrete means
∑
k(X, h) (see (1.3)) and its continuous
counterpart Ik(X, h) (see (1.4)), precisely in the range where we expect the
∆-functions to be close to one another. This is
Theorem 2. For 1  h = h(X) 6 12
√
X we have
(2.5)
∑
2
(X, h) = I2(X, h) + O(h5/2 log5/2 X),
while, for a fixed integer k > 3,
(2.6)
∑
k
(X, h) = Ik(X, h) + Oε(Xεh3).
3. Proof of Theorem 1
We start from Perron’s classical inversion formula (see e.g., the Appendix
of [2]). Since dk(n) ε nε, this yields
(3.1)
1
2πi
∑
n6x
dk(n) =
∫ 1+ε+iT
1+ε−iT
xs
s
ζk(s) ds + Oε(X1+εT−1),
where X 6 x 6 2X, and T is parameter satisfying 1  T  X that will
be suitably chosen a little later. We replace the segment of integration by
the contour joining the points
1 + ε− iT, σ(k)− iT, σ(k) + iT, 1 + ε + iT.
In doing this we encounter the pole of ζk(s) at s = 1 of order k, and
the residue at this point will furnish xPk−1(log x), the main term in (1.1).
Hence by the residue theorem (3.1) gives, applied once with x and once
with x + h,
∆k(x + h)−∆k(x) =
1
2πi
∫ σ(k)+iT
σ(k)−iT
(x + h)s − xs
s
ζk(s) ds(3.2)
+ Oε(X1+εT−1) + O(Rk(x, T )),
where
(3.3) Rk(x, T ) :=
1
T
∫ 1+ε
σ(k)
xα|ζ(α + iT )|k dα.
Mean square of the divisor function in short intervals 257
By using the Cauchy-Schwarz inequality for integrals, (2.1) and (2.2), it
follows that
∫ 2T0
T0
Rk(x, T ) dT 
1
T0
∫ 1+ε
σ(k)
xα
(∫ 2T0
T0
|ζ(α + iT )|k dT
)
dα
(3.4)
ε
X1+ε
T0
sup
σ(k)6α61+ε
(
T0
∫ 2T0
T0
|ζ(α + iT )|2k dT
)1/2
ε X1+ε.
Therefore (3.4) implies that there exists T ∈ [T0, 2T0] such that
(3.5) Rk(x, T ) < cX1+εT−10
for a suitable c > 0, uniformly in X 6 x 6 2X. It is this T that we initially
take in (3.2)–(3.3), and using
(x + h)s − xs
s
=
∫ h
0
(x + v)s−1 dv (s 6= 0)
we obtain from (3.2) and (3.5)
(3.6)
∆k(x+h)−∆k(x) =
1
2πi
∫ σ(k)+iT
σ(k)−iT
∫ h
0
(x+v)s−1 dv ζk(s) ds+Oε(X1+εT−10 ).
On squaring (3.6) and integrating over x, we obtain
(3.7)
∫ 2X
X
(∆k(x + h)−∆k(x))2 dx
ε
∫ 2X
X
∣∣∣∫ T
−T
∫ h
0
(x + v)σ(k)−1+itζk(σ(k) + it) dv dt
∣∣∣2 dx + X3+εT−20 .
Let now ϕ(x) (> 0) be a smooth function supported in [X/2, 5X/2], such
that ϕ(x) = 1 when X 6 x 6 2X and ϕ(r)(x) r X−r (r = 0, 1, 2, . . . ). In
the integrals under the absolute value signs in (3.7) we exchange the order
of integration and then use the Cauchy-Schwarz inequality for integrals.
We infer that the integral on the right-hand side of (3.7) does not exceed
h
∫ 5X/2
X/2
ϕ(x)
∫ h
0
∣∣∣∫ T
−T
(x + v)σ(k)−1+itζk(σ(k) + it) dt
∣∣∣2 dv dx
= h
∫ h
0
∫ T
−T
∫ T
−T
ζk(σ(k) + it)ζk(σ(k)− iy)J dy dt dv,
say, where
J = Jk(X; v, t, y) :=
∫ 5X/2
X/2
ϕ(x)(x + v)2σ(k)−2(x + v)i(t−y) dx.
258 Aleksandar Ivić
Integrating by parts we obtain, since ϕ(X/2) = ϕ(5X/2) = 0,
J =
−1
i(t− y) + 1
∫ 5X/2
X/2
(x + v)2σ(k)−1+i(t−y)
(
ϕ′(x) +
2σ(k)− 2
x + v
ϕ(x)
)
dx.
By repeating this process it is seen that each time our integrand will be
decreased by the factor of order
 X
|t− y|+ 1
· 1
X
ε X−ε
for |t−y| > Xε. Thus if we fix any ε, A > 0, the contribution of |t−y| > Xε
will be  X−A if we integrate by parts r = r(ε, A) times. For |t− y| 6 Xε
we estimate the corresponding contribution to J trivially to obtain that the
integral on the right-hand side of (3.7) is
ε h2X2σ(k)−1
∫ T
−T
∫ T
−T,|t−y|6Xε
|ζ(σ(k) + it)ζ(σ(k) + iy)|k dy dt
ε h2X2σ(k)−1
∫ T
−T
|ζ(σ(k) + it)|2k
(∫ t+Xε
t−Xε
dy
)
dt
ε h2X2σ(k)−1+εT,
where we used (2.1) and the elementary inequality
|ab| 6 12(|a|
2 + |b|2).
Since T0 6 T 6 2T0, it is seen that the left-hand side of (3.7) is
(3.8) ε Xε(h2T0X2σ(k)−1 + X3T−20 ).
If σ(k) = 12 , then
h2T0 + X3T−20 = 2Xh
4/3
with the choice T0 = Xh−2/3, which clearly satisfies 1  T0  X. Therefore
(2.3) follows from (3.7) and (3.8).
If 12 < σ(k) < 1, then we choose first
T0 = X1+εh−1,
so that the bound in (3.8) becomes
hX2σ(k)+2ε + X1−εh2  X1−εh2
for
h > X2σ(k)−1+3ε.
Therefore (2.4) follows if 0 < ε1 < 13(θ(k)− 2σ(k) + 1). This completes the
proof of Theorem 1.
Mean square of the divisor function in short intervals 259
4. Proof of Theorem 2
We may suppose that X > h (> 2) are integers, for otherwise note that
by replacing X with [X] in (2.5) and (2.6) we make an error which is, by
trivial estimation, ε h2Xε, and likewise for h. Write
(4.1) Ik(X, h) =
∫ 2X
X
(
∆k(x + h)−∆k(x)
)2
dx =
∑
X6m62X−1
Ik,h(m),
say, where for m ∈ N we set
Ik,h(m) :=
∫ m+1−0
m
(
∆k(x + h)−∆k(x)
)2
dx.
Recall that
∑
n6y
dk(n) =
∑
n6[y]
dk(n) (y > 1),
so that
∑
x<n6x+h
dk(n) =
∑
m<n6m+h
dk(n) (m 6 x < m + 1; m,h ∈ N).
If Qk−1 := Pk−1 + P ′k−1 (see (1.1)), then we have (for some 0 6 θ 6 1)
∆k(x + h)−∆k(x) =
∑
x<n6x+h
dk(n)
+ xPk−1(log x)− (x + h)Pk−1(log(x + h))
=
∑
m<n6m+h
dk(n)− hQk−1(log(x + θh))(4.2)
=
∑
m<n6m+h
dk(n)− hQk−1(log m)
+ O(h2X−1 logk−1 X)
= ∆k(m + h)−∆k(m) + O(h2X−1 logk−1 X),
260 Aleksandar Ivić
where we used the mean value theorem. By using (4.2) we see that the
left-hand side of (4.1) becomes
∑
X6m62X−1
Ik,h(m)
(4.3)
=
∑
X6m62X−1
m+1−0∫
m
(
∆k(m + h)−∆k(m) + O(h2X−1 logk−1 X)
)2
dx
=
∑
X6m62X
(∆k(m + h)−∆k(m))2 + Oε(Xεh2)
+ O
( ∑
X6m62X−1
|∆k(m + h)−∆k(m)|h2X−1 logk−1 X
)
+ O(h4X−1 log2k−2 X).
If k > 3, we use the trivial bound (1.5) to obtain∑
X6m62X−1
|∆k(m + h)−∆k(m)|h2X−1 logk−1 X ε h3Xε,
and (2.5) follows, since
h4X−1 log2k−2 X  h2 log2k−2 X (1  h = h(X) 6 12
√
X),
and the last term is smaller than all the error terms in (2.5) and (2.6).
If k = 2, then we apply the Cauchy-Schwarz inequality to the sum in the
first O-term in (4.3) and use (1.7). We obtain∑
X6m62X−1
|∆(m + h)−∆(m)|h2X−1 log X(4.4)
 X−1/2h2 log X
 ∑
X6m62X
(∆(m + h)−∆(m))2
1/2
 h5/2 log5/2 X,
and (2.5) follows from (4.3) and (4.4). This ends the proof of Theorem 2.
Mean square of the divisor function in short intervals 261
References
[1] G. Coppola, S. Salerno, On the symmetry of the divisor function in almost all short
intervals. Acta Arith. 113(2004), 189–201.
[2] A. Ivić, The Riemann zeta-function. John Wiley & Sons, New York, 1985 (2nd ed., Dover,
Mineola, N.Y., 2003).
[3] A. Ivić, On the divisor function and the Riemann zeta-function in short intervals. To
appear in the Ramanujan Journal, see arXiv:0707.1756.
[4] M. Jutila, On the divisor problem for short intervals. Ann. Univer. Turkuensis Ser. A I
186 (1984), 23–30.
[5] P. Shiu, A Brun-Titchmarsh theorem for multiplicative functions. J. Reine Angew. Math.
313 (1980), 161–170.
[6] E.C. Titchmarsh, The theory of the Riemann zeta-function (2nd ed.). University Press,
Oxford, 1986.
[7] W. Zhang, On the divisor problem. Kexue Tongbao 33 (1988), 1484–1485.
Aleksandar Ivić
Katedra Matematike RGF-a
Universitet u Beogradu, -Dušina 7
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Two remarks on the inverse Galois problem for intersective polynomials
Tome 21, no 2 (2009), p. 435-437.
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Journal de Théorie des Nombres
de Bordeaux 21 (2009), 435-437
Two remarks on the inverse Galois problem for
intersective polynomials
par Jack SONN
Résumé. Un polynôme (unitaire) f(x) ∈ Z[x] est dit intersectif
si la congruence f(x) ≡ 0 mod m a des solutions pour tout entier
positif m. On dit que f(x) est non-trivialement intersectif s’il est
intersectif et n’a pas de racine rationnelle. L’auteur a prouvé que
tout groupe fini G, résoluble non cyclique, peut être réalisé comme
groupe de Galois sur Q d’un polynôme non-trivialement intersec-
tif (non cyclique est une condition nécessaire). Notre premiere
remarque est l’observation que le résultat correspondant pour les
groupes non-résolubles G se ramène au problème de Galois inverse
ordinaire pour G sur Q. La seconde remarque concerne la rareté
d’exemples explicites de polynômes non-trivialement intersectifs
de groupe de Galois donné, et nous donnons le premier exemple
connu pour le groupe dihédral d’ordre dix.
Abstract. A (monic) polynomial f(x) ∈ Z[x] is called inter-
sective if the congruence f(x) ≡ 0 mod m has a solution for all
positive integers m. Call f(x) nontrivially intersective if it is in-
tersective and has no rational root. It was proved by the author
that every finite noncyclic solvable group G can be realized as
the Galois group over Q of a nontrivially intersective polynomial
(noncyclic is a necessary condition). Our first remark is the ob-
servation that the corresponding result for nonsolvable G reduces
to the ordinary inverse Galois problem for G over Q. The second
remark has to do with the scarcity of explicit examples of non-
trivial intersective polynomials with given Galois group, and gives
the first known example for the dihedral group of order ten.
A (monic) polynomial f(x) ∈ Z[x] is called intersective if the congru-
ence f(x) ≡ 0 mod m has a solution for all positive integers m. Call f(x)
nontrivially intersective if it is intersective and has no rational root. It was
proved by the author that every finite noncyclic solvable group G can be
realized as the Galois group over Q of a nontrivially intersective polyno-
mial (noncyclic is a necessary condition) [4, Thm 2.2]. This note makes the
observation that the corresponding result for nonsolvable G reduces to the
ordinary inverse Galois problem for G over Q.
436 Jack Sonn
Proposition. Let G be a finite nonsolvable group. The following are equiv-
alent:
(1) G is realizable as the Galois group of a nontrivially intersective
polynomial over Q
(2) G is realizable as a Galois group over Q.
Proof. (1)=⇒ (2) is trivial. Assume (2). As noted in [4], a monic polynomial
f(x) ∈ Z[x] has a root mod n for all n if and only if f(x) has a root in Qp
for all (finite) primes p. Let G = G(K/Q). By [4, Prop. 2.1], the following
are equivalent:
(i) K is the splitting field of a product f = g1 · · · gm of m irreducible
polynomials of degree greater than 1 in Q[x] and f has a root in Qp for all
primes p.
(ii) G is the union of the conjugates of m proper subgroups A1, ..., Am,
the intersection of all these conjugates is trivial, and for all primes p of K,
the decomposition group G(p) is contained in a conjugate of some Ai.
It therefore suffices to show that G has such a covering A1, ..., Am. Since
any decomposition group G(p) is necessarily solvable, hence a proper sub-
group of G, we can take for example A1, ..., Am to be the set of all proper
subgroups of G (obviously not the most economical choice). Clearly their
intersection is trivial, since for example a nonsolvable group has elements
of prime order for at least two distinct primes. The result then follows. 
An Example. In [2], (see also [1]), the polynomials (xr − 2)Φr(x), r ≥ 3
a prime (Φr(x) is the rth cyclotomic polynomial), with Galois groups the
Frobenius groups of order r(r − 1), are given as examples of polynomials
with no rational roots and roots mod p for all p. Examples of nontrivial
intersective polynomials are still rare. For instance the above examples do
not include the dihedral group D10 of order ten, which, like the above
examples, is a Frobenius group and in particular is the union of conjugates
of two proper subgroups, and where the desired polynomial should be a
product of two irreducible factors. The results in [4] are purely existence
theorems and do not give explicit polynomials. We now give an example
of a nontrivial intersective polynomial with Galois group D10 which is a
product of two irreducible factors.
In [3, p. 171], the polynomial f(x) = x5 +x4−5x3−4x2 +3x+1 is given
as an example of a polynomial with Galois group D10 over Q, with all real
roots. Its discriminant is 4012 (401 is a prime) hence the quadratic subfield
of its splitting field is Q(
√
401). We claim that all decomposition groups are
cyclic. Since 401 ≡ 1 (mod 8), the prime 2 splits completely in Q(
√
401),
so the only ramified prime in Q(
√
401) is 401. From the structure of D10
we see that the decomposition group at 401 is cyclic (of order 2). Secondly,
Two remarks on the inverse Galois problem for intersective polynomials 437
since the prime 2 splits completely in Q(
√
401), the decomposition group
at 2 is cyclic as well. In fact, f is irreducible mod 2, so 2 is unramified with
decomposition group cyclic of order 5. All other primes are unramified, so
all other decomposition groups are also cyclic. We may now apply condition
(ii) in the proof of the proposition above, with A1 the subgroup of order 5
and A2 of order 2. It follows that f(x)(x2 − 401) has a root in Qp for all p.
References
[1] D. Berend and Y. Bilu, Polynomials with roots modulo every integer. Proc. AMS 124
(1996), 1663–1671.
[2] R. Brandl, Integer polynomials with roots mod p for all primes p. J. Alg. 240 (2001),
822–835.
[3] C. Jensen, A. Ledet, and N. Yui, Generic Polynomials. Cambridge Univ. Press,
Cambridge-New-York, 2002.
[4] J. Sonn, Polynomials with roots in Qp for all p. Proc. AMS 136 (2008), 1955–1960.
Jack Sonn
Department of Mathematics
Faculty of Mathematics
Technion–Israel Institute of Technology
Haifa, Israel
E-mail : sonn@tx.technion.ac.il


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