INTEGERS 15 (2015)
#A4
ELEMENTARY ESTIMATES FOR THE NUMBER OF PURE
NUMBER FIELDS OF DEGREE p
Yusuke Fujisawa
Graduate School of Mathematics, Nagoya University, Furocho, Chikusa-ku,
Nagoya, Japan
fujisawa.gifu@gmail.com
Received: 11/27/13, Revised: 11/9/14, Accepted: 12/13/14, Published: 1/19/15
Abstract
Let p be an odd prime number, and X a large real number. In this note, we consider
the lower and upper bounds of the number of pure number fields of degree p with
the absolute values of discriminants at most X by elementary methods.
1. Introduction
Let n be a positive integer, X a large positive number, and Nn (X) the number of
number fields F of degree n with |d(F )|  X. Here d(F ) is the discriminant of a
number field F . A well-known conjecture asserts that
Nn (X) ⇠ cn X
for some cn . This conjecture has been proved for n = 2, 3, 4, and 5 ([3], [1], [2]).
However, this problem is very difficult and deep. In this paper, we consider the
distribution of pure number fields of odd prime degree by elementary methods.
Let p be an odd prime number, and F a number field of degree p. If there exists
p
a p-free positive integer n > 1 such that F = Q( p n), then we shall call F a pure
number field of degree p. (If there is no prime number l such that lk |n, then n is
said to be k-free.) For X > 0, we denote the number of pure number fields F of
degree p with |d(F )|  X by Pp (X).
Theorem 1. For an odd prime number p, we have
Bp p 1 1
Ap
X
 Pp (X) 
X
⇣(2)
⇣(p)
where ⇣(s) is the Riemann zeta function,
Ap =
pp+1
pp
(pp
1
+ pp
1)pp
2
1
,
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and
Bp =
1
(p + 1)p
p
p
2
1
+
p
(p + 1)p p
.
p
1
For example, P3 (X) is the number of pure cubic fields F with |d(F )|  X, and
p
1
37
p
X  P3 (X) 
X.
351⇣(3)
2 3⇣(2)
To show this theorem, we use two important lemmas.
First, we explain a result of Cohen and Robinson [4]. Let k, q > 1 be positive
integers, a 2 Z/qZ, and X a real positive large number. We set
Qk (X; a, q) = ]{n : k-free  X, n ⌘ a (mod q)}.
If there is a k-free positive integer n such that n ⌘ a (mod q), then n = cq + a for
some c 2 Z and gcd(a, q) must be k-free. Thus, since Qk (X; a, q) = 0 when gcd(a, q)
is not k-free, we assume gcd(a, q) is k-free.
A divisor d > 0 of q is called a unitary divisor when gcd(d, q/d) = 1. If d is a
unitary divisor of q, we write d|⇤ q. The largest unitary divisor of q which is a divisor
of a is denoted by (a, q)⇤ . Moreover, we denote the core of H by H0 . Namely, H0
is the largest square-free divisor of H.
In [4], Cohen and Robinson proved the following result.
Lemma 1. Let H := (a, q)⇤ . We have
Qk (X; a, q) =
p
q k 1 '⇤ (H0k /H) 1
k
X + O( X)
k
Jk (q) H0 /H ⇣(k)
where
k
Jk (n) = n
Y ✓
1
1
lk
l|n
l:prime
◆
and
Y ✓
' (n) = n
1
1
le
⇤
le |⇤ n
l:prime
◆
.
For example, if H = 1, then
Qk (X; a, q) =
✓
1 Y
1
q l|q
l:prime
1
lk
◆
1
p
1
k
X + O( X).
⇣(k)
Next lemma computes the discriminants of pure number fields.
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Lemma 2 (Fujisaki [5], p.133). Let p be an odd prime number, n a p-free positive
p
integer, and n0 the core of n. Put F = Q( p n). We have
(
p 1
( 1) 2 pp 2 (n0 )p 1 if np 1 ⌘ 1 mod p2 ,
d(F ) =
p 1
( 1) 2 pp (n0 )p 1 if np 1 6⌘ 1 mod p2 .
Since the author doesn’t find the proof of this lemma in the literature, and the
original proof was published in Japanese, we shall sketch the proof here. Every
Qp 1
p-free positive integer n can be put uniquely into the form n = i=1 aii where
Qp 1
ai 2 Z>0 and n0 = i=1 ai . Put
↵j =
pY1
ij [ ij ]p
ai p
i=1
! p1
for j = 0, 1, · · · , p 1. Let n be the Z-module generated by ↵0 , · · · , ↵p 1 . We can
check d(n) = (OF : n)2 d(F ) where OF is the ring of integers of F , and d(n) is the
discriminant of n as Z-module. By algebraic argument, we obtain
d(n) = ( 1)
p
1
2
pp (n0 )p
1
and
(
p if np
(OF : n) =
1 if np
1
1
⌘ 1 mod p2 ,
6⌘ 1 mod p2 .
For more details, refer to [5].
2. Proof
First, we consider the upper bound. It is obvious that
p
Pp (X)  ]{n : p-free > 1, |d(Q( p n))|  X}
2
=
p
X
a=1
p
]{n : p-free > 1, n ⌘ a(mod p2 ), |d(Q( p n))|  X}
2
=:
p
X
Pp,a (X).
a=1
If ap
1
⌘ 1(mod p2 ), we have
Pp,a (X) = ]{n : p-free > 1, n ⌘ a(mod p2 ), pp
2 p 1
n0
 X}
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by Lemma 2. Since n  np0
1
for p-free number n, we see
✓
◆
X
2
Pp,a (X)  Qp
; a, p
pp 2
in this case. Similarly, if ap
1
6⌘ 1(mod p2 ), we obtain
✓
◆
X
2
Pp,a (X)  Qp
; a, p
pp
by Lemma 2.
We can estimate Qp (X; a, p2 ) for a = 1, 2, · · · , p2 by using Lemma 1. Note that
(a, p2 )⇤ = p2 if a = p2 and (a, p2 )⇤ = 1 otherwise. Thus, we have
pp 2 1
X,
pp 1 ⇣(p)
Qp (X; a, p2 ) ⇠
when a = 1, · · · , p2
1, and
pp 2 1 1
X.
pp 1 ⇣(p)
Qp (X; p2 , p2 ) ⇠
Therefore, for a large number X, we obtain that
X
X
Pp (X) =
Pp,a (X) +
Pp,a (X)
ap

⇠
ap
ap
1 ⌘1(mod
X
1 ⌘1(mod
X
1 ⌘1(mod
p2 )
Qp
p2 )
✓
ap
X
pp
1 6⌘1(mod
; a, p2
2
◆
p2 )
ap
p2 )
pp 2 1 1 X
pp 1 ⇣(p) pp
✓
pp 2
= (p 1) p
(p
1)pp
✓ p+1
p 1
p
p
+ pp
=
(pp 1)pp
1 X
+
1 ⇣(p) pp 2
1 6⌘1(mod
X
p 2
p
pp
X
+
ap
Qp
p2 )
1 6⌘1(mod p2 )
a6=p2
✓
X
; a, p2
pp
◆
pp 2 1 X
pp 1 ⇣(p) pp
+
2
2
2
pp
+ (p
p) p
(p
◆
1
X
.
⇣(p)
2
pp
+
1)pp
(pp
2
1
1)pp
◆
X
⇣(p)
Thus, we have proved that Pp (X)  Ap X/⇣(p).
p
p
Finally, we consider the lower bound. Since Q( p n) 6= Q( p m) for distinct square-
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free numbers m and n, it is obvious that
p
]{n : square-free > 1, d(Q( p n))  X}
Pp (X)
2
=
p
X
a=1
p
]{n : square-free > 1, n ⌘ a(mod p2 ), d(Q( p n))  X}
2
=:
p
X
0
Pp,a
(X).
a=1
If ap
1
⌘ 1(mod p2 ), we have
0
Pp,a
(X) = ]{n : square-free > 1, n ⌘ a(mod p2 ), pp
by Lemma 2. Since np
np0
1
1
1
 X}
, we see
0
Pp,a
(X)
in this case. Similarly, if ap
2 p 1
n0
✓
Q2
X
pp
2
◆ p1 1
2
; a, p
6⌘ 1(mod p2 ), we obtain
0
Pp,a
(X)
Q2
✓
X
pp
◆ p1 1
; a, p2
!
!
by Lemma 2.
Since gcd(p2 , p2 ) = p2 is not square-free, we see that
Q2 (X; p2 , p2 ) = 0.
By Lemma 1, we have
Q2 (X; a, p2 ) ⇠
(p2
1
X
1)⇣(2)
for a = 1, 2, · · · , p2 1.
Therefore, for a large number X, we obtain that
!
✓
◆ p1 1
X
X
2
Pp (X)
Q2
; a, p +
pp 2
p 1
2
a
⇠ (p
=
1)
ap
⌘1(mod p )
1
1)⇣(2)
(p2
1
p
(p + 1)p p
2
1
+
✓
X
pp
2
◆ p1 1
p
(p + 1)p p
p
1
!
+ (p2
1
Xp 1
.
⇣(2)
p)
(p2
X
Q2
1 6⌘1(mod p2 )
a6=p2
1
1)⇣(2)
✓
X
pp
◆ p1 1
✓
X
pp
◆ p1 1
2
; a, p
!
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Thus, it holds that Pp (X)
tion.
Bp X p
1
1
/⇣(2). This completes the proof of the asser-
Acknowledgements. The author would like to thank the anonymous referee and
Professor Hiroshi Suzuki for carefully reading the first version of this paper and for
making helpful comments and suggestions.
References
[1] M. Bhargava, The density of discriminants of quartic rings and fields, Ann. of Math. 162
(2005), 1031–1063.
[2] M. Bhargava, The density of discriminants of quintic rings and fields, Ann. of Math. 172
(2010), 1559–1591.
[3] H. Davenport and H. Heilbronn, On the density of discriminants of cubic fields. II, Proc.
Royal Soc. London Ser. A 322 (1971), 405–420.
[4] E. Cohen and R. L. Robinson, On the distribution of the k-free integers in residue classes,
Acta Arith. 8 (1962/1963), 283–293. errata, ibid. 10 (1964/1965), 443.
[5] G. Fujisaki, daisuteki seisuron nyumon ge, (in Japanese) shokabo, 1975.