Math 627
Homework #4
November 7, 2014
Due Monday, November 24
1. Let ζ = e2πi/7 , and let K = Q(ζ).
(a) Show that the free part of O× has rank 2.
(b) For j = 1, . . . , 6, let
uj =
1 − ζj
.
1−ζ
×
Show that uj ∈ OK
.
×
(c) Show that huj | 1 ≤ j ≤ 6i ⊆ OK
is a subgroup of finite index.
2. Let K be a number field with r1 real embeddings and r2 pairs of complex conjugate
embeddings, and let r = r1 + r2 . Suppose that {ε1 , . . . , εr−1 } forms a basis of the free
×
part of the unit group of OK
. I.e.,
m
×
r−1
1
OK
= {uεm
1 · · · εr−1 | u is a root of unity, m1 , . . . , mr−1 ∈ Z}.
Form the (r − 1) × r matrix,


log |σ1 (ε1 )| · · · log |σr1 (ε1 )|
2 log |σr1 +1 (ε1 )| · · · 2 log |σr (ε1 )|
 log |σ1 (ε2 )| · · · log |σr (ε2 )|
2 log |σr1 +1 (ε2 )| · · · 2 log |σr (ε2 )| 
1


M =
,
..
..
..
..


.
.
.
.
log |σ1 (εr−1 )| · · · log |σr1 (εr−1 )| 2 log |σr1 +1 (εr−1 )| · · · 2 log |σr (εr−1 )|
where as usual σ1 , . . . , σr1 are the real embeddings of K and σr1 +1 , . . . , σr are representatives of the pairs of complex conjugate embeddings of K.
(a) Let M 0 be the (r − 1) × (r − 1) matrix obtained by deleting any of the columns of
M . Show that RK := | det(M 0 )| is independent of the choice of deleted column.
(Hint: use the fact that the columns add up to 0.)
(b) Show that RK is independent of the choice of generators ε1 , . . . , εr−1 .
×
(c) Show that RK 6= 0. (Hint: show that it is the covolume of the lattice L(OK
)⊆
r−1
∼
W = R .)
√
√
3. Let K = Q( 3 2), and let OK = Z[ 3 2]. Factor (2), (3), (7), (29), and (31) into prime
ideals in OK .
1
Problems 4–7 are focused on number theoretic functions: let R be the set of all functions
f : N → C, where N = {1, 2, 3, . . .} is the set of positive integers.
4. Show that R is a commutative ring under the following operations + and ∗,
(f + g)(n) := f (n) + g(n),
X
(f ∗ g)(n) :=
f (n/d)g(d).
d|n
P
(Here ‘ d|n ’ means that we take the sum over all positive divisors d of n.) Show that
the function I : N → C defined by I(1) = 1 and I(n) = 0 for n 6= 1 is an identity
element with respect to ∗, so R is a ring with 1.
5. Let u : N → C be the function defined by u(n) = 1 for all n, and let µ : N → C be the
function defined as follows:


if n = 1,
1
r
µ(n) = (−1) if n = p1 · · · pr , with p1 , . . . , pr distinct primes,


0
if there is a prime p with p2 | n.
(a) Show that u and µ are multiplicative inverses with respect to ∗. The function µ
is also known as the Möbius function.
(b) Use this to recover the Möbius inversion formula: for f ∈ R, if we set
X
g(n) =
f (d),
d|n
then
f (n) =
X
µ(n/d)g(d).
d|n
6. Let D be the ring of formal Dirichlet series. That is,
)
(∞
X an
: an ∈ C .
D=
ns
n=1
Here “formal” means that we do not worry about whether or not the series converges,
so although s is supposed to be a complex number, we think of it purely as a variable
quantity. (And ns obeys all of the familiar rules of exponentiation: (ns )(ms ) = (nm)s ,
etc.)
(a) Convince yourself that D is a ring under addition and multiplication of series.
(b) Show that R is isomorphic to D as a ring under the map R → D given by
f 7→ Df (s) :=
∞
X
f (n)
n=1
2
ns
.
(c) Show that if ζ(s) =
P∞
1
n=1 ns
is the Riemann zeta function, then
∞
X µ(n)
1
=
.
ζ(s) n=1 ns
7. Recall that a function f ∈ R is multiplicative if f (1) = 1 and whenever gcd(m, n) = 1
we have f (mn) = f (m)f (n). Show that f is multiplicative if any only if Df (s) has an
Euler product:
Df (s) =
∞
X
f (n)
n=1
ns
=
Y
1 + f (p)p−s + f (p2 )p−2s + · · · ,
p
where the product runs over all primes p.
3