MAT 322: Algebra with Galois Theory
Professor Claus Sorensen
Problem Set 3
Rik Sengupta
October 9, 2009
1.
Show that
Q
/
Z
is isomorphic to
μ
∞
(
C
)
, the group of
z
∈
C
*
such that
z
n
= 1
for some positive integer
n
. Is it a ﬁnitely generated abelian group?
What is the torsion subgroup of
Q
/
Z
?
Solution.
We deﬁne a homomorphism from
Q
to
μ
∞
(
C
) by
ϕ
:
Q
→
μ
∞
(
C
), with the following mapping:
p
7→
e
2
πip
. This is obviously a ho-
momorphism, because
ϕ
(
p
+
q
) =
e
2
πi
(
p
+
q
)
=
e
2
πip
e
2
πiq
=
ϕ
(
p
)
ϕ
(
q
). In
fact, the image space of the homomorphism has been chosen because we
can simply raise the image of the homomorphism to the denominator of
the rational number in question, and then we will get 1, as follows:
[
ϕ
(
p
q
)]
q
=
e
2
πi
p
q
q
=
e
2
πip
= 1
Thus the image space is well-deﬁned, in the sense that all the images of
the map lie in
μ
∞
(
C
).
Note that the kernel of this homomorphism is the set of points
p
∈
Q
for
which
ϕ
(
p
) = 1, i.e.
e
2
πip
= 1. But this happens exactly when
p
∈
Z
. So,
the kernel of the homomorphism is simply
Z
.
Furthermore, the map is surjective, because given any element
z
∈
μ
∞
(
C
),
we know
z
n
= 1 for some
n
, and hence,
z
is one of the
n
th roots of unity.
So we can simply write
z
as
e
2
πik
for some rational
k
. Then,
k
is in
Q
, so
it is a preimage in the homomorphism, i.e. our map is surjective.
So then, from the First Isomorphism Theorem, we immediately get
Q
/
Z
is isomorphic to
μ
∞
(
C
).
It is not ﬁnitely generated. Suppose it is in fact a ﬁnitely generated group,
generated by
p
1
Z
,p
2
Z
,...p
r
Z
for ﬁnitely many rationals
p
i
. Then we can
simply take the denominators of all the
p
i
and ﬁnd some natural number
that is coprime to them all. Then, take any rational
s
whose denominator
in its lowest form is the natural number we just found. Then,
s
Z
is not
1