Serge Ballif
MATH 535 Homework 7
October 19, 2007
1. This exercise shows (part of) the fact every ring R can be considered as a subring of a ring with identity element.
We mimic putting the subring
i
into the full ring of Gaussian integers: On
S
:=
×
, define the operations
(
m, r
) + (
n, s
) := (
m
+
n, r
+
s
)
(
m, r
)(
n, s
) := (
mn, nr
+
ms
+
rs
)
.
a) Show associativity of multiplication.
b) Show one of the distributivity laws.
c) Find the identity element.
d) Give the embedding of
R
into
S
and check that it is a homomorphism.
(a)
(
(
m, r
)(
n, s
)
)
(
o, t
) = (
mn, nr
+
ms
+
rs
)(
o, t
)
= (
mno, o
(
nr
+
ms
+
rs
) +
mn
(
t
) + (
nr
+
ms
+
rs
)
t
)
= (
mno, onr
+
oms
+
ors
+
mnt
+
nrt
+
mst
+
rst
)
= (
mno, no
(
r
) +
m
(
os
+
nt
+
st
) +
r
(
os
+
nt
+
st
)
= (
m, r
)(
no, os
+
nt
+
st
)
= (
m, r
)
(
(
n, s
)(
o, t
)
)
.
(b)
(
(
m, r
) + (
n, s
)
)
(
o, t
) = (
m
+
n, r
+
s
)(
o, t
)
=
(
(
m
+
n
)
o, o
(
m
+
n
) + (
m
+
n
)
t
+ (
r
+
s
)
t
)
= (
mo
+
no, om
+
mt
+
rt
+
on
+
nt
+
st
)
= (
mo, om
+
mt
+
rt
) + (
no, on
+
nt
+
st
)
= (
m, r
)(
o, t
) + (
n, s
)(
o, t
)
.
(c) We claim that the element (1
,
0) is the multiplicative identity of
S
.
We
verify this claim with the calculation (1
,
0)(
m, r
) = (1(
m
)
, m
(0) + 1(
r
) + 0(
r
)) =
(
m
(1)
,
1(
r
) +
m
(0) +
r
(0) = (
m, r
)(1
,
0).
(d) Let
j
:
R
→
S
be given by
j
(
r
) = (0
, r
).
Claim 1.
The map
j
is a ring homomorphism.
Proof.
j
(
rs
) = (0
, rs
) = (0
,
0(
r
) + 0(
s
) +
rs
) = (0
, r
)(0
, s
) =
j
(
r
)
j
(
s
).
j
(
r
+
s
) = (0
, r
+
s
) = (0 + 0
, r
+
s
) = (0
, r
) + (0
, s
) =
j
(
r
) +
j
(
s
).
Claim 2.
The map
j
is injective.
Proof.
Suppose that (0
, r
) =
j
(
r
) =
j
(0) = (0
,
0).
Then (0
, r
) = (0
,
0) is the
unique additive identity of
S
, so
r
= 0. Hence
j
is injective.
2. a) Let
a
1
, a
2
∈
A
, an (additively written) abelian group. Let
a
1
, a
2
have relatively prime (finite) orders
n
1
, n
2
.
Show that
a
1
+
a
2
has order
n
1
n
2
.
b) Conclude that if
G
1
and
G
2
are cyclic groups of relatively prime (finite) orders, then
G
1
×
G
2
is also cyclic.
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