Proof
It is clear from the definition of the relation
∼
S
that (
r, s
)
∼
S
(
r, s
)
for all (
r, s
)
∈
R
×
S
. Moreover elements (
r, s
) and (
r
0
, s
0
) of
R
×
S
satisfy
(
r, s
)
∼
S
(
r
0
, s
0
) if and only if (
r
0
, s
0
)
∼
S
(
r, s
). Thus the relation
∼
S
on
R
×
S
is both reflexive and symmetric.
Let (
r, s
), (
r
0
, s
0
) and (
r
00
, s
00
) be elements of
R
×
S
. Suppose that (
r, s
)
∼
S
(
r
0
, s
0
) and (
r
0
, s
0
)
∼
S
(
r
00
, s
00
). Then there exist elements
u
and
v
of
S
such
that
us
0
r
=
usr
0
and
vs
00
r
0
=
vs
0
r
00
. Then
(
uvs
0
)(
s
00
r
)
=
(
vs
00
)(
us
0
r
) = (
vs
00
)(
usr
0
)
=
(
us
)(
vs
00
r
0
) = (
us
)(
vs
0
r
00
)
=
(
uvs
0
)(
sr
00
)
,
and
uvs
0
∈
S
, and therefore (
r, s
)
∼
S
(
r
00
, s
00
).
Thus the relation
∼
S
on
R
×
S
is transitive. It follows that this relation is an equivalence relation on
R
×
S
.
Lemma 2.39
Let
S
be a multiplicatively closed subset of a unital commu-
tative ring
R
, and let
∼
S
be the equivalence relation on
R
×
S
defined so
that elements
(
r, s
)
and
(
r
0
, s
0
)
of
R
×
S
satisfy
(
r, s
)
∼
S
(
r
0
, s
0
)
if and only if
there exists some element
u
of
S
for which
us
0
r
=
usr
0
. Let
(
r
1
, s
1
)
,
(
r
0
1
, s
0
1
)
,
(
r
2
, s
2
)
and
(
r
0
2
, s
0
2
)
be elements of
R
×
S
satisfying
(
r
1
, s
1
)
∼
S
(
r
0
1
, s
0
1
)
and
(
r
2
, s
2
)
∼
S
(
r
0
2
, s
0
2
)
. Then
(
s
2
r
1
+
s
1
r
2
, s
1
s
2
)
∼
S
(
s
0
2
r
0
1
+
s
0
1
r
0
2
, s
0
1
s
0
2
)
and
(
r
1
r
2
, s
1
s
2
)
∼
S
(
r
0
1
r
0
2
, s
0
1
s
0
2
)
.
Proof
There exist elements
u
1
and
u
2
of
S
such that
u
1
s
0
1
r
1
=
u
1
s
1
r
0
1
and
u
2
s
0
2
r
2
=
u
2
s
2
r
0
2
. Then
(
u
1
u
2
)(
s
0
1
s
0
2
)(
s
2
r
1
+
s
1
r
2
)
=
(
u
2
s
2
s
0
2
)(
u
1
s
0
1
r
1
) + (
u
1
s
1
s
0
1
)(
u
2
s
0
2
r
2
)
=
(
u
2
s
2
s
0
2
)(
u
1
s
1
r
0
1
) + (
u
1
s
1
s
0
1
)(
u
2
s
2
r
0
2
)
=
(
u
1
u
2
)(
s
1
s
2
)(
s
0
2
r
0
1
+
s
0
1
r
0
2
)
,
and
u
1
u
2
∈
S
, and therefore
(
s
2
r
1
+
s
1
r
2
, s
1
s
2
)
∼
S
(
s
0
2
r
0
1
+
s
0
1
r
0
2
, s
0
1
s
0
2
)
.
40