Aaron Goldsmith
Homework 1 Section 3.4
Algebra II
3
a)
Let
E
=
{
2
n

n
∈
Z
+
}
. If 2
n
1
,
2
n
2
∈
E
, then the product
(2
n
1
)(2
n
2
) = 2(2
n
1
n
2
)
∈
E
holds, and
E
is multiplicative. Now, we see that
E

1
Z
is a com
mutative ring (Thm 4.3) and the subring of
Q
trademarked by
even denominators. For the reverse inclusion, let
p/q
∈
Q
. Then,
p/q
= 2
p/
2
q
∈
E

1
Z
.
b)
In addition to 0
6∈
S
and of course multiplicativity,
S
must in
tersect every nonzero ideal. Equivalently, for every prime
p
, there
must be an
s
∈
S
such that
p
divides
s
. This is necessary because
if there were a prime
p
that divided no
s
∈
S
, then 1
/p
6∈
S

1
Z
. It
is suﬃcient because we can take the prime factorization of any de
nominator, replace the primes
p
with elements of
S
that
p
divides,
and choose a suitable numerator to cancel all that’s extra.
10
⊃
Let
S
=
R

M
. By deﬁnition,
R
M
=
S

1
R
. To see that 1
∈
S
,
note that
R
is an integral domain containing 1, and if 1
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 Spring '09
 GELLER
 Algebra, Ring, Prime number, Integral domain, Ring theory, Commutative ring

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