Topics in algebra Chapter IV: Commutative rings and modules I - 1

Proof. (a) If S contains a non-zero zero-divisor, say s, then there is a nonzero
element a ∈ R such that as = 0, it follows that a ∈ ker(iS) as a
= 0 in RS.
1
If iS is not a monomorphism, then there is a non-zero element a ∈ R such
that a
1 = 0 in RS, or equivalently, there is an element s ∈ S such that as = 0.
It follows that s is a zero-divisor.
(b) If IS = RS, then 1 a
= for some a ∈ I and s ∈ S, so that there is an t ∈ S
1 s
such that t(a − s) = 0, it follows that st ∈ I and then I ∩ S = ∅. Conversely,
if I ∩ S = ∅, then there is an element a ∈ I ∩ S, so that 1 a
=
1 a ∈ IS, it
follows that RS = IS.
(c) It is clear that J ce ⊆ J. Let a a
∈ J; then
s 1 ∈ J so that a ∈ J c , it follows
that a 1 a
= ·
s s 1 ∈ J ce .
(d) It is clear that P ⊆ P ec . Conversely, let a ∈ P ec ; then a b
= for some
1 s
b ∈ P and s ∈ S, so that there is an t ∈ S such that t(as − b) = 0, it follows
that ast ∈ P . However, P ∩ S = ∅, we conclude that a ∈ P and P ec ⊆ P .
(e) If P ∩ S = ∅ then PS = RS by (b). So, we may assume that P ∩ S = ∅.
Let a b
·
s t ∈ PS; then ab ∈ P ec = P by (d), so that a ∈ P or b ∈ P , it follows
that a
s ∈ PS or b
∈ PS.
t
From Proposition 3.9, we see that if I is an ideal of R with I ∩ S = ∅ then
I = I ec does not hold in general. However, if P is a prime ideal of R with
P ∩ S = ∅, then P ec = P . Therefore we may ask that what kind of ideals in
R will satisfy I ec = I. The answer is primary ideals.
Definition 3.10 Let Q be an ideal of a ring R; Q is said to primary if
ab ∈ Q then either a ∈ Q or b n ∈ Q for some positive integer n.
Lemma 3.11 Let Q be a primary ideal of a ring R; then √ Q is a prime
ideal.
Proof. Let P = √ Q and ab ∈ P with b /∈ P ; then b n /∈ Q for every n, so that
by the definition of primary ideal a ∈ Q ⊆ P .
From now on, if Q is a primary ideal of a ring R with √ Q = P , then we say
that Q is a P -primary ideal.
Corollary 3.12 The following are equivalent for an ideal of a ring R:
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