Hungerford: Algebra
III.4. Rings of Quotients and Localization
1. Determine the complete ring of quotients of the ring Zn for each n ≥ 2.
Proof: Denote Zn = {0, 1, · · · , n − 1}, and the set of all non zero divisors of Zn by S(Zn ).
(1.1) Claim: S(Zn ) = {m : (m, n) = 1}.
Suppose that 0 < m < n and (m, n) = d. If m 6∈ S(Zn ). Then there exists an x with 0 < x < n such
that mx = 0. Thus n|(mx). If (m, n) = 1, then n|x, contrary to the assumption that 0 < x < n.
Thus d > 1, and so S(Zn ) ⊇ {m : (m, n) = 1}.
Conversely, assume that d > 1. Then for some positive integers m0 and n0 with 0 < m0 < n and
0 < n0 < n, we have m = m0 d and n = n0 d. Therefore
mn0 = mn0 = nm0 = 0 ∈ Zn ,
and so m 6∈ S(Zn ). Thus S(Zn ) ⊆ {m : (m, n) = 1}. This proves the claim.
By the claim, we can write
S −1 Zn = {m/s : (n, s) = 1}.
Example: For n = 4,
S −1 Z4 = {1/3, 2/3} ∪ Z4 .
2. Let R be a multiplication subset of a commutative ring with identity and let T be a multiplicative
subset of the ring S −1 R. Let S∗ = {r ∈ R : r/s ∈ T for some s ∈ S}. Then S∗ is a multiplicative
subset of R and there is a ring isomorphism S∗−1 R ∼
= T −1 (S −1 R).
Proof: We may assume that 0 6∈ S.
For any r, r0 ∈ S∗ , there exist s, s0 ∈ S such that r/s, r0 /s0 ∈ T . Since T is multiplicative,
0
(rr )/ss0 ) ∈ T ; and since S is multiplicative, ss0 ∈ S. Thus rr0 ∈ S∗ , and so S∗ is also multiplicative.
We now define a map f : S∗−1 R 7→ T −1 (S −1 R). For any r/w ∈ S∗−1 R with r ∈ R and w ∈ S∗ ,
there exists s ∈ S such that w/s ∈ T . Thus we define
f (r/w) = (r/s)/(w/s) ∈ T −1 (S −1 R).
(2.1) f is well-defined.
Suppose that r/w = r0 /w0 in S∗−1 R. As w, w0 ∈ S∗ , there exist s, s0 ∈ S such that w/s, w0 /s0 ∈ T .
Since r/w = r0 /w0 in S∗−1 R, we have, for some r00 ∈ S∗ , r00 (rw0 − r0 w) = 0 in R, and for some s00 ∈ S,
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r00 /s00 ∈ T . It follows that in R, we also have
r00 (rw0 ss0 − r0 wss0 ) = 0.
Hence in S −1 R, we have (r00 /s00 )((r/s)(w0 /s0 )−(rw0 )/(ss0 )) = 0, and so (r/s)(w0 /s0 ) = (rw0 )/(ss0 ) =
(r0 w)/(ss0 ) = (r0 /s0 )(w/s). This, in turn, implies that in T −1 (S −1 R),
f (r/w) = (r/s)/(w/s) = (r0 /s0 )/(w0 /s0 ) = f (r0 /w0 ).
(2.2) f : S∗−1 R 7→ T −1 (S −1 R) is a ring homomorphism.
Let r1 /w1 , r2 /w2 ∈ S∗−1 R. Then ∃s1 , s2 ∈ S such that w1 /s1 , w2 /s2 ∈ T , which implies that
(w1 w2 )/(s1 s2 ) ∈ T .
f (r1 /w1 + r2 /w2 )
(addition in S∗−1 R)
= f ((r1 w2 + r2 w1 )/(w1 w2 ))
=
((r1 w2 + r2 w2 )/(s1 s2 ))/((w1 w2 )/(s1 s2 ))
(definition of f )
=
((r1 /s1 )(w2 /s2 ) + (r2 /s2 )(w1 /s1 ))/((w1 /s1 )(w2 /s2 ))
(addition and multiplication in T −1 (S −1 R)
=
(r1 /s1 )/(w1 /s1 ) + (r2 /s2 )/(w2 /s2 )
= f (r1 /w1 ) + f (r2 /w2 )
(addition in T −1 (S −1 R)
(definition of f )
Let 1 denote the identity of R. Since S is multiplicative, 1/1 = s/s in S −1 R for any s ∈ S.
For any w/s ∈ T , (w/s)((1/1)(w/s) − (w/s)(1/1)) = 0/s and so (1/1)/(1/1) = (w/s)/(w/s) in
T −1 (S −1 R).
(2.3) For any w ∈ S∗ and for any s ∈ S, sw ∈ S∗ .
In fact, if w ∈ S∗ , then ∃s0 ∈ S such that w/s0 ∈ T . For any s ∈ S, since 1/1 = s/s ∈ S −1 R,
(sw)/(ss0 ) = (1/1)(w/s) ∈ T,
and so as ss0 ∈ S, sw ∈ S∗ .
(2.4) Let r ∈ R. If for some w0 ∈ S∗ , rw0 = 0, then for any w ∈ S∗ , r/w = (rw0 w)/(ww0 ) =
0/(ww0 ) = 0/w. In particular, in S∗−1 R,
{0/w} = {r/w : r ∈ R, w ∈ S∗ , and for one w ∈ S∗ , rw = 0 in R}.
(2.5) f is a bijection.
For any (r/s)(w/s) ∈ T −1 (S −1 R), we have r ∈ R, s ∈ S and w/s ∈ T . Thus w ∈ S∗ and so
f (r/w) = (r/s)(w/s). Thus f is surjective.
By (2.3) and (2.4)
Kerf
= {r/w ∈ S∗−1 R : f (r/w) = 0 ∈ T −1 (S −1 R)}
= {r/w ∈ S∗−1 R : for some w ∈ R, s ∈ S with w/s ∈ T, (r/s)/(w/s) = (0/s)/(w/s) ∈ T −1 (S −1 R)}
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= {r/w ∈ S∗−1 R : for some w ∈ R, s ∈ S with w/s ∈ T, (rw)/s = 0/s ∈ S −1 R}
= {r/w ∈ S∗−1 R : for some w ∈ R, s ∈ S with w/s ∈ T, rws = 0 ∈ R}
= {0/w ∈ S∗−1 R}.
3. (a) The set E of positive even integers is a multiplicative subset of Z such that E −1 Z is the field
of the rational numbers.
(b) State and prove condition(s) on a multiplicative subset S of Z which insure that S −1 Z is the
field of rational numbers.
Proof: (a) Since the product of two positive even numbers is still a positive even number, E is a
multiplicative subset of Z. Let Q = (Z − {0})−1 Z denote the field of the rational numbers. Then
∀p/q ∈ Q, p, q ∈ Z and q 6= 0. Without loss of generality, we may assume that q > 0, and so
p/q = (2p)/2q) ∈ E −1 bf Z. This proves that Q ⊆ E −1 bf Z, and so equality must hold.
(b) We claim that for a multiplicative subset S ⊆ Z − {0}, S −1 Z = Q if and only if for any
prime p, S ∩ pZ 6= ∅.
First assume that S −1 Z = Q. For any prime p, as 1/p ∈ Q = S −1 Z, ∃a ∈ Z and b ∈ S such that
a/b = 1/p, or in Z, ap = b ∈ S. HenceS ∩ pZ = ∅.
Conversely, we assume that for any prime p, S ∩ pZ 6= ∅. Let a/b ∈ Q. Then as Z is a PID (and
so a UFD, and every irreducible element is a prime), for some primes p1 , p2 , · · · , pm , and positive
integers n1 , n2 , · · · , nm ,
b = pn1 1 pn2 2 · · · pnmm .
Since pi Z ∩ S 6= ∅, for each i, ∃si ∈ Z such that si pi ∈ S. Let s = s1 s2 · · · sm . Then sb =
(s1 p1 )(s2 p2 ) · · · (sm pm ) ∈ S, and so a/b = (sa)/sb) ∈ S −1 Z. This proves Q ⊆ S −1 |bf Z, and
completes the proof.
4. If S = {2, 4} and R = Z6 , then S −1 R ∼
= Z3 .
Proof: For the simplicity of notation, we denote Z6 = {0, 1, 2, 3, 4, 5}. Therefore, |S × R| = 12.
As 23 = 34 = 02 = 04 , and so the equivalence class in S −1 R represented by 02 contains 4 elements;
as 41 = 52 = 22 = 44 , and so the equivalence class in S −1 R represented by 22 contains 4 elements; as
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1
2
4
−1
R represented by 42 contains 4 elements. Thus
4 = 2 = 4 = 2 , and so the equivalence class in S
S × R has exactly 3 equivalence classes.
Define φ : S −1 R 7→ Z3 by φ( 20 ) − 0, φ( 22 ) = 1 and φ( 42 ) = 2. Then one can verify that this is a
ring epimorphism (verification is routine and so omitted here). Since |S −1 R| = Z3 | = 3, φ is also a
monomorphism and so φ is an isomorphism.
5. Let R be an integral domain with quotient field F . It T is an integral domain such that
R ⊆ T ⊆ F , then F is (isomorphic to) the quotient field of T .
Proof: Let E be the field of quotients of T . Then there is an monomorphism f : T 7→ E. By
Corollary III-4.6 (Let R be an integral domain considered as a subring of its quotient field F . If
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E is a field and f : R 7→ E is a monomorphism of rings, then there is a unique monomorphism of
fields f˜ : F 7→ E such that the restriction of f˜ to R is f . We here apply Corollary III-4.6 with T
replacing R.), there is a field monomorphism f˜ : E 7→ F extending f . View this isomorphism as an
embedding, we may assume that
T ⊆ E ⊆ F.
Now apply Corollary III-4.6 to R with E replacing F in Corollary III-4.6, then we have
R ⊆ F ⊆ E.
8. Let R be a commutative ring with identity, I an ideal of R and π : R 7→ R/I the canonical
projection.
(a) If S is a multiplicative subset of R, then πS = π(S) is a multiplicative subset of R/I.
(b) The mapping θ : S −1 R 7→ (πS)−1 (R/I) given by r/s 7→ π(r)/π(s) is a well defined function.
(c) θ is a ring epimorphism with kernel S −1 I and hence induces a ring isomorphism S −1 R/S −1 I ∼
=
−1
(πS) (R/I).
Proof: (a) ∀s, s0 ∈ S, as S is a multiplicative subset, ss0 ∈ S. Thus π(s)π(s0 ) = π(ss0 ) ∈ π(S) =
πS, and so πS is a multiplicative subset.
(b) Suppose that r/s = r0 /s0 ∈ S −1 R. Then rs0 = r0 s. Since π : R 7→ R/I is a homomorphism,
π(r)π(s0 ) = π(rs0 ) = π(r0 s) = π(r)π(s0 ).
It follows that θ(r/s) = π(r)/π(s) = π(r0 )/π(s0 ) = θ(r0 /s0 ) in (πS)−1 (R/I).
(c) For any r/s, r0 /s0 ∈ S −1 R,
θ(r/s + r0 /s0 )
= θ((rs0 + r0 s)/ss0 ) = π(rs0 + r0 s)/π(ss0 ) = (π(r)π(s0 ) + π(r0 )π(s))/(π(s)π(s0 ))
= π(r)/π(s) + π(r0 )/π(s0 ) = θ(r/s) + θ(r0 /s0 ).
θ((r/s)(r0 /s0 ))
= θ((rr)/(ss0 )) = π(rr0 )/π(ss0 ) = (π(r)π(r0 ))/)π(s)π(s0 ))
=
(π(r)/π(s))(π(r0 )/π(s0 )) = θ(r/s)θ(r0 /s0 ).
For any a/b ∈ (πS)−1 (R/I) with a ∈ R/I and b ∈ πS, since π is an epimorphism, ∃r ∈ R, s ∈ S
such that a = π(r) and b = π(s). Therefore, θ(r/s) = a/b and so θ is also an epimorphism.
Kerθ
= {r/s ∈ S −1 R : θ(r/s) = 0 ∈ (πS)−1 (R/I)}
= {r/s ∈ S −1 R : π(r)/π(s) = 0/π(s) ∈ (πS)−1 (R/I)}
= {r/s ∈ S −1 R : π(r) = 0 ∈ R/I and π(s) ∈ π(S)}
= {r/s ∈ S −1 R : r ∈ I and s ∈ S} = S −1 I.
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9. Let S be a multiplicative subset of a commutative ring with identity. If I is an ideal in R, then
S −1 (Rad I) = Rad (S −1 (I)).
Proof: It is routine to show that S −1 ( Rad I) ⊆ Rad (S −1 I). In fact, if ab ∈ S −1 ( Rad I), then
a ∈ Rad (I) and b ∈ S. Thus for some integer n > 0, an ∈ I. As S is multiplicative, bn ∈ S. It
n
follows that ( ab )n = abn ∈ S −1 I, and so S −1 ( Rad I) ⊆ Rad (S −1 I).
Conversely, let ab ∈Rad(S −1 I). Then for some integer m > 0,
m
for some c ∈ I and d ∈ S, abm = dc . Thus ∃s ∈ S, such that
am
bm
= ( ab )m ∈ S −1 I. Therefore,
s(am d − bm c) = 0, or equivalently, am sd = bm sc.
Since c ∈ I and since I is an ideal, am sd = bm sc ∈ I. Since S is a multiplicative set, and since
b, s, d ∈ S, we have bsd ∈ S and so
m
asd
m
m
m−1
m
∈ S −1 I.
(asd) = a sd(sd)
∈ I, (bsd) ∈ S, and so
bsd
Thus
asd
bsd
∈ Rad(S −1 I). But then for any s0 ∈ S, s0 (absd − absd) = 0, and so
a
asd
=
∈ Rad(S −1 I).
b
bsd
This proves that Rad (S −1 (I)) ⊆ S −1 (Rad(I)).
12. A commutative ring with identity is local if and only if for all r, s ∈ R, r + s = 1R implies r or
s is a unit.
Proof: Let R be a commutative ring with identity 1R .
Suppose first that R is local, and so R has exactly one maximal ideal M . Suppose that r, s ∈ R
with r + s = 1. If neither r nor s are units, then (r) and (s) are principal ideals. Let M1 and M2 be
maximal ideals of R containing (r) and (s), respectively. Then since M is the only maximal ideal
of R, we have M1 = M2 = M , and so r, s ∈ M . Since r + s = 1R , we must have 1R ∈ M , and so
M = R, contrary to the assumption that M 6= R.
Conversely, assume that ∀r, s ∈ R, that r + s = 1R implies that r or s is a unit. By contradiction,
we assume that R has at least two distinct maximal ideals M1 and M2 . Since M1 and M2 are distinct
maximal ideals, the ideal M1 + M2 must be R itself. Therefore, since 1R ∈ R, ∃r ∈ M1 and s ∈ M2
such that r + s = 1R . Therefore, either r or s is a unit, and so either M1 or M2 equals R, contrary
to the fact that a maximal ideal is not equal to R.
13. The ring R consisting of all rational numbers with denominators not divisible by some (fixed)
prime p is a local ring.
Proof: Consider the ring Z of the integers. For a fixed prime p, (p) is a prime ideal of Z and
so S = Z − (p) is a multiplicative subset. By Theorem III-4.11(ii) (Let P be a prime idea in a
commutative ring R with identity, the ideal PP = S −1 P is the unique maximal ideal of the ring
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RP = S −1 R, the localization of R at P ), Z(p) , having S −1 (p) as its only maximal ideal, is a local ring.
Fact: The structure of Z(p) . As (p) = pZ, S = Z − (p) = {n ∈ Z − {0} : p 6 |n}. Thus
Z(p) = {n/m ∈ Q : p 6 |m}.
In particular.
Z(2) = {n/m ∈ Q : m is odd }.
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