3.1 Examples and Basic Properties
1.
a. Not an additive group; only 0 has an inverse.
c. It is an additive abelian group (as in Example 4), unity is 
, multiplication is associative (it is function composition). But h 
(f + g) = h f + h g fails: h (f + g)(x) = h[f(x) + g(x)] while
, multiplication is associative (it is function composition). But h (f + g) = h f + h g fails: h (f + g)(x) = h[f(x) + g(x)] while
These may not be equal. Note that (f + g) h = f h + g h does hold.
h = f h + g h does hold.
3.
a. Clearly 
and 
. If 
and 
are in S then 
and 
because
because
and . If and are in S then and because
because
c. Clearly I, 0 
S and A, B S ⇒ A + B and −A S. To check multiplication
S and A, B S ⇒ A + B and −A S. To check multiplication
5. We have 1, 0 C(X) as 1x = x = x1 and 0x = 0 = x0 for all x X. If c, d C(x) then
C(X) as 1x = x = x1 and 0x = 0 = x0 for all x X. If c, d C(x) then
show that −c, c + d, and cd are in C(x).
6.
a. If ab = 0 and a ≠ 0 then a−1 exists so b = a−1ab = 0.
7. If 
then 
that is 
. Hence c = 0 and a = d, so 
. Similarly the fact that A commutes with 
forces b = 0. Finally 
shows ar = ra for all r R; that is a Z(R). Conversely if 
. Thus
then that is . Hence c = 0 and a = d, so . Similarly the fact that A commutes with forces b = 0. Finally shows ar = ra for all r R; that is a Z(R). Conversely if . Thus
8.
a. If (a + b)(a − b) = a2 − b2. Then a2 + ba − ab − b2 = a2 − b2; whence ba − ab = 0, ab = ba. Conversely if ab = ba then
9.
Hence a + b + a + b = a + a + b + b, so b + a = a + b (additive cancellation on the left and on the right).
10.
a. ab + ba = 1 gives aba + ba2 = a, so ba2 = a − aba. Similarly a2b = a − aba so a2b = ba2. Then ab = a3b = a(a2b) = aba2 = (1 − ba)a2 = a2 − ba. Hence a2 = ab + ba = 1.
11. If a2 = 0 implies a = 0, let an = 0, n ≥ 1. If a ≠ 0, let am = 0, am−1 ≠ 0. Then m ≥ 2 and so (am−1)2 = a2m−2 = 0 since 2m − 2 ≥ m (m ≥ 2). This contradicts the hypothesis. So a = 0. The converse is clear.
13. By the hint: 
. Hence
. Hence
is a unit, being a product of units [Theorem 2.5]. Thus
15. (3) ⇒ (1). Let e be the unique right unity. Given b R,
R,
for all r R. Hence e + eb − b is a right unity too, so e = e + eb − b by uniqueness. Thus eb = b for all b, that is e is also a left unity.
R. Hence e + eb − b is a right unity too, so e = e + eb − b by uniqueness. Thus eb = b for all b, that is e is also a left unity.
16.
a. This is by the subring test since 0 = 01R; 1R = 11R ; k1R + m1R = (k + m)1R; −(k1R) = (− k)1R and (k1R)(m1R) = (km)1R. To see that 
is central, let s R. Then (k1R)s = (k1R)(1s) = (k1)(1Rs) = ks by Theorem 2, and s(k1R) = ks in the same way.
is central, let s R. Then (k1R)s = (k1R)(1s) = (k1)(1Rs) = ks by Theorem 2, and s(k1R) = ks in the same way.
c. If char R = 0 define 
by σ(k) = k1R. Then σ(km) = σ(k) · σ(m) and σ(k + m) = σ(k) + σ(m) as in (b), and σ(k) = σ(m) ⇒ (k − m)1R = 0 ⇒ k = m because o(1R) =∞. Since σ is clearly onto, it is an isomorphism.
by σ(k) = k1R. Then σ(km) = σ(k) · σ(m) and σ(k + m) = σ(k) + σ(m) as in (b), and σ(k) = σ(m) ⇒ (k − m)1R = 0 ⇒ k = m because o(1R) =∞. Since σ is clearly onto, it is an isomorphism.
17. If R has characteristic 1 then 1 · 1R = 0, 1R = 0. Hence R = 0, the zero ring.
18.
a. 
. The unity is (1, 1) so k(1, 1) = 0 
k1 = 0 in 
and k1 = 0 in 
and m|k. So the characteristic of 
. The unity is (1, 1) so k(1, 1) = 0 k1 = 0 in and k1 = 0 in and m|k. So the characteristic of
c. 
. k(1, 1) = 0 k1 = 0 in 
and k1 = 0 in 
. So the characteristic of 
is 0.
. k(1, 1) = 0 k1 = 0 in and k1 = 0 in . So the characteristic of is 0.
19. If u R∗, 
, then k1 = 0 in R ⇒ ku = 0. But
R∗, , then k1 = 0 in R ⇒ ku = 0. But
So o(u) = o(1) = char R (since char R< ∞).
21.
a. (1 − 2e)2 = (1 − 2e)(1 − 2e) = 1 − 2e − 2e + 4e2 = 1.
22.
a. If a = (1 − e)re then ea = 0 and ae = a. Hence a2 = (ae)a = a(ea) = 0. Similarly if b = er(1 − e), then eb = b, be = 0, so b2 = 0.
c. This follows from (a) and Example 17.
23. (2) ⇒ (1). If r R, f = e + (1 − e)re is an idempotent, so ef = fe by (2). Hence re = ere. Similarly re = ere, so re = re and e is central.
R, f = e + (1 − e)re is an idempotent, so ef = fe by (2). Hence re = ere. Similarly re = ere, so re = re and e is central.
24. (1) ⇒ (2). If an = 0 then 1 + a is a unit [(1 + a)−1 = 1 − a + a2 − 
]. Hence r(1 + a) = (1 + a)r for all r R, that is ra = ar for all r.
]. Hence r(1 + a) = (1 + a)r for all r R, that is ra = ar for all r.
25. If r3 = r then e = r2 is idempotent. Iterating r3 = r: r9 = r3 = r, r27 = r3 = r, and in general 
for k ≥ 1. Hence if rn = 0 then 
for any k such that if 3k ≥ n. Thus R has no nonzero nilpotents. Hence idempotents in R are central by Example 18, so r2 is central for all r. Finally, if r, s R, rs = (rs)3 = (rs)2rs = r(rs)2s = r2srs2 = s2srr2 = s3r3 = sr.
for k ≥ 1. Hence if rn = 0 then for any k such that if 3k ≥ n. Thus R has no nonzero nilpotents. Hence idempotents in R are central by Example 18, so r2 is central for all r. Finally, if r, s R, rs = (rs)3 = (rs)2rs = r(rs)2s = r2srs2 = s2srr2 = s3r3 = sr.
26.
a. If ab = 1 and R = {r1, r2, . . ., rn} then br = bs ⇒ r = s in R. But then br1, br2, . . ., brn are all distinct, so {br1, br2, . . ., brn} = R. In particular bc = 1 for some c. Then a = a(bc) = (ab)c = c so 1 = bc = ba.
27.
a. If am = am+n then
Then am+t = am+t+kn for all t ≥ 0, so ar+kn = ar for all r ≥ m and for all k ≥ 1. We want r and k such that r + kn = 2r; that is kn = r. Since n ≥ 1 choose k such that kn ≥ m, and take r = km. Then ar is an idempotent.
28.
a. 
. Units: ±1; idempotents: 0,1; nilpotents: 0.
. Units: ±1; idempotents: 0,1; nilpotents: 0.
c. 
29. It is clearly an additive abelian group and 
. The associative and distributive laws hold for all matrices. Units: 
; nilpotents: 
; idempotents: 
.
. The associative and distributive laws hold for all matrices. Units: ; nilpotents: ; idempotents: .
31. If m is odd then m2 − m = (m − 1)m is a multiple of 2m. Thus 
in 
in
33. If r2 = r for all r R, then
R, then
Hence 0 = r + r = 2r. Since R ≠ 0, this shows that R has characteristic 2. In particular, −r = r for all r R. If r, s R, then
R. If r, s R, then
Thus rs + sr = 0, so sr = − rs = rs. Thus R is commutative.
35. Define
It is clear that σ is a bijection and preserves addition. As to multiplication
Hence σ is an isomorphism.
36.
a. If 
is an isomorphism let 
. Then
is an isomorphism let . Then
which is impossible for 
. So no such isomorphism exists.
. So no such isomorphism exists.
c. If 
is an isomorphism and 
, then
is an isomorphism and , then
a contradiction in 
So no such isomorphism exists.
So no such isomorphism exists.
37. Put σ(1) = e. Given r′ R′, write r′ = σ(r), r R. Then
R′, write r′ = σ(r), r R. Then
Similarly r′e = r′, so e is the unity of R′.
38. Let 
be an isomorphism.
be an isomorphism.
a. If z Z(R) let s S, say s = σ(r). Then
Z(R) let s S, say s = σ(r). Then
so σ(z) Z(S). If z′ Z(S) and 
, then for any
Z(S). If z′ Z(S) and , then for any
Since σ is one-to-one, wr = rw. Thus 
. It follows that σ : Z(R) → Z(S) is onto. It is clearly one-to-one and so is an isomorphism of rings.
. It follows that σ : Z(R) → Z(S) is onto. It is clearly one-to-one and so is an isomorphism of rings.
39. It is a routine matter to verify that α + β and αβ are again endomorphisms. The distributive law α(β + γ) = αβ + αγ follows because, for all x X:
X:
The other distributive law is similar, as are the rest of the axioms. The zero is the zero endomorphism θ : X → X where θ(x) = 0 for all x X, and 1X is the unity.
X, and 1X is the unity.
41. If a (eRe)∗, let ab = e, b (eRe)∗. Write f = 1 − e, so ef = 0 = fe and e + f = 1. Then
(eRe)∗, let ab = e, b (eRe)∗. Write f = 1 − e, so ef = 0 = fe and e + f = 1. Then
because af = (ae)f = 0 and fb = f(eb) = 0. Thus σ : (eRe)∗ → R∗ is a mapping, and (∗) shows it is a group homomorphism. It is one-to-one because σ(a) = σ(b) means a + f = b + f, so a = b.
42.
a. Let 
, pi distinct primes, ni ≥ 1. If 
is nilpotent in 
then km ≡ 0 (mod n) so n|km. Hence pi|km for all i so pi|k. Conversely, if pi|k for all i then p1p2 pr|k because the pi are relatively prime in pairs. Hence if m = max {n1, . . ., nr} then km = 0.
, pi distinct primes, ni ≥ 1. If is nilpotent in then km ≡ 0 (mod n) so n|km. Hence pi|km for all i so pi|k. Conversely, if pi|k for all i then p1p2 pr|k because the pi are relatively prime in pairs. Hence if m = max {n1, . . ., nr} then km = 0.
c. Let 
in 
. Hence n|e(1 − e). Then we can write n = ab, a|e, b|(1 − e). [Exercise 35 §1.2], say e = xa, 1 − e = yb. Thus 1 = xa + yb and e = xa.
in . Hence n|e(1 − e). Then we can write n = ab, a|e, b|(1 − e). [Exercise 35 §1.2], say e = xa, 1 − e = yb. Thus 1 = xa + yb and e = xa.
43. Let |R| = 4. This is an additive group of order 4, so o(1) = 4 or o(1) = 2. If o(1) = 4 then R = {0, 1, 2, 3} is isomorphic to 
via the obvious map. So assume o(1) = 2; that is the characteristic of R is 2. Hence r + r = 0 for all r in R. If a ≠ 0, 1 then 1 + a ≠ 0, 1, a. Hence R = {0, 1, a, 1 + a}. Thus
via the obvious map. So assume o(1) = 2; that is the characteristic of R is 2. Hence r + r = 0 for all r in R. If a ≠ 0, 1 then 1 + a ≠ 0, 1, a. Hence R = {0, 1, a, 1 + a}. Thus
the addition table is as shown (it is the Klein group). All but four entries in the multiplication table are prescribed as shown. The rest of the table is determined by the choice of a2.
1. a2 = 1 + a. The table is then as shown. This is clearly a field if it is associative. (See Section 4.3).
2. a2 = a. The table is as shown. This is isomorphic to 
with a = (1, 0) and 1 + a = (0, 1).
with a = (1, 0) and 1 + a = (0, 1).
3. a2 = 1. The table is as shown. This is isomorphic to
with 
and 
and
4. a2 = 0. If b = 1 + a then 1 + b = a and b2 = 1 + a2 = 1. Hence this is the same as Case 3.
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