3. If e2 = e in a domain, then e(1 − e) = 0 so e = 0 or e = 1. If an = 0, n ≥ 1, then a = 0. For if a ≠ 0 then aan−1 = 0 gives an−1 = 0, . . ., and eventually a = 0, a contradiction.
5. Let
. Then
A2 = 0 but
A ≠ 0. So
Mn(
R) is not a domain.
7. If ab = 0 then (ba)2 = b(ab)a = 0, so ba = 0 by hypothesis.
9. In
, 1
2 + 2
2 = 0; in
, let
a2 +
b2 = 0. If either
a = 0 or
b = 0, the other is 0 (
x2 = 0 ⇒
x = 0 in a field). If
a ≠ 0,
b ≠ 0 then
a,
b {1, 2}. But 1
2 + 1
2 ≠ 0, 1
2 + 2
2 = 2 ≠ 0, 2
2 + 2
2 = 3 ≠ 0.
11. The group F∗ = F {0} has order q − 1 so aq−1 = 1 for all a ≠ 0 (by Lagrange's theorem). Thus aq = a if a ≠ 0; this also holds if a = 0.
13. Since |
F| =
p is prime, (
F, +) is cyclic and is generated by 1 (or any nonzero element) by Lagrange's theorem. Hence the map
given by
is an isomorphism of additive groups. It is a ring isomorphism because (
km)1 = (
k1)(
m1) in
F.
15. Let
Z denote the center of a division ring
D. If 0 ≠
z Z we have
zd = 1 =
dz for some
d D ; we must show that
d Z. Given
r D we have (
rz)
d =
r(
zd) =
r and (
zr)
d =
d(
zr) = (
dz)
r =
r. Hence (
rz)
d = (
zr)
d so
rz =
zr.
16.
a. If
K is a subfield, let 0 ≠
a K. If
a′ is the inverse of
a in
K then
aa′ = 1. But
aa−1 = 1 in
F so
a′ =
a−1 by cancellation. Hence
a−1 K. Conversely, if the condition holds, then the inverse of
a in
F serves as its inverse in
K.
c. Here |K| = 1, 2, 4, 8, 16, and |K∗| divides 15 so |K∗| = 1, 3, 5, 15. Thus |K| = 2, 4, 8, 16. The common values are |K| = 2, 4, 16. So K = {0, 1}, |K| = 4, or K = F.
17. It is clearly a subring of
. If
a =
r +
si ≠ 0 then
(one of
r ≠ 0 or
s ≠ 0). Since
we have
too, so
18.
a. It is clearly a subring of
. As in Example 4, if
define
and
N(
a) =
aa∗ =
r2 + 5
s2. If
a ≠ 0 then
N(
a) ≠ 0 in
so
. Thus
.
19. is a subfield of
by Example 4, and it contains
If
F is any subfield of
then
(because
), and hence
(because
for all
n,
m ≠ 0 in
). If also
F, this means
for all
. Thus
.
21.
a. is a subring of
, and so is an integral domain by Example 3.
c. r∗∗ =
r is obvious. If
r =
n +
mw and
then
e. If
r is a unit in
then
rr−1 = 1 in
so, by (d)
N(
r)
N(
r−1) =
N(1) = 1 in
. It follows that
N(
r) = ± 1. Conversely, if
N(
r) = ± 1 then
rr∗ = ± 1 so
r−1 = ±
r∗.
23. Let
R = {
r1,
r2, . . .,
rn} be a domain with
n elements. If 0 ≠
a R, then the elements of
aR = {
ar1,
ar2, . . .,
arn} are distinct (because
a can be cancelled) so |
aR| =
n = |
R|. Hence
aR =
R so
ab = 1 for some
b R. Similarly
Ra = {
r1a, . . .,
rna} =
R so
ca = 1 for some
c R. Thus
c =
c(
ab) = (
ca)
b =
b, and this element is the inverse of
a.
24.
a.
c. If
F is a field of characteristic
p, the map
σ :
F →
F with
σ(
a) =
ap satisfies
σ(1) = 1,
σ(
ab) = (
ab)
p =
apbp =
σ(
a) ·
σ(
b), and, using (b),
σ(
a +
b) = (
a +
b)
p =
ap +
bp =
σ(
a) +
σ(
b). Hence
σ is a homomorphism. We claim
σ is one-to-one. Let
a ker
σ, that is
σ(
a) = 0. Then
ap = 0 so
a = 0 because
F is a field). Since
F is finite,
σ is also onto, and so is an automorphism of
F.
25. Given
σ :
R →
R, we have
Q = {
ru−1 r R, 0 ≠
u R}. Since
u ≠ 0 implies
σ(
u) ≠ 0, if
exists it must be given by
So define
by this formula. If
ru−1 =
sv−1 then
rv =
su so
that is
. Hence
is well defined. Now
Similarly
preserves multiplication. If
then
σ(
r)[
σ(
u)]
−1 = 0 so
σ(
r) = 0. Hence
r = 0,
ru−1 = 0;
is one-to-one. Finally let
sv−1 Q. Let
r =
σ−1(
s) and
. Then
u ≠ 0 and
σ(
ru−1) =
σ(
r) · [
σ(
u)]
−1 =
sv−1. Thus
is onto.
26.
a. If
and
then
ru′ =
ur′ and
sv′ =
vs′, and so
This shows
c.
27. Let
R ⊆
F where
F is a field, and let
. Define
σ :
Q →
F by
. Then
, so
σ is well defined and one-to-one. We have
Hence
R ≅
σ(
R) = {
ru−1 r,
u R,
u ≠ 0} and
σ(
R) is a subring of
F. It is a subfield because, if
ru−1 ≠ 0 then
r ≠ 0 so (
ru−1)
−1 =
ur−1 σ(
R).
29.
a. If
r =
i and
s = 1 in
, consider
a =
r +
sω in
. Then
aa∗ =
r2 +
s2 = 0, but
a ≠ 0 and
a∗ ≠ 0 in
. Thus
is not a field. In
let
a = 1 + 2
ω. Then
aa∗ = 1
2 + 2
2 = 0, and
a ≠ 0 ≠
a∗. So
is not a field. However
is a field. If
a =
r +
si ≠ 0 in
then
aa∗ =
r2 +
s2 and it suffices to show
r2 +
s2 ≠ 0 in
. Suppose
r2 +
s2 = 0. If
r = 0 or
s = 0 then
a = 0, contrary to hypothesis. Thus
r ≠ 0 ≠
s. Then 0 =
s−1(
r2 +
s2) = (
s−1r)
2 + 1 so (
s−1r)
2 = − 1 in
. This is not the case because 0
2 = 0, 1
2 = 1 = 6
2, 2
2 = 4 = 5
2, 3
2 = 2 = 4
2 in
.
c. Let
a =
r +
si ≠ 0 in
p ≡ 3 (mod 4). Then
aa∗ =
r2 +
s2 so it suffices to show
r2 +
s2 ≠ 0 (than
a−1 = (
r2 +
s2)
−1a∗). Suppose
r2 +
s2 = 0. Now
r ≠ 0 or
s ≠ 0 (because
a ≠ 0). If
s ≠ 0 then 0 =
s−2(
r2 +
s2) = (
s−1r)
2 + 1. Thus
x =
s−1r satisfies
x2 = − 1 in
, contrary to the Corollary to Theorem 8 §1.3. Similarly if
r ≠ 0.
e. If
e =
r +
sω and
e2 =
e then
r2 −
s2 =
r and 2
rs =
s. If
s = 0 then
r2 =
r;
r = 0, 1 ;
e = 0, 1. If
s ≠ 0 then 2
r = 1,
,
, (2
s)
2 = − 1.
30.
a. and c. These are routine calculations.
e.
31.
a. and c. are routine verifications.
e. If
then (b) shows that
where
denotes the conjugate of
z. In this form
H is easily verified to be a subring of
. Since (a) and (d) determine the multiplication, we have
.
32. Since
R is commutative, Lemma 1 holds in
; the proof is the same.
a. If
q is a unit in
then 1 =
N(1) =
N(
qq−1) =
N(
q)
N(
q−1), so
N(
q) is a unit in
R. Conversely
qq∗ =
N(
q) shows
q−1 =
N(
q)
−1q∗ if
N(
q)
R∗.
c. Let
q =
a +
bi +
cj +
dk in
. Then
Similarly
qj =
jq 2
a = 2
c = 0 and
qk =
kq 2
b = 2
c = 0. Since
,
qj =
jq and
qk =
kq, the result follows. If
R has characteristic 2, then 2
r = 0 for all
r R, so
, that is
is commutative. Conversely, if
is commutative, (1 +
ai)
j =
j(1 +
ai) for all
a R, so
ak = −
ak, 2
a = 0. Thus
R has characteristic 2. Finally,
A2(
Z6) = {0, 3} = 3
Z6, so