3.2 Integral Domains and Fields
1.
a. 1, − 4.
c. 0, 1.
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.
. 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 
, 12 + 22 = 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 12 + 12 ≠ 0, 12 + 22 = 2 ≠ 0, 22 + 22 = 3 ≠ 0.
, 12 + 22 = 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 12 + 12 ≠ 0, 12 + 22 = 2 ≠ 0, 22 + 22 = 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.
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.
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.
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
. 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 + 5s2. If a ≠ 0 then N(a) ≠ 0 in 
so 
. Thus 
.
. As in Example 4, if define and N(a) = aa∗ = r2 + 5s2. 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 
.
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.
is a subring of , and so is an integral domain by Example 3.
c. r∗∗ = r is obvious. If r = n + mw and 
then
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∗.
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.
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.
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
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
by this formula. If ru−1 = sv−1 then rv = su so
that is 
. Hence 
is well defined. Now
. 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.
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
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
. 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).
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 + 2ω. Then aa∗ = 12 + 22 = 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 02 = 0, 12 = 1 = 62, 22 = 4 = 52, 32 = 2 = 42 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 + 2ω. Then aa∗ = 12 + 22 = 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 02 = 0, 12 = 1 = 62, 22 = 4 = 52, 32 = 2 = 42 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.
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 2rs = s. If s = 0 then r2 = r; r = 0, 1 ; e = 0, 1. If s ≠ 0 then 2r = 1, 
, 
, (2s)2 = − 1.
, , (2s)2 = − 1.
30.
a. and c. These are routine calculations.
e. 
31.
a. and c. are routine verifications.
e. If 
then (b) shows that
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 
.
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.
; 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∗.
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
. Then
Similarly qj = jq 
2a = 2c = 0 and qk = kq 2b = 2c = 0. Since 
, qj = jq and qk = kq, the result follows. If R has characteristic 2, then 2r = 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, 2a = 0. Thus R has characteristic 2. Finally, A2(Z6) = {0, 3} = 3Z6, so
2a = 2c = 0 and qk = kq 2b = 2c = 0. Since , qj = jq and qk = kq, the result follows. If R has characteristic 2, then 2r = 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, 2a = 0. Thus R has characteristic 2. Finally, A2(Z6) = {0, 3} = 3Z6, so
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