114 Groups and Rings 2008–09
Suggested solutions to exercise set 8
April 27, 2009
33.12 As in class, we adopt the convenient notation
eiθ = cos θ + i sin θ.
The fifth complex roots of unity are then
1, e2πi/5 , e4πi/5 , e6πi/5 , e8πi/5 .
Geometrically:
e2πi/5
e4πi/5
1
e6πi/5
e8πi/5
1
33.21 Let C′ be the multiplicative group of the ring C; that is, C′ = (C \ {0}, ·).
Let G = {e2πik/n : k = 0, 1, . . . , n − 1}, which is the set of complex nth roots
of unity. It suffices to show that G is a cyclic subgroup of C′ .
Since e2πik/n = (e2πi/n )k , if we write x = e2πi/n then xn = 1 and
G = {xk : k = 0, 1, . . . , n − 1} = {xk : k ∈ Z} = hxi,
so G is the cyclic subgroup of C′ generated by x.
33.22 Let x = e2πi/n . By Example C.1, the sum of the nth complex roots of unity
is
1 − xn
1 + x + x2 + · · · + xn−1 =
= 0.
1−x
33.23 Let x = e2πi/n . The product of the nth complex roots of unity is
1
2πi 1
1 · x · x2 · · · · · xn−1 = x0+1+2+···+(n−1) = x 2 n(n−1) = e n · 2 n(n−1)
(
−1
if n is even,
= (eπi )n−1 = (−1)n−1 =
1
if n is odd.
34.9 (a) False. For example, in R[x] we have (1 + x) + (−x) = 1, which has
| {z } | {z }
degree 1
degree 1
degree 0, not 1 + 1 = 2.
(b) False. For example, in Z4 [x] we have |{z}
2 · (1 + 2x) = 2+4x = 2, which
| {z }
degree 0
degree 1
has degree 0, not 0 + 1 = 1. (Note that the statement is true in R[x] if R is
an integral domain).
34.11 If R has characteristic zero then for every integer n > 0 there is a ∈ R with
na 6= 0, which implies that the constant polynomial c(x) = a then satisfies
nc(x) = na 6= 0. So R[x] also has characteristic zero.
If R has characteristic 1 then a = 1a = 0 for all a ∈ R so R = {0}. So R[x]
only contains the zero polynomial, so R[x] has characteristic 1.
If R has characteristic n with n > 1 then na = 0 for all a ∈ R, and there is
b ∈ R with (n − 1)b 6= 0. Now if p(x) = a0 + a1 x + · · · + am xm is in R[x] then
np(x) = na0 + na1 x + · · · + nam xm = 0 + 0x + · · · + 0xm = 0.
So char R[x] ≤ n. Moreover, the constant polynomial c(x) = b satisfies
(n − 1)c(x) = (n − 1)b 6= 0 in R[x], so char R[x] > n − 1. So char R[x] = n.
(Note that this solution can be shortened if R is unital, since then char R is
the smallest integer n with ne = 1 and the constant polynomial u(x) = e is
the unit of R[x], so we simply have to examine the equation nu(x) = 0.)
2
34.12 Let θ : R → S be an isomorphism of rings. Let φ : R[x] → S[x] be given by
φ(a0 + a1 x + · · · + an xn ) = θ(a0 ) + θ(a1 )x + · · · + θ(an )xn .
We have to show that φ is a well-defined ring isomorphism.
If a0 + a1 x + · · · + an xn = b0 + b1 x + · · · + bm xm in R[x] then these two
expressions differ only by some terms of the form 0xk . Since θ(0) = 0, the
expressions θ(a0 ) + θ(a1)x + · · · + θ(an )xn and θ(b0 ) + θ(b1 )x + · · · + θ(bm )xm
will differ only by some terms of the form 0xk , so they are equal. So φ is
well-defined.
Since θ is an isomorphism, we have θ(a) = 0 ⇐⇒ a = 0. So we can reverse
the argument of the last paragraph to see that φ is one-to-one. Moreover,
since θ is onto, φ is clearly onto.
Let p(x) = a0 + a1 x + · · · + an xn and q(x) = b0 + b1 x + · · · + bm xm be two
polynomials in R[x], and let r(x) = p(x)+q(x). Let us write N = max{m, n}
and define aj = 0 if j > n and bj = 0 if j > m. Then
φ(p(x) + q(x)) = φ((a0 + b0 ) + (a1 + b1 )x + · · · + (aN + bN )xN )
= θ(a0 + b0 ) + θ(a1 + b1 )x + · · · + θ(aN + bN )xN
= θ(a0 ) + θ(b0 ) + (θ(a1 ) + θ(b1 ))x + · · · + (θ(aN ) + θ(bN ))xN
= θ(a0 ) + θ(a1 )x + · · · + θ(aN )xN + θ(b0 ) + θ(b1 )x + · · · + θ(bN )xN
= φ(p(x)) + φ(q(x)).
Consider the product p(x)q(x) = c0 + c1 x + · · · + cn+m xn+m where
ck = a0 bk + a1 bk−1 + · · · + ak−1 b1 + ak b0 .
[Here we retain the convention that aj = 0 if j > n and bj = 0 if j > m.]
Since θ is a ring isomorphism, we have
θ(ck ) = θ(a0 )θ(bk ) + θ(a1 )θ(bk−1 ) + · · · + θ(ak−1 )θ(b1 ) + θ(ak )θ(b0 ),
which is the coefficient of xk in the product φ(p(x))φ(q(x)). So
φ(p(x)q(x)) = θ(c0 ) + θ(c1 )x + · · · + θ(cn+m )xn+m = φ(p(x))φ(q(x)).
So φ : R[x] → S[x] is a ring isomorphism, so R[x] ≈ S[x].
3
35.4 In Z3 [x], we have x2 + 2 = x2 − 1 = (x − 1)(x + 1), so q(x) = x + 1 and
r(x) = 0. If we didn’t spot this, here’s the long division:
x+1
x − 1 x +0x+2
x2 −x
x+2
x−1
0
2
35.5 In Z5 [x] we have f (x) = 3x4 + 2x2 + 4. We compute:
4x2 +2x +2
2x2 + 4x 3x4 +0x3 +2x2 +0x+4
3x4 +x3
4x3 +2x2 +0x+4
4x3 +3x2
4x2 +0x+4
4x2 +3x
2x+4
So g(x) = 4x2 + 2x + 2 and q(x) = 2x + 4.
4