James Pringle
June 15, 2011
Abstract Algebra
Dummit and Foote
Third Edition
Section 13.1
Problems: 1, 2, 3, 4
[1] Show that p(x) = x3 + 9x + 6 is irreducible in Q[x]. Let θ be a root of p(x). Find
the inverse of 1 + θ in Q(θ).
According to Eisenstein’s Criterion for Z[x] (chapter 9, corollary 14, page 310), since
f (x) = x3 + 0x2 + 9x + 6 ∈ Z[x] and prime p = 3 divides 6, 9, and 0, but p2 = 9 - 6,
it follows f (x) is irreducible in Q[x]. Therefore, by the Euclidean Algorithm, we
find polynomials a(x) and b(x) in Q[x] such that a(x)(1 + x) + b(x)(x3 + 9x + 6) = 1.
By long division (x3 + 9x + 6 divided by 1 + x), we have
x3 + 9x + 6 = (1 + x)(x2 − x + 10) − 4
−(1 + x)(x2 − x + 10) + x3 + 9x + 6 = −4
1
1 2
(x − x + 10)(1 + x) + (− )(x3 + 9x + 6) = 1
4
4
Now, in Q(θ), p(θ) = 0 since θ is a root of p(x). So, when θ is plugged in for x, the
equation above becomes
1
1
1 2
(θ − θ + 10)(1 + θ) + (− )(0) = (θ2 − θ + 10)(1 + θ) = 1
4
4
4
The two factors are multiplicative inverses of each other. Hence, (1 + θ)−1 =
1 2
(θ − θ + 10).
4
[2] Show that x3 − 2x − 2 is irreducible over Q[x] and let θ be a root. Compute
(1 + θ)(1 + θ + θ2 ) and (1 + θ)/(1 + θ + θ2 ) in Q(θ).
Once again by Eisenstein’s Criterion, choosing p = 2, x3 − 2x − 2 is irreducible over
Q[x]. Since θ3 − 2θ − 2 = 0, it follows θ3 = 2θ + 2. Expanding, we see (1 + θ)(1 + θ +
θ2 ) = 1+2θ +2θ2 +θ3 = 3+4θ +2θ2 . To divide, we need to find (1+θ +θ2 )−1 . As in
problem 1, we find a(x) and b(x) such that a(x)(1 + x + x2 ) + b(x)(x3 − 2x − 2) = 1
using the Euclidean Algorithm.
x3 − 2x − 2 = (x2 + x + 1)(x − 1) + (−2x − 1)
1
1
3
x2 + x + 1 = (−2x − 1)(− x − ) +
2
4
4
4
1
1
4
− (−2x − 1)(− x − ) + (x2 + x + 1) = 1
3
2
4
3
1
James Pringle
June 15, 2011
Solving the first equation for −2x − 1 and substituting that into the third equation
we have
4
1
1
4
− ((x3 − 2x − 2) − (x2 + x + 1)(x − 1))(− x − ) + (x2 + x + 1) = 1
3
2
4
3
After combining terms
(x2 + x + 1)(−2x2 /3 + x/3 + 5/3) + (x3 − 2x − 2)(2x/3 + 1/3) = 1
Therefore (1 + θ + θ2 )(−2θ2 /3 + θ/3 + 5/3) = 1, and (1 + θ + θ2 )−1 = (−2θ2 /3 +
θ/3 + 5/3). Computing,
(1 + θ)/(1 + θ + θ2 ) = (1 + θ)(−2θ2 /3 + θ/3 + 5/3)
= −2θ3 /3 − θ2 /3 + 2θ + 5/3
= −2(2θ + 2)/3 − θ2 /3 + 2θ + 5/3
= −θ2 /3 + 2θ/3 + 1/3
[3] Show that x3 + x + 1 is irreducible over F2 and let θ be a root. Compute the powers
of θ in F2 .
Since the polynomial has degree 3, if it is reducible, it will have a linear factor.
So, it is sufficient to show that it has no roots. The possible roots are 0 and 1,
the only elements in the field. Notice 03 + 0 + 1 = 1 and 13 + 1 + 1 = 1. Thus
x3 + x + 1 is irreducible over F2 . To compute the powers of θ, we use θ3 = θ + 1
(since θ3 + θ + 1 = 0).
θ0
θ1
θ2
θ3
θ4
θ5
θ6
θ7
=1
=θ
= θ2
=θ+1
= θ(θ3 ) = θ(θ + 1) = θ2 + θ
= θ(θ2 + θ) = θ + 1 + θ2 = θ2 + θ + 1
= θ(θ2 + θ + 1) = θ3 + θ2 + θ = θ2 + 1
= θ(θ2 + 1) = θ3 + θ = 1
Thus the powers of θ, θj , are unique for 0 ≤ j ≤ 6.
√
√
√
[4] Prove directly that the map a + b 2 7→ a − b 2 is an isomorphism of Q( 2) with
itself.
√
√
First√we show this
map,
call
it
ϕ,
is
a
homomorphism:
ϕ(a
+
b
2)
+
ϕ(c
+
d
2) =
√
√
√
a − b 2 + c − d 2√= a + c −√
(b + d) 2 = √
ϕ(a + c +√
(b + d) 2), showing
the
addition
√
√
part, and ϕ(a + b √
2)ϕ(c + d 2) = (a − b 2)(c
− bc 2 −√
ad 2 + 2bd =
√− d 2) = ac √
ac+2bd−(bc+ad) 2 = ϕ(ac+2bd+(bc+ad) 2) = ϕ((a+b 2)(c+d 2)), showing
the multiplication part. Hence ϕ is a homomorphism.
kernel of ϕ =
√ The √
√ {0},
and therefore
ϕ is injective. For any element a + b 2 ∈ Q( 2), ϕ(a − b 2) =
√
a + b 2, and therefore
√ ϕ is surjective. These individual parts show the map ϕ is
an isomorphism of Q( 2) with itself.
2