抽象代数Chapter11(1)习题答案

Math 5286H: Fundamental Structures of Algebra II
HW 1 Solutions, (February 3rd, 2012)
Problems from Chapter 11 of Artin’s Algebra:
√ 1.1 To show that α = 7 + 3 2 is an algebraic number, we use the Binomial Theorem and note that
S
is
also
closed
under multiplication.
1.8 (a) The units in Z/12Z are {1, 5, 7, 11}. We note that 1−1 = 1, 5−1 = 5 (25 ≡ 1 mod 12), 7−1 = 7 (49 ≡ 1 mod 12), and 11−1 = 11 (121 ≡ 1 mod 12). The other eight elements are not units since they are a factor of (or share a common factor with) zero, i.e. 12. These are called zero divisors.
1
The kernel of ϕ : Z[x] → R where ϕ(x) → γ is a principal ideal, namely (x4 − 10x2 + 1) using methods similar to Problem 1.1 and Example 11.3.23. Hence Z[γ] ∼= Z[x]/(x4 − 10x2 + 1), which means that Z[γ]
closed
under
additive
inverses,
i.e.
subtraction.
S
contains
the
fraction
1 1
,
the
multiplicative
identity.
Finally
a b
·
c d
=
ac bd
,
which
again
is
reduced
if
possible,
and
so
√
√
√
√
√√
Furthermore, γ3 = 2 2 + 3(2 3) + 3(3 √2) + 3 √3 = 11 2 + 9 3 ∈ Q[γ]. Thus by taking appropriate
Q-linear combinations of γ and γ3 yields 2 and 3 by themselves:
Using the same logic as Example 11.3.23 (i.e. using Lemma 11.3.24), we see that this is also the kernel of
the map from Z[x] → R.
d) By the sam√e log√ic as (c)√, the k√ernel is √the p√rincipal i√deal g√enerated by x4 − 10x2 + 1 which equals the product (x − 2 − 3)(x − 2 +√ 3)(x√+ 2 − 3)(x + 2 + 3), i.e. the lowest degree monic polynomial with integer coefficients having 2 + 3 as a root.
1.9 Let a, b ∈ R, we wish to show commutativity of addition. Since R is a ring, it contains a multiplicative identity element 1, and let −1 denote its additive inverse (note that it is possible, e.g. in characteristic 2, that −1 = 1 in R). Then by distributivity, (−1)(b + a) = −b − a and by associativity of addition we have
For instance 8 · 3 ≡ 0 mod 12.
(b) Similarly, the units in Z/8Z are {1, 3, 5, 7}. See part (c) for the proof.
(c) In general, in Z/nZ, the units are the subset of {1, 2, . . . , n − 1} of numbers that are relatively prime to n. Firstly, if a ∈ {1, 2, . . . , n − 1} is not relatively prime to n, i.e. gcd(a, n) = d > 1, then all integral multiples ka mod n will still be divisible by d. Hence, no such integral multiple ka can be 1. On the other hand, if gcd(a, n) = 1, then there exists r, s ∈ Z such that ra + sn = 1, i.e. there exists integer r such that ra ≡ 1 mod n. This is the definition of a unit.
also in the basis {1, γ, γ2, γ3} with integer coefficients, and hence we would again get a integer linear
combination of {1, γ, γ2, γ3} equal to α, a contradiction.) Hence, Z[γ] ⊇ Z[α, β] and so Z[γ] = Z[α, β].
1.6
(a)
Let
S
=
{
a b
:
a, b
∈
Z
and b is not divisible by 3}. To show that S is a subring of Q, we note the
following:
Firstly,
a b
+
c d
=
ad+bc bd
(where
we
then
reduce
into
1 2
γ
3
−
9 2
γ
=
Байду номын сангаас
√ 2
and
−1 2
γ
3
+
11 2
γ
=
√ 3.
Hence we have the other inclusion and we conclude that Q[α, β] = Q[γ]. However, these inclusions required rational (not int√egral) coefficients. We wish to show that there is not a Z-lin√ear combination of powers of γ equal to α = 2, i.e. an integer polynomial g(x) ∈ Z[x] such that g(γ) = 2.
has Z-basis {1, γ, γ2, γ3}. So if α ∈ Z[γ, then α = a + bγ + cγ2 + dγ3 for some integers a, b, c, d. However,
Q[x]/(x4 − 10x2 + 1)
similarly
has
Q-basis
{1, γ, γ2, γ3},
having 2 + i as a root. This ideal is principal since R[x] is a P.I.D.
√
√
c) The kernel of the map Q[x] → R is the principal ideal generated by x2−2x−1 = (x−1− 2)(x−1+ 2).
√
√
√
√
√
(7 + 3 2)3 = 343 + 3 · 49 3 2 + 3 · 7 3 4 + 2 = 345 + 147 3 2 + 21 3 4,
√
√√
and (7 + 3 2)2 = 49 + 14 3 2 + 3 4.
√ Consequently, α3 − 21α2 = −684 − 147 3 2, hence
and
we
already
know
that
α=
1 2
γ
3
−
9 2
γ
.
Thus
there cannot be an expression for α involving integer linear combinations of powers of γ. (Note: if there