MATH 6/71052
Homework #8
Solutions
Notation: If k is a positive integer, denote by k a (complex) primitive k th root of unity.
13.5 #1: If m and n are relatively prime positive integers, then m n is a primitive mnth
root of unity.
Sketch of Proof. It is e
MATH 6/71052
Homework #7
Solutions
13.5 #3: If d and n are positive integers, then xd 1 | xn 1 if and only if d | n.
Proof. By the Division Algorithm, we have n = dq + r with 0
r < d, and so
xn 1 = xdq xr xr + xr 1
= xr (xd )q 1 + (xr 1)
= xr (xd 1) (xd )
MATH 6/71052
Homework #9
Solutions
General Observations (applicable to several problems and life in general):
Terminology: R C, so real numbers are complex numbers. A complex number that
is not real has nonzero imaginary part, but is not necessarily imag
MATH 6/71052
Homework #12
Solutions
14.6 #7: The Galois group of f (x) = x4 + 2x2 + x + 3 over Q is S4 .
Sketch of Proof. First note that f (x) is irreducible over Q. Indeed, the reduction of f (x)
modulo 2 is f (x) = x4 + x + 1, which has no roots in F2
MATH 6/71052
Homework #11
Solutions
I. Let p be an odd prime, p a primitive pth root of unity, and K = Q(p ). If is any pth
1 if is primitive
root of unity, then TK ( ) =
Q
p 1 if is not primitive.
Proof. The Galois group of K over Q is Gal(K/Q) = cfw_a |
MATH 6/71052
Homework #10
Solutions
14.5 #2: Let = 8 be a primitive 8th
root of unity. The subelds of Q( ) generated by
the periods of are Q, Q( 2), Q(i 2), and Q( ). The subeld Q(i) does not have a
period as a primitive element.
Sketch of Proof. Recall f
MATH 6/71052
Homework #6
Solutions
1. An example of a chain 1
2
3
of algebraically closed elds.
Example. Let 1 = Q, the eld of algebraic numbers, which is the algebraic closure of Q in
C and hence is algebraically closed. For i 1, let xi be an indetermin
MATH 6/71052
Homework #5
Selected Solutions and Notes
13.1 #4: The map a + b 2 a b 2 is an isomorphism of Q( 2) to itself.
Note on Proof. It is straightforward to
show directly that the map is a ring homomorphism
and is surjective. Recall that since Q( 2)
MATH 6/71052
Homework #4
Selected Solutions and Notes
9.2 #1: If F is a eld and f (x) F [x] is of degree n 1, then for each g (x) F [x]/(f (x)
there is a unique polynomial g0 (x) F [x] of degree at most n 1 such that g (x) = g0 (x).
Proof. Let g (x) F [x]
MATH 6/71052
Homework #3
Selected Solutions and Notes
8.2 #1: In a PID R, two ideals (a) and (b) are comaximal if and only if a and b are
relatively prime.
Note on Proof. We have shown that if d is a GCD for a and b in a PID R, then d = ax + by
for some x
MATH 6/71052
Homework #2
Selected Solutions and Notes
7.6 #1: Let R be a ring with 1 and e R a central idempotent of R (that is, e2 = e and
re = er for all r R). Then 1 e is also a central idempotent, Re and R(1 e) are 2-sided
ideals of R and R Re R(1 e).
MATH 6/71052
Homework #1
Selected Solutions and Notes
The following results are useful for various approaches to Problem #IV and 7.4 #13. The
proofs are left as (highly recommended and straightforward) exercises. Some parts were
previously done as homewor