Homework #11 Solutions
p 348, #12 Since 3 F , is a root of the polynomial x3 3 F [x]. This polynomial is
irreducible over F since the only possible root in F would be itself, and it is easy to show
that F (if it were, would be algebraic over Q). Therefore
Homework #10 Solutions
p 348, #10 The keys to this exercise are the following.
Lemma 1. Let V be a vector space over a eld F . If cfw_v1 , . . . , vn is a linearly independent
set in V and w v1 , . . . , vn then cfw_v1 , . . . , vn , w is linearly indepe
Homework #9 Solutions
Handout, #1 As suggested, we induct on m. When m = 1 we must prove the following
statement: if p1 , q1 , . . . , qn D (n Z+ ) are primes and p1 = q1 q2 qn then n = 1. So,
suppose we have the stated hypotheses and assume that n 2. Sin
Homework #7 Solutions
#1 Let I = f (x), g (x) . Since F [x] is a PID there is an h(x) F [x] so that I = h(x) .
But this implies that h(x) divides both f (x) and g (x). As f (x) and g (x) are relatively prime
this can only happen if h(x) is a unit in F [x]
Homework #6 Solutions
p 315, #4 Let f (x) = xn + an1 xn1 + a0 Z[x] and suppose that x r divides f (x) for
some r Q. Then we must have f (r) = 0. We will use this fact to prove that in fact r Z.
If r = 0 then there is nothing to prove. So assume r = 0 and
Homework #4 Solutions
p 286, #8 Let : Zn Zn be a ring homomorphism. Let a = (1). Then for any
0 = r Zn = cfw_1, 2, . . . , n 1 we have
(r) = (1 + 1 + + 1) = (1) + (1) + + (1) = r (1) = r a = ra mod n.
r times
r times
Since
a = (1) = (1 1) = (1)(1) = a2
we
Homework #3 Solutions
pp 268-271: 12, 14, 18, 42, 44 pp 268-271: 6, 30, 32, 36, 52, 56
p 268, #6 We will nd the maximal ideals in the general case of Zn only. The ideals of Zn
are, rst of all, additive subgroups of Zn . These we know to all have the form
Homework #2 Solutions
pp 254-257: 18, 34, 36, 50, 54
p 241, #18 We apply the subring test. First of all, S = since a 0 = 0 implies 0 S .
Now let x, y S . Then a(x y ) = ax ay = 0 0 = 0 and a(xy ) = (ax)y = 0 y = 0 so
that x y, xy S . Therefore S is a subr
Homework #1 Solutions
p 241, #2 The identity element is easily seen to be 6. Indeed, in Z10 we have
26
46
66
86
=
=
=
=
12 = 2
24 = 4
36 = 6
48 = 8.
p 241, #4 There are many possible examples. Probably the simplest occurs in Z4 , where
both 1 and 3 are so
Modern Algebra II
Spring 2007
Factorization
Exercise 1. Let D be an integral domain. Prove that if p1 , p2 , . . . , pm , q1 , q2 , . . . , qn
D (m, n Z+ ) are primes and p1 p2 pm = q1 q2 qn then m = n and, after possibly
reordering, pi and qi are associ