Sample Exam 1 Solutions
1. (a) This IS a group. We quickly verify the 4 necessary properties.
Closure: It suces to verify that a b = 1 (mod 13) for all a, b G since 1
12 (mod 13) and G contains all other congruence classes modulo 13.
a b = 12 a + b + ab
Homework 12 Solutions
9.26 Assume F and K are elds with F K and f, g K [x] with f, g irreducible and not
associates in K . Notice that f and g are relatively prime in K [x]. If h K [x] is such that h|f
and h|g then f = ha and g = hb for some a, b K [x]. S
21-373 Exam 1 Version B Solutions
March 1, 2012
1.
(a) We claim that this is a group.
Closure: Let a, b be two nonzero rational numbers and consider ab . It should be
2
clear that this product is still rational. Moreover, it is zero if and only if one of
Exam 2B Solutions
1. Z/6Z. We know modular arithmetic is commutative, so this is a commutative ring under the
operations of addition and multiplication modulo 6. However, its not an integral domain
because it has zero divisors. 3 2 0 (mod 6) but 3 0 and 2
Homework 1 Solutions
1.13 Case 1: I = cfw_0. Let d = 0 so I = dZ as desired.
Case 2: I = cfw_0. Then a I such that a = 0. Property 2 implies that a I as well and
one of these 2 must be positive. Let S be the set of positive integers in I . We know S =
so
Homework 1 Solutions
3.6 (1) Q( 2) is a group.
Closure For x, y Q( 2) , xy = 0 since x, y = 0. Also xy = (a + b 2)(c + d 2) =
(ac + 2bd) + (ad + bc) 2 Q( 2) .
Associativity Follows from the associativity of multiplication on R.
Identity 1 Q( 2) and 1 is t
Homework 3 Solutions
3.22 () If H G and a, b H then ab H since subgroups are closed under the binary
operation inherited from G (subgroups are groups themselves).
() Assume H is a non-empty subset of G having the property that if a, b H then ab H .
We wis
Homework 4 Solutions
4.16 Assume K H G. We want to show that [G : K ] = [G : H ][H : K ]. (Note that we can not
assume that any of these groups are nite.). We break this up into 2 parts. First we show
that if both sides of the equation are nite then this
Homework 5 Solutions
4.48
1) Let N G and assume gN g 1 N for all g G. We will show N G by showing that
gN = N g for all g G by double containment.
Let g G and n N . By assumption, gng 1 N so gng 1 = n for some n N . Thus
gn = n g and therefore gN N g . Si
Homework 6 Solutions
5.12 We assume G and X are both nite and F (g ) = |cfw_x X | gx = x|. Consider the set
S = cfw_(g, x) G X | gx = x. We count the size of this set in two ways (by counting over
the g s and by counting over the xs).
By counting over the
Homework 7 Solutions
6.8 Assume G is a nite p-group and H G is nontrivial. We know |G| = pn for some n 1
by Proposition 6.4 and the fact that it has a nontrivial subgroup. Thus |H | = pk for some
1 k n by Lagranges theorem. Let G act on H by conjugation.
Homework 8 Solutions
6.28 Let G be a group with |G| = p2 q . First we suppose that p and q are distinct primes. Clearly
if np = 1 or nq = 1 were done, since then one of the Sylow subgroups would be a nontrivial,
proper normal subgroup. BWOC, suppose that
Homework 9 Solutions
7.7 (): By contrapositive. Suppose m is composite. Let m = ab for 1 < a, b < m. We claim
a Z/mZ has no multiplicative inverse modulo m. To see this, suppose ax 1 (mod m).
Then m|ax 1 which implies ax 1 = mn for some n Z which in turn
Homework 10 Solutions
8.3 Dene X = C ([0, 1]) and M = cfw_f X | f (1) = 0. We claim M is a maximal ideal.
First we show that M is an ideal using the ideal test. Clearly M = because the
constant 0 function, f (x) = 0 x, is in M . Let f, g M . Then (f g )(
Homework 11 Solutions
8.27 Assume R is a UFD and P R is a nonzero prime ideal. Let a P with a = 0. Note that a
is necessarily not a unit since otherwise P = R. Since R is a UFD we can write a as
a = p1 p2 pn
with each pi prime (equivalently, each is irred
Homework 11 Solutions
8.27 Assume R is a UFD and P R is a nonzero prime ideal. Let a P with a = 0. Note that a
is necessarily not a unit since otherwise P = R. Since R is a UFD we can write a as
a = p1 p2 pn
with each pi prime (equivalently, each is irred
Homework 10 Solutions
8.3 Dene X = C ([0, 1]) and M = cfw_f X | f (1) = 0. We claim M is a maximal ideal.
First we show that M is an ideal using the ideal test. Clearly M = because the
constant 0 function, f (x) = 0 x, is in M . Let f, g M . Then (f g )(
Sample Exam 2 Solutions
1. M atn (R) and the quaternions are two of the standard examples. There are, of course, many
more possibilities.
2. (a) Irreducible by Eisensteins Criterion. 3 does not divide the leading coecient, but 3
divides all of the other c
Exam 1A Solutions
1. (a) (Z, #) is a group. We quickly verify that it satises the 4 necessary properties.
Closure: Since Z is closed under addition we know that for x, y Z, x#y = x+y +2 Z.
Associativity: Let x, y, z Z
(x#y )#z = (x + y + 2)#z
= (x + y + 2
21-373 Exam 1 Version B Solutions
March 1, 2012
1.
(a) We claim that this is a group.
Closure: Let a, b be two nonzero rational numbers and consider ab . It should be
2
clear that this product is still rational. Moreover, it is zero if and only if one of
Exam 2A Solutions
1. R must be a commutative, have a multiplicative identity and have no zero divisors.
2. We prove this with a chain of is
g CG (x1 ax) gx1 ax = x1 axg
xgx1 a = axgx1
xgx1 CG (a)
g x1 CG (a)x
3. (a) Irreducible by the Mod p irreducibil
Exam 2B Solutions
1. Z/6Z. We know modular arithmetic is commutative, so this is a commutative ring under the
operations of addition and multiplication modulo 6. However, its not an integral domain
because it has zero divisors. 3 2 0 (mod 6) but 3 0 and 2
Homework 1 Solutions
1.13 Case 1: I = cfw_0. Let d = 0 so I = dZ as desired.
Case 2: I = cfw_0. Then a I such that a = 0. Property 2 implies that a I as well and
one of these 2 must be positive. Let S be the set of positive integers in I . We know S =
so
Homework 1 Solutions
3.6 (1) Q( 2) is a group.
Closure For x, y Q( 2) , xy = 0 since x, y = 0. Also xy = (a + b 2)(c + d 2) =
(ac + 2bd) + (ad + bc) 2 Q( 2) .
Associativity Follows from the associativity of multiplication on R.
Identity 1 Q( 2) and 1 is t
Homework 3 Solutions
3.22 () If H G and a, b H then ab H since subgroups are closed under the binary
operation inherited from G (subgroups are groups themselves).
() Assume H is a non-empty subset of G having the property that if a, b H then ab H .
We wis
Homework 4 Solutions
4.16 Assume K H G. We want to show that [G : K ] = [G : H ][H : K ]. (Note that we can not
assume that any of these groups are nite.). We break this up into 2 parts. First we show
that if both sides of the equation are nite then this
Homework 5 Solutions
4.48
1) Let N G and assume gN g 1 N for all g G. We will show N G by showing that
gN = N g for all g G by double containment.
Let g G and n N . By assumption, gng 1 N so gng 1 = n for some n N . Thus
gn = n g and therefore gN N g . Si
Homework 6 Solutions
5.12 We assume G and X are both nite and F (g ) = |cfw_x X | gx = x|. Consider the set
S = cfw_(g, x) G X | gx = x. We count the size of this set in two ways (by counting over
the g s and by counting over the xs).
By counting over the
Homework 7 Solutions
6.8 Assume G is a nite p-group and H G is nontrivial. We know |G| = pn for some n 1
by Proposition 6.4 and the fact that it has a nontrivial subgroup. Thus |H | = pk for some
1 k n by Lagranges theorem. Let G act on H by conjugation.
Homework 8 Solutions
6.28 Let G be a group with |G| = p2 q . First we suppose that p and q are distinct primes. Clearly
if np = 1 or nq = 1 were done, since then one of the Sylow subgroups would be a nontrivial,
proper normal subgroup. BWOC, suppose that