Math 0430
04-3
Sample Final Exam Solutions
1. See the textbook.
2. (a) If (a,c) = 1 and (b,c) = 1, then there are integers s, t, u and v such that 1 = ua + vc and
1 = sb + tc. So 1 = usab + vcsb +uatc +vctc = usab+ c (vsb + aut +vct). Therefore, 1 is
a linear combination of ab and c, so (ab, c) = 1.
(b) Let d = (a,b) and c = sa + tb where a, b, c, d, s, t  . Then there are integers x and y
such that a = xd and b = yd. So c = sxd +tyd = d(xs +ty) and d | c.
3. (a) Let n = 2k where k is an odd integer. Then there is an integer x such that k = 2x + 1,
n = 4x + 2, and k2 = 4x2 + 4x + 1. So k2- k = 4x2 + 4x + 1 – (2x + 1) = 4x2 + 2x =
x(4x + 2) = xn and n | k2- k . Therefore, k2 k(mod n).
(b) Let n be an odd integer. Then there is some integer x with n = 2x+ 1. n 2 = 4x2 + 4x + 1=
4x(x + 1) + 1 and n2 - 1= 4x(x + 1). Now, 2|x or 2|(x + 1), so 8| (n2 - 1) and n21(mod 8).
4. (a) Let f(x)  F[x]. Then f(x) – f(x) = 0F, and p(x) |( f(x) – f(x)). So f(x)  f(x)(mod p(x)) and
the relation of “congruence modulo p(x)” is reflexive.
(b) Let f(x)  g(x)(mod p(x). Then p(x) | (f(x) – g(x)), so p(x) | (g(x) – f(x)) and
g(x)  f(x)(mod p(x)). Therefore, the relation of “congruence modulo p(x)” is symmetric.
(c) Let f(x)  g(x)(mod p(x)) and g(x)  h(x)(mod p(x)). Then p(x) | (f(x) – g(x)) and
p(x) | (g(x) – h(x)). So there are polynomials a(x) and b(x) in F[x] such that f(x) – g(x) =
a(x)p(x) and g(x) – h(x) = b(x)p(x). Therefore, f(x) – h(x) = (f(x) –g(x)) + (g(x) – h(x)) =
a(x)p(x) + b(x)p(x) = p(x)(a(x) + b(x)) and p(x) | (f(x) - h(x)). So f(x)  h(x)(mod p(x)) and
the relation of “congruence modulo p(x)” is transitive.
5. x5 +1 = x2 ( x3 +1) + (x2 +1).
x3 +1 = x (x2 +1).+ (x + 1)
x2 +1 = (x + 1)(x + 1) + 0, so (x5 +1, x3 +1) = x +1
x + 1 = x(x5 +1) + ( x3 +1)) ( x3 +1) in 2[x].
6. The ring
Z 2 [ x]
3
( x  x  1)
has 23 = 8 elements:
[0], [1], [x], [x + 1], [ x 2 ], [ x 2  1], [ x 2  x ], and [ x 2  x  1]. x3 + x +1,is irreducible in Z2[x],
so the ring is a field and has no zero divisors. Every nonzero element is a unit, so the
units are: [1], [x], [x + 1], [ x 2 ], [ x 2  1], [ x 2  x ], and [ x 2  x  1].
[x3 + x2 +1] [ x 2  x  1] = [x5+ x4+ x2+ x4+ x3+ x + x3+ x2+ 1] = [x5+ x + 1] =
[x2(x + 1) + x + 1] = [x3+ x2+ x+1] = [x2].
a b
a b
 | a, b  R}. Show that S is an ideal of the ring T = { 
 | a, b, c  R}
7. Let S = { 
0 0
0 c
0 0
 a b   c d
 r s
 ,  T . Then
  S. Let 
 , 
  S and 
Proof: S is nonempty since 
0 t 
0 0
0 0 0 0
 a b   c d   a  c b  d
 a b   r s   ar as  bt 
 r s a b
  S, 
 = 
 · S, and 
 

 - 
 = 
 
 =
0 
0 
 0 t  0 0
0 0 0 0  0
0 0  0 t   0
 ra rb 

  S. So S is an ideal of T
0 0
8. Show that 5 = (5) is a prime ideal of , but 10 = (10) is not.
Proof. If a, b   such that ab  (5), then 5 | ab. Since 5 is prime, 5|a or 5|b. This implies
that a  (5) or b  (5). Therefore, (5) is a prime ideal of .
2·5 10, but 2 10 and 5  10 , so 10 is not a prime ideal of .
9. A homomorphism :   2 is defined by (z) = [r] where z = 2q + r, 0  r < 2 as in the
Division Algorithm. Find the kernel of  and express it as a principal ideal of .
Proof: Note that [z] = [r] here. Let K be the kernel of . Then K = { z | (z) = [0]} =
{z | [r] = [0]} = { z | [z] = [0]} = {z | (2 | z)} = (2).
10. Define multiplication on Z3  Z3 by (a,b)  (c,d) = (ac – bd, bc + ad) Complete the
multiplication table for Z3  Z3 , noting that the multiplication is commutative. Is this the
multiplication table of an integral domain? Of a field? Justify your answers.

(0,0)
(0,0)
(0,0)
(0,1)
(0,0)
(0,2)
(0,0)
(1,0)
(0,0)
(1.1)
(0,0)
(1,2)
(0,0)
(2,0)
(0,0)
(2,1)
(0,0)
(2,2)
(0,0)
(0,1)
(0,0)
(2,0)
(1,0)
(0,1)
(2,1)
(1,1)
(0,2)
(2,2)
(1,2)
(0,2)
(1,0)
(0,0)
(0,0)
(1,0)
(0,1)
(2,2)
(0,2)
(0,2) (1,2)
(1.0) (1,1)
(2.2) (0,1)
(1,2) (2,0)
(1,1)
(2,1)
(2,1)
(2,2)
(1,1)
(0,0)
(2,1)
(1,2)
(1,1)
(2,0)
(2,2)
(1,0)
(0,1)
(1,2)
(0,0)
(1,1)
(2,2)
(1.2) (2.0) (0,1)
(2,1)
(0,2)
(1,0)
(2.0) (0,0)
(2.1) (0,0)
(0,2)
(2,2)
(0,1)
(1,1)
(2.0) (2,2)
(2.1) (1,0)
(2,1)
(1,2)
(1,0)
(1,2)
(1,2)
(0,1)
(1,1)
(2,0)
(2,2)
(1,2)
(2,1)
(2,2)
(1,0)
(1,1)
(2,0)
(0,1)
(0,0)
(0,2)
(0,1)
This is the multiplication table of an integral domain because there are no zero divisors. It is
the multiplication table of a field because a finite integral domain is a field or because every
nonzero element has a multiplicative inverse.