Math 0430 Exam #2
2067
Solutions
1 - 3 See the textbook.
4. 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. The polynomials of degree 2 in Z2[x] are x 2 , x 2  1, x 2  x , and x 2  x  1. By the Factor
Theorem, we can get all polynomials of degree 3 with 1 as a root by multiplying these by
(x + 1). So (x + 1) x 2 = x3 + x2, (x + 1)( x 2  1) = x3 + x2+ x +1, (x + 1)( x 2  x ) = x3 + x
and (x + 1)( x 2  x  1) = x3 + 1 are all the polynomials of degree 3 with 1 as a root.
6. x4 + x +1 = (x2 +1)( x2 +2) + (x + 2).
x2 +1 = (x + 2)(x + 1) + 2
x + 2 = 2(2x + 1) + 0, so (x4 + x +1, x2 +1) = 1 (note that it must be monic).
1 = (x + 1)(x4 + x +1) + (2x3 + 2x2+ x )( x2 +1) in Z3[x].
7.
Z 5 [ x]
( x 2  1)
has 52 = 25 elements. It contains zero divisors because x2 +1 is reducible
in Z5[x]. [x + 2] and [x + 3] are zero divisors. There are units because [1], [2], [3], [4] have
inverses. Also [2x +3][x + 1] = [1] shows that [2x +3] and [x + 1] are units.
8. To find all associates of x3 + 2x + 3 in Z12[x], we must multiply it by the units of Z12[x], the
constants that are relatively prime to 12, i.e., 1,5,7,and 11. So the associates are
x3 + 2x + 3, 5x3 + 10x + 3, 7x3 + 2x + 9, and 11x3 + 10x + 9.
9. Show that Q( 2 ) is isomorphic to Q[x]
Proof: Define : Q( 2 )  Q[x]
Q[x]
(x 2 - 2)
(x 2 - 2)
(x 2 - 2)
.
by (r + s 2 ) = [r + sx]. Since every element of
is of the form [r + sx] where r, s  Q,  is surjective. Let (a + b 2 ), (r + s 2 ) 
Q( 2 ) with (a + b 2 ) = (r + s 2 ). Then [a + bx] = [r + sx] and
(x 2 - 2) | ((a + bx) – (r + sx)). Because the degree of the polynomial on the left is 2 and the
polynomial on the on the right is either the zero polynomial or of degree less than 2, we must
conclude that ((a + bx) – (r + sx)) is the zero polynomial, and (a + bx) = (r + sx). This means
that  is injective.
Let (a + b 2 ), (r + s 2 )  Q( 2 ). Then ((a + b 2 ) + (r + s 2 )) = ((a +r) + (b + s) 2 ) =
[(a +r) + (b + s)x] = [a + bx] + [r + sx] = ((a + b 2 )) +
 (r + s 2 )). Also, ((a + b 2 )·(r + s 2 )) = ((ar + 2bs) +(as + br) 2 ) =
[(ar + 2bs) +(as + br)x] = [ar +(bs) x 2 +(as + br)x] [(a + bx)(r + sx)] =
[a + bx][r + sx] = ((a + b 2 )) · ((r + s 2 )). So  is an isomorphism.
10.
Z 3 [ x]
3
( x  2x  1)
is an integral domain that has 27 elements, since x3 + 2x + 1 is
irreducible in Z3[x].
Extra Credit: By the Remainder Theorem, this is f(1) = 1 + 2 + 1 + 7 + 2 +1 = 14 = 2 in Z12.