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MAT 3110 - HW 6
12. 02 In the ring {0, 2, 4, 6, 8} under addition and multiplication mod 10,
6 · 0 = 0, 6 · 2 = 2, 6 · 4 = 4, 6 · 6 = 6, and 6 · 8 = 8. So 6 is the unity
element in the ring.
13.02 (1) Z is not a field, because 2 does not have a multiplicative inverse.
(2) Z[i] is not a field, because 2 does not have a multiplicative inverse.
(3) Z[x] is not a field, because 2x does not have a multiplicative inverse.
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(4) Z[ 2] is not a field, because 2 does not have a multiplicative inverse.
(5) Zp is a field.
13.43 Using the Cayley table for Z3 [i] on page 258, we can determine that
Z3 [i] has an element of order 8 (one of these is 1 + i). So Z3 [i] ≈ Z8 .
16.14 Let f (x) and g(x) be cubic polynomials with integer coefficients such
that f (a) = g(a) for four integer values of a. Prove that f (x) = g(x).
Let f (x) = a1 x3 + b1 x2 + c1 x + d1 and g(x) = a2 x3 + b2 x2 + c2 x + d2 .
Suppose f (xi ) = g(xi ) for some x1 , x2 , x3 , x4 ∈ Z. Then each xi is a
zero of (f − g)(x) = (a1 − a2 )x3 + (b1 − b2 )x2 + (c1 − c2 )x + (d1 − d2 ).
So (f − g)(x) is a polynomial of degree at most 3, with four distinct
roots. This implies (f − g)(x) is the zero polynomial and so a1 − a2 =
b1 − b2 = c1 − c2 = d1 − d2 = 0, thus f (x) = g(x) in Z[x].
16.16 No. Suppose for a contradiction that f (x) = an xn + · · · + a0 ∈ Z[x]
with n > 0 has a multiplicative inverse. Then there exists g(x) =
bm xm + · · · + b0 ∈ Z such that f (x)g(x) = an bm xn+m + · · · + a0 b0 = 1,
so in particular there exists bm 6= 0 ∈ Z such that bm an = 0. But
an 6= 0 and Z has no zero divisors, a contradiction. So f (x) has no
multiplicative inverse.
17.15 Let f (x) = x3 + 6 ∈ Z7 [x]. Then the roots of f (x) are 1, 2 and 4 so
f (x) = (x − 1)(x − 2)(x − 4) = (x + 6)(x + 5)(x + 3) ∈ Z7 [x].
17.16 Let f (x) = x3 + x2 + x + 1 ∈ Z2 [x]. The only root of f (x) in Z2 is
1, so x − 1 = x + 1 is a factor of f (x). Using polynomial division
we obtain f (x) = (x + 1)(x2 + 1). But then x2 + 1 is not irreducible,
since it has 1 as a root, so using polynomial division again we obtain
x2 + 1 = (x + 1)(x + 1). So then f (x) = (x + 1)3 .
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