114 Groups and Rings 2008–09
Suggested solutions to exercise set 9
April 27, 2009
35.8 Let a(x) = 2x5 − 3x3 + 2x + 1. Then
a(−3) = 2 · (−3)5 − 3 · (−3)3 + 2 · (−3) + 1 = −410,
which, by the Remainder Theorem, is the remainder when a(x) is divided by
x + 3 = x − (−3).
35.10 Let a(x) = ix9 + 3x7 + x6 − 2ix + 1. Then
a(−i) = i · (−i)9 + 3 · i7 + i6 − 2i · i + 1 = 1 − 3i − 1 + 2 + 1 = 3 − 3i,
which, by the Remainder Theorem, is the remainder when a(x) is divided by
x + i = x − (−i).
35.13 By the Factor Theorem, x − 2 is a factor of x4 + x3 + x2 + x in Zp if and only
if [24 + 23 + 22 + 2] = [0] in Zp , i.e. p|30 = 2 · 3 · 5. Since p an odd prime, this
gives p = 3 or p = 5.
35.15 Let a(x) = 1 and b(x) = 2 in Z[x]. Suppose that a(x) = b(x)q(x)+r(x) where
q(x), r(x) ∈ Z[x] and either r(x) = 0 or deg r(x) < deg b(x). Since deg b(x) =
0, we must have r(x) = 0, so 1 = 2q(x); if q(x) = q0 + q1 x + · · · + qn xn
where q0 , . . . , qn ∈ Z then by comparing constant terms we obtain 1 = 2q0 ,
which has no solution q0 ∈ Z. So we cannot write a(x) in this way and the
conclusion of the Division Algorithm fails.
35.17 f (x) = x(x − 1)(x − 2)(x − 3)(x − 4) will do, since for any c ∈ Z5 we see
that x − c divides f (x), so by the Factor Theorem, f (c) = 0, so c is a root
of f (x).
36.10 We have
x4 + x2 + 1 = (x2 + 1)2 − x2 = (x2 + 1 + x)(x2 + 1 − x).
So this polynomial is reducible.
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If we didn’t spot this trick, we could argue as follows: x4 + x2 + 1 has
no roots in Z5 , since it never takes the value [0] when you substitute x =
[0], [1], [2], [3], [4]. So it has no factor of degree 1, so it cannot have a factor of
degree 3. So we should look for two factors of degree 2, and we may assume
that they are both monic. Now write x4 + x2 + 1 = (x2 + bx + c)(x2 + βx + γ),
expand and equate coefficients, and you should find the factors above.
36.12 We have x5 + x4 + x2 + 2x = x(x4 + x3 + x + 2). Let a(x) = x4 + x3 + x + 2.
Since a(0) = 2, a(1) = 2 and a(2) = 1 in Z3 , a(x) has no linear factors by
the Factor Theorem. However, it does have quadratic factors, since we can
solve
a(x) = (x2 + bx + c)(x2 + βx + γ)
for b, β, c, γ ∈ Z5 . Equating the constant coefficients gives 2 = cγ, so one of
c and γ is 2 and the other is 1. Without loss of generality, c = 1 and γ = 2.
Then equating the other coefficients gives:
x1 :
x2 :
x3 :
1 = 2b + β
0 = bβ
1=b+β
By inspection, one (and in fact the only) solution is b = 0 and β = 1. We
obtain the factorisation
x4 + x3 + x + 2 = (x2 + 1)(x2 + x + 2).
Since this polynomial has no linear factors, the two quadratic factors must
be irreducible. So
x5 + x4 + x2 + 2x = x(x2 + 1)(x2 + x + 2)
is the required factorisation.
36.19 (a) If f (x) = a0 + a1 x + · · · + an xn then f (1) = a0 + a1 + · · · + an is the sum
of the coefficients of f (x). If f (x) ∈ Z2 [x] then every non-zero aj is equal
to 1 = [1]. So if m is the number of nonzero coefficients, then f (1) = [m]
which is equal to zero if and only if m is even. By the Factor Theorem,
x − 1|f (x) ⇐⇒ f (1) = 0 ⇐⇒ m is even.
(b) If f (x) is irreducible over Z2 and deg f (x) > 1 then x − 1 6 |f (x), so the
number of nonzero coefficients of f (x) is odd. Also, x 6 |f (x), so the constant
coefficient of f (x), which is f (0), is not zero (by the Factor Theorem). So it
is 1.
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(c) Following the rule from (b), the only possible irreducible polynomials of
degree 4 or less over Z2 are:
degree
degree
degree
degree
1:
2:
3:
4:
x, x + 1
x2 + x + 1
x3 + x2 + 1, x3 + x + 1
x4 + x3 + x2 + x + 1, x4 + x3 + 1, x4 + x2 + 1, x4 + x + 1.
Which of these actually are irreducible? Any linear (degree one) polynomial
is irreducible. Moreover, by following the rule in (b) we have ensured that
no linear polynomial divides any of these, so the degree 2 and degree 3
polynomials we’ve listed are irreducible. Since the degree 4 polynomials
above have no linear factor, if they are not irreducible they can only have
an irreducible degree 2 factor, i.e. x2 + x + 1, and the quotient must also be
irreducible of degree 2, so we just have to eliminate (x2 +x+1)2 = x4 +x2 +1:
this is not irreducible but the others are.
(d) Using the table above, we obtain the following factorisations into irreducible polynomials:
x3 + 1 = (x + 1)(x2 + x + 1)
x3 + x = x(x2 + 1) = x(x + 1)(x + 1)
x3 + x + 1 is irreducible
x3 + x2 = x · x · (x + 1)
x3 + x2 + 1 is irreducible
x3 + x2 + x = x(x2 + x + 1)
x3 + x2 + x + 1 = (x + 1)(x + 1)(x + 1).
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