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4.2
4.2
J.A.Beachy
1
Factors
from A Study Guide for Beginner’s by J.A.Beachy,
a supplement to Abstract Algebra by Beachy / Blair
21. Over the field of rational numbers, use the Euclidean algorithm to show that
2x3 − 2x2 − 3x + 1 and 2x2 − x − 2 are relatively prime.
Let 2x3 − 2x2 − 3x + 1 = f (x) and 2x2 − x − 2 = g(x).
Solution: Using the division algorithm to divide f (x) by g(x), we obtain f (x) =
(x − 21 )g(x) − 23 x as the first step. At the next step we can use x rather than 32 x, and
then dividing g(x) by x gives us g(x) = (2x − 1) · (x) − 2. The constant remainder at
this step shows that gcd(f (x), g(x)) = gcd(g(x), x) = 1.
22. Over the field of rational numbers, find the greatest common divisor of x3 − 1 and
x4 + x3 + 2x2 + x + 1, and express it as a linear combination of the given polynomials.
Solution: Let x4 + x3 + 2x2 + x + 1 = f (x) and x3 − 1 = g(x). From the division
algorithm we first obtain f (x) = (x + 1)g(x) + 2(x2 + x + 1), and then the next step
yields g(x) = (x − 1)(x2 + x + 1), so gcd(f (x), g(x)) = x2 + x + 1. Solving the first
equation for the remainder shows that (x2 + x + 1) = 21 f (x) − 21 (x + 1)g(x).
23. Find the greatest common divisor of x3 − 2x + 1 and x2 − x − 2 in Z5 [x], and express
it as a linear combination of the given polynomials.
Solution: Rather than using the standard congruence classes 0, 1, 2, 3, 4 in Z5 , we will
use 0, ±1, ±2, and in the computations we will reduce coefficients at each step. (For
simplicity, we are omitting the brackets in the congruence class notation.)
Let x3 − 2x + 1 = f (x) and x2 − x − 2 = g(x). Using the division algorithm to divide
f (x) by g(x), we get f (x) = (x + 1)g(x) + x − 2.
x2 − x − 2
x3
x3
x
+1
−2x
−2x
+1
−x2
x2
x2
−x
+1
−2
x
−2
The next step yields g(x) = (x − 1)(x − 2), so gcd(f (x), g(x)) = x − 2. Solving the
first equation for the remainder shows that x − 2 = f (x) − (x + 1)g(x).
24. (a) Express x4 + x as a product of polynomials irreducible over Z5 .
Solution: In general, we have x4 + x = x(x3 + 1) = x(x + 1)(x2 − x + 1). The factor
p(x) = x2 − x + 1 is irreducible over Z5 since it can be checked that it has no roots in
Z5 . (We get p(0) = 1, p(1) = 1, p(−1) = 3, p(2) = 3, p(−2) = 2.)
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(b) Show that x3 + 2x2 + 3 is irreducible over Z5 .
Solution: For p(x) = x3 + 2x2 + 3, we have p(0) = 3, p(1) = 1, p(−1) = −1, p(2) = 4,
and p(−2) = 3, so p(x) is irreducible over Z5 since it has no roots in Z5 .
25. Express 2x3 + x2 + 2x + 2 as a product of polynomials irreducible over Z5 .
Solution: We first factor out 2, using (2)(−2) = −4 ≡ 1 (mod 5). This reduces the
question to factoring p(x) = x3 − 2x2 + x + 1. (We could also multiply each term by
3.) Checking for roots shows that p(0) = 1, p(1) = 1, p(−1) = −3, p(2) = 3, and
p(−2) ≡ −2, so p(x) itself is irreducible over Z5 since it has no roots in Z5 .
26. Factor x4 + 2 over Z3 .
Solution: If we replace 2 with the congruent number −1, we get x4 + 2 = x4 − 1 =
(x2 −1)(x2 +1) = (x−1)(x+1)(x2 +1). Since x2 +1 has no root in Z3 , it is irreducible
over Z3 . Using the standard congruence classes for Z3 , we have the factorization
x4 + 2 = (x + 1)(x + 2)(x2 + 1).
27. Factor x4 + 1 over Z2 , over Z5 , over Z7 , and over Z11 .
Solution: Over Z2 , we have x4 + 1 = (x2 + 1)(x2 − 1) = (x + 1)4 , and so 1 is a root
with multiplicity 4.
Over Z5 , check that there are no roots. For the factorization, we get (x2 −2)(x2 +2) =
x4 − 4 ≡ x4 + 1.
Over Z7 , there are no roots. We get (x2 + 3x + 1)(x2 − 3x + 1) = x4 − 7x2 + 1 ≡ x4 + 1.
Over Z11 , there are no roots. We get (x2 +3x−1)(x2 −3x−1) = x4 −11x2 +1 ≡ x4 +1.
Note that since x4 + 1 has no roots in Z5 , Z7 , or Z11 , it’s irreducible factors over
those fields must be quadratic, and thus the quadratic factors given above must be
the irreducible factors of x4 + 1.
Comment: Yes, there is a pattern to these factorizations. In Problem 40 you are
asked to find a method to factor x4 + 1 over Zp .
28. Find a polynomial q(x) such that (a + bx)q(x) ≡ 1 (mod x2 + 1) over Z3 .
Solution: A quick calculation shows that x2 + 1 has no roots in Z3 , so it is irreducible
in Z3 [x]. This means that we can use the Euclidean algorithm to find q(x), since it
follows that gcd((a + bx), x2 + 1) = 1.
If both a and b are zero, there is no solution. If b = 0 and a 6= 0, then we can take
q(x) to be the constant polynomial a−1 . Thus we can assume that b 6= 0. Then we
need to find a linear combination q(x)(a + bx) + s(x)(x2 + 1) of a + bx and x2 + 1 that
is equal to 1. The answer q(x) will be a multiplicative inverse in the set of congruence
classes of Z3 [x] modulo x2 + 1. The first step is to divide bx + a into x2 + 1.
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b−1 x
bx + a
x2
x2
3
−ab−2
+1
+ab−1 x
−ab−1 x
−ab−1 x
+1
−a2 b−2
1 + a2 b−2
Thus we get x2 +1 = (b−1 x−ab−2 )(bx+a)+(1+a2 b−2 ). Multiplying both sides of the
equation by b2 and rearranging the terms gives us (a+bx)(a−bx)+b2 (x2 +1) = a2 +b2 .
In Z3 , we must have b2 = 1, since b 6= 0, and then for all possible values of a we have
a2 + b2 6= 0, so (a2 + b2 )−1 exists in Z3 . Multiplying both sides by (a2 + b2 )−1 , we get
(a + bx)(a2 + b2 )−1 (a − bx) + (a2 + b2 )−1 b2 (x2 + 1) = 1. We finally have the answer:
the multiplicative inverse of a + bx is (a2 + b2 )−1 (a − bx).
Note: Compare this answer with the one you get when you compute (a + bi)−1 in C.
Alternate solution: We know that x2 + 1 ≡ 0 (mod x2 + 1), so the easiest way to
calculate the product of two congruence classes a + bx and c + dx modulo x2 + 1
(over the field Z3 ) is to take the product (a + bx)(c + dx) = ac + (ad + bc)x + bdx2
and substitute −1 for x2 , since they represent congruent polynomials. This give
(a + bx)(c + dx) ≡ (ac − bd) + (ad + bc)x (mod x2 + 1).
Now we can solve the equation (a + bx)(c + dx) ≡ 1 (mod x2 + 1). (Can we really
guarantee that the answer has to be a linear polynomial? Yes, since every polynomial
is congruent to its remainder on division by x2 + 1, and these never have degree larger
than 1.) This leads to two equations in the two unknowns c and d. We get bc + ad = 0
and ac − bd = 1; multiplying the first by b and the second by a allows us to solve for
c. As above we can reduce to the case b 6= 0, and then we get c = a(a2 + b2 )−1 and
d = −b(a2 + b2 )−1 .
29. Let F be a field, and let f (x), g(x) ∈ F [x], with deg(f (x) = n and deg(g(x) = m,
where m < n ∈ Z+ . Write n = qm + r, where r = 0 or r < m. Show that there exist
polynomials r0 (x), r1 (x), . . . , rq (x) such f (x) = rq (x)g(x)q + . . . + r1 (x)g(x) + r0 (x),
where deg(ri (x)) < m for 0 ≤ i ≤ q.
Solution: (Outline only) Use the division algorithm to write f (x) = q(x)g(x)q + r(x),
where r(x) = 0 or deg(r(x)) < mq. It follows that deg(q(x)) < m since deg(q(x)) =
r < m. Continue by induction, repeating this process with r(x) in place of f (x).
30. Let F be a field, and let f (x) ∈ F [x]. Prove that f (x) is irreducible over F if and
only if f (x + c) is irreducible over F .
Solution: Suppose that f (x) is irreducible, but f (x + c) = p(x)q(x) can be factored
into a product of polynomials of degree less than deg(f (x)). Then substituting x − c
for x gives a proper factorization f (x) = p(x − c)q(x − c), a contradiction. Thus
f (x + c) is irreducible.
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A similar proof shows the converse. Both directions depend on the fact that you get
the same answer when you first substitute and then multiply as when you multiply
first and then substitute.
ANSWERS AND HINTS
33. Over Z3 , find gcd(x5 −x4 +x3 −x2 , x3 −x2 +x−1) and write it as a linear combination
of the given polynomials.
Hint: The gcd is x3 − x2 + x − 1.
34. Over Z5 , find gcd(x5 + x4 − 2x3 − x2 + 2x − 2, x3 − x2 + x − 1) and write it as a linear
combination of the given polynomials.
Hint: The gcd is x2 + 1.
35. Over Z7 , find gcd(x5 + 3x4 − 2x3 − 3x − 3, x3 − x2 + x − 1) and write it as a linear
combination of the given polynomials.
Hint: The gcd is x2 + 1.
36. Over Q, find gcd(x5 − 8x4 + 25x3 − 38x2 + 28x − 8, x5 − x4 − 2x3 + 2x2 + x − 1) and
write it as a linear combination of the given polynomials.
Hint: The gcd is x2 − 2x + 1.
39. Factor x3 + 6 over Z7 .
Hint: x3 + 6 ≡ x3 − 1 (mod 7)
40. Show that x4 + 1 has a proper factorization over Zp , for all primes p.
2
4
2
2
Hint: If there exists b ∈ Z×
p with b = −1, then x + 1 = (x − b)(x + b). If there
2
4
2
2
×
exists a ∈ Zp with a = 2, then x + 1 = (x + ax + 1)(x − ax + 1). If there exists
2
4
2
2
a ∈ Z×
p with a = −2, then x + 1 = (x + ax − 1)(x − ax − 1). Use Problem 4.1.31
to show that one of these cases must occur. (Compare the answers to Problem 27.)