Mathematics 351 – Abstract Algebra 1 Solutions for Problem Set 3 September 22, 2007

Section 4.3

5. ⇒: If f and g are associates, then f = ug, where u is a unit in F [x]. By Corollary 4.9, this says f = ug where u ≠ 0 is a constant polynomial. The equation f = ug shows g|f by definition. Since u^{−1} exists in F, we also have u^{−1}f = g, which shows f|g.

⇐: Assume f|g and g|f. Then we have equations g = fk and f = gℓ for some k, ℓ ∈ F [x]. Substituting from the second equation into the first and using the associative law for multiplication in F [x], we have g = (kℓ)g. By considering degrees, we see that deg(kℓ) = 0, so both k and ℓ must be nonzero constants. Hence f, g are associates in F [x].

6. If x^{2} + 1 = (ax + b)(cx + d) for a, b, c, d ∈ Q, then by equating coefficients on both sides, ac = 1, bd = 1 and ad + bc = 0. This Shows d = \frac{1}{b} and c = \frac{1}{a}, so \frac{a}{b} + \frac{b}{a} = 0. Multiplying both sides of the last equation by ab, a^{2} + b^{2} = 0, which is only possible if a = b = 0, which is absurd since ac = bd = 1. This contradiction shows x^{2} + 1 must be irreducible in ℚ[x].

14. The two ways are x^{2}+x = x(x+1) and x^{2}+x = (x+3)(x+4) (note x^{2}+7x+12 ≡ x^{2}+xmod 6). Comment: This problem shows that unique factorization for polynomials (as in Theorem 4.12) can fail in R[x] when R is not a field!

16. ⇒: Assume that p is irreducible in F [x] and p does not divide g. We must show that (p, g) = 1. Suppose not. Then There is some polynomial d of positive degree with d|p and d|g. Since p is irreducible, the first divisibility statement can only be true if d is an associate of p. But then p|g as well since p and d differ only by a nonzero constant multiple. This is a contradiction to the assumption that d does not divide g. Hence (p, g) = 1.

⇐: Assume that p is a polynomial such that for every g ∈ F [x], either p|g or (p, g) = 1. We want to show that p must be irreducible. Suppose not. Then p = hk for two polynomials h, k of positive degrees. Apply the hypothesis with g = h. we get that either p|h or (p, h) = 1. But both of these lead to contradictions. First, p cannot divide h, since p = hk and deg(k) ≥ 1. Second, (p, h) = 1 is also impossible since h is a common divisor of h and p of positive degree. Therefore, p satisfying the given condition must be irreducible.

20. The hypothesis is that p, q are both irreducible, but p ≠ cq for any nonzero constant c ∈ F. By Exercise 16 above, since p is irreducible, either p|q or (p, q) = 1. However, p|q is not possible since q irreducible says that the only factors of q are units and associates of q. But we are given that p and q are not associates. Hence p and q are relatively prime.

21. (a) The reducible monic polynomials of degree 2 in ℤ_{p}[x] come in two “flavors:” (x + a)(x + b) with a, b ∈ F distinct, and (x + a)^{2} for a ∈ F. There are p of the second type because a can be any one of the p elements of ℤ_{p}. There are \frac{p(p−1)}{2} of the second type because:

• a can be any one of the p elements ℤ_{p},

• b can be any one of the p − 1 elements of ℤ_{p} different from a, but

• every such polynomial is obtained in exactly two different ways by this construction.

The total number is p + \frac{p(p−1)}{2} = \frac{p(p+1)}{2}.

(b) Every monic polynomial of degree 2 is either reducible or irreducible (these are mutually exclusive properties). There are p^{2} total monic polynomials f = x^{2} + αx + β of degree 2 in ℤ_{p}[x], since α can be any one of the p elements of ℤ_{p} and for each α, β can be any one of the p elements of ℤ_{p}. Hence the number of monic irreducibles of degree 2 is

p^{2} − \frac{p(p + 1)}{2} = \frac{p(p − 1)}{2}.

22. (a) The quick way to see this is to note that for all a ∈ Z_{3}, since Z_{3}[x] is a commutative ring:

(x + a)^{3} = x^{3} + 3ax^{2} + 3a^{2}x + a^{3} ≡ x^{3} + amod 3.

This shows x^{3} + a is reducible.

(b) Similar to (a):

(x + a)^{5} = x^{5} + 5ax^{4} + 10a^{2}x^{3} + 10a^{3}x^{2} + 5a^{4}x + a^{5} ≡ x^{5} + amod 5.

(where we used the fact a^{5} ≡ amod 5 for all a, easy to check, and mentioned in class).

23. (a) Since x^{2} + 2 is quadratic, using some results (Corollary 4.18 in particular) from the next section, it suffices to show that it has no roots in ℤ_{5}. This is clear since 0^{2} + 2 ≡ 2, 1^{2} + 2 ≡ 3, 2^{2} + 2 ≡ 1, 3^{2} + 2 ≡ 1, 4^{2} + 1 ≡ 3.

(b) x^{4} − 4 = (x^{2} + 2)(x^{2} − 2). Both factors can be seen to be irreducible. The first one is by part (a), the other one is by a similar argument.

Section 4.4

4. (a) x − 2 is a factor of the polynomial if and only if 2 is a root, or 16 − 40 + 20 + 6 + k = 0, which says k = −2.

(b) x + 1 is a factor if and only if −1 ≡ 4 is a root in ℤ_{5}. This says 1 − 2 − 3 + k + 1 = 0, or k = −3 ≡ 2mod 5.

5. By Theorem 4.15, x − 1_{F} divides f(x) = a_{n}x^{n} + ⋯ a_{1}x + a_{0} if and only if 1_{F} is a root of f. This is equivalent to

0_{F} = f(1_{F}) = a_{n} + a_{n−1} + ⋯ + a_{1} + a_{0},

which is what we wanted to show.

8. (e) Since this is a cubic polynomial, we can test for irreducibility by seeing if there are roots in ℤ_{11}. With x = 4, we have 4^{3} − 9 = 64 − 9 = 55 ≡ 0mod 11. Hence x − 4 is a factor of x^{3} − 9 in ℤ_{11}, and this polynomial is reducible. (f) If we substitute x = 1 we get 1^{4} + 1^{2} + 1 ≡ 0mod 3, so 1 is a root and this polynomial is reducible too.

9. (See Exercise 21 from Section 4.3 above!) There are precisely \frac{3(3−1)}{2} = 3 monic irreducibles of degree 2 in ℤ_{3}[x]:

x^{2} + 1, x^{2} + x + 2, x^{2} + 2x + 2.

The systematic ways to find them are either

• to list all monic polys of degree 2, then cross off the ones that have roots in ℤ_{3}, or

• to make all of the reducible monic polys of degree 2 in ℤ_{5} as in Exercise 21 from Section 4.3, multiply them out, and take all the polynomials not obtained that way.

Similarly, in ℤ_{5}[x], we get \frac{5(5−1)}{2} = 10 such polynomials:

x^{2} + 2, x^{2} + 3, x^{2} + 2x + 4, x^{2} + x + 1, x^{2} + 3x + 3,
x^{2} + 3x + 4, x^{2} + 4x + 1, x^{2} + 4x + 2, x^{2} + x + 2, x^{2} + 2x + 3.

12. If c_{n}a^{n} + ⋯ + c_{1}a + c_{0} = 0_{F}, then multiply both sides by (a^{−1})^{n} to obtain c_{n} + c_{n−1}a + ⋯ + c_{1}a^{n−1} + c_{0}a^{n}. This shows a^{−1} is a root of the “reversed” polynomial c_{n} + c_{n−1}x + ⋯ + c_{0}x^{n}.

13. (a) If f, g are associates in F [x], then g = uf for some u ≠ 0 ∈ F. Then for all a ∈ F, g(a) = uf(a). If a is a root of f, then f(a) = 0_{F}, so g(a) = u · 0_{F} = 0_{F} as well. Conversely, if a is a root of g, then g(a) = 0_{F}. Hence 0_{F} = uf(a), so f(a) = u^{−1}0_{F} = 0_{F}, and a is a root of f.

(b) The answer is no, since for instance g could equal f times a polynomial in F [x] that has no roots in F. An explicit counterexample is f(x) = x − 1 and g(x) = (x − 1)(x^{2} + 1) in ℚ[x]. These have the same roots in F (namely x = 1), but f, g are not associates.

16. Consider the polynomial h = f − g. By the given information deg(h) = deg(f − g) ≤ n. But h(c_{i}) = f(c_{i}) − g(c_{i}) = 0 for i = 0, …, n, so h has n + 1 roots in F. Therefore, by Corollary 4.16, h = 0 (the zero polynomial), and f = g.

18. Let f = c_{n}x^{n} + ⋯ + c_{1}xc_{0} ∈ ℚ[x]. Apply φ to both sides of the equation 0 = f(r):

0 = φ(f(r)) = φ(c_{n}r^{n} + ⋯ + c_{1}r + c_{0})
= φ(c_{n}r^{n}) + ⋯ + φ(c_{1}r) + φ(c_{0}) (sum property of homomorphisms)
= φ(c_{n})φ(r)^{n} + ⋯ + φ(c_{1})φ(r) + φ(c_{0}) (mult. property of homomorphisms)
= c_{n}φ(r)^{n} + ⋯ + c_{1}φ(r) + c_{0} (since c_{i} ∈ ℚ)

This shows that φ(r) is also a root of f.

19. (a) ⇐: If a is a multiple root of f, then f(x) = (x − a)^{k}q(x) for some k ≥ 2, and some q ∈ ℝ[x]. Hence by the product rule for derivatives

f′(x) = k(x − a)^{k−1}q(x) + (x − a)^{k}q′(x).

Since k ≥ 2, k − 1 ≥ 1, so f′(a) = 0 and a is also a root of f′. ⇒: Conversely, assuming a is a root of f and f′, write f(x) = (x − a)^{k}q(x) where q(a) ≠ 0 (i.e. k is as large as possible). Then by the equation above,

0 = f′(a) = k(a − a)^{k−1}q(a) + (a − a)^{k}q′(a) = k(a − a)^{k−1}q(a).

Since q(a) ≠ 0, k − 1 ≥ 1, and hence k ≥ 2. This shows a is a multiple root of f.

(b) We show the contrapositive: If f has a multiple root, then (f, f′) ≠ 1. But this is clear from the above, since if k ≥ 2, then (x − a)|f and (x − a)|f′.

24. Consider the mapping

φ_{a} : F [x] → F
f ↦ f(a).

Then φ_{a} is a homomorphism of rings since for all f, g ∈ F [x],

φ_{a}(f + g) = (f + g)(a) = f(a) + g(a) = φ_{a}(f) + φ_{a}(g)
φ_{a}(f · g) = (f · g)(a) = f(a) · g(a) = φ_{a}(f) · φ_{a}(g).

It is surjective (onto), since for all b ∈ F there are f ∈ F [x] such that φ_{a}(f) = f(a) = b. For a very simple example, take the constant polynomial f = b (!)