Mathematics 351 – Abstract Algebra 1 Solutions for Problem Set 4 September 28, 2007

Section 4.5

1 e. By the Rational Root Test, the possibilities for rational roots of f(x) = 2x^{4}+7x^{3}+5x^{2}+7x+3 = 0 are

x = ±1, ±3, ±\frac{1}{2}, ±\frac{3}{2}.

Since all of the coefficients in f(x) are positive, note that only the negative numbers here are actually candidates as roots. Testing them one by one, we find f(−1) = −4, f(−3) = 0, f (\frac{−1}{2}) = 0 and f (\frac{−3}{2}) = −\frac{39}{4}. Hence

f(x) = (2x + 1)(x + 3)(x^{2} + 1),

(The x^{2} + 1 is found by dividing (2x + 1)(x + 3) = 2x^{2} + 7x + 3 into f(x) and taking the quotient. The remainder must be zero by the Factor Theorem.) Each factor is clearly irreducible in ℚ[x].

3. By the Rational Roots Test, if f(x) = x^{n} + a_{n−1}x^{n−1} + … + a_{1}x + a_{0} ∈ ℤ[x], then any rational root of f must have the form x = \frac{p}{q} where p, q ∈ ℤ and p|a_{0} and q|1. This means q = ±1, so x ∈ ℤ.

4 a. First, by the Rational Roots Test, the only possible roots of the polynomial f(x) = x^{4} + 2x^{3} + x+1 in ℚ are x = ±1. But neither is a root. This means that no ax+b ∈ ℤ[x] divides f(x). We will use Theorem 4.22 now to say that it is enough to consider factorizations of f(x) into polynomials with integer coefficients. However, it is still possible that we have f(x) = (x^{2} + ax + b)(x^{2} + cx + d) for some a, b, c, d ∈ ℤ. However, this would say x^{4} + 2x^{3} + x + 1 =

(x^{2} + ax + b)(x^{2} + cx + d) = x^{4} + (a + c)x^{3} + (b + ac + d)x^{2} + (ad + bc) + bd,

so a + c = 2, b + ac + d = 0, ad + bc = 1 and bd = 1. The last equation says b = d = 1 or b = d = −1 since b, d ∈ ℤ. In the first case the other equations reduce to a + c = 2, ac + 2 = 0, and a + c = 1 This is a contradiction since a + c cannot be 1 and 2(!) The other case is similar. We get a + c = 2, ac − 2 = 0, and −a − c = 1. This is also impossible. Hence f(x) must be irreducible.

5. a. Eisenstein applies using p = 2 (2 |̸ 1, 2|4 and 2|22, but 4 |̸ 22). So the polynomial is irreducible in ℚ[x]. b. Eisenstein applies using p = 5 so the polynomial is irreducible in ℚ[x]. c. Eisenstein applies using either p = 2 or p = 3 so the polynomial is irreducible in ℚ[x]. Comment: Note that part c shows it is possible for several different primes to satisfy the coditions in Eisenstein’s Criterion.

6. By the Eisenstein Criterion, x^{9} + 12x^{5} − 21x + k is irreducible whenever p = 3 divides k, but 9 |̸ k. There are infinitely many k that satisfy these requirements: we can take k = 3ℓ for any ℓ ∈ ℤ satisfying ℓ ≡ 1 or 2 mod 3.

12. Following the hint to prove the contrapositive form, assume f(x) is reducible in F [x]. Then we can factor it as f(x) = g(x)h(x) with deg(g), deg(h) ≥ 1. From this equation, it follows that for all c ∈ F, f(x + c) = g(x + c)h(x + c). If g(x) = a_{r}x^{r} + ⋯ + a_{1}x + a_{0}, then g(x + c) = a_{r}(x + c)^{r} + ⋯ + a_{1}(x + c) + a_{0} = a_{r}x^{r} + · · ·. So deg(g(x)) = deg(g(x + c)). Similarly for h(x). Hence the equation f(x + c) = g(x + c)h(x + c) shows f(x + c) is also reducible.

13. Let f(x) = x^{4} + 4x + 1. Then f(x + 1) = x^{4} + 4x^{3} + 6x^{2} + 8x + 6 (using the binomial theorem to expand (x + 1)^{4}). Hence by the Eisenstein Criterion with p = 2, f(x + 1) is irreducible in ℚ[x]. By Exercise 12, f(x) is also irreducible.

14. Let f(x) = x^{4} + x^{3} + x^{2} + x + 1. Then

f(x + 1) = (x^{4} + 4x^{3} + 6x^{2} + 4x + 1) + (x^{3} + 3x^{2} + 3x + 1) + (x^{2} + 2x + 1) + x + 1 + 1
= x^{4} + 5x^{3} + 10x^{2} + 10x + 5

The Eisenstein Criterion with p = 5 shows that this polynomial is irreducible in ℚ[x]. Hence f(x) is irreducible in ℚ[x] as well by Exercise 12. Comment: You can also derive the formula above for f(x + 1) by noticing that f(x) = \frac{x^{5}−1}{x−1}. Hence

f(x + 1) = \frac{(x + 1)^{5} − 1}{x} = x^{4} + 5x^{3} + 10x^{2} + 10x + 5.

which explains why we are seeing the binomial coefficients \binom{5}{ℓ} in the expansion of f(x + 1).