Math 307 Abstract Algebra Homework 12 Sample solution 1. (a) Write x3 + 6 ∈ Z7 [x] as a product of irreducible polynomials over Z7 . (b) Write x3 + x2 + x + 1 ∈ Z2 [x] as a product of irreducible polynomials over Z2 . Solution. (a) Let f (x) = x3 + 6, then f (1) = f (2) = f (4) = 0. Thus, (x − 1), (x − 2), (x − 4) are factors, and one easily checks that f (x) = (x − 1)(x − 2)(x − 4). (b) Let g(x) = x3 + x2 + x + 1. Then g(1) = 0 and we have g(x) = (x + 1)(x2 + x + 1). Note that x2 + x + 1 has no linear factor, and is irreducible. 2. (a) Find all units in Z4 [x] with degree at most 1. (b) Let p be a prime. Find all units in Zp [x]. (c) Show that 1 is the only solution of x25 − 1 = 0 in Z37 . Solution. (a) Suppose f (x) = ax + b is a unit, and g(x) = c0 + · · · + cr xr is its inverse. Then f (x)g(x) = 1. Thus, acr = 0 and c0 b = 1. Because cr 6= 0, it follows that a ∈ {0, 2} and b ∈ {1, 3}. If a = 0, then f (x) = b is a unit if and only if a ∈ {1, 3}. Suppose a = 2 and b ∈ {1, 3}. Then f (x) = 2x + b has inverse f (x). Thus, {1, 3, 2x + 1, 2x + 3} are the required set. (b) If f (x) ∈ Zp [x] has positive degree, then so is f (x)g(x) for any nonzero g(x) ∈ Zp [x], and thus f (x)g(x) 6= 1. Hence, units in Zp [x] must have zero degree. One sees that the units are 1, . . . , p − 1. (c) Note that (F∗ , ·) is a group with 36 elements. If x ∈ F satisfies x25 = 1, then the order of x is a factor of 25 in F∗ . Now, x ∈ F∗ implies that the order of x is a factor of 36. Thus, x has order 1. 3. Determine which of the polynomials below is (are) irreducible over Q. (a) x3 + x2 + x + 1. (b) x4 + x + 1. (c) x5 + 5x2 + 1. Solution. (a) Let f (x) be the given polynomial. By factor theorem, −1 is a zero as f (−1) = 0. Thus, (x + 1) is a factor of f (x), and it is reducible. (b) Let f (x) be the given polynomial. If (ax + b) is a factor, then a and b are factors of 1. We may assume that a = 1 and b ∈ {1, −1}. But f (1) 6= 0 and f (−1) 6= 0. Thus, f has no linear factor. Now, suppose x4 + x + 1 is a product of two degree 2 polynomials. Then we may assume that the leading coefficients are 1 and the constant terms are both 1 or −1. Thus, f (x) = (x2 + ax + c)(x2 + bx + c) = x4 + (a + b)x3 + (ab + 2c)x2 + c(a + b)x + c2 . So, a + b = 0 and 2c + ab = 1, which is impossible. Hence f (x) is irreducible. (c) Let f (x) be the given polynomial. Because f (1) 6= 0 and f (−1) 6= 0, f (x) has no linear factor. Assume ax2 + bx + c is a degree 2 factor. Then we may assume that a = 1 and c = ±1, and f (x) = (x2 + bx + c)(x3 + dx2 + f x + g) = x5 + (d + b)x4 + (c + bd + f )x3 + (g + bf + cd)x2 + (bg + f c)x + cg. Thus, g = c, b + d = 0, 0 = gb + cf = c(b + f ), and c + b − b2 = c + bd + f = 0. So, c = b(b − 1), which is impossible. Hence f (x) is irreducible. 1 4. Show that f (x) = x3 + 2x + 1 ∈ Z3 [x] is irreducible. Let F = Z3 [x]/A with A = hf (x)i. (a) Let a = (x2 + 1) + A and b = (x2 + x + 1) + A in F. Compute ab, a2 , a−2 . (b) Find an element in F satisfying y 3 + 2y + 1 = 0. (c) Find a ∈ F∗ such that F∗ = hai under multiplication. Solution. One easily checks that f (k) 6= 0 for k = 0, 1, 2 ∈ Z3 . Thus, f (x) has no linear factor and is irreducible. (a) ab = (x2 + 1)(x2 + x + 1) + A = x4 + x3 + 2x2 + x + 1 + A = x(x3 + 2x + 1) + A = x3 + 1 + A = −2x + A = x + A. a2 = (x4 + 2x2 + 1) + A = x(x3 + 2x + 1) − x + 1 + A = 2x + 1 + A. Suppose a−1 = cx2 + dx + e. Then aa−1 = 1 implies that (x2 + 1)(cx2 + dx + e) = cx4 + dx3 + (c + e)x2 + dx + e = (x3 + 2x + 1)h(x) + 1 for some h(x) ∈ Z3 [x]. Applying long division, we see that h(x) = cx + d, and thus 2c = c + e, d = c + 2dc, 1 + d = e. It follows that d = 1 = −e = −c. One can check that (2x2 + x + 2) + A is the inverse of a. (b) Let y = x + A. Then y 3 + 2y + 1 = x3 + 2x + 1 + A = 0 + A. (c) Note that a2 = x + A 6= 1 + A. Also, a12 = x6 + A = (−2x − 1)2 + A = (x2 + x + 1) + A, which is not the inverse of a. So, a13 6= 1 + A. Note that the order of a must be a factor of 27 − 1 = 26. So, a has order 26 and is a generator of F∗ . 5. Let F be a field. Show that I = {a0 + · · · + an xn : ai ∈ F, a0 + · · · + an = 0} is an ideal, and find f (x) ∈ F[x] such that I = hf (x)i. Solution. Clearly, g(x) ∈ I if and only if g(1) = 0. Thus, h(x − 1)i. 6. Let p be a prime. (a) Show that the number of reducible polynomials over Zp of the form x2 +ax+b is p(p+1)/2. (b) Determine the number of irreducible quadratic polynomials over Zp . Solution. (a) For a reducible polynomials of the form x2 + ax + b = (x − u)(x − v), correspond to distinct choices of u, v ∈ Zp allowing repetition. So, there are p(p + 1)/2 pairs, namely, 0 ≤ u ≤ v ≤ q. (b) There are p2 polynomials of the form x2 + ax + b, and p(p + 1)/2 are reducible. So, there are p(p − 1)/2 irreducible monic polynomial. For every such polynomial g(x), we can get p − 1 polynomials of the form rg(x) with r ∈ {1, . . . , p − 1}. Thus, there are (p − 1)2 p/2 reducible polynomials. 2 Optional 7. Suppose F is a finite field with n elements. (a) Suppose the finite Abelian group (F∗ , ·) is isomorphic to Zpr1 ⊕ · · · ⊕ Zprk . 1 Show that if k lcm(pr11 , . . . , prk ) k = m ≤ n, then every element in F∗ satisfies xm − 1 = 0. (b) Recall that a degree s polynomial in F[x] can have at most s distinct roots. Show that m = n − 1 in (a). (c) Deduce from (a) and (b) that (F∗ , ·) is a cyclic group. Solution. (a) Since every element y in Zpr1 ⊕ · · · ⊕ Zprk satisfies y + · · · + y = (1, . . . , 1), and 1 | {z } k (F∗ , ·) is isomorphic to this group, we have am = 1 for every a ∈ F∗ . m (b) By (a), every element a ∈ F∗ satisfies xm −Q1 = 0. But this polynomial can have at most r m distinct roots. So, m ≥ n − 1. But n − 1 = pj j ≥ m. So, n − 1 = m. Q r (c) Note that lcm(pr11 , . . . , prkk ) = pj j = n − 1 if and only if all the primes p1 , . . . , pk are distinct. In such a case, we see that (1, . . . , 1) has order n − 1. So, Zpr1 ⊕ · · · ⊕ Zprk is cyclic, 1 k and so is F∗ . 8. (8 points) Let F be a field. For f (x) = a0 + · · · + an xn ∈ F[x] let D(f (x)) = a1 + 2a2 x + 3a3 x2 + · · · + nan xn−1 , where kak = ak + · · · + ak (k times), and Dr (f (x)) = D(Dr−1 (f (x))) for 2 ≤ r ≤ n. (a) Suppose f1 (x) = a0 + · · · + am xm and f2 (x) = b0 + · · · + bn xn . Show that f1 (x)f2 (x) = m+n X ck xk k=0 with ck = X ai bj , i+j=k D(f1 (x) + f2 (x)) = D(f1 (x)) + D(f2 (x)) and D(f1 (x)f2 (x)) = D(f1 (x))f2 (x) + f1 (x)D(f2 (x)). (b) Show that a ∈ F satisfies f (x) = (x − a)2 g(x) and g(a) 6= 0 if and only if f (a) = D(f (a)) = 0 and D2 (f (a)) 6= 0. (c) Suppose r ≥ 3. Show that a satisfies f (x) = (x − a)r g(x) and g(a) 6= 0 if and only if f (z) = 0, Dj (f (a)) = 0 for j = 1, . . . , r − 1 and Dr (f (a)) 6= 0. Solution. (a) Routine checking. (b) If f (x) = (x−a)2 g(x), then D(f (x)) = 2(x−a)g(x)+(x−a)2 Dg(x). So, f (a) = D(f (a)) = 0. Now, D2 (f (x)) = 2g(x)+2(x−a)D(g(x))+(x−a)2 D2 (g(x)). Thus, D3 (f (a)) = 2g(a) 6= 0. Conversely, suppose f (a) = Df (a)) = 0 and D2 (f (a)) 6= 0. By factor theorem, f (a) = 0 implies that f (x) = (x − a)h(x). Then D(f (x)) = h(x) + (x − a)Dh(x) so that 0 = D(f (a)) = h(a). Hence, h(x) = (x − a)g(x) and f (x) = (x − a)2 g(x). If g(a) = 0, we will have D2 (f (a)) = 0, which is a contradiction. So, D2 (f (a)) 6= 0. (c) By induction. 3 9. (10 points) Let F be a field, f (x), g(x) ∈ F[x] be nonzero polynomials, and A = {f (x)u(x) + g(x)v(x) : u(x), v(x) ∈ F[x]}. Let d(x) = d0 + · · · + dk xk with dk = 1 be a monic (non-zero) polynomial in A with minimum degree. (a) Show that A = {d(x)p(x) : p(x) ∈ F[x]}. (b) Show that d(x) is a factor of f (x) and g(x). (c) Show that h(x)|d(x) for any h(x) such that h(x)|f (x) and h(x)|g(x). (d) Show that if f (x) and g(x) has no common factor q(x) ∈ F[x] of degree at least 1, then there is u(x), v(x) ∈ F[x] such that 1 = f (x)u(x) + g(x)v(x). (e) Let f (x) = x5 + x2 + 1, g(x) = x3 + x + 1 ∈ Z3 [x]. Determine u(x), v(x) ∈ Z3 [x] so that gcd(f (x), g(x)) = d(x) = f (x)u(x) + g(x)v(x). Solution. (a) If there is a polynomial h(x) in A which is not a multiple of d(x), then we can get h(x) = d(x)q(x) + r(x), where r(x) has lower degree in A. (b) Since f (x), g(x) ∈ A, the result follows. (c) This follows from (b). (d) This follows from the fact that d(x) = 1. (e) Routine calculation. [Note that the polynomial d(x) constructed is known as gcd(f (x), g(x)) that satisfies (1) d(x)|f (x) and d(x)|g(x), (2) h(x)|d(x) for any h(x) such that h(x)|f (x) and h(x)|g(x).] 4