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391 Homework 8 solutions • Exercises 2.3: 18. Show that Z[i] is an integral domain, describe its field of fractions and find the units. There are two ways to show it is an integral domain. The first is to observe: Any subring of a field is an integral domain. (Proof: Suppose R is a subring of a field F . Take x, y ∈ R with xy = 0. We need to show that one of x, y is zero. Suppose y 6= 0. Then multiplying both sides by y −1 in the big field F , you get x = 0. Done.) The second way is to use some tricks. Suppose wz = 0 for w, z ∈ Z[i]. Then wzwz = 0 hence wwzz = 0. Hence |w||z| = 0. Hence either |w| = 0 or |z| = 0, i.e. either w = 0 or z = 0. The field of fractions of Z[i] is isomorphic to Q[i], i.e. the complex numbers of the form ab + dc i for a, b, c, d ∈ Z with b, d 6= 0. Finally to find the units in Z[i], suppose w is a unit, so wz = 1 for some z ∈ Z[i]. Then wzwz = 1 hence |w||z| = 1 hence ww = |w|2 = 1. But if w = x + iy then |w|2 = x2 + y 2 = 1. That means either x = ±1, y = 0 or x = 0, y = ±1. Hence the units are ±1, ±i. • Exercises 2.4: 1(a)(b)(c). (a) z 2 − (6 + i)z + (8 + 2i) = 0. By the quadratic formula, √ 6 + i ± 1 + 4i z= . 2 You can simplify and put the answer in Cartesian form x + iy if you like, but there is no need. (b) z 2 − iz + 2 = 0. By the quadratic formula, √ i ± −9 i ± 3i z= = . 2 2 so z = 2i or z = −i are the solutions. (c) z 4 + 2z 2 + 4 = 0. By the quadratic formula, √ √ −2 ± −12 2 z = = −1 ± 3i. 2 Let me simplify this time just for fun... first into polar form z 2 = 2e±2π/3 1 2 Now square root once more... √ z = ± 2e±π/3 . If you like you can convert back into cartesian form, but there is no need. If you do, you get √ √ √ 1 3 1 3 z = ± 2( ± i ) = ± √ ± √ i. 2 2 2 2 (There are four different answers!) • Exercises 3.1: 1(e), 2(a)(b)(c), 5, 6, 8, 9(a)(c), 10, 13, 14, 20(c). 1(e) f (x) = x7 + x6 + x4 + x + 1, g(x) = x3 + x + 1, F = Z2 . Long division gives x7 + x6 + x4 + x + 1 = (x3 + x + 1)(x4 + x3 + x2 + x) + 1. VERY IMPORTANT: if you don’t know how I was able to just write that answer down with no working, COME AND SEE ME IN AN OFFICE HOUR!!!!! 2(a)(b)(c) (a) f (x) = x3 − 1, g(x) = x4 + x3 − x2 − 2x − 2, F = Q. Long division gives... x4 + x3 − x2 − 2x − 2 = (x3 − 1)(x + 1) − (x2 + x + 1) x3 − 1 = (x2 + x + 1)(x − 1) + 0. Hence the GCD is x2 + x + 1 and x2 + x + 1 = (x4 + x3 − x2 − 2x − 2) − (x + 1)(x3 − 1). √ √ (b) f (x) = x2 + (1 − 2)x − 2, g(x) = x2 − 2, F = R. Long division and a little scratch work gives √ √ √ √ x2 + (1 − 2)x − 2 = (x2 − 2).1 + (1 − 2)x + (2 − 2) √ √ √ √ (x2 − 2) = ((1 − 2)x + (2 − 2))((−1 − 2)x + (−2 − 2)) + 0. √ √ So the GCD is (1 − 2)x + (2 − 2) made monic, which is √ x − 2. Moreover, √ √ √ 1 1 √ (x2 + (1 − 2)x − 2) − √ (x2 − 2). x− 2= 2− 2 2− 2 (c) f (x) = x2 + 1, g(x) = x2 − i + 2, F = C. Long division gives x2 + 1 = (x2 − i + 2).1 + (i − 1). Since i − 1 is a unit, the GCD is i = 1 made monic, which is 1. Moreover, 1 1 1= (x2 + 1) + (x2 − i + 2). i−1 1−i 3 5 Suppose deg(f (x)) = n and deg(g(x)) = m with m ≥ n. Prove of give a counterexample... (a) deg(f (x) + g(x)) = m. FALSE. Take g(x) = −f (x) – then f (x) + g(x) = 0!!! (b) deg(f (x)g(x)) = m + n. FALSE. Take the coefficient ring to be Z4 and f (x) = g(x) = 2x + 1. Then f (x)g(x) = 4x2 + 4x + 1 = 1 since we can reduce the coefficients mod 4. (BUT this is true if you assume that f (x) and g(x) are not zero and the coefficient ring is an integral domain). 6 Prove that if F is a field, f (x) ∈ F [x] and deg f (x) = n ≥ 1 then f (x) has at most n roots in F . Proof. Proceed by induction on n. The case n = 1 is clear: since f (x) is linear and F is a field it has exactly one root in this case. Induction step. Assume true if deg f (x) = k. Suppose deg f (x) = k + 1. If f (x) has no roots in F , then we are certainly okay. So assume that f (x) has at least one root c ∈ F . Then, (x − c) is a factor, i.e. f (x) = (x − c)g(x) where deg g(x) = k. By induction g(x) has at most k roots. So f (x) = (x − c)g(x) has at most (k + 1) roots. 8 Let F be a field. Prove that if f (x) ∈ F [x] is a polynomial of degree 2 or 3 then f (x) is irreducible in F [x] if and only if f (x) has no root in F . Proof. I’ll show f (x) is reducible in F [x] if and only if f (x) has a root. (⇒). Suppose f (x) is reducible, then f (x) = g(x)h(x) where – because f (x) is of degree 2 or 3 – g(x) must be a linear factor. But linear factors always have roots. So f (x) has a root too. (⇐). Suppose f (c) = 0. Then (x − c) is a linear factor of f (x), so f (x) is reducible. 9(a)(c) (a) Show that unique factorization fails horribly in R[x] when R is not an integral domain. Consider x2 − 1 over Z8 . You can factor it in two different ways: (x − 1)(x + 1) or (x − 3)(x + 3). So unique factorization fails when the coefficient ring is not an integral domain. (c) How many roots does f (x) = 2x − 4 in Z6 [x] have? Search... x = 2, 5 are both roots, so it has two roots! 10 Decide whether each of the following is irreducible. (a) f (x) = x2 + 1 over Z5 . REDUCIBLE: x = 2 is a zero. (b) f (x) = x2 + 1 over Z7 . IRREDUCIBLE: it doesn’t have any zeros so its irreducible (as it is of degree 2 or 3). (c) f (x) = x2 + 1 over Z1 9. 4 IRREDUCIBLE: it doesn’t have any zeros (note it is enough to check x = 0, 1, 2, . . . , 9 to see this!). (d) f (x) = x3 − 9 over Z1 1. x = 4 gives 55 = 0 so it is REDUCIBLE. (e) f (x) = x3 + x + 1 over Z2 . IRREDUCIBLE as no zeros (and it is of degree 2 or 3). (f) f (x) = x4 + x2 + 1 over Z2 . Since it has no zeros it cannot have a LINEAR factor. But it could still possible factor as a product of two irreducible quadratics. There is only one irreducible quadratic over Z2 , namely, x2 + x + 1. So if x4 + x2 + 1 factors it must factor as (x2 + x + 1)(x2 + x + 1) – and it does! So its REDUCIBLE. 13 List all the irreducible polynomials in Z2 [x] of degree ≤ 4. Factor x7 + 1 as a product of irreducibles in Z2 [x]. Linear polynomials are always irreducible, so x, x + 1 count. Then for quadratics there is only x2 +x+1 – the other 3 possible quadratics all have zeros. For cubics there are 8 possibilities at first: x3 , x3 +1, x3 +x, x3 + x + 1, x3 + x2 , x3 + x2 + 1, x3 + x2 + x, x3 + x2 + x + 1. Obviously we must have +1 at the end else x = 0 is a root. Also there must be an odd number of non-zero terms else x = 1 is a root. So this leaves just x3 + x + 1 and x3 + x2 + 1. Since these have no roots and are cubics they ARE irreducible. For quartics, similar reasoning narrows the candidates down to x4 + x + 1, x4 + x2 + 1, x4 + x3 + 1, x4 + x3 + x2 + x + 1. Now which of these are irreducible? They have no linear factors, but they could factor as a product of two irreducible quadratics, i.e. as (x2 + x + 1)2 . That rules out x4 + x2 + 1. The remaining three MUST be irreducible. Answer: x, x + 1, x2 + x + 1, x3 + x + 1, x3 + x2 + 1, x4 + x + 1, x4 + x3 + 1, x4 + x3 + x2 + x + 1. 14 For each of the following numbers c find an irreducible polynomial in √ Q[x] that has c as a root. (a) 1 + √3. c = 1 + 3, so (c − 1)2 = 3 so c is a √ root of x2 − 2x − 2. That is irreducible over Q as its zeros 1 ± 3 are not rational... (b) 2 + 21/3 . c is a solution of (x−2)3 = 2, hence a root of the cubic x3 −6x2 + 12x − 10. To see that the latter is irreducible, use Eisenstein’s criterion with p = 2 (see Theorem 3.5 on p.109). (c) 2 + i. It is a solution of (x − 2)2 = −1, so a root of x2 − 4x + 5. That is irreducible over Q as it has no zeros. p √ (d) 1 + 3. 5 √ c2 = 1 + 3 so (c2 − 1)2 = 3. So c is root of x4 − 2x2 − 2. To see that really is irreducible, use Eisenstein with p = 2. 20(c) Solve f (x) ≡ x2 + 1 (mod x3 + x + 2), f (x) ≡ 2x + 1 (mod x2 + x + 2) in Z3 [x]. First we must find the GCD... x3 + x + 2 = (x2 + x + 2)(x + 2) + 1 So the GCD is 1 (so we can use the Chinese Remainder Theorem directly) and 1 = (x3 + x + 2) − (x + 2)(x2 + x + 2). Now to solve the equation, one solution is f (x) = (2x + 1)(x3 + x + 2) − (x2 + 1)(x + 2)(x2 + x + 2). The general solution is therefore f (x) ≡ (2x+1)(x3 +x+2)−(x2 +1)(x+2)(x2 +x+2) (mod (x3 +x+2)(x2 +x+2)).