z = i ± so z = 2i or z = i are the solutions. (c) z 4 + 2z = 0. By the quadratic formula,


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1 91 Homework 8 solutions Exercises.: 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 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 a b + c di for a, b, c, d Z with b, d 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 = 1. But if w = x + iy then w = x + y = 1. That means either x = ±1, y = 0 or x = 0, y = ±1. Hence the units are ±1, ±i. Exercises.4: 1(a)(b)(c). (a) z (6 + i)z + (8 + i) = 0. z = 6 + i ± 1 + 4i. You can simplify and put the answer in Cartesian form x + iy if you like, but there is no need. (b) z iz + = 0. z = i ± 9 = i ± i. so z = i or z = i are the solutions. (c) z 4 + z + 4 = 0. z = ± 1 = 1 ± i. Let me simplify this time just for fun... first into polar form z = e ±π/ 1
2 Now square root once more... z = ± e ±π/. If you like you can convert back into cartesian form, but there is no need. If you do, you get z = ± ( 1 ± i ) = ± 1 ± i. (There are four different answers!) Exercises.1: 1(e), (a)(b)(c), 5, 6, 8, 9(a)(c), 10, 1, 14, 0(c). 1(e) f(x) = x 7 + x 6 + x 4 + x + 1, g(x) = x + x + 1, F = Z. Long division gives x 7 + x 6 + x 4 + x + 1 = (x + x + 1)(x 4 + x + x + 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!!!!! (a)(b)(c) (a) f(x) = x 1, g(x) = x 4 + x x x, F = Q. Long division gives... x 4 + x x x = (x 1)(x + 1) (x + x + 1) x 1 = (x + x + 1)(x 1) + 0. Hence the GCD is x + x + 1 and x + x + 1 = (x 4 + x x x ) (x + 1)(x 1). (b) f(x) = x + (1 )x, g(x) = x, F = R. Long division and a little scratch work gives x + (1 )x = (x ).1 + (1 )x + ( ) (x ) = ((1 )x + ( ))(( 1 )x + ( )) + 0. So the GCD is (1 )x + ( ) made monic, which is x. Moreover, x 1 = (x + (1 )x 1 ) (x ). (c) f(x) = x + 1, g(x) = x i +, F = C. Long division gives x + 1 = (x i + ).1 + (i 1). Since i 1 is a unit, the GCD is i = 1 made monic, which is 1. Moreover, 1 = 1 i 1 (x + 1) i (x 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 Z 4 and f(x) = g(x) = x + 1. Then f(x)g(x) = 4x + 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 or 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 or 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 x 1 over Z 8. You can factor it in two different ways: (x 1)(x + 1) or (x )(x + ). So unique factorization fails when the coefficient ring is not an integral domain. (c) How many roots does f(x) = x 4 in Z 6 [x] have? Search... x =, 5 are both roots, so it has two roots! 10 Decide whether each of the following is irreducible. (a) f(x) = x + 1 over Z 5. REDUCIBLE: x = is a zero. (b) f(x) = x + 1 over Z 7. IRREDUCIBLE: it doesn t have any zeros so its irreducible (as it is of degree or ). (c) f(x) = x + 1 over Z 1 9.
4 4 IRREDUCIBLE: it doesn t have any zeros (note it is enough to check x = 0, 1,,..., 9 to see this!). (d) f(x) = x 9 over Z 1 1. x = 4 gives 55 = 0 so it is REDUCIBLE. (e) f(x) = x + x + 1 over Z. IRREDUCIBLE as no zeros (and it is of degree or ). (f) f(x) = x 4 + x + 1 over Z. 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 Z, namely, x + x + 1. So if x 4 + x + 1 factors it must factor as (x + x + 1)(x + x + 1) and it does! So its REDUCIBLE. 1 List all the irreducible polynomials in Z [x] of degree 4. Factor x as a product of irreducibles in Z [x]. Linear polynomials are always irreducible, so x, x + 1 count. Then for quadratics there is only x +x+1 the other possible quadratics all have zeros. For cubics there are 8 possibilities at first: x, x +1, x +x, x + x + 1, x + x, x + x + 1, x + x + x, x + x + x + 1. Obviously we must have +1 at the end else x = 0 is a root. Also there must be an odd number of nonzero terms else x = 1 is a root. So this leaves just x + x + 1 and x + x + 1. Since these have no roots and are cubics they ARE irreducible. For quartics, similar reasoning narrows the candidates down to x 4 + x + 1, x 4 + x + 1, x 4 + x + 1, x 4 + x + x + 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 (x + x + 1). That rules out x 4 + x + 1. The remaining three MUST be irreducible. Answer: x, x + 1, x + x + 1, x + x + 1, x + x + 1, x 4 + x + 1, x 4 + x + 1, x 4 + x + x + x For each of the following numbers c find an irreducible polynomial in Q[x] that has c as a root. (a) 1 +. c = 1 +, so (c 1) = so c is a root of x x. That is irreducible over Q as its zeros 1 ± are not rational... (b) + 1/. c is a solution of (x ) =, hence a root of the cubic x 6x + 1x 10. To see that the latter is irreducible, use Eisenstein s criterion with p = (see Theorem.5 on p.109). (c) + i. It is a solution of (x ) = 1, so a root of x 4x + 5. That is irreducible over Q as it has no zeros. (d) 1 +.
5 c = 1 + so (c 1) =. So c is root of x 4 x. To see that really is irreducible, use Eisenstein with p =. 0(c) Solve f(x) x + 1 (mod x + x + ), f(x) x + 1 (mod x + x + ) in Z [x]. First we must find the GCD... x + x + = (x + x + )(x + ) + 1 So the GCD is 1 (so we can use the Chinese Remainder Theorem directly) and 1 = (x + x + ) (x + )(x + x + ). Now to solve the equation, one solution is f(x) = (x + 1)(x + x + ) (x + 1)(x + )(x + x + ). The general solution is therefore f(x) (x+1)(x +x+) (x +1)(x+)(x +x+) 5 (mod (x +x+)(x +x+)).
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