Chapter 4. Remember: F will always stand for a field.

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1 Chapter 4 Remember: F will always stand for a field Take f(x) = x F [x]. Could there be a polynomial g(x) F [x] such that f(x)g(x) = 1 F? Could f(x) be a unit? 19. Compare with Problem #21(c). 20. For D : R[x] R[x] to be a homomorphism, it would have to be true that D(f(x)g(x)) = D(f(x))D(g(x)) for all polynomials f(x), g(x) R[x]. What do you think? Either prove it or find a specific example where it fails to be true This problem can be done in many different ways. The easiest approach is to compute the gcd(x + a, x + b) using the Euclidean Algorithm. Another would be to notice that every common divisor of x + a and x + b must divide 1 (x + a) + ( 1) (x + b). Yet another, but less elegant approach would be to show that the only monic divisor of x + a of degree greater than zero is x + a, which does not divide x + b; consequently, all common monic divisors of x + a and x + b are constant. 4. Use the definition on top of Page 96 and consider the degrees of all polynomials involved. (Alternatively, you could use the fact that F [x] has no zero divisors and therefore allows for cancellation.) 6. Follow the steps of the example in class. 7. Use the given hint. Compare with Problem It is given that gcd(f(x), g(x)) = 1. Show that if d(x) f(x)h(x) and d(x) g(x), then d(x) h(x) and d(x) g(x). The converse is obviously true. Recall from class that d(x) h(x) gcd(d(x), f(x)). So, you want to argue that the gcd(d(x), f(x)) = As a point of clarification, they meant to state: show that besides the obvious factorization x 2 + x = x(x + 1), there are polynomials f(x) and g(x) in Z 6 [x] with x 2 + x = f(x)g(x) such that f(x) and g(x) are not units and not associates of x or x + 1.

2 15. (a) First explain why every monic polynomial f(x) = x 2 +a 1 x+a 0 in Z p [x] which is not irreducible can be factored as f(x) = (x + a)(x + b) for some a, b Z p. (See hint in the back of your textbook.) In how many ways can you choose two elements a and b from the set Z p, with possible repetition but disregarding the order? (How many have a b and how many have a = b?) Why do different choices for a and b lead to different polynomials f(x)? (Use either #3 of 4.2 or Thm 4.14 and the fact that every polynomial in Z p [x] of degree 1 is irreducible.) (b) How many monic polynomials of degree 2 are left? 21. This problem shows that in Z 9 [x] there would be no end to the usual idea of factoring: f(x) = g(x)h(x) = g 1 (x)g 2 (x)h 1 (x)h 2 (x) =. (a) Look for non-zero elements a, b Z 9 such that (ax + 1)(bx + 1) = 0x 2 + 0x + 1 = 1. (b) If c is a constant polynomial in Z 9 [x] with c 0, c 3, and c 6, then c = c 1 = c(ax + 1)(bx + 1) = (cax + c)(bx + 1). (Do not neglect to explain why ca 0.) Note that for c = 0, you have instead c = (ax)(bx) since ab = 0. That covers the constant polynomials (other than 3 and 6). Given a non-constant polynomial f(x) Z 9 [x], you can write f(x) = f(x) 1 = f(x)(ax + 1)(bx + 1) = p(x)q(x) with p(x) = f(x)(ax + 1) and q(x) = (bx + 1). Caution: Since the problem requires you to find two nonconstant polynomials p(x) and q(x), you have to deal with the exceptional case that you might have f(x)(ax + 1) = c (1) for some constant c Z 9. If that is so, then multiply both sides of Equation (1) by (bx + 1) to learn that the polynomial

3 f(x) = c(bx + 1) must have been linear. Say, f(x) = rx + s. Show that there are exactly six linear polynomials for which the exceptional Equation (1) holds. For each of these six polynomials f(x) = rx + s, show that f(x)(bx + 1) is not a constant. That is, in these six exceptional cases, you can regroup your original strategy as follows: ( ) f(x) = f(x) 1 = f(x)(bx + 1) (ax + 1). This is now a product of two non-constant polynomials. 22. (a) Show that the complete factorization of the polynomial x 3 +a in Z 3 is given by (x 3 + a) = (x + a)(x + a)(x + a). In order to check this, all you need to do is multiply it out: (x + a) 3 = x 3 + 3x 2 a + 3xa 2 + a 3. Finally verify that a 3 = a for all a Z 3. (b) Built on the idea of Part (a). While we are at it, let s mention the following two important theorems: Theorem. [The Freshman s Dream] Let p be a positive prime and R a commutative ring in which pr = 0 for every r R, where pr = r + r + + r denotes repeated addition of p terms. Then for every a, b R, (a + b) p = a p + b p. Proof. From the Binomial Theorem we know that ( ) ( p p (a + b) p = a p + a p 1 b However, so long as 1 n p 1 each ( ) p = n ) a p 2 b p! n!(p n)! ( ) p ab p 1 + b p. p 1 is a multiple of p. This is because p is a prime so that none of the integers in the denominator, all whose prime factors are less than p, can cancel p in the numerator. Consequently, all terms except for the first and the last, are equal to zero.

4 Theorem. [Fermat s Little Theorem] Let p be a positive prime and a Z p. Then a p = a in Z p. Proof. We will prove this theorem later in the course Use Corollary 4.19 for the polynomials in Parts (a) through (e). For Part (f) use Corollary There are quadratic polynomials in Z 6 [x] with exactly four roots in Z 6. There are also quadratics with exactly six roots. 19. (a) If a is a multiple root of f(x), then f(x) = (x a) k g(x) for some k 2. Now check f (a) after applying the product rule of differentiation. Conversely, if f(a) = f (a) = 0, then (x a) is a factor of both f(x) and f (x). Suppose, to the contrary, that a is not a multiple root of f(x), then f(x) = (x a)g(x) and g(a) 0. Show that f (a) 0 to arrive at a contradiction. (b) Suppose, to the contrary, that f(x) did have a multiple root. Use Part (a) to find a linear common factor of f(x) and f (x). What s wrong with that? 28. The ring T of all polynomial functions from Z 3 to Z 3 consists of all polynomials f(x) in Z 3 [x] when regarded as functions f : Z 3 Z 3 rather than polynomials. An element f : Z 3 Z 3 of T is equal to the zero element 0 T exactly if f(a) = 0 for all a Z 3. Show that neither f(x) = x + 1 nor g(x) = x 2 +2x is the zero element of T, but that their product is the zero element of T. So, there are zero divisors in T. Does Z 3 [x] have zero divisors? Explain why there are exactly 3 3 = 27 different functions from Z 3 to Z 3. Thus, the ring T cannot have more than 27 elements. How many elements does Z 3 [x] have? Although it is not important here, you might want to think about how many elements T actually has. Experiment with the formula f(x) = 2a(x 1)(x 2) + 2b(x 0)(x 2) + 2c(x 0)(x 1) (c) Take p = 3. Show that the polynomial is irreducible in Z p [x]. Do this in two steps: first show that it has no root in Z 3 [x]. This

5 rules out linear factors. Then show that it does not have an irreducible monic quadratic factor in Z 3 [x] by listing all irreducible monic quadratic polynomials in Z 3 [x] and calculating the remainders when you divide by them. Finally, explain in detail why you are done. (Hint: when making your list of all irreducible monic quadratic polynomials in Z 3 [x], start with x 2 + ax + b and let a, b {0, 1, 2}. Of these nine polynomials, three have no roots in Z 3 but six of them do have roots in Z 3. The ones without roots in Z 3 are your irreducible quadratics in Z 3 [x].) (c) Since z 1 = 3 + 2i is given to be a root, you will automatically know a second root z 2 by Lemma So, f(x) = (x z 1 )(x z 2 )g(x) for some quadratic g(x). Rather than dividing out the two linear polynomials (x z 1 ) and (x z 2 ) in two separate divisions (which would be very ugly), find their product (x z 1 )(x z 2 ) = x 2 +ax+b (which is a quadratic with real coefficients) and divide f(x) by x 2 + ax + b to get q(x). Since q(x) is again a quadratic, you can find its roots with the quadratic formula. 2. The key here is Lemma (c) First find a rational root using the rational root test. How much further you can factor now depends on whether you are in Q[x] or in C[x].