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Math 412. Polynomial Rings Over a Field

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Math 412. Polynomial Rings Over a Field (c)Karen E. Smith 2018 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Math 412. Polynomial Rings over a Field. For this worksheet, F always denotes a fixed field, such as Q; R; C; or Zp for p prime. THEOREM 4.6: THE DIVISION ALGORITHM FOR POLYNOMIALS. Let f; d 2 F[x], where d 6= 0. There there exist unique polynomials q; r 2 F[x] such that f = qd + r where r = 0 or deg(r) < deg(d): DEFINITION: Fix f; g 2 F[x]. We say f divides g (and write fjg) if there exists an h 2 F[x] such that g = fh. 1 DEFINITION: Fix f; g 2 F[x]. The greatest common divisor of f and g is the monic polynomial of largest degree which divides both f and g. THEOREM 4.8 In F[x], the greatest common divisor of two polynomials f and g is the monic polynomial of smallest degree which is an F[x]-linear combination of f and g. DEFINITION: A polynomial f 2 F[x] is irreducible if its only divisors are units in F[x] and polynomials of the form uf where u is a unit. THEOREM 4.14 Every non-constant polynomial in F[x] can be factored into irreducible polynomials. This factorization is essentially unique in the sense that if we have two factor- izations into irreducibles f1 ··· fr = g1 ··· gs; then r = s, and after reordering, each fi is a unit multiple of gi for all i. A. THE GCD AS A LINEAR COMBINATION IN F[x]. Suppose f and g are polynomials in Z5[x] such that (x3 + 3x + 5)f + (x17 + 4x4 + 3x2 + 3x + 1)g = x. What are the possible values of the gcd of f and g? The possibilities are x or 1, since the gcd is the smallest degree monic polynomial which is a linear combination of f and g. Since x is a monic polynomial of degree 1 which is linear combination, the gcd is either x, or a lower degree polynomial. The only lower degree monic polynomial is 1. [Note: it is implicit in the statement of Theorem 4.8 that the lowest degree monic polynomial that is a linear combination of both f and g is unique! Otherwise, we could not say the gcd is “the” linear combination of lowest degree. To prove the uniqueness, suppose there are two, d1 and d2. Write d1 = a1f + b1g and d2 = a2f +b2g. But then d1 −d2 = (a1 −a2)f +(b1 −b2)g is also a linear combination of f and g, and since d1 and d2 are monic of the same degree, the difference d1 − d2 is of strictly smaller degree (or is zero). The −1 leading coefficient (call it λ) may not be 1, but we can multiply d1 − d2 by λ to get a monic polynomials of degree strictly less than deg d1 which is a linear combination of f and g, Since we assume there is not lower degree linear combination, it must be that d1 − d2 = 0; that is, that d1 = d2.] B. THE REMAINDER THEOREM AND THE FACTOR THEOREM. Fix f 2 F[x]. (1) Remainder Theorem: Prove that for any λ 2 F; the remainder when f is divided by (x − λ) is f(λ). (2) Factor Theorem: Prove that (x − λ) divides f if and only if f(λ) = 0. 4 (3) Show that 1; 2; 3 and 4 are all roots of f(x) = x − 1 in Z5[x] by evaluating f. 2 (4) Use the factor theorem and (3) to quickly find the factorization of x5 − x completely into irreducibles in the ring Z5[x] as guaranteed by Theorem 4.14. 5 (5) Find a different factorization of x − x into irreducibles in Z5[x] and explain why it is “essentially the same.” (6) Find the factorization of x5 − x completely into irreducibles as guaranteed by Theorem 4.14 in the ring Z7[x]. You might want to start by factoring the way you would in high school (but remember the exotic setting). (1) The Remainder Theorem was proved last time. Read this in the book, too: it is Theorem 4.15. (2) The Factor Theorem was proved last time. Read this in the book, tooit is Theorem 4.16. (3) Check: f(1) = (1)4 − 1 = 0, f(2) = (2)4 − 1 = (4)2 − 1 = (−1)2 − 1 = 1 − 1 = 0, f(3) = (−2)4 − 1 = (4)2 − 1 = (−1)2 − 1 = 1 − 1 = 0, f(4) = (−1)4 − 1 = 1 − 1 = 0. (4) Since 0; 1; 2; 3; 4 are all roots, we know x; (x − 1); (x − 2); (x − 3); (x − 4) are all factors. So 5 x − x = ux(x − 1)(x − 2)(x − 3)(x − 4) where u 2 Z5[x]. Looking at the degrees, we see the degree of u must be zero, so u is a non-zero constant, hence a unit and this must be the complete factorization. But then u = 1 since the leading coefficient on the left is x5 and on the right is ux5. (5) 2x(3x−3)(x−2)(x−3)(x−4). We multiplied one factor by 2, and one by its multiplicative inverse 3. This is the same up to the other, up to unit multiples. 5 4 2 2 2 (6) In Z7[x], we have x − x = x(x − 1) = x(x − 1)(x + 1) = x(x − 1)(x + 1)(x + 1). We only need to check finally that (x2 + 1) can not be factored. If it could, then one would have the form 2 (x − a) for some a 2 Z7[x], which would mean that a is a root of (x + 1). But there is no element 2 in Z7[x] satisfying (x + 1): just check each element: 0; ±1; ±2; ±3 are not roots! C. IRREDUCIBILITY. Let F be any field. (1) Show that if a polynomial g 2 F[x] has degree 2 or 3, then g is irreducible if and only if g has no roots. (2) Show that (1) is false for polynomials of degree 4. (1) Suppose g is irreducible of any degree. Then in particular, we can’t factor g as (x − a)h for any a 2 F and any h 2 F[x]. So g has no root (according to the Factor Theorem). Conversely, say g has degree two or three. If g factors as fh in a non-trivial way, then deg f+deg h = deg f = 2or3 so at least one of f or h must have degree 1, say (without loss of generality) that c f = bx − c for some coefficients b; c 2 F with b 6= 0. This means that g = fh = b(x − b )h, so c c (x − b ) is a factor of g. So b is a root of g. 2 2 2 (2) In Z7[x], we already showed x + 1 is has no roots. So (x + 1) is an non-irreducible polynomial with no roots. Similarly, (x2 + 1)2 is irreducible in R[x] as well (but not in C[x]; since i is a root). D. EUCLIDEAN ALGORITHM IN F[x]: Fix a field F. (1) Use the Euclidean Algorithm to compute gcd(f; g), where f = x5 + x2 + x + 1 and 2 g = x + x in Z2[x]. (2) Write the gcd as a linear combination of f and g in Z2[x]. (1) Arithmetic in Z2[x] is great because 1 = −1 so no sign errors! Dividing, we get a quotient of x3 + x2 + x and a remainder of x + 1, so that f = g(x3 + x2 + x) + (x + 1). So gcd(f; g) = gcd (g; x + 1). Dividing again, the remainder is zero, so the gcd is x + 1. (2) (x − 1) = 1 · f + (x3 + x2 + x) · g. 3 E. CONGRUENCE IN F[x]: Fix a polynomial f(x) 2 F[x]. Define two polynomials g; h 2 F[x] to be congruent modulo f if fj(g − h). We write g ≡ h mod f. The set of all polynomials congruent to g modulo f is written [g]f . (1) Prove that [g]f = fg + kf j k 2 F[x]g. (2) Prove that Congruence is an equivalence relation: (a) reflexive: for all g, we have g ≡ g mod f; (b) symmetric: g ≡ h mod f implies h ≡ g mod f for all g; h 2 F[x]. (c) transitive: g ≡ h mod f and h ≡ k mod f implies g ≡ k mod f for all g; h; k 2 F[x]. (3) Prove that if h 2 [g]f , then [g]f = [h]f . (4) Explain why, for any two polynomials g; h 2 F[x], either [g]f = [h]f or [g]f \ [h]f = ;: (1) Take arbitrary h 2 [g]f . This means fj(g − h). So we can write g − h = fk for some k 2 F[x]. So h = g + (−k)f 2= fg + kf j k 2 F[x]g. For the other direction, take arbitrary g + kf 2 fg + kf j k 2 F[x]g. Then g − (g + kf) = −kf is divisible by f, so g + kf 2 [g]f . QED. (2) Easy: Reflexive follows from g − g is divisible by f. Symmetric follows from fj(g − h) implies fj(h − g). Transitive: fj(g − h) and fj(h − k) implies fj((g − h) + (h − k)) so fj(g − k). QED. 0 (3) Say h 2 [g]f .
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