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Math 412. Quotient Rings of Polynomial Rings Over a Field

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Math 412. Quotient Rings of 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. Quotient Rings of Polynomial Rings over a Field. For this worksheet, F always denotes a fixed field, such as Q; R; C; or Zp for p prime. DEFINITION. 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]f for the set fg + kf j k 2 F[x]g of all polynomials congruent to g modulo f. We write F[x]=(f) for the set of all congruence classes of F[x] modulo f. The set F[x]=(f) has a natural ring structure called the Quotient Ring of F[x] by the ideal (f). CAUTION: The elements of F[x]=(f) are sets so the addition and multiplication, while induced by those in F[x], is a bit subtle. A. WARM UP. Fix a polynomial f(x) 2 F[x] of degree d > 0. (1) Discuss/review what it means that congruence modulo f is an equivalence relation on F[x]. (2) Discuss/review with your group why every congruence class [g]f contains a unique polynomial of degree less than d. What is the analogous fact for congruence classes in Z modulo n? 2 (3) Show that there are exactly four distinct congruence classes for Z2[x] modulo x + x. List them out, in set-builder notation. For each of the four, write it in the form [g]x2+x for two different g. (4) How many distinct congruence classes are there for Zp[x] modulo f, where f is a poly- nomial of degree d? (1) This means that it is reflexive (g is congruent to g), symmetric (if g is congruent to h then h is congruent to g, and transitive (if g is congruent to h and h is congruent to k, then g is congruent to k.) In this case, the equivalence relation partitions the set F[x] up into non-overlapping equivalence (congruence) classes. (2) Using the division algorithm, writing g = qf + r, we see that the remainder r is in the class [g]f , and we know deg r < deg f = d. There can’t be another polynomial r0 of degree less than d in 0 0 0 [g]f , because that r = g + q f for some q 2 F[x], which would violate the uniqueness part of the division algorithm. (3) These are [0] = [x2 + x] = f(x2 + x)h ; j h 2 F[x]g; [1] = [1 + x2 + x] = f1 + (x2 + x)h ; j h 2 F[x]g; [x] = [x2] = fx + (x2 + x)h ; j h 2 F[x]g; [x + 1] = [x2 + 1] = fx + 1 + (x2 + x)h ; j h 2 F[x]g. d d−1 d−2 (4) p , since we are counting the number of polynomials of the form ad−1x + ad−2x + ··· + a1x + a0, where there are p choices for each of the ai. B. RING STRUCTURE ON THE SET OF CONGRUENCE CLASSES F[x]=(f(x)): Fix a polynomial f(x) 2 F[x] of degree d > 0. 0 0 0 0 (1) Consider arbitrary g; h 2 F[x]. Suppose g 2 [g]f and h 2 [h]f . Prove that g + h 2 [g + h]f . (2) Explain how (1) allows us to define a well-defined binary operation (call it “addition”) on the set F[x]=(f(x)) of congruence classes. Note that the elements of F[x]=(f(x)) are sets and the notation [g]f is biased— you can represent this set by many different g—so you need to be careful. 2 0 0 1 0 0 (3) Suppose g 2 [g]f and h 2 [h]f . Prove that g h 2 [gh]f . (4) Explain how (3) allows us to define a well-defined binary operation (call it “multiplication”) on the set F[x]=(f(x)) of congruence classes. Why is this non-obvious? (5) Find elements in F[x]=(f(x)) that serve as additive and multiplicative identities for your operations in (2) and (4). (6) Discuss what needs to be checked to ensure that the operations defined in (2) and (4) make F[x]=(f(x)) into a commutative ring. Now, illustrate how to check one or two of the axioms, pointing out how they follow from the corresponding axioms in F[x]. Wiithout writing it all out, make sure everyone is confident how to prove that F[x]=(f(x)) is commutative ring. (1) Write g−g0 = fq for some q 2 F[x] and h−h0 = fq0 for some q0 2 F[x]. Then (g−g0)+(h−h0) = 0 0 0 0 0 0 f(q + q ). So (g + h) − (g + h ) = f(q + q ), so that g + h 2 [g + h]f . (2) We define [g]f + [h]f to be [g + h]f . Problem (1) ensure that no matter which element I use to stand 0 0 for the congruence classes, we get the same result. That is, if [g ]f = [g]f and [h ]f = [h]f , then 0 0 [g + h]f = [g + h ]f . (3) Assume g−g0 and h−h0 are divisible by f. We need to show gh−g0h0 is divisible by f. Equivalently, that gh+(g0h−g0h)−g0h0 is divisible by f. Write gh+(g0h−g0h)−g0h0 = (g −g0)h+(h−h0)g0, which is a sum of two things both divisible by f. So the sum, gh − g0h0; is divisible by f QED. (4) Similar to (2), we define [g]f ·[h]f to be [gh]f . Problem (3) ensure that no matter which element I use 0 0 to stand for the congruence classes, we get the same result. That is, if [g ]f = [g]f and [h ]f = [h]f , 0 0 then [gh]f = [g h ]f . (5) The additive identity is [0]f which is the set of all multiplies of f. The multiplicative identity is [1]f . (6) We need to check both operations are associative, commutative, have identiities, and every element has an additive inverse. Also that the distributive property holds. These all follow from the corre- sponding facts in F[x]. For example, [g] + [h] = [g + h] = [h + g] = [h] + [g]. 2 C. EXAMPLES. Consider an arbitrary monic degree two polynomial x + ax + b 2 Z2[x]. We will investigate the 2 quotient rings of the form Z2[x]=(x + ax + b). There are four cases to consider, depending on the values of a and b: 2 R1 = F[x]=(x ); 2 R2 = F[x]=(x + 1); 2 R3 = F[x]=(x + x); 2 R4 = F[x]=(x + x + 1). (1) Abusing notation slightly, the elements of each Ri can be denoted [0]; [1]; [x]; and [x + 1]. Why? In what sense is this an “abuse”? These notations have completely different meanings in each Ri. Explain. (2) Write down the addition table for R1. How do the addition tables for the other Ri look? (3) For each of the four rings Ri, write out the multiplication tables. (4) For each of the four rings Ri, identify all nilpotent elements. (5) For each of the four rings Ri, identify all idempotent elements. (6) For each of the four rings Ri, identify all zero divisors. (7) For each of the four rings Ri, identify all units. (8) One of these Ri is a field (with four elements!). Which one? (9) None of the three rings R2;R3; or R4 can be isomorphic to any other. Explain why, using your answers to (4)–(7). (10) Find an isomorphism φ : R1 ! R2. [Hint: Your answers to (4)–(7) are a clue.]. (11) Find an isomorphism : R3 ! Z2 × Z2. [Hint: Your answers to (4)–(7) are a clue.] There are two correct answers.] (12) Explain why the ring Z4 is not isomorphic to any of the rings Ri. [Hint: Consider 1 + 1.] 1Hint: you need to show fj(g0h0 − gh). Use the trick of “cleverly adding zero”: equivalently, you need to show fj(g0h0 + (gh0 − gh0) − gh). 3 2 (1) The notation drops the polynomial x + ax + b, but the set [g]x2+ax+b depends strongly on the 2 2 polynomial x + ax + b. For example [0]x2 and [0]x2+1 are two different set: one is multiples of x and the other is multiplies of x2 + 1. For example, x2 is in the former but not the latter. + [0] [1 ] [x] [x+1] [0] [0] [1 ] [x] [x+1] (2) The addition table of R1 is [1 ] [1 ] [0] [x+1] [x] . The addition tables for all [x] [x] [x+1] [0] [1] [x+1] [x+1] [x] [1] [0] four look ”the same”. ·R1 [0] [1 ] [x] [x+1] ·R3 [0] [1 ] [x] [x+1] [0] [0] [0 ] [0] [0] [0] [0] [0 ] [0] [0] (3) [1 ] [0 ] [1] [x] [x+1] [1 ] [0 ] [1] [x] [x+1] [x] [0] [x] [0] [x] [x] [0] [x] [x] [0] [x+1] [0] [x+1] [x] [1] [x+1] [0] [x+1] [0] [x+1] ·R2 [0] [1 ] [x] [x+1] ·R4 [0] [1 ] [x] [x+1] [0] [0] [0 ] [0] [0] [0] [0] [0 ] [0] [0] [1 ] [0 ] [1] [x] [x+1] [1 ] [0 ] [1] [x] [x+1] [x] [0] [x] [1] [x+1] [x] [0] [x] [x+1] [1] [x+1] [0] [x+1] [x+1] [0] [x+1] [0] [x+1] [1] [x] (4) The zero element [0] is (trivially) nilpotent in all.
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