(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) ∈ F[x]. Define two polynomials g, h ∈ F[x] to be congruent modulo f if f|(g − h).

We write [g]f for the set {g + kf | k ∈ F[x]} 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) ∈ 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 ∈ F[x], which would violate the uniqueness part of the division algorithm. (3) These are [0] = [x2 + x] = {(x2 + x)h , | h ∈ F[x]}; [1] = [1 + x2 + x] = {1 + (x2 + x)h , | h ∈ F[x]}; [x] = [x2] = {x + (x2 + x)h , | h ∈ F[x]}; [x + 1] = [x2 + 1] = {x + 1 + (x2 + x)h , | h ∈ F[x]}. 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 STRUCTUREONTHESETOF CONGRUENCE CLASSES F[x]/(f(x)). Fix a polynomial f(x) ∈ F[x] of degree d > 0. 0 0 0 0 (1) Consider arbitrary g, h ∈ F[x]. Suppose g ∈ [g]f and h ∈ [h]f . Prove that g + h ∈ [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 ∈ [g]f and h ∈ [h]f . Prove that g h ∈ [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 ∈ F[x] and h−h0 = fq0 for some q0 ∈ 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 ∈ [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 ∈ 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 ψ : 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 f|(g0h0 − gh). Use the trick of “cleverly adding zero”: equivalently, you need to show f|(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. In R1, [x] is nilpoent and in R2, [x+1] is nilpotent. (5) [0] and [1] are idempotent in all the rings. There are no further idempotents in R1 or R2 or R4. In R3, all elements are idempotent. (6) In R4, only [0] is a zero-divisor. In R3, all elements except [1] are zero divisors. In R1, [x] is an additional zero divisor and in R2, [x + 1] is. (7) In all rings, [1] is a unit. In R4, all element except [0] are units. In R3, only 1 is a unit. In R2, x is a unit and in R1, x + 1 is a unit. (8) R4 is a field! (9) If these rings were isomorphic, there would be a bijiection between their units. But they all have different numbers of units: 3 in R4, one in R3 and two in R2. (10) φ([0]) = [0], φ([1]) = [1], φ([x]) = [x + 1], φ([x + 1]) = [x]. We know that the additive identities must correspond, and that the multiplicative identities must correspond. Since nilpotents also corre- spond, this is the only possible map. Comparing the tables, we see that this is in fact an isomorphism. (11) ψ([0]) = (0, 0), ψ([1]) = (1, 1), ψ([x]) = (1, 0), ψ([x + 1]) = (0, 1), (12) In all the Ri, we have [1] + [1] = [0], but in Z4 [1] + [1] 6= [0]. So there is no isomorphism from Z4 to any Ri.

D.RINGSWITHTWOORTHREEELEMENTS: (1) Show that any two rings with two elements are isomorphic. That is, up to isomorphism, there is only one ring with two elements. What is this ring, up to isomorphism? (2) Show that any two rings with three elements are isomorphic. That is, up to isomorphism, there is only one ring with three elements. What is this ring, up to isomorphism? [Hint: you can call the elements 0, 1, and a. Start making addition and multiplication tables and see what happens.]

(1) There must be a zero and a 1. So a ring with two elements must consist only of {0R, 1R}. (Note: we can’t have 0R = 1R because we have proved that in any ring 0R × x = 0 for all x ∈ R, so if 0 = 1, then 0 × x = 1 × x = x = 0, and R = {0R}. Any other ring with two elements is isomorphic, by just matching the zero’s and the ones. Note that in both rings, we must have 1 + 1 = 0 because it 1 + 1 = 1, then adding −1 to both sides, we would have 1 = 0. (2) Suppose R is a ring with three elements {0, 1, a}. What can 1 + a be? If 1 + a = a, then 1 = 0, which is impossible. If 1 + a = 1, then a = 0, which is impossible. So we must have 1 + a = 0. Likewise, if 1 + 1 = 1, then 1 = 0, a contradiction. If 1 + 1 = 0, then 1 + 1 + a = a, so using 4

the fact that 1 + a = 0, we have 1 = a, a contradiction. So the only possibility is 1 + 1 = a. This determines the addition table completely, which we see is a re-naming of the addition table for Z3: + 0 1 a 0 0 1 a 1 1 a 0 a a 0 1 The multiplication table is also completely determined, since a = 1 + 1, so we can use repeated uses of the distributive property to figure out any product: eg a × a = (1 + 1)(1 + 1) = 1 + 1 + 1 + 1 = a + 1 + 1 = 1. We get the multiplication table is · 0 1 a 0 0 0 0 1 0 1 a a 0 a 1

By inspection, we see that this is isomorphic to Z3.

E.RINGSWITH FOUR ELEMENTS: (1) You found a field with four elements! What is it? It is a finite field but not of the form Zp for any p (why not?). This is very exotic! (2) You showed in C that up to isomorphism, there are at least four different rings with four elements. Ex- plain. List out representatives of each isomorphism type. (3) BONUS: Show that any ring with four elements is isomorphic to one on your list above.

You can turn in a complete answer for this by Wednesday October 17 for extra credit. Scan and email it to me.