# 4 Unique Factorization and Applications

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1 Number Theory (part 4): Unique Factorization and Applications (by Evan Dummit, 2014, v. 1.00) Contents 4 Unique Factorization and Applications Integral Domains Euclidean Domains and Division Algorithms Unique Factorization Domains Modular Arithmetic in Euclidean Domains Euler's and Fermat's Theorems The Chinese Remainder Theorem Factorization in F [x] Modular Arithmetic in F [x] The Polynomial x p x over F p The Polynomial x pn x over F p Shamir's Secret Sharing, Secured Voting Primitive Roots Modular Arithmetic in Z[i] and Sums of Two Squares Residue Classes in Z[i] Prime Factorization in Z[i] Unique Factorization and Applications In this chapter, we dene a notion of a division algorithm in a general ring and then formalize the idea of when a ring possesses unique factorization. This ring theory underlies much of what we do in the rest of the course: one of our central goals is to generalize number-theoretic properties of Z to other number systems in order to reap the benets of applying the abstract machinery in those settings. Toward this end, we generalize the notion of modular arithmetic into a the setting of general commutative rings, and provide generalizations of Euler's Theorem, Fermat's Little Theorem, and the Chinese Remainder Theorem. We then apply our results in two rings relevant to number theory: the polynomial ring F [x] and the Gaussian integers Z[i]. In particular, we study the structure of the polynomials with coecients in Z/pZ, we characterize those m for which a primitive root exists modulo m, and we make an in-depth study of the structure of the Gaussian integers, including giving a description of the modular arithmetic and prime factorization in Z[i]. Along the way, we will also study nite elds from several perspectives, and prove Fermat's characterization of the integers that can be written as the sum of two squares. 4.1 Integral Domains Recall that in a general commutative ring R, we say that a b if there exists k R such that b = ak. (Reminder) Denition: If R is a commutative ring, we say that x R is a zero divisor if x 0 and there exists a nonzero y R such that xy = 0. (Note in particular that 0 is not a zero divisor!) We originally dened zero divisors when discussing the ring structure of Z/mZ. 1

2 In Z/6Z, since 2 3 = 4 3 = 0, the residue classes represented by 2, 3, and 4 are zero divisors. As a general philosophy, zero divisors can be a bit troublesome (at least, to a novice ring theorist), since they behave counter to one's natural intuition that products of nonzero elements should be nonzero. We recall a few important properties of zero divisors: An integer a is a zero divisor modulo m i 1 < gcd(a, m) < m. In particular, Z/mZ contains zero divisors if and only if m is composite. The ring Z/pZ is a eld (which, in particular, contains no zero divisors). In a commutative ring with 1, a unit can never be a zero divisor. Denition: If R is a commutative ring with 1 that contains no zero divisors, R is called an integral domain (or often, just a domain). Example: Any eld is an integral domain. More generally, any ring that is a subset of a eld is an integral domain: hence, the integers Z and the ring Z[ D] for any D are integral domains (since they are all subsets of the complex numbers C). Example: The ring of polynomials F [x] where F is any eld is also an integral domain, because the degree of p(x) q(x) is the degree of p(x) times the degree of q(x), since the product of the two leading terms cannot be zero. More generally, the ring of polynomials R[x] where R is any integral domain is also an integral domain, for the same reason. Integral domains are nice, in the sense that they obey more of the laws of arithmetic that are familiar from Z. In an integral domain, cancellation of a nonzero element is allowed. Explicitly, if c 0 and ac = bc, then a = b. Proof: Rearrange to write the equality as (a b)c = 0. Since R contains no zero divisors, the only way this product can equal zero is if a b = 0 or c = 0. Since we assumed c 0, we see a b = 0 so that a = b. If a b and b a and a, b are nonzero, then a = bu for some unit u. (Reminder) Denition: Two elements a and b are called associates if there exists a unit u with a = bu. Proof: Since a b, there is some u with a = bu. Since b a, there is some w with b = aw. Multiplying the two equations gives ab = abuw, so ab(1 uw) = 0. Since a and b are nonzero and R is a domain, we can cancel to see that 1 uw = 0, so that u is a unit. For any m 0, a b is equivalent to (ma) (mb). Proof: Follows directly from the cancellation property. Denition: Let a, b R where R is a domain. We say d is a common divisor if d a and d b, and we say that a common divisor d R is a greatest common divisor of a and b if d 0 and for any other common divisor d, it is true that d d. Example: 2 is a greatest common divisor of 14 and 20 in Z, and 2 is also a greatest common divisor of 14 and 20. (In particular, note that the gcd is no longer unique.) Example: In the polynomial ring C[x], a greatest common divisor of x 2 4 and x 3 8 is x 2, as can be seen by factoring these two polynomials into linear terms. Important Warning: In arbitrary domains, there can exist pairs (a, b) that do not possess a greatest common divisor. For example, in the ring Z[ 5], the elements 2(1 + 5) and 6 do not possess a greatest common divisor. To see this, observe that 2 and are both common divisors of 2(1 + 5) and 6. However, there is no element d that divides 2(1+ 5) and 6 that is also itself divisible by 2 and This can be seen by taking norms: necessarily N(d) would divide N( ) = 24 and N(6) = 36, hence divide 12. 2

3 Also, N(d) would also necessarily be a multiple of N(2) = 4 and N(1 + 5) = 6, hence be a multiple of 12. The only possibility is N(d) = 12, but there are no elements of norm 12, since there are no integer solutions to a 2 + 5b 2 = 12. Proposition: If d 1 and d 2 are both greatest common divisors of a and b, then there exists a unit u such that d 1 = ud 2. Conversely, if d 1 is a gcd of a and b, then so is ud 1 for any unit u. Another way of phrasing this result is: any two gcds of a and b are associates. Proof: Since d 1 is a gcd, and d 2 is a common divisor, we see d 1 d 2, and similarly d 2 d 1. By one of the facts about divisibility in domains above, this implies d 1 = ud 2 for a unit u. For the other statement, if d 1 a then there exists c R with a = d 1 c. a = (ud 1 ) (cu 1 ), so that ud 1 a. Similarly, we see ud 1 b. But then we can also write Also, if d is any common divisor, then d d 1. Clearly this implies d ud 1, so ud 1 is also a gcd of a and b. 4.2 Euclidean Domains and Division Algorithms We now discuss what it means for an integral domain to possess a division algorithm. Denition: If R is a domain, any function N : R N {0} such that N(0) = 0 is called a norm on R. Observe that this is a rather weak property, and that any given domain may possess many dierent norms. Denition: A Euclidean domain (or domain with a division algorithm) is an integral domain R that possesses a norm N with the property that, for every a and b in R with b 0, there exist some q and r in R such that a = qb + r and either r = 0 or N(r) < N(b). The purpose of the norm function is to allow us to compare the size of the remainder to the size of the original element. Example: Any eld is a Euclidean domain, because any norm will satisfy the dening condition. This follows because for every a and b with b 0, we can write a = qb + 0 with q = a b 1. Example: The integers Z are a Euclidean domain, because if we set N(n) = n, then, as we have already proven, the standard division algorithm allows us to write a = qb + r with either r = 0 or r < b. Before we give additional examples, we will remark that the reason Euclidean domains have that name is that we can perform the Euclidean algorithm in such a ring (in precisely the same manner as in R): Denition: If R is a Euclidean domain, then for any a, b R with b 0, the Euclidean Algorithm in R consists of repeatedly applying the division algorithm to a and b as follows, until a remainder of zero is obtained: a = q 1 b + r 1 b = q 2 r 1 + r 2 r 1 = q 3 r 2 + r 3. r k 1 = q k r k + r k+1 r k = q k+1 r k+1. By the construction of the division algorithm, we know that N(r 1 ) > N(r 2 ) >, and since N(r i ) is a nonnegative integer for each i, this sequence must eventually terminate with the last remainder equalling zero (else we would have an innite decreasing sequence of nonnegative integers). Theorem (Bézout): If R is a Euclidean domain and a and b are arbitrary elements with b 0, then the last nonzero remainder d arising from the Euclidean Algorithm applied to a and b is a greatest common divisor of a and b. Furthermore, there exist elements x, y R such that d = ax + by. 3

4 Corollary: Any two elements in a Euclidean domain always possess at least one gcd. The ideas in the proof are the same as for the proof over Z. Proof: By an easy induction (starting with r k = q k+1 r k+1 ), d = r k+1 divides r i for each 1 i k. Thus we see d a and d b, so the last nonzero remainder is a common divisor. Suppose d is some other common divisor of a and b. By another easy induction (starting with d (a q 1 b) = r 1 ), it is easy to see that d divides r i for each 1 i k + 1, and therefore d d. Hence d is a greatest common divisor. For the existence of x and y with d = ax + by, we simply observe (by yet another easy induction starting with r 1 = a q 1 b) that each remainder can be written in the form r i = x i a + y i b for some x i, y i R. Proposition: The Gaussian integers Z[i] are a Euclidean domain, under the norm N(a + bi) = a 2 + b 2. Remark: For the rings Z[ D] in general, the function N(a + b D) = a 2 Db 2 is a norm, but it does not in general give a division algorithm. Explicitly, given a + bi and c + di in Z[i], we will describe how to produce q, r Z[i] such that a + bi = q(c + di) + r, and N(r) 1 N(c + di). This is even stronger than is needed (once we note that the only 2 element of norm 0 is 0). Proof: We need to describe the algorithm for producing q and r when dividing an element a + bi by an element c + di. If c + di 0, then we can write a + bi c + di = x + iy where x = (ac + bd)/(c2 + d 2 ) and y = (bc ad)/(c 2 + d 2 ) are real numbers. Now we dene q = s + ti where s is the integer closest to x and t is the integer closest to y, and set r = (a + bi) q(c + di). Clearly, (a + bi) = q(c + di) + r. All we need to do now is show N(r) 1 r N(c+di): rst observe that 2 c + di = a + bi q = (x s)+(y t)i. c + di Then because x s 1 2 and y t 1 2 by construction, the triangle inequality implies r 2 c + di 2. Squaring both sides and rearranging yields N(r) 1 N(c + di), as desired. 2 Proposition: If F is any eld, the ring of polynomials F [x] in the variable x with coecients in F is a Euclidean domain, under the norm given by N(p(x)) = deg(p). The idea is simply to show the validity of polynomial long division. The reason we require F to be a eld is that we need to be able to divide by arbitrary nonzero coecients to be able to perform the divisions. (Over Z, for instance, we cannot divide x 2 by 2x and get a remainder that is a constant polynomial.) Explicitly, we will show that if a(x) and b(x) are polynomials with b(x) 0, then there exist q(x) and r(x) such that a(x) = q(x)b(x) + r(x), and either r(x) = 0 or deg(r) < deg(b). Proof: We prove this by induction on the degree n of a(x). The base case is trivial, as we may take q = r = 0 if a = 0. Now suppose the result holds for all polynomials a(x) of degree n 1. If deg(b) > deg(a) then we can simply take q = 0 and r = a, so now also assume deg(b) deg(a). Write a(x) = a n x n + a n 1 x n a 0 and b(x) = b m x m + + b 0, where b m 0 since b(x) 0. Observe that the polynomial a (x) = a(x) a n b(x) has degree less than n, since we have cancelled the b m leading term of a(x). (Here we are using the fact that F is a eld, so that a n also lies in F.) b m By the induction hypothesis, a (x) = q (x)b(x)+r (x) for some q (x) and r (x) with r = 0 or deg(r ) < deg(b). Then a(x) = [ q (x) + a ] n b(x) + r (x), so that q(x) = q (x) + a n b m b m and r(x) = r (x) satisfy all of the requirements. Remark: It is also straightforward to see that the quotient and remainder are unique, by observing that if a = qb + r = q b + r, then r r has degree less than deg(b) but is also divisible by b(x), hence must be zero. 4

5 4.3 Unique Factorization Domains Now that we have given the denition of a general ring having a division algorithm, we turn our attention to dening when a ring has unique factorization. In order to do this, we rst need a notion generalizing the idea of a prime number in Z: namely, an element which does not have any nontrivial divisors. (Remainder) Denition: If R is an integral domain, a nonzero element a R is irreducible if it is not a unit and, for any factorization a = bc with b, c R, one of b and c must be a unit. Example: In the integers, the irreducible elements are the prime numbers. Example: In the ring of polynomials F [x], for any eld F, the irreducible elements are the irreducible polynomials: namely, the polynomials which cannot be factored into a product of polynomials of smaller degree. (Thus, for example, the polynomial x is an irreducible element in the ring R[x].) Important Warning: Whether a given number is irreducible depends on the ring R we are considering it as an element of. For example, the number 5 is irreducible in the ring Z, but it is not irreducible in the ring Z[i] because in this ring we can write 5 = (2 + i)(2 i) and neither of these elements is a unit. An irreducible element behaves much like a prime number in Z. However, there is a separate notion of a prime element in a general domain: Denition: If R is an integral domain, a nonzero element p R is prime if it is not a unit and, for any a, b R such that p ab, it must be the case that p a or p b. Frequently, irreducible elements and prime elements are related. Prime elements are always irreducible, as we will show in a moment, but the converse is not always true. (It is true when R is a Euclidean domain, as we will show in a moment.) (Non-)Example: In the ring R = Z[ 5], the element 2 is irreducible, but not prime. To see 2 is irreducible, observe that N(2) = 4, and so if there were any nontrivial factorization of 2, it would necessarily be the product of 2 elements of norm 2. However, there are clearly no such elements, because there are no integer solutions to 2 = N(a + b 5) = a 2 + 5b 2. To see 2 is not prime, observe that 2 clearly divides 6 = (1 + 5) (1 5), but 2 divides neither of the individual factors, since and 1 5 are not elements of R. 2 2 Alternatively, we could have observed that N(1 ± 5) = 6, and N(2) = 4 does not divide 6. Proposition: If R is an integral domain and p R is a prime element, then p is irreducible. Proof: Suppose p is prime and has a factorization p = bc. By denition, it must be the case that p b or p c; by relabeling assume p b, with b = pu for some u. Then p = puc so p(1 uc) = 0. Cancelling p yields uc = 1, so c is a unit. Thus, p is irreducible. Proposition: If R is a Euclidean domain and p R is irreducible, then p is prime. Proof: Suppose p ab. If p a, we are done, so suppose p a, and let d be a gcd of p and a (which exists since R is a Euclidean domain). By hypothesis, d divides p, so (since p is irreducible) either d is a unit, or d = up for some unit u however, the latter cannot happen, because then up (hence p) would divide a. Hence d is a unit, say with inverse e. By the Euclidean algorithm, we see that there exist c and d such that cp + da = d. Multiplying by be yields (bce)p + de(ab) = (de)b = b, so since p divides both terms on the LHS, we conclude p b. We would like to say a ring possesses unique factorization if every nonzero element can be written uniquely as the product of irreducibles. However, there are some issues with this attempted denition. 5

6 In the Gaussian integers, we can write 5 = (2 i)(2 + i) = (1 + 2i)(1 2i). It would seem that these are two dierent factorizations, but we should really consider them the same, because all we have done is moved some units around: (2 i) i = 1 + 2i and (2 + i) ( i) = 1 2i. We should declare that two factorizations are equivalent if the only dierences between them are by moving units around which is equivalent to replacing elements with associates. Denition: An integral domain R is a unique factorization domain (UFD) if (i) every nonzero nonunit r R can be written as a nite product of irreducibles r = p 1 p 2 p k, and (ii) such a factorization is unique up to associates: if r = q 1 q 2 q d is some other factorization, then d = k and there is some reordering of the factors such that p i is associate to q i for each 1 i k. Our main result is the following: Theorem: If R is a Euclidean domain, then R is also a unique factorization domain. The main ideas of the proof are the same as those over Z, but they are a bit obfuscated by some of the technical diculties. The existence portion of the proof contains precisely the same ideas as in our characterization of the gcd of a collection of integers as the minimal positive linear combination of those integers. The uniqueness portion is essentially the same as for Z: an induction argument on the number of irreducible terms in a factorization. Proof (Existence): Let R be a Euclidean domain and r a nonzero nonunit. If r is irreducible, we are done. Otherwise, by denition we can write r = r 1 r 2 where neither r 1 nor r 2 is a unit. If both r 1 and r 2 are irreducible, we are done: otherwise, we can continue factoring (say) r 1 = r 1,1 r 1,2 with neither term a unit. If r 1,1 and r 1,2 are both irreducible, we are done: otherwise, we factor again. We claim that this process must terminate eventually: otherwise, there necessarily exists an innite chain of elements x 1, x 2, x 3,..., such that x 1 r, x 2 x 1, x 3 x 2, and so forth, where no two elements are associates. Consider the set I of all (nite) R-linear combinations I = {r 1 x 1 +r 2 x 2 + +r k x k : k 1, r i R}, and let y I be a nonzero element in I of minimal norm. We claim that every element in I is divisible by y: otherwise, if there were some element s I with y not dividing s, applying the division algorithm to write s = qy + r would yield the element r = (s qy) I of smaller norm than y, contradiction. But now since y I, we can write y = r 1 x r d x d for some d. Since x d x d 1 x 1, the LHS is a multiple of x d, meaning x d y. But now since x d y and y x d+1 we conclude x d x d+1. But by assumption, x d+1 x d, meaning that they are associates; this is a contradiction. Hence the factoring process must terminate, as claimed. Proof (Uniqueness): Let R be a Euclidean domain and r be a nonzero nonunit. We prove the factorization of r is unique by induction on the number of irreducible factors of r = p 1 p 2 p d. If d = 0, then r is a unit. If r had some other factorization r = qc with q irreducible, then q would divide a unit, hence be a unit (impossible). Now suppose d 1 and that r = p 1 p 2 p d = q 1 q 2 q k has two factorizations into irreducibles. Since p 1 (q 1 q k ) and p 1 is irreducible, repeatedly applying the fact that p irreducible and p ab implies p a or p b shows that p 1 must divide q i for some i. Then q i = p 1 u for some u: then since q i is irreducible (and p 1 is not a unit), u must be a unit, so p 1 and q i are associates. Cancelling then yields the equation p 2 p d = (uq 2 ) q k, which is a product of fewer irreducibles. By the induction hypothesis, such a factorization is unique up to associates. This immediately yields the desired uniqueness result for r as well. 6

7 Remark (for those who like ring theory): The set I from the existence proof is an ideal. The underlying idea of the proof given above is to show that any ideal in a Euclidean domain is actually principal (generated by a single element) in other words, that Euclidean domains are principal ideal domains. In fact, it is actually true that a Euclidean domain is a principal ideal domain, and that any principal ideal domain is a unique factorization domain. Corollary: The Gaussian integers Z[i] are a unique factorization domain, as is the polynomial ring F [x] for any eld F. Although most of the unique factorization domains we will discuss are also Euclidean domains, there exist UFDs that are not Euclidean domains. (We will not pursue this topic much here.) One example is the ring C[x, y] of polynomials in the two variables x and y, with coecients in C. There is no Euclidean division algorithm in this ring (the degree map is not a norm since, e.g., there is no way to divide y 2 by x and obtain a remainder of degree zero). However, polynomials in C[x, y] can still be factored into a product of irreducible terms. We often prefer to speak of prime factorization, rather than factorization into irreducibles. Over a unique factorization domain, these are equivalent: Proposition: If R is a unique factorization domain, an element p R is prime if and only if it is irreducible. Proof: Prime elements are always irreducible, as we showed earlier. For the other direction, suppose p R is an irreducible element and p ab: thus, there exists c R with ab = pc. Write a = q 1 q 2 q j, b = r 1 r 2 r k, and c = s 1 s l as products of irreducible elements. Then the two factorizations q 1 q j r 1 r k = ab = pc = ps 1 s l, must be equivalent up to associates. Since p is irreducible, we see that it must be associate to one of the q i (in which case p a) or to one of the r i (in which case p b): hence, p is a prime element. 4.4 Modular Arithmetic in Euclidean Domains In previous chapters, we described the division algorithm over Z and used it to study modular arithmetic in Z. The goal of this section is to show that there is a meaningful extension of the notion of modular arithmetic modulo a general element r in a Euclidean domain R. Our primary interest is when R is Z[i] or F [x], for F a eld. However, many of the notions will hold in general, and so we will work in the general setting whenever possible. We will see that almost all of the proofs are exactly the same as over Z. We will also, when possible, remark when the results we prove hold (or fail to hold) for more general classes of rings. Remark (for those who like ring theory): All of what we do here is subsumed by the theory of ideals in a general ring R, and our construction of modular arithmetic is a special case of the quotient of a ring by an ideal. (Specically, we are studying the quotient R/I, where I is a principal ideal. Every ideal is principal in a Euclidean domain, so we do not lose too much by studying the quotient without speaking of ideals explicitly.) Our underlying denition of modular congruences and residue classes are exactly the same as over Z: Denition: Let R be a commutative ring with 1. If a, b, r R, we say that a is congruent to b modulo c, written a b (mod c), if c (b a). The residue class of a modulo r, denoted a is the set S = {a + dr : d R} of all elements in R congruent to a modulo r. Example: In Z[i], it is true that 13 3i 2 i modulo 3+4i, because (13 3i) (2 i) = (1 2i)(3+4i). Example: In F 2 [x], it is true that x 3 + x x + 1 modulo x 2 + x + 1, because (x 3 + x) (x + 1) = (x + 1) (x 2 + x + 1). 7

8 Proposition: The set R/rR consisting of all residue classes in R modulo r forms a ring (under the addition and multiplication from R). The proof of this fact is exactly the same as over Z. Proof: The most dicult part is showing that the addition and multiplication operations are well-dened: that if we choose dierent elements a ā and b b, the residue class of a + b is the same as that of a + b, and similarly for the product. Since a ā there exists k 1 with a = a + k 1 r and similarly there exists k 2 with b = b + k 2 r. Then a +b = (a+b)+r(k 1 +k 2 ), and since these dier by a multiple of r, we see that a + b = a + b. Similarly, a b = (a + k 1 r)(b + k 2 r) = ab + r(k 1 b + k 2 a + k 1 k 2 r), so a b = ab. Hence the operations are well-dened. For the ring axioms (R1)-(R6), we observe that associativity, commutativity, and the distributive law already hold in R. The additive identity in R/rR is 0, the additive inverse of a is a, and the multiplicative identity is 1. Over Z, we usually work with a specic collection of representatives for the residue classes modulo m, generally the integers 0 through m 1. In a general ring, there is not usually a natural choice for residue class representatives. Or, if there does happen to be a good choice, it is not always obvious what that choice is Euler's and Fermat's Theorems One of our foundational results in Z/mZ was Euler's theorem. There is a natural generalization of the Euler ϕ-function and of Euler's theorem that holds in the case where there are nitely many units in R/rR. First, we give a characterization of the units in R/rR if R is a Euclidean domain: Proposition: If R is a Euclidean domain, an element s R is a unit in R/rR if and only if 1 is a gcd of r and s. (In such a case, we say that r and s are relatively prime.) Proof: If r and s are relatively prime, then since R is a Euclidean domain, there exist a, b R such that ar + bs = 1. Then, modulo r, we have b s = 1, meaning that s is a unit in R/rR. Conversely, suppose that s is a unit in R/rR: then there exists some b such that b s = 1 in R/rR: hence there exists some a R with bs = 1 ar i.e., ar + bs = 1. Since any gcd of r and s divides ar + bs, we conclude that any gcd must be a unit. Since all gcds are associates, we conclude 1 is a gcd of r and s. Theorem (Generalization of Euler's Theorem): If R is a commutative ring with 1 and r R, let ϕ(r) denote the number of units in the ring R/rR, assuming this number is nite. Then if a is any unit in R/rR, we have a ϕ(r) 1 (mod r). The proof is the same as over Z/mZ: the point is that if a is a unit and u 1,, u k are the units in R, then the elements au 1,, au k are the same as u 1,, u k, just in a dierent order. Proof: Let the set of all units in R/rR be u 1, u 2,..., u ϕ(r), and consider the elements a u 1, a u 2,, a u ϕ(r) : we claim that (modulo r) they are simply the elements u 1, u 2,..., u ϕ(r) again (possibly in a dierent order). Since there are ϕ(r) elements listed and they are all still units, it is enough to verify that they are all distinct. So suppose a u i a u j (mod r). Since a is a unit, multiply by a 1 : this gives u i u j (mod r), whence i = j. Hence modulo r, the elements a u 1, a u 2,, a u ϕ(r) are simply u 1, u 2,..., u ϕ(r) in some order. Therefore we have (a u 1 )(a u 2 ) (a u ϕ(r) ) u 1 u 2 u ϕ(r) (mod r), whence, upon cancelling u 1 u 2 u ϕ(r) from both sides, we obtain as desired. a ϕ(r) 1 (mod r), 8

9 One of the other nice properties of Z/mZ is that if m is prime, then Z/mZ is actually a eld. This remains true if we replace Z with an arbitrary Euclidean domain: Proposition: If R is a Euclidean domain, the element p R is a prime element (equivalently, irreducible) if and only if R/pR is a eld. Proof: Suppose p is a prime element. If a p, then a 0 (mod p). Otherwise, because p is prime (hence irreducible), we see that 1 is a gcd of a and p, so by the Euclidean algorithm in R, there exist b and c such that ab + cp = 1. This implies ab 1 (mod r), meaning a is a unit in R/pR. We conclude any nonzero element in R/pR is a unit, so R/pR is a eld. Conversely, suppose R/pR is a eld. If a, b R are such that p ab, then ab 0 (mod p). Since R/pR is a eld, it has no zero divisors, meaning that a 0 (mod p) or b 0 (mod p) hence, p a or p b. Thus, p is a prime element of R. Remark: This proposition is not true if R is only assumed to be a general commutative ring with 1, or even a unique factorization domain. (The unique factorization domain R = C[x, y] is a counterexample: the element x is prime, but the ring R/xR is not a eld, since y has no multiplicative inverse there.) The correct equivalence in that case is the element p R is a prime element i R/pR is an integral domain. The above proposition gives us a very easy way to construct new elds, which we will explore more in a moment. Corollary (Generalization of Fermat's Little Theorem): If R is a Euclidean domain and p R is a prime element, and the number of elements in R/pR is n, then a n a (mod p) for every a R. Proof: Since R/pR is a eld, the only nonunit is zero, so ϕ(p) = n 1. By the generalization of Euler's theorem, we know that a ϕ(p) 1 (mod p) for every a that is a unit modulo p, so a n = a ϕ(p)+1 a (mod p) for such a. Since a n a (mod p) is also true when p a, we see that it holds for every a R The Chinese Remainder Theorem Another foundational result of arithmetic in Z was the Chinese Remainder Theorem. This result generalizes to arbitrary Euclidean domains, with essentially the same statement. We rst start with the analogous proposition on solving a single linear congruence. Proposition: Let R be a Euclidean domain, with a, b R, and let d any gcd of a and r. Then the equation ax b (mod r) has a solution for x R if and only if d b. In this case, if b = b d and r = r d, and x = k is any solution, then the set of all solutions is given by x k (mod r ). The proof is the same as over Z. Proof: If x is a solution to the congruence ax b (mod r), then there exists an s R with ax rs = b. Then since d divides the left-hand side, it must divide b. Now if we set a = a/d, b = b/d, and r = r/d, our original equation becomes a dx b d (mod r d). Solving this equation is equivalent to solving a x b (mod r ), by one of our properties of congruences. But since a and r are relatively prime, a is a unit modulo r, so we can simply multiply by its inverse to obtain x b (a ) 1 (mod r ). As over Z, the above proposition converts a problem of solving a general system of congruences in the variable x to a system of the form x a i (mod r i ). 9

10 Theorem (Chinese Remainder Theorem): Let R be a Euclidean domain and r 1, r 2,..., r k be pairwise relatively prime elements of R, and a 1, a 2,..., a k be arbitrary elements of R. Then the system x a 1 (mod r 1 ) x a 2 (mod r 2 )... x a k (mod r k ) has a solution x 0 R. Furthermore, the set of all solutions for x is precisely the residue class of x 0 modulo r 1 r 2 r k. The proof is the same as over Z. Proof: Since we may repeatedly convert two congruences into a single one until we are done, by induction it suces to prove the result for two congruences x a 1 (mod r 1 ) x a 2 (mod r 2 ). For existence, the rst congruence implies x = a 1 +kr 1 for some k R; plugging into the second equation then yields a 1 + kr 1 a 2 (mod r 2 ). Rearranging yields kr 1 (a 2 a 1 ) (mod r 2 ). Since by hypothesis r 1 and r 2 are relatively prime, by our proposition above we see that this congruence has a unique solution for k modulo r 2, and hence a solution for x. For uniqueness, suppose x and y are both solutions. Then x y is 0 modulo r 1 and 0 modulo r 2, meaning that r 1 (x y) and r 2 (x y). But since r 1 and r 2 are relatively prime, their product must therefore divide x y, meaning that x is unique modulo r 1 r 2. Finally, it is obvious that any other element of R congruent to x modulo r 1 r 2 also satises the system. Example: In R = C[x], solve the system q(x) 1 (mod x 1), q(x) 3 (mod x). Since x 1 and x are relatively prime polynomials, by the Chinese Remainder Theorem all we have to do is nd one polynomial satisfying the system. If we take the solution q(x) = 3 + ax to the second equation and plug it into the rst equation, we need to have 3 + ax 1 (mod x 1). Since 3 + ax (3 + a) mod (x 1) by the remainder theorem, we can take a = 2. Hence the polynomial q(x) = 3 2x is a solution to the system. 3 2x + x(x 1) s(x) for an arbitrary polynomial s(x) R. The general solution is therefore 4.5 Factorization in F [x] We now use our ring theory machinery to study the structure of polynomials whose coecients lie in a eld F. As we observed in the previous sections, F [x] is a Euclidean domain, hence also a unique factorization domain, so we may freely speak of irreducible polynomials, factorizations, and so forth. For ease of notation, we will write F p = Z/pZ for p a prime, since (as we have discussed) the ring Z/pZ is a eld, and the notation (Z/pZ)[x] is rather ungainly in comparison to F p [x]. Frequently, F p is called the eld with p elements. It is also occasionally denoted GF (p), for the Galois Field with p elements, since these elds were rst extensively studied by Galois. The uses of the word the are justied, as it can be proven that up to relabeling the elements, there is only one eld with p elements for any prime p. We rst start with an observation that is likely familiar from elementary algebra (where it is sometimes called the remainder theorem): 10

11 Proposition: If q(x) F [x] is a polynomial and a F, then the remainder upon dividing q(x) by x a is q(a). In particular, x a divides q(x) if and only if q(a) = 0. (In this case we say a is a zero of q(x).) Proof: Suppose q(x) = d c i x i. Observe rst that (x i a i ) = (x a)(x i 1 + x i 2 a + + a i 2 x + a i 1 ), i=0 so in particular, x a divides x i a i for all i. d Now we simply write q(x) = q(x) f(a) = c i (x i a i ), whence we see that x a divides q(x) q(a). i=0 Since q(a) is a constant, it is therefore the remainder after dividing q(x) by x a. The other statement is immediate. Proposition: If q(x) F [x] is a polynomial of degree d, then q(x) has at most d distinct zeroes in F. Proof: We induct on the degree d. For d = 1, the polynomial is of the form ax + c for a 0, which has exactly one root, namely c/a. Now suppose the result holds for all polynomials of degree d and let q(x) be a polynomial of degree d + 1. If q has no zeroes we are obviously done, so suppose q(a) = 0. We can then factor to write q(x) = (x a)r(x) for some polynomial r(x) of degree d. By the induction hypothesis, r(x) has at most d roots, so q(x) has at most d+1 roots, because (x a)r(x) = 0 only when x = a or r(x) = 0 (since F is a eld). The above results, while seemingly obvious, fail completely if the coecients do not form a eld. Here are some especially distressing examples: The quadratic polynomial q(x) = x 2 1 visibly has four solutions modulo 8: x 1, 3, 5, and 7. Furthermore, q(x) can be factored in two dierent ways: as (x 1)(x 7) and as (x 3)(x 5). The linear polynomial q(x) = x, despite having degree 1, is not irreducible modulo 6: it can be written as the product (2x + 3)(3x + 2). Furthermore, q(x) = x has one zero (namely x = 0), even though its two factors 2x + 3 and 3x + 2 each have no zeroes modulo 6. In general, it is not easy to determine when an arbitrary polynomial is irreducible. If the degree is small, however, this task can be done by examining all possible factorizations. The following result is frequently useful: Proposition: If F is a eld and q(x) F [x] has degree 2 or 3 and has no zeroes in F, then q(x) is irreducible. Proof: If q(x) = a(x)b(x), taking degrees shows 3 = deg(q) = deg(a) + deg(b). Since a and b both have positive degree, one of them must have degree 1. Then its root is also a root of q(x). Taking the contrapositive gives the desired statement. Example: Over R, the polynomial x 2 +x+11 has no roots (since it is always positive), so it is irreducible. Example: Over F 5, the polynomial q(x) = x 3 + x + 1 has no roots, since q(0) = 1, q(1) = 3, q(2) = 1, q(3) = 1, and q(4) = 4. Another property that we can fruitfully study in a general eld is the presence of repeated factors (when we factor a polynomial into irreducibles). Example: Over C, the polynomial x 3 + x 2 x 1 factors into irreducibles as (x 1) 2 (x + 1), and thus has the repeated factor x 1 (appearing with multiplicity 2). Example: Over F 2, the polynomial x 4 + x factors into irreducibles as (x 2 + x + 1) 2, which has repeated factors. Denition: If q(x) = d a n x n is an element of F [x], the derivative is q (x) = n=0 d n a n x n 1. n=1 11

12 This, of course, agrees with the usual denition of the derivative of a real-valued polynomial. However, in a general eld, there is no notion of limit (and thus, no analogy, so instead we dene the derivative explicitly, as above. It is easy to see that the derivative is a linear operator. As would be expected, Leibniz's relation (often called the product rule) also holds: (fg) = f g + fg. This relation holds for powers of x by a straightforward computation, and then holds for general polynomials by the linearity of the derivative. Over the real numbers, the derivative allows us to detect multiple roots: If q(x) R[x] is a polynomial, then (x a) is a repeated factor of q(x) if and only if q(a) = q (a) = 0. This also holds over arbitrary elds, in a fairly general fashion. Proposition: Let q(x) F [x]. If r(x) F [x] is an irreducible polynomial, then [r(x)] 2 divides q(x) if and only if r(x) divides q(x) and q (x). Thus, q(x) will have a repeated (irreducible) factor if and only if q(x) and q (x) have a nontrivial gcd. Proof: For clarity, we will suppress the argument of the polynomials. Suppose r 2 divides q. Then there exists a polynomial s(x) with q = r 2 s. By the product rule, we have q = r [2r s + rs ], so r divides q. Now suppose r divides q and q. Then there exists a polynomial t such that q = rt. By the product rule, we have q = r t + rt. Then r divides r t, but since r is irreducible and does not divide its derivative (by degree considerations), we conclude that r divides t: then r 2 divides q, as claimed. Since we can rapidly compute the gcd of q(x) and q (x) using the Euclidean algorithm in F [x], we can quickly determine if a given polynomial has a repeated factor. Example: Determine whether q(x) = x 4 + 3x 3 + 3x 2 + 3x + 1 has a repeated factor in F 5 [x]. We have q (x) = 4x 3 + 4x 2 + x + 3. Now we perform the Euclidean algorithm: dividing q by q yields a quotient of 4x + 3 and a remainder of 2x 2 + 3x + 2. Dividing q by 2x 2 + 3x + 2 yields a quotient of 2x + 4 and a remainder of 0. Hence, we see that 2x 2 + 3x + 2 is a nontrivial repeated factor. It is associate to the monic polynomial x 2 + 4x + 1. Indeed we can see that, in F 5 [x], q(x) = (x 2 + 4x + 1) Modular Arithmetic in F [x] In the ring F [x], where F is a eld, we have a natural collection of residue class representatives that arise from the division algorithm. Proposition: If R = F [x] and q(x) R is a polynomial of degree d, then the polynomials in F [x] of degree d 1 are a full set of residue class representatives for R/qR. Proof: By the division algorithm, every polynomial is congruent modulo q(x) to some polynomial of degree less than d, namely, to the remainder after dividing by q(x). Conversely, each of these residue classes is distinct, because two distinct polynomials in F [x] of degree less than d cannot be congruent modulo q(x): if they were, their dierence would be a multiple of q(x) of degree less than d, but the only such multiple is 0. Example: Describe the multiplication in the ring R/qR, where q(x) = x

13 By our previous results, we know that q(x) is irreducible over the real numbers, since it has degree 2 and no real roots. Thus, R/qR is a eld. We also know that the elements of this ring are of the form a + bx where a, b R. The multiplication is simply multiplication of polynomials, subject to the relation x = 0. Thus, in general we can write (a + bx) (c + dx) = ac + (bc + ad)x + bdx 2 = (ac bd) + (bc + ad)x. This multiplication should look very familiar: in fact, it is exactly the same as the multiplication of complex numbers (a + bi) (c + di) = (ac bd) + (bc + ad)i. There is an obvious reason for this: the ring R/qR is really just the complex numbers C, where instead of using i 2 = 1, we say x 2 = 1 instead. (In the language of algebra, we would say that R/qR and C are isomorphic rings, since their ring structures are exactly the same: we have just labeled the elements dierently.) We can fruitfully apply our results to the case where F = F p = Z/pZ is a nite eld with p elements: Theorem: If q(x) F p [x] is an irreducible polynomial of degree d, then the ring R/qR is a nite eld with p d elements. Proof: We simply invoke our previous results: by the above, the residue classes in the ring R/qR are given by the polynomials in F p [x] of degree d 1. Such a polynomial has the form a 0 +a 1 x+ +a d 1 x d 1, where the coecients a i are arbitrary elements of F p. There are clearly p d such polynomials. Now by our earlier results, we know that if q(x) is irreducible, R/qR is an integral domain. Since it is also nite, it is a eld. Example: The ring R/qR, where R = F 2 [x] and q(x) = x 2 + x + 1, is a eld with 4 elements. This follows because x 2 +x+1 is irreducible modulo 2, since (if it had a nontrivial factorization) it would necessarily have a root (which it does not). Explicitly, the four elements of this eld are 0, 1, x, and x+1. Addition is taken with coecients modulo 2, and multiplication is performed under the convention that x 2 + x + 1 = 0 (i.e., x 2 = x + 1, since coecients are taken modulo 2). Even more explicitly, we can write out the addition and multiplication tables as follows: x x x x x + 1 x x x x x + 1 x + 1 x x x x x + 1 x 0 x x x x x We can see, for example, that x and x + 1 are both primitive roots in this eld, since their powers yield all of the nonzero elements. (We will prove shortly that any nite eld possesses a primitive root.) Example: The ring R/qR, where R = F 3 [x] and q(x) = x 2 + 1, is a eld with 9 elements. This follows because x 2 + x + 1 is irreducible modulo 3, since, if it had a nontrivial factorization, it would necessarily have a root (which it does not). Explicitly, the nine elements of this eld are 0, 1, 2, x, x + 1, x + 2, 2x, 2x + 1, and 2x + 2. Addition is taken with coecients modulo 3, and multiplication is performed under the convention that x = 0 (i.e., x 2 = 2, since coecients are taken modulo 3). Here is the multiplication table for the units in this eld: 1 2 x x + 1 x + 2 2x 2x + 1 2x x x + 1 x + 2 2x 2x + 1 2x x 2x + 2 2x + 1 x x + 2 x + 1 x x 2x 2 x + 2 2x x + 1 2x + 1 x + 1 x + 1 2x + 2 x + 2 2x 1 2x x x + 2 x + 2 2x + 1 2x x x + 1 2x 2 2x 2x x 1 2x + 1 x + 1 2x + 2 2x + 1 d + 2 2x + 1 2x + 1 x + 2 x x 2x + 2 x 1 2x + 2 2x + 2 x + 1 2x + 1 x 2 x x 13

14 One can check, for example, that x + 1, x + 2, 2x + 1, and 2x + 2 are all primitive roots in this eld. For posterity, we also write down the analogue of Fermat's Little Theorem in an arbitrary nite eld: Proposition: If F is a nite eld, then a F = a for any a F. Remark: The notation F denotes the number of elements of F. Proof: Apply the generalization of Euler's Theorem with R = F and r = 0: this yields the statement that if a is any unit in F, we have a F 1 = 1. Hence a F = a for every nonzero a F, but this clearly holds when a = 0 as well The Polynomial x p x over F p The polynomial x p x in F p [x] has a number of interesting properties. By Fermat's Little Theorem, a p a (mod p), so if q(x) = x p x, then q(a) = 0 for every a F p. In other words, this polynomial x p x has the rather strange property that its value is always zero, yet it is not the zero polynomial. Proposition: The factorization of x p x in F p [x] is x p x = a F p (x a). Proof: As noted above, q(x) = x p x is such that q(a) = 0 for every a F p. Hence, x a is a divisor of q(x) for every a F p. However, because this polynomial has at most p roots, and we have exhibited p roots, the factorization of q(x) must be q(x) = a F p (x a), since the leading terms agree. We can also show that x p x is essentially the only polynomial which is identically zero. Explicitly: Proposition: If q(a) = 0 for all a F p, then q(x) is divisible by x p x. Proof: If q(a) = 0 for all a F p, then x a is a divisor of q(x) for all a F p [x]. But by the above factorization, this means a F p (x a) = x p x divides q(x). Another immediate application of this factorization is an easy proof of Wilson's Theorem. By dividing through by x, we see that x p 1 1 = (x a). a F p,a 0 Now examine the constant term of the product: it is ( 1) p 1 a F p,a 0 (a) = ( 1)p 1 (p 1)!. But (modulo p) the constant term is also equal to 1, so we deduce (p 1)! ( 1) p 2 1 (mod p) The Polynomial x pn x over F p As we observed above, the polynomial x p x has a nice factorization in F p [x]. factorization of the polynomial x pn x in F p [x]. Let us now consider the Example: For n = 2 and p = 2, we nd the irreducible factorization x 4 x = x(x + 1)(x 2 + x + 1). Example: For n = 3 and p = 2, we nd the irreducible factorization x 8 x = x(x + 1)(x 3 + x 2 + 1)(x 3 + x + 1). Example: For n = 4 and p = 2, we nd the irreducible factorization x 16 x = x(x + 1)(x 2 + x + 1)(x 4 + x 3 + 1)(x 4 + x + 1)(x 4 + x 3 + x 2 + x + 1). 14

15 Example: For n = 2 and p = 3, we nd the irreducible factorization x 9 x = x(x+1)(x+2)(x 2 +2)(x 2 + x + 2)(x 2 + 2x + 2). Example: For n = 2 and p = 5, the list of irreducible factors of x 25 x is x, x + 1, x + 2, x + 3, x + 4, x 2 + 2, x 2 + 3, x 2 + x + 1, x 2 + x + 2, x 2 + 2x + 3, x 2 + 2x + 4, x 2 + 3x + 3, x 2 + 3x + 4, x 2 + 4x + 1, and x 2 + 4x + 2. We notice (especially in the p = 5 example) that the irreducible factors all appear to be of small degree, and that there are no repeated factors. In fact, it seems that the factorization of x pn x over F p contains all of the irreducible polynomials of degree n, or of degree dividing n. Theorem: The polynomial x pn x factors in F p [x] as the product of all monic irreducible polynomials over F p of degree dividing n. Proof: Let q(x) = x pn x and R = F p [x]. We prove the result in the following way: rst, we show that there are no repeated factors. Second, we show that every irreducible polynomial of degree dividing n divides q(x). Finally, we show that no other irreducible polynomial can divide q(x). For the rst part, we have q (x) = p n x pn 1 1 = 1, so q(x) and q (x) are relatively prime. Thus, by our earlier results, we know that q(x) has no repeated irreducible factors. Before starting the rest of the proof, we rst show a simple lemma: Lemma: If p is any prime, the gcd of p n 1 and p d 1 is p gcd(n,d) 1. Write n = qd + r, and let a = p r (p (q 1)d + p (q 2)d + + p d + 1). Some arithmetic will show that p n 1 = (p d 1)a + (p r 1). Then gcd(p n 1, p d 1) = gcd(p d 1, p r 1). But this means we can perform the Euclidean algorithm on the exponents without changing the gcd. The end result is p gcd(n,d) 1, so this is the desired gcd. For the second part, now suppose that s(x) F p [x] is an irreducible polynomial of degree d, where n = ad. We know that R/sR is a nite eld F having p d elements, so by Euler's Theorem in F, we see that x pd 1 1 (mod s). But, by the lemma, p d 1 divides p n 1, so raising to the appropriate power modulo s shows x pn 1 1 (mod s). We conclude that s divides x pn x, as desired. For the nal part, suppose s(x) F p [x] is an irreducible polynomial that divides x pn x and has degree d not dividing n. Since s(x) x, we can assume s divides x pn 1 1. As above, R/sR is a nite eld F having p d elements, so by Euler's Theorem in F, we see that a pd 1 1 (mod s) for every nonzero a F. Since a pn 1 1 (mod s) holds for every nonzero a F by the above assumptions, we conclude that a pgcd(d,n) 1 1 (mod s). But this is impossible, because q(t) = t pgcd(d,n) 1 1 is then a polynomial of degree p gcd(d,n) 1 which has p d 1 roots over the eld F p. As a corollary, the above theorem allows us to count the number of monic irreducible polynomials in F p [x] of any particular degree n. Let f p (n) be the number of monic irreducible polynomials of exact degree n in F p [x]. The theorem above says that p n = df p (d), since both sides count the total degree of the product of d n all irreducible polynomials of degree dividing n. Using this recursion, we can compute the rst few values: n f p (n) p 2 (p2 p) 3 (p3 p) 4 (p4 p ) 5 (p5 p) 6 (p6 p 3 p ) 7 (p7 p) 8 (p8 p 4 ) 15

16 For example, the formula says that there are 2 irreducible polynomials of degree 3 over F 2, which there are: x 3 + x and x 3 + x + 1. In fact, we can essentially write down a general formula. { 0 if n is divisible by the square of any prime Denition: The Möbius function is dened as µ(n) =. ( 1) k if n is the product of k distinct primes In particular, µ(1) = 1. Proposition (Möbius inversion): If f(n) is any sequence satisfying a recursive relation of the form g(n) = f(d), for some function g(n), then f(n) = µ(d)g(n/d). d n d n Proof: First, consider the sum d n µ(d): we claim it is equal to 1 if n = 1 and 0 if n 0. To see this, if n = p a1 1 pa k k, the only terms that will contribute to the sum µ(d) are those values d n of d = p b1 1 pb k k where each b i is 0 or 1. If k > 0, then half of these 2 k terms will have µ(d) = 1 and the other half will have µ(d) = 1, so the sum is zero. Otherwise, k = 0 means that n = 1, in which case the sum is clearly 1. Now we prove the desired result by (strong) induction. It clearly holds for n = 1, so now suppose the result holds for all k < n. By hypothesis and induction, we have µ(d)g(n/d) d n = µ(d) d n but this last sum is simply f(n), because d (n/d ) d (n/d) = dd n µ(d)f(d ) = d n f(d ) d (n/d ) f(d ) µ(d) µ(d) is zero unless n/d is equal to 1. By applying Möbius inversion to our particular function f p (n), we immediately obtain the following: Corollary: If f p (n) is the number of monic irreducible polynomials of degree n in F p [x], then f p (n) = 1 p n/d µ(d). n d n From this corollary, we see that f p (n) = 1 n pn + O(p n/2 ), where the big-o notation means that the error is of size bounded above by a constant times p n/2. This has the following interesting reinterpretation: let X be the number of polynomials in F p [x] of degree less than n. Clearly, X = p n. Now we ask: of all these X polynomials, how many of them are prime (i.e., irreducible)? This is simply f p (n) = 1 n pn + O(p n/2 X ) = log p (X) + O( X). In other words: the number of primes less than X is equal to X, up to a bounded error term. log p (X) 16

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