1/30: Polynomials over Z/n.

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1 1/30: Polynomials over Z/n. Last time to establish the existence of primitive roots we rely on the following key lemma: Lemma 6.1. Let s > 0 be an integer with s p 1, then we have #{α Z/pZ α s = 1} = s. We now discuss the idea leading to the lemma, which is of great interest by its own. Let n > 0 be a fixed integer. Recall that we denote by Z/n (will specializes to n = p a prime later) the set of residue classes modulo n. We can add, subtract, and multiply residue classes, and there are 0 and 1, the additive identity and the multiplicative identity, following the common rules: For any ā, b, c Z/n we have 1. ā + b = b + ā. 2. (ā + b) + c = ā + ( b + c). 3. ā + 0 = ā. 4. b := 0 b, ā b = ā + ( b). 5. ā b = b ā. 6. (ā b) c = ā ( b c). (We will frequently omit the for multiplication from now on.) 7. ā( b + c) = ā b + ā c. 8. ā 1 = 1ā. In Math 120, we say Z/n is a commutative ring with (multiplicative) identity. We typically just refer to 0 (resp. 1) as zero (reps. one), since it has all the desired properties. When n = p is a prime, every non-zero element in Z/n is invertible, in which case we say Z/p is a field, the main object of study in Math 121. Now there is another object in Math 120 which is our lead today: Definition 6.2. (i) A polynomial over Z/n is a formal object of the form α k x k +α k 1 x k α 1 x 1 + α 0 x 0 where k 0 is some non-negative integer and all α i Z/n. We can add, subtract and multiply two polynomials using the addition, subtraction and multiplication in Z/n and the usual rule for polynomials. Two polynomials are considered the same, if after omitting those terms with 0 (i.e. 0) coefficients they have identical terms. (ii) We denote by (Z/n)[x] the collection of all polynomials over Z/n. 1

2 (iii) In (i) we may thus assume P (x) = α k x k + α k 1 x k α 1 x 1 + α 0 x 0 has either P (x) = 0 (i.e. all coefficients are 0), or α k 0 by omitting those before. In the latter case, say k is the degree of q and write deg(q) = k, otherwise deg(q) =. When deg(q) 0 we also say α k is the leading coefficient, and that q is monic if the leading coefficient α k = 1. (iv) For any β Z/n, we may plug β into P (x) and get P (β) Z/n. Warning 6.3. A polynomial is NOT a function! For example, say n = 3. Then q 1 (x) = x 3 and q 2 (x) = x satisfy q 1 (β) = q 2 (β) for all β Z/3. (This is Fermat s little Theorem.) Still, we consider q 1 and q 2 to be different polynomials over Z/3. Remark 6.4. You might have realized that to write a polynomial over Z/n, we can first write a polynomial over Z and then take it modulo n. Even better, addition, subtraction and multiplication of polynomials over Z and over Z/n are compatible; one may either add two polynomials over Z first and then take them modulo n, or take them modulo n first and then add them, and get the same result in both routes. Because of this flexibility, we frequently describe a polynomial over Z/n as a polynomial over Z. Definition 6.5. We call β Z/n is a root of P (x) if P (β) = 0 (i.e. = 0). Under this definition, we see that Lemma 6.1 can be rephrased as: Lemma 6.6. Suppose n = p is a prime. Let s > 0 be such that s p 1. Then x s 1 has s roots in Z/p. In high school algebra, with polynomials over real numbers we have that if β is a root of P (x), then (x β) is a factor of P (x) and we can write P (x) = (x β)q(x). Do we have the same thing here? Ideally we would like polynomials in (Z/n)[x] to enjoy unique factorizations. Following our experience in Z, we want to be able to divide polynomials: Definition 6.7. Let P 1 (x), P 2 (x) (Z/n)[x]. We say P 2 (x) P 1 (x) if there exists q(x) (Z/n)[x] with P 1 (x) = P 2 (x)q(x). We also write P 2 P 1 and say P 2 is a divisor of P 1, that P 1 is a multiple of P 2, etc. Proposition 6.8. Let P 1 (x), P 2 (x) (Z/n)[x]. Suppose P 2 (x) 0 and its leading coefficient is invertible. Then we can divide P 1 (x) by P 2 (x); there exists q(x), r(x) (Z/n)[x] such that P 1 (x) = P 2 (x)q(x) + r(x) and deg(r) < deg P 2. We have P 2 (x) P 1 (x) iff r(x) = 0. Proof. Proved by classical long division algorithm. multiple of P 2 only when r = 0. Since deg(r) < deg(p 2 ), it can be a 2

3 Begin with P 1 (x), P 2 (x) (Z/n)[x], and apply Euclid s algorithm: by swapping assume deg(p 1 ) deg(p 2 ). Divide P 1 by P 2 as in Proposition 6.8 unless (1) P 2 does not have invertible leading coefficient or (2) P 2 = 0. As long as we can divide, we replace P 1 by r and re-do the process. Since deg(r) < deg(p 2 ) deg(p 1 ), the total degree deg(p 1 ) + deg(p 2 ) decreases by at least 1 in each loop, so by some step the algorithm must stop. Proposition 6.9. Suppose we are lucky enough that the Euclid s algorithm stops due to situation (2) (but not to situation (1) where the leading coefficient of P 2 is not invertible). Say we stop at P 1 (x) = d(x) and P 2 (x) = 0. Let α be the leading coefficient of d(x) and d(x) := α 1 d(x) is monic. We have 1. d P 1 and d P If d is another common divisor of P 1 and P 2, then d d. Such a monic polynomial d with the above two properties is called the greatest common divisor of P 1 and P 2, and written gcd(p 1, P 2 ) := d. Moreover, there exists polynomials g 1 (x), g 2 (x) (Z/n)[x] that d = P 1 g 1 + P 2 g 2. Proof. Same proof as in the integer case for d. See Lemma 1.4 on the lecture note dated 1/10. There is a little difference between d and d. But as α is invertible we have d d and d d, so that anything divisible by d can be divided by d and vise versa, and anything that divides d will also divides d and vise versa. Now the highlight is when n = p is a prime number: Corollary Suppose n = p. Then any two polynomials P 1 (x) and P 2 (x) have a greatest common divisor d(x). Moreover, there exists polynomials g 1 (x), g 2 (x) (Z/p)[x] that d = P 1 g 1 + P 2 g 2. Proof. When n = p so Z/n = Z/p, we have that any non-zero residue class modulo p is invertible, and therefore any non-zero polynomial in Z/p has an invertible leading coefficient. Thus Euclid s algorithm always work to our desire, and the corollary follows from Proposition 6.9. From now on we work with polynomials over Z/p, where p is a fixed prime. It is also common to write F p := Z/p, to emphasize that it is a field, and also F p = (Z/p). Some people write Z p for F p, but in number theory we will avoid doing so 1. Now with Euclid s algorithm for polynomials over F p working as before, we expect a unique factorization result. First we have to define the analogue of prime numbers. 1 The reason behind is to reserve the notation for p-adic integers. See P-adic_number 3

4 Definition A monic polynomial P (x) F p [x] is called irreducible, if we cannot write P (x) = P 1 (x)p 2 (x) with deg(p 1 ), deg(p 2 ) > 0. Example By definition, a monic polynomial of degree 1 is always irreducible, as when we write P = P 1 P 2 we have deg(p ) = deg(p 1 )+deg(p 2 ). In other words, x α is irreducible for any α F p. Corollary Suppose P (x) F p [x] is irreducible, and A(x), B(x) are such that P AB. Then either P A or P B. Proof. Same proof as for integers, see the proof of Lemma 1.2. Theorem Let P (x) F p [x] be any non-zero polynomial. Then we can write P (x) = c P 1 (x) P 2 (x)... P k (x) (1) for some irreducible monic polynomials P 1,..., P k and c F p. This expression is unique up to re-ordering the irreducible factors. Proof. Same as Theorem 1.1. Now recall our original purpose was to discuss the roots of x s 1 F p [x]. We first give a general observation. Lemma Let P (x) (Z/n)[x] (n does not need to be a prime for this lemma). Then α Z/n is a root of P (x) if and only if (x α) P (x). Proof. We divide P (x) by (x α) to have P (x) = Q(x)(x α) + R(x). As deg R(x) < deg(x α) = 1, we have either deg R(x) = 0 or deg R(x) =. In either case R(x) = r for some r Z/n. Now we plug in α to get P (α) = Q(α)(α α) + r = P (α) = r. This shows that α is a root iff r = 0, i.e. if (x α) divides P (x). Corollary We have x p x = α F p (x α) Proof. Firstly, some factorization of x p x exists. Secondly, any α F p is a root of x p x; this is Fermat s little theorem. By Lemma 6.15, all (x α), α F p are irreducible factors of P (x), which must appear in the factorization of x p x. These are p factors. If P (x) := x p x is going to have more factors in (1), then P (x) is going to have its degree larger than p, which is not the case. Thus all the linear (i.e. degree 1) monic polynomials are precisely all factors of P. Since P is monic, we also have c = 1 in (1). This proves the Corollary. Now we can easily prove Lemma 6.1: 4

5 Proof of Lemma 6.1. When s p 1. We note that x s 1 divides x p 1 1 and thus also x p x. Thus the factorization of x s 1 is part of that of x p x. In particular, x s 1 is a product of distinct monic linear polynomials. Since deg(x s 1) = s, x s 1 has s many such factors, i.e. s roots. This shows #{α Z/pZ α s = 1} = s. Let us also mention another useful corollary: Corollary For any P (x) F p [x] a non-zero polynomial over F p, the number of roots of P (x) is always no larger than deg(p ). Proof. P (x) has a unique factorization as in (1), and its roots correspond to those linear factors in (1). By degree consideration, the number of such cannot exceed deg(p ). Remark A fundamental analogy we work with here is Z versus F p [x]. There are way, way much more to say about this analogy. We would like to mention two points without going into the detail: Firstly, R[x] can be think of as polynomial functions on the real line. Likewise, F p [x] can be thought of as polynomial functions on the F p -line, this is the starting point of arithmetic geometry. Secondly, beginning from Z we observe the properties of prime numbers (which we really defined to be irreducible numbers) and go into the field Z/p. We can likewise begin with an irreducible polynomial P (x) and go into the field F p [x]/p (x); that is, similar arguments as we had done will show that all elements in F p [x]/p (x) are invertible. For example, in Problem 1 of Problem Set 2 we were concerned with whether x 2 5 has a root (modulo p) or not, and drew some conclusion (e.g. p a p 1 ) if it has a root. Things become more complicated when x 2 5 does not have a root: in this case P (x) := x 2 5 is irreducible, and one can work with the finite field F p [x]/p (x) (First half of Math 121 is devoted to that), get generalizations of Fermat s little theorem and a lot more, and in particular have corollaries like p a p+1 dropping out. 5