1. Algebra 1.5. Polynomial Rings

Transcription

1 1. ALGEBRA Algebra 1.5. Polynomial Rings Lemma Let R and S be rings with identity element. If R > 1 and S > 1, then R S contains zero divisors. Proof. The two elements (1, 0) and (0, 1) are different from the zero element and their product is (0, 0). The above lemma shows that the direct product of two fields is never a field. In order to construct finite fields we will have introduce some more definitions. Let R be a commutative ring with identity element and 0 1. A polynomial in one variable X with coefficients in R is an expression f(x) = a n X n + a n 1 X n a 1 X + a 0, where X is the variable and a n, a n 1,..., a 1, a 0 R. We will often write f instead of f(x). The set of polynomials in one variable X with coefficients in R will be denoted by R[X]. If a n 0, then n will be called the degree of the polynomial, in this case n will be denoted by deg(f) and a n will be called the leading coefficient of f. Let f be a polynomial of degree n. If a n = 1, then the polynomial will be called monic, if a n 0 and a n 1 = = a 1 = a 0 = 0, then f will be called monomial. The polynomial 0 has no degree and a polynomial has degree 0 if and only if it is an element of R different from 0. s The polynomial 2X 3 + X + 1 Z [X] has degree 3; The polynomial X + 5 Z [X] has degree 1 and is monic; The polynomial 3 Z [X] has degree 0 and is monomial; The polynomial X 6 Z [X] has degree 6 and is monomial and monic; Notation If R = Z /m Z, with m 2, and f R[X], then the coefficients of f will not be written as [a i ] m but as integers. Among those there is a unique one with 0 r < m which will be the preferred notation. The polynomial 2X 3 + X + 1 Z /2 Z [X] has degree 1, for 2X 3 + X + 1 = X + 1. Let R be a commutative ring with identity element. Let f, g R[X] with f(x) = a n X n + a n 1 X n a 1 X + a 0 and g(x) = b m X m + b n 1 X m b 1 X + b 0, with n m. We define the sum + of f and g, denoted by f + g, as the polynomial where b n = = b m+1 = 0. (a n + b n )X n + (a n 1 + b n 1 )X n (a 1 + b 1 )X + (a 0 + b 0 ),

2 1. ALGEBRA 20 We define the product of f and g, denoted by f g, as the polynomial c n+m X n+m + c n+m 1 X n+m c 1 X + c 0, where c k = k a i b k i, and 0 k n + m, a j = 0 if j > n and b j = 0 if j > m. i=0 The polynomial 0 is neutral with respect to the sum and the polynomial 1 is neutral with respect to the product. Lemma (R[X], +, ) is a commutative ring with identity element. Proof. Exercise. s Let f(x), g(x) Z [X], with f(x) = X 2 + X + 1 and g(x) = X 3 + 2X 2 + 3X + 2. f + g = X 3 + 3X 2 + 4X + 3. Then Let f(x), g(x) Z [X], with f(x) = X 2 + X + 1 and g(x) = X 3 + 2X. Then f g = (X 2 + X + 1)(X 3 + 2X) = X 5 + X 4 + X 3 + 2X 3 + 2X 2 + 2X = X 5 + X 4 + 3X 3 + 2X 2 + 2X. Let f(x), g(x) Z /4 Z [X], with f(x) = X 2 + X + 1 and g(x) = X 3 + 2X 2 + 3X + 2. Then f + g = X 3 + 3X Let f(x), g(x) Z /3 Z [X], with f(x) = X 2 + X + 1 and g(x) = X 3 + 2X. Then f g = (X 2 + X + 1)(X 3 + 2X) = X 5 + X 4 + X 3 + 2X 3 + 2X 2 + 2X = X 5 + X 4 + 2X 2 + 2X. Let f(x), g(x) Z /6 Z [X] with f(x) = 2X 2 + X and g(x) = 3X + 1. Then the degree of f is 2 and the degree of g is 1. But f g = (2X 2 + X)(3X + 1) = 6X 3 + 3X 2 + 2X 2 + X = 5X 2 + X, has degree 2! Moreover, 2X 2, 3X Z /6 Z [X] are both different from 0, but (2X 2 ) (3X) = 6X 3 = 0, so Z /6 Z [X] has zero divisors. To avoid this type of behavior we need to put some extra conditions on R. Lemma Let K be a field. (i) The ring K[X] has no zero divisors; (ii) If f, g K[X] with f 0 and g 0, then deg(f g) = deg(f) + deg(g). Proof. Let f, g K[X] with f 0 and g 0. Write f(x) = a n X n + a n 1 X n a 1 X + a 0, where a n 0 and g(x) = b m X m + b n 1 X m b 1 X + b 0, where b m 0. Now f g = (a n b m )X n+m + monomials of lower degree. Since K is a field, and so has no zero divisors, we have a n b m 0. Whence f g 0 and deg(f g) = n + m = deg(f) + deg(g). For the remainder we will focus our attention on the ring K[X], with K a field. So far the only finite fields we encountered arose from modular arithmetic on the ring Z when we calculate modulo a prime. We will now mimic this construction with K[X] taking the role of Z and certain polynomials taking the role of prime numbers. Theorem (Division with remainder) Let K be a field, f, g K[X], with g 0. There exist unique q, r K[X], with f = qg + r and either r = 0 or deg(r) < deg(g).

3 1. ALGEBRA 21 Proof. Existence: If f = 0 we can take r = q = 0. So we may suppose that f 0. If deg(f) < deg(g), then we can take q = 0 and r = f. So we may assume that deg(f) deg(g). Let deg(f) = n and deg(g) = m, so n m. We will use induction on n. If n = 0, then m = 0 and f, g K, so we can take q = fg 1 and r = 0. Suppose now that the theorem is true for all polynomials of K[X] of degree less then n. Write f(x) = a n X n +a n 1 X n a 1 X + a 0, where a n 0 and g(x) = b m X m + b n 1 X m b 1 X + b 0, where b m 0. Let f 1 = f a n b 1 m X n m g. Then f 1 = 0 or deg(f 1 ) < deg(f). If f 1 = 0, we can take q = a n b 1 m X n m and r = 0. If f 1 0, then there exist q 1, r 1 K[X], with f 1 = q 1 g + r 1 and either r 1 = 0 or deg(r 1 ) < deg(g). Now f = f 1 + a n b 1 m X n m g = (q 1 g + r 1 ) + a n b 1 m X n m g = (q 1 + a n b 1 m X n m )g + r 1. Taking q = q 1 + a n b 1 m X n m and r 1 = r, we obtain f = qg + r and either r = 0 or deg(r) < deg(g) as wanted. Uniqueness: Suppose f = q 1 g + r 1 with either r 1 = 0 or deg(r 1 ) < deg(g) and f = q 2 g + r 2 with either r 2 = 0 or deg(r 2 ) < deg(g). Then 0 = (q 1 g + r 1 ) (q 2 g + r 2 ) = (q 1 q 2 )g + (r 1 r 2 ), whence (q 1 q 2 )g = (r 2 r 1 ). If r 1 = r 2, then (q 1 q 2 )g = 0. But since K[X] has no zero divisors and g 0, we must have (q 1 q 2 ) = 0. Hence q 1 = q 2. If r 1 r 2, then r 2 r 1 0 and thus also (q 1 q 2 ) 0. Now deg((q 1 q 2 )g) = deg(q 1 q 2 ) + deg(g) deg(g) and deg(r 2 r 1 ) < deg(g). So deg(g) deg((q 1 q 2 )g) = deg(r 2 r 1 ) < deg(g), a contradiction. Observe that the proof give an algorithm of computing q and r. This is essentially the long division you are used to. The polynomial q will be called the quotient, and the polynomial r will be called the remainder after division by g. Consider in the ring Z /2 Z and the polynomials X 3 +X +1 and X 2 +X. The quotient of X 3 +X +1 divided by X 2 + X is X + 1, the remainder 1. Let K be a field and f, g K[X]. We say that g divides f if there exists a q K[X] with f = q g. In this case we say that g is a divisor of f, or f is a multiple of g. We will write g f. In the ring Z /5 Z [X] X + 1 is a divisor of 3X 2 + 4X + 1, for 3X 2 + 4X + 1 = (3X + 1)(X + 1); 2X + 2 is a divisor of 3X 2 + 4X + 1, for 3X 2 + 4X + 1 = (4X + 3)(2X + 2). Lemma Let K be a field and f, g K[X]. If g divides f, then kg divides f, and g divides kf, for any k K. Proof. Since g divides f there exists a q K[X] with f = q g. Since f = (qk 1 ) (kg), and qk 1 K[X] we have that kg divides f. Since kf = (kq)g and kq K[X] we have that g divides kf. Let K be a field and f, g K[X] not both 0. A divisor d of g and f is said to be a greatest common divisor if any divisor of f and g is also a divisor of d.

4 1. ALGEBRA 22 Note that if d is a common divisor of f and g, then f = q 1 d and g = q 2 d, for certain q 1, q 2 K[X]. Since f and g are not both 0, we have d 0. In particular, the degree of d is defined. Moreover, if d is a greatest common divisor of f and g, then so is kd, for any k K. The following lemma shows that a greatest common divisor is unique, up to multiplying with K. Lemma If c and d are two greatest common divisors of f and g, then there exists a k K, with c = kd. Proof. Since c is a greatest common divisors of f and g, and d a divisor of f and g, we have d c. So there exists a q 1 K[X] with q 1 0 and c = q 1 d. In particular, deg(c) = deg(q 1 )+deg(d) deg(d). Since d is a greatest common divisors of f and g, and c a divisor of f and g, we have c d. So there exists a q 2 K[X] with q 2 0 and d = q 2 c. In particular, deg(d) = deg(q 2 ) + deg(c) deg(c). It follows that deg(d) = deg(c) and thus deg(q 1 ) = deg(q 2 ) = 0. Hence q 1, q 2 K. Let K be a field and f, g K[X] not both 0. The greatest common divisor of f and g is a greatest common divisor of f and g which is monic. The division with remainder gives rise to the Euclidean algorithm. The algorithm stops when the remainder becomes 0. Lemma Let K be a field, f, g K[X] with g 0. Let q, r K[X] such that f = qg + r with r = 0 or deg(r) < deg(g). Let A = {h K[X] h is a common divisor of f and g} and B = {h K[X] h is a common divisor of g and r}. Then A = B. Proof. Let q, r K[X] such that f = qg + r with r = 0 or deg(r) < deg(g). Let h A, then h divides f and g, thus also qg. Hence also f qg. It follows that h divides r and g, whence h B. Let h B, then h divides g and r, thus also qg. Hence also qg + r. It follows that h divides f and g, whence h A. Thus A = B. If the Eucledian algorithm starts with two polynomials f and g and ends with r n 1 = r n q n+1 +0, where r n 1, r n K[X] are the two previous remainders and r n 0. The previous lemma shows that the set of common divisors of f and g is equal to the set of common divisors of r n and 0. But the set of common divisors of r n and 0 is just the set of all divisors of r n, and in particular, r n is on of them. Hence all common divisors of f and g divide r n and r n divides f and g. Hence r n is a greatest common divisor of f and g. Thus the Eucledian algorithm proves that a greatest common divisor of f and g always exists if f and g are not both 0. The Euclidean algorithm gives rise to the extended Euclidean algorithm, which allows you to write the geatest common divisor as a sum of multiples of f and g. Consider the ring Z /5 Z [X] and the polynomials X 2 1 and X 3 + X 2 + X + 1. X 3 + X 2 + X + 1 = (X 2 1)(X + 1) + (2X + 2) X 2 1 = (2X + 2)(3X + 2) + 0 So 2X + 2 is a greatest common divisor of X 2 1 and X 3 + X 2 + X + 1, and X + 1 is the greatest common divisor of X 2 1 and X 3 + X 2 + X + 1. The extended Eucledian algorithm gives 2X + 2 = (X 3 + X 2 + X + 1) (X 2 1)(X + 1). After multiplying with 2 1 we obtain X +1 = 3(X 3 +X 2 +X +1) (X 2 1)(3X +3) and so X +1 = 3(X 3 +X 2 +X +1)+(2X +2)(X 2 1). Consider the ring Z /2 Z [X] and the polynomials X 4 + X 2 and X X 4 + X 2 = (X 3 + 1)(X) + (X 2 + X)

5 1. ALGEBRA 23 X = (X 2 + X)(X + 1) + (X + 1) X 2 + X = (X + 1)(X) + 0 So X +1 is the greatest common divisor of X 4 X 2 and X Now we use the extended Euclidean algorithm to write X + 1 as a sum of multiples of X 4 + X 2 and X X + 1 = (X 3 + 1) (X 2 + X)(X +1) = (X 3 +1) ((X 4 +X 2 ) (X 3 +1)(X))(X +1) = (X 3 +1)(1+X 2 +X) (X 4 +X 2 )(X +1). That is X + 1 = (X 3 + 1)(X 2 + X + 1) + (X 4 + X 2 )(X + 1). Let R be a commutative ring with identity element and 0 1. Let f(x) R[X] and r R. If f(x) = a n X n + a n 1 X n a 1 X + a 0, then f(r) = a n r n + a n 1 r n a 1 r + a 0 is called the value of f at r. If f(r) = 0, then r will be called a root of the polynomial. The polynomial X Z /2 Z [X] has 1 as a root since = 0. The polynomial 3X + 2 Z /6 Z [X] has no roots since = 2, = 5, = 2, = 5, = 2 and = 5. Lemma Let K be a field. If f K[X] has degree 1, then f has roots. Proof. If f K[X] has degree 1, then f(x) = ax +b, for some a, b K. Since a 0 it is invertible. Let r = a 1 b, then r is a root. Lemma Let K be a field, f K[X] with f 0. Then k K is a root of f if and only if (X k) is a divisor of f. Proof. We will show that k K is a root of f if and only if f = (X k)q, for some q K[X]. Let k K. Consider the polynomials f and X k of K[X]. There exist q, r K[X], with f = (X k)q + r and r = 0 or deg(r) < 1. The condition on r implies that r K. Now if k is a root of f, then f(k) = 0, so r = 0. If r = 0, then f(k) = (k k)q = 0. Consider the ring Z /5 Z [X] the polynomial 3X 2 X +1. Then 3 and 4 are roots and 3X 2 X +1 = (X 3)(3X + 3) and 3X 2 X + 1 = (X 4)(3X + 1). Indeed 3X 2 X + 1 = (X 4)(X 3)3. Consider the ring Z /2 Z [X] the polynomials X and X They both have only one root, namely 1. X = (X 1)(X 1) and X = (X 1)(X 2 + X + 1). Observe that (X 2 + X + 1) has no roots. Lemma Let K be a field, f K[X], with f 0. The polynomial has at most deg(f) distinct roots. Proof. Let n = deg(f). We will use induction on n. If n = 0, then f K and so f has no roots. Suppose n > 0 and any polynomial of degree m with m < n has at most m distinct roots. If f has no roots then it has at most n roots. If f has a root k, then f = (X k)q for some q K[X] with deg(q) = n 1. Any root l of f different of k has to be a root of q, for 0 = f(l) = (l k)q(l) and K has no zero divisors. Since q has at most n 1 distinct roots, f can have at most n distinct roots. Consider the ring Z /2 Z [X]. X 2 + X has two roots: 0 and 1; X has one root: 1; X 2 + X + 1 has no roots; X has one root: 1.

6 1. ALGEBRA 24 Let f K[X] with deg(f) 1. The polynomial f is said to be reducible if there exist g, h K[X] with f = g h and 0 < deg(g) < deg(f) and 0 < deg(h) < deg(f). Otherwise the polynomial is said to be irreducible. Consider the ring Z /2 Z [X]. The polynomials X 2, X 2 + X and X are reducible for X 2 = XX, X 2 + X = X(X + 1) and X = (X + 1)(X + 1). The polynomial X 2 + X + 1 is irreducible for if it would be reducible there would exist g, h K[X] with X 2 + X + 1 = g h and 0 < deg(g) < 2 and 0 < deg(h) < 2. Hence deg(g) = deg(h) = 1, but then g and h have a root and so f has roots too. A contradiction. We want to establish if the polynomial X 4 +X 2 +1 is reducible or irreducible. First observe that it has no roots. Suppose X 4 +X 2 +1 is reducible. Then there exist g, h K[X] with X 4 +X 2 +1 = g h and 0 < deg(g) < 4 and 0 < deg(h) < 4. Since any root of g or h would be a root of X 4 + X 2 + 1, we deduce that g and h have no roots. In particular, deg(g) 1 deg(h). Since deg(g) + deg(h) = 4, we must have deg(g) = deg(h) = 2. Let ax 2 + bx + c Z /2 Z [X] be a polynomial of degree 2 with no roots. Then a = 1 and, since 0 is not a root, c 0, and since 1 is not a root, we also have 1 + b + c 0. It follows that c = 1 and b = 1. Thus X 2 + X + 1 is the only polynomial of degree 2 without roots. It follows that g = h = X 2 + X + 1. Since (X 2 + X + 1)(X 2 + X + 1) = X 4 + X we see that X 4 + X is reducible. Lemma Let K be a field and f K[X] with deg(f) 1. (i) If deg(f) = 1, then f is irreducible and has one root. (ii) If deg(f) = 2 or deg(f) = 3, then f is irreducible if and only if it has no roots. (iii) If deg(f) 4, then if f has a root, then f is reducible. Proof. Exercise. If deg(f) 4 and has no roots, then nothing can be said in general. It might be irreducible or the product of polynomials of degree at least 2 which have no root. Irreducible polynomials in K[X] behave similarly as prime numbers in Z. Lemma Let K be a field and f, g K[X] with f irreducible and g f. Then g K or g = kf, for some k K. Proof. Suppose g divides f. Then f = qg for some q K[X]. Since f 0 we have g 0 and q 0. Now deg(f) = deg(q) + deg(g). If 0 < deg(g) < deg(f), then 0 < deg(q) < deg(f) too and f is reducible. A contradiction. Hence deg(g) = 0 or deg(g) = deg(f). If deg(g) = 0, then g K. If deg(g) = deg(f), then deg(q) = 0 and so q K. Let k = q 1 K, then g = kf. Lemma Let K be a field and f, g, h K[X] with f irreducible and f gh. Then f g or f h. Proof. Suppose f divides gh but f does not divide g. Then, since f is irreducible, we have gcd(f, g) = 1 and, by the Euclidean algorithm, there exist a, b K[X] with 1 = af + bg. In particular, h = afh + bgh and thus f h.