Chapter 7 Rings ring. ring integral domain, ring The Ring of Integers ring Z., Z,,. Euclid s Algorithm,.,. Theorem (Euclid s Algorithm). n

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2 Chapter 7 Rings ring. ring integral domain, ring The Ring of Integers ring Z., Z,,. Euclid s Algorithm,.,. Theorem (Euclid s Algorithm). n, m Z, h, r Z, 0 r < n, m = h n + r. Proof.,. ring,. W = {m t n t Z}. t, W. r W, r W, h Z r = m h n. 0 r < n. r, r n ( r )., r r = n + r, r 0. m = h n + r = h n + (n + r ) = (h + 1) n + r, r = m (h + 1) n W. 0 r < r, r W.. 119

3 Rings Theorem well-ordering principle,, ring., ring Euclid s Algorithm. integral domain Euclid s Algorithm. integral domain : Euclidean domain. Theorem 7.1.1! Theorem Z ideal principle ideal. Proof. : I Z ideal, I a I = ( a ) = {h a h Z}, I a.,?! Z trivial ideal Z {0}, 1 0, principle ideal. Z nontrivial proper ideal. I Z nontrivial proper ideal, I {0}, b 0, b I. I ideal, b I, I. a I I, I = ( a ). a I, h Z h a I, ( a ) I. I ( a ), I a. m I m a? ( m a.) Theorem 7.1.1, h, r Z, 0 r < a r = m h a. m I h a I, I ideal r = m h a I. a I, r = 0, m = h a ( a ). I ( a )., ring ideal principle ideal. integral domain ideal principle ideal, integral domain principle ideal domain. Z principle ideal domain (Theorem 7.1.2), Z Euclidean domain (Theorem 7.1.1).,,. Definition a, b Z. (1) d Z h Z a = h d, d a divisor, d a. (2) c Z, c a c b, c a, b common divisor. (3) d Z a, b common divisor, d a, b greatest common divisor.

4 7.1. The Ring of Integers 121 greatest common divisor, Theorem greatest common divisor. Proposition a, b Z, d N ( d ) = ( a ) + ( b ) d a, b greatest common divisor Proof. Lemma ( a ) + ( b ) = {r a + s b r, s Z} Z ideal. Theorem d Z ( d ) = ( a ) + ( b ). d, 1 Z unit Lemma ( d ) = ( d ). d N a, b greatest common divisor. d a, b common divisor. a ( a ) ( a ) + ( b ) = ( d ), r Z a = r d. d a., b ( d ) d b. d a, b common divisor. d a, b common divisor? d ( d ) = ( a ) + ( b ), m, n Z d = m a + n b. c a, b common divisor, c a c b, r, s Z a = r c b = s c. d = m (r c) + n (s c) = (m r + n s) c. c d. d a, b common divisor. Proposition greatest common divisor, greatest common divisor. Corollary a, b Z d a, b greatest common divisor, d : (1) m, n Z d = m a + n b. (2) c a c b, c d. :. p 1. p a b p a p b,. ring,. Definition Z p. (1) d p d Z d = ±1 d = ±p, p irreducible element. (2) p a b a, b Z p a p b, p prime element.

5 Rings,. (prime). Proposition Z p irreducible element, p prime element., p prime element, p irreducible element. Proof. p irreducible p prime. p irreducible. p a b : p a p b. p a b r Z a b = r p. p a, p a. p, a greatest common divisor d. d p p irreducible d = 1 d = p. d p, d p, a common divisor p = d a: p a. d = 1, Corollary n, m Z 1 = n p + m a. b p b. b = (n b) p + m (a b) = (n b) p + m (r p) = (n b + m r) p,, p prime element p irreducible. d p, d = ±1 d = ±p. d p r Z p = d r, p d r. p prime, p d p r. p d, d p d = ±p. p r, s Z r = s p. p = d r = d (s p) d s = 1. d, s Z, d s = 1 d = ±1..,. Theorem a N a > 1, p 1,..., p r, p i prime, a = p n 1 1 pnr r, n i N, i {1,..., r}. a a = q m 1 1 qs ms, q i prime, r = s p i = q i, n i = m i, i {1,..., r}. Proof.,. : 1 ( ) prime. a prime, a = p 1 ( r = 1, n 1 = 1),. a prime? Proposition a irreducible, a 1, b 1 N a 1 1, b 1 1 a = a 1 b 1. a 1, b 1 prime. prime, prime.. a prime.,,. a = 2 2 prime,

6 7.2. Ring of Polynomials over a Field a 1. a prime,, a prime, Proposition a = a 1 b 1 a 1, b 1 N 1 < a 1 < a 1 < b 1 < a. a 1 b 1 prime, a prime., a = p n 1 1 pn r r = q m 1 1 q m s s, p 1,..., p r prime, q 1,..., q s prime. p 1 prime, p 1 a = q m 1 1 q m s s j {1,..., s} p 1 q j. p 1 q 1. q 1 prime, Proposition q 1 irreducible., q 1 divisor ±1 ±q 1. p 1 q 1 p 1 = q 1. a = p n p n r r = q m q m s s. p 1 a/p 1 < a, r = s p 1 = q 1,..., p r = q r n 1 = m 1, n 2 = m 2,..., n r = m r,. integral domain Z irreducible element, integral domain unique factorization domain Ring of Polynomials over a Field,. field. polynomial ring.. F field. F f(x) = a 0 + a 1 x + + a n 1 x n 1 + a n x n, a i F i = 0,..., n F [x]. F [x] : f(x) = a 0 + a 1 x + a n x n g(x) = b 0 + b 1 x + + b m x m F [x], f(x) + g(x) = c 0 + c 1 x + + c r x r, i {1,..., r}, c i = a i + b i r = max{m, n}. f(x) g(x) = d 0 + d 1 x + + d m+n x m+n, i {1,..., m + n}, d i = a 0 b i + a 1 b i a i 1 b 1 + a i b 0., j > n a j = 0 k > m b k = 0. : ;. F [x] commutative ring with 1,. F [x] ring identity 0 0

7 Rings, 0 ( 0 F 0). identity 1 1, 1 ( 1 F 1) 0. F [x] the ring of polynomials in x over F. ring, zero divisor unit? polynomial ring, degree. Definition f(x) = a 0 + a 1 x + + a n x n F [x] a n 0, f(x) degree n deg(f(x)) = n. 0, 0 0, 0 degree 0. 0 degree ( deg(0) = ). degree. Lemma f(x) g(x) F [x] 0, deg(f(x) g(x)) = deg(f(x)) + deg(g(x)). Proof. deg(f(x)) = n deg(g(x)) = m f(x) = a 0 + a 1 x + + a n x n g(x) = b 0 +b 1 x+ +b m x m, a n 0 b m 0. f(x) g(x) = c k x k, c k = i+j=k a i b j. k > n + m c k = 0. i n j m, i + j n + m. k > n + m i + j = k, i > n j > m. a i = 0 b j = 0. k > n + m c k = 0. k = n + m, i + j = k i = n j = m a i 0 b j 0. c n+m = a n b m. F field, F zero divisor, a n 0 b m 0 c n+m 0. deg(f(x) g(x)) = n + m. Lemma 7.2.2, F [x] zero divisor unit. Proposition F field. (1) F [x] zero divisor, F [x] integral domain. (2) F [x] unit 0. Proof. (1) f(x), g(x) F [x] 0. deg(f(x)) = n deg(g(x)) = m, Lemma deg(f(x) g(x)) = n + m., f(x) g(x) x n+m 0. f(x) g(x) 0. F [x] zero divisor. (2) f(x) F [x] unit, g(x) F [x] f(x) g(x) = 1. 1 degree 0, Lemma deg(f(x) g(x)) = deg(f(x)) + deg(g(x)) = 0. deg(f(x)) 0 deg(g(x)) 0, deg(f(x)) = 0 f(x). 0 unit, f(x) 0., f(x) = c 0, c F c 0. F field,

8 7.2. Ring of Polynomials over a Field 125 F c inverse c 1. g(x) = c 1 F [x], f(x) g(x) = 1. f(x) = c unit. polynomial ring. Theorem (Euclid s Algorithm). F field, polynomials f(x), g(x) F [x], g(x) 0, h(x), r(x) F [x] f(x) = h(x) g(x)+r(x), r(x) = 0 deg(r(x)) < deg(g(x)). Proof., r(x) 0, 0 degree, deg(r(x)) < deg(g(x)), r(x) = 0. Theorem W = {f(x) l(x) g(x) l(x) F [x]}. 0 W, h(x) F [x] f(x) h(x) g(x) = 0, r(x) = 0. 0 W, r(x) W W degree polynomial. deg(r(x)) = m deg(g(x)) = n, m < n. m n, r(x) x m a, g(x) x n b. b F b 0, s(x) = r(x) ((a b 1 )x m n ) g(x). r(x) ((a b 1 )x m n ) g(x) x m a, deg(s(x)) < m = deg(r(x)). r(x) W l(x) F [x] r(x) = f(x) l(x) g(x). s(x) = f(x) l(x) g(x) ((a b 1 )x m n ) g(x) = f(x) (l(x)+(a b 1 )x m n ) g(x) W. s(x) W r(x) degree polynomial, r(x) W degree. m < n h(x) F [x] r(x) = f(x) h(x) g(x) deg(r(x)) < deg(g(x).. Remark , Theorem F field ( g(x) b inverse b 1 ). Theorem ring polynomials. Z[x]. f(x) = x 2, g(x) = 2x h(x) f(x) h(x) g(x) = 0 deg(f(x) h(x) g(x)) < deg(g(x)). (Theorem 7.1.1) Z ideal principle ideal (Theorem 7.1.2). (Theorem 7.2.4),. Theorem F field, F [x] ideal principle ideal. Proof. F [x] ideal, I. I g(x) ( g(x) ) = I. g(x) I degree polynomial, ( g(x) ) = I. g(x) I ( g(x) ) I., I ( g(x) ) f(x) I h(x) F [x] f(x) = h(x) g(x). Theorem h(x), r(x) F [x] f(x) = h(x) g(x) + r(x) r(x) = 0 deg(r(x)) < deg(g(x)). g(x), f(x) I, r(x) = f(x) h(x) g(x) I.

9 Rings r(x) 0, r(x) I g(x) degree polynomial, g(x). r(x) = 0, f(x) = h(x) g(x) ( g(x) ). F [x].,. Definition f(x), g(x) F [x]. (1) d(x) F [x] h(x) F [x] f(x) = h(x) d(x), d(x) f(x) divisor, d(x) f(x). (2) l(x) F [x], l(x) f(x) l(x) g(x), l(x) f(x), g(x) common divisor. (3) d(x) F [x] f(x), g(x) common divisor degree polynomial, d(x) f(x), g(x) greatest common divisor. greatest common divisor. greatest common divisor common divisor degree 1 polynomial, greatest common divisor. greatest common divisor, Theorem greatest common divisor. Proposition f(x), g(x) F [x], d(x) F [x] ( d(x) ) = ( f(x) ) + ( g(x) ) d(x) f(x), g(x) greatest common divisor Proof. Theorem d(x) F [x] ( d(x) ) = ( f(x) ) + ( g(x) ). d(x) F [x] f(x), g(x) greatest common divisor. d(x) f(x), g(x) common divisor. f(x) ( f(x) ) ( f(x) ) + ( g(x) ) = ( d(x) ), h(x) F [x] f(x) = h(x) d(x). d(x) f(x)., g(x) ( d(x) ) d(x) g(x). d(x) f(x), g(x) common divisor. d(x) f(x), g(x) common divisor degree? d(x) ( d(x) ) = ( f(x) ) + ( g(x) ), m(x), n(x) F [x] d(x) = m(x) f(x) + n(x) g(x). l(x) f(x), g(x) common divisor, l(x) f(x) l(x) g(x), r(x), s(x) F [x] f(x) = r(x) l(x) g(x) = s(x) l(x). d(x) = m(x) (r(x) l(x)) + n(x) (s(x) l(x)) = (m(x) r(x) + n(x) s(x)) l(x). l(x) d(x). d(x) f(x), g(x) common divisor degree. Proposition greatest common divisor, greatest common divisor.

10 7.2. Ring of Polynomials over a Field 127 Corollary f(x), g(x) F [x] d(x) f(x), g(x) greatest common divisor, d(x) : (1) m(x), n(x) F [x] d(x) = m(x) f(x) + n(x) g(x). (2) l(x) f(x) l(x) g(x), l(x) d(x). ring, unit. Z 1 1. F [x] units 0 (Proposition 7.2.3), divisor. (irreducible element). Definition F [x] p(x). (1) d(x) p(x) d(x) F [x], d(x) = c d(x) = c p(x), 0 c F, p(x) irreducible element. (2) p(x) f(x) g(x) f(x), g(x) F [x] p(x) f(x) p(x) g(x), p(x) prime element. irreducible element degree polynomial. irreducible prime, F [x] polynomial. Proposition F [x] p(x) irreducible element, p(x) prime element., p(x) prime element, p(x) irreducible element. Proof. p(x) irreducible p(x) prime. p(x) irreducible. p(x) f(x) g(x) : p(x) f(x) p(x) g(x). p(x) f(x) g(x) r(x) F [x] f(x) g(x) = r(x) p(x). p(x) f(x), p(x) f(x). p(x), f(x) greatest common divisor d(x). d(x) p(x) p(x) irreducible d(x) = c d(x) = c p(x), 0 c F. d(x) c p(x), d(x) p(x), f(x) common divisor p(x) = c 1 d(x) f(x) ( c F [x] unit). p(x) f(x). d(x) = c, Corollary n(x), m(x) F [x] c = n(x) p(x) + m(x) f(x). c 1 g(x) g(x) = c 1( n(x) g(x) ) p(x) + c 1( m(x) (f(x) g(x)) ) = c 1( n(x) g(x) + m(x) r(x) ) p(x), p(x) g(x)., p(x) prime element p(x) irreducible. d(x) p(x), d(x) = c d(x) = c p(x). d(x) p(x)

11 Rings r(x) F [x] p(x) = r(x) d(x), p(x) r(x) d(x). p(x) prime, p(x) d(x) p(x) r(x). p(x) d(x), s(x) F [x] d(x) = s(x) p(x). p(x) = r(x) d(x) d(x) = (s(x) r(x)) d(x). d(x) (s(x) r(x) 1) = 0, F [x] zero divisor (Proposition 7.2.3) d(x) 0, s(x) r(x) = 1, s(x) unit. s(x) c, d(x) = s(x) p(x) = c p(x). p(x) r(x), s(x) F [x] r(x) = s(x) p(x). p(x) = d(x) r(x) = d(x) (s(x) p(x)) d(x) s(x) = 1. d(x) F [x] unit, d(x) = c., Z F [x]. F [x] Z. unit, Z. F [x] d(x) f(x) divisor, h(x) F [x] f(x) = d(x) h(x), F [x] 0 c F [x] unit, c 1 h(x) F [x]. f(x) = (c d(x)) (c 1 h(x)) c d(x) f(x) divisor. 0 c F, d(x) c d(x) f(x) divisor. c d(x) f(x) divisor. c c d(x) 1,. Definition f(x) F [x] f(x) 1 f(x) monic polynomial. Lemma monic polynomial. Lemma p(x), q(x) F [x] monic irreducible element p(x) q(x), p(x) = q(x). Proof. q(x) irreducible, q(x) divisor c c q(x). p(x) ( irreducible) p(x) q(x) c F p(x) = c q(x). p(x), q(x) monic polynomial, 1. c = 1, p(x) = q(x). F [x]. Theorem f(x) F [x] deg(f(x)) 1, c F p 1 (x),..., p r (x), p i (x) monic irreducible elements, f(x) = c p 1 (x) n1 p r (x) n r, n i N, i {1,..., r}. f(x) f(x) = d q 1 (x) m1 q s (x) ms, d F q i (x) monic irreducible elements, c = d, r = s p i (x) = q i (x), n i = m i, i {1,..., r}.

12 7.2. Ring of Polynomials over a Field 129 Proof. Theorem Theorem 7.1.8,, degree F [x], degree induction. ( f(x) ): deg(f(x)) = 1 f(x) = ax + b, 0 a F, f(x) a (x + b a 1 ). x + b a 1 degree 1 polynomial, x + b a 1 monic irreducible element.. degree 1 n 1 polynomials. deg(f(x)) = n. f(x) irreducible a, a 1 f(x) monic irreducible element, f(x) = a (a 1 f(x)),. f(x) irreducible, f(x) = g(x) h(x) g(x), h(x) F [x] 1 deg(g(x)) < n 1 deg(h(x)) < n. g(x) = c 1 p 1 (x) n1 p u (x) n u h(x) = c 2 p 1 (x) m1 p v (x) m v, p i (x), p j (x) monic irreducible elements, monic irreducible elements, f(x). : deg(f(x)) = 1, f(x) = ax + b,. degree 1 n 1 polynomials, deg(f(x)) = n. f(x) = c p 1 (x) n1 p r (x) n r = d q 1 (x) m1 q s (x) m s, c, d F, p i (x), q j (x), p i (x), q j (x) monic irreducible element., p i (x), q j (x) monic, c d f(x). polynomial, c = d. p 1 (x) irreducible Proposition prime, p 1 (x) f(x) = cq 1 (x) m1 q s (x) ms j {1,..., s} p 1 (x) q j (x). p 1 (x) q 1 (x), p 1 (x) q 1 (x) monic irreducible element Lemma p 1 (x) = q 1 (x). f(x) f(x) = c p 1 (x) n1 p 2 (x) n2 p r (x) n r = c p 1 (x) m1 q 2 (x) m2 q s (x) m s. c p 1 (x), c p 1 (x) (p 1 (x) n 1 1 p 2 (x) n2 p r (x) n r p 1 (x) m 1 1 q 2 (x) m2 q s (x) m ) s = 0. c p 1 (x) 0 F [x] integral domain, p 1 (x) n 1 1 p 2 (x) n2 p r (x) n r p 1 (x) m 1 1 q 2 (x) m2 q s (x) m s = 0. g(x) = p 1 (x) n 1 1 p 2 (x) n2 p r (x) nr. deg(g(x)) = deg(f(x)) deg(p 1 (x)) < deg(f(x)) = n

13 Rings g(x) = p 1 (x) n 1 1 p 2 (x) n2 p r (x) n r = p 1 (x) m 1 1 q 2 (x) m2 q s (x) m s g(x), r = s p 1 (x) = q 1 (x),..., p r (x) = q r (x) n 1 = m 1, n 2 = m 2,..., n r = m r, Polynomials over the Integers polynomials, polynomials. polynomials. polynomial ring, polynomial ring. Q[x] polynomials Z[x] polynomials. Q[x] ring, polynomial ring over Q. Z[x] ring, polynomial ring over Z. Z[x] 0 1 Q[x] 0 1. Z[x] degree ( Z[x] Q[x] ). Lemma 7.2.3, Z[x] integral domain. Z[x] Q[x] Q[x] 0 unit, Z[x] ±1 unit. Lemma Z[x] unit degree 0, Z[x] unit,, Z unit Z[x] unit, ±1., Z[x]. Remark Z[x], Q[x] ideal principle ideal (Theorem 7.2.6) Z[x]. Z[x] ( ) ideal principle ideal. Example Z[x] I = (2) + (x) principle ideal. I principle ideal, f(x) Z[x] I = ( f(x) ). 2 I, 2 ( f(x) ), h(x) Z[x] 2 = h(x) f(x). degree deg(f(x)) = 0, f(x) c Z. x I = ( c ) g(x) Z[x] x = c g(x). c g(x) c ( g(x) Z[x], g(x) ). x = c g(x) x c. x x 1, c 1, c = ±1. c unit, Lemma I = ( c ) = Z[x], 1 I = ( 2 ) + ( x ). ( 2 ) + ( x ) n(x), m(x) Z[x] 1 = 2 n(x) + x m(x). x m(x), 2 n(x) 2, 2 n(x) + x m(x) 1. n(x), m(x) Z[x] 1 = 2 n(x) + x m(x). I principle ideal, I = ( 2 ) + ( x ) Z[x] principle ideal.

14 7.3. Polynomials over the Integers 131 Z[x] ideal principle ideal Proposition Z[x] irreducible element prime element.,! Z[x] irreducible element prime element,. Z[x] Q[x] Z[x]. f(x) = a 0 + a 1 x + + a n x n Z[x] f(x) degree polynomials, ( ±1 Z[x] ).? a 0, a 1,..., a n!. Lemma f(x) Z[x] 0 polynomial, f(x) f(x) = c f (x), c N, f (x) Z[x] f (x) 1. Proof. : f(x) = a 0 +a 1 x+ +a n x n, d = gcd(a 0, a 1,..., a n ). a 0 = d b 0, a 1 = d b 1,..., a n = d b n gcd(b 0, b 1,..., b n ) = 1. f(x) d (b 0 + b 1 x + + b n x n ). : f(x) = c f (x), c N f (x) Z[x]. c f (x), f(x) a 0, a 1,..., a n c. c a 0, a 1,..., a n. c d = gcd(a 0, a 1,..., a n ), f (x) d/c 1, f (x) 1. d = c, d f (x) = d (b 0 + b 1 x + + b n x n ). Z[x] integral domain, f (x) = b 0 + b 1 x + + b n x n. Lemma 7.3.2,. Definition f(x) Z[x] f(x) = c f (x), c N, f (x) Z[x] f (x) 1. c f(x) content, c(f). f(x) Z[x] c(f) = 1, f(x) primitive polynomial. c(f) f(x). Lemma f(x) Z[x] content primitive polynomial. Lemma Q[x]. Proposition f(x) Q[x] 0 polynomial, f(x) f(x) = c f (x), c Q, c > 0 f (x) Z[x] primitive polynomial. Proof. : f(x) = a 0 + a 1 x + + a n x n, a i Q. m m f(x) Z[x] ( m a i ). m f(x) Z[x] Lemma a f (x) Z[x]

15 Rings f (x) primitive polynomial, m f(x) = a f (x). f(x) = a m f (x). f(x) = d f (x) = d g(x) d, d f (x), g(x) Z[x] primitive polynomials. d d a/b a /b, a, a, b, b N. (a b ) f (x) = (a b) g(x). (a b ) f (x), (a b) g(x) Z[x] a b, a b N f (x), g(x) primitive polynomial, Lemma : a b = b a ( d = d ) f (x) = g(x).. Proposition 7.3.4, content Q[x], f(x) Q[x] f(x) = c(f) f (x), 0 < c(f) Q f(x) content, f (x) Z[x] primitive polynomial. f(x), g(x) Q[x], f(x) g(x) content,. polynomial,,. f(x) g(x) content f(x) g(x) contents. f(x) g(x) contents 1. Lemma (Gauss Lemma). f(x), g(x) Z[x] primitive polynomials, f(x) g(x) primitive polynomial. Proof. f(x) = a n x n + + a 1 x + a 0, g(x) = b m x m + + b 1 x + b 0, c(f) = c(g) = 1, c(f g) = 1. c(f g) = d 1, p p d, p f(x) g(x). c(f) = c(g) = 1, a i, b j p a i p b j. r p a r ( p a r, i < r, p a i ), s p b s. f(x) g(x) x r+s : a i b j. i+j=r+s a r b s, a i b j i < r j < s. i > r j > s i + j > r + s i + j = r + s. i < r r p a i, p a i b j., j < s p a i b j., f(x) g(x) x r+s a r b s a i b j p. p a r p b s, p a r b s. f(x) g(x) x r+s p. p f(x) g(x). c(f g) 1, f(x) g(x) primitive polynomial. Gauss Lemma f(x), g(x) Q[x], c(f g).

16 7.3. Polynomials over the Integers 133 Proposition f(x), g(x) Q[x] 0 polynomial, c(f g) = c(f) c(g). Proof. Lemma f(x) g(x) f(x) = c(f) f (x) g(x) = c(g) g (x), f (x) g (x) primitive polynomials. f(x) g(x) = ( c(f) c(g) ) (f (x) g (x) ). Lemma f(x) g(x) c(f g) h(x) h(x) primitive polynomial. Lemma f (x) g (x) primitive polynomial, f (x) g (x) = h(x) c(f) c(g) = c(f g). Z[x], Z[x] Q[x]. f(x), g(x) Z[x], f(x) g(x) in Z[x] h(x) Z[x] g(x) = h(x) f(x). f(x) g(x) in Q[x] l(x) Q[x] g(x) = l(x) f(x). h(x) Z[x], l(x) Q[x]. f(x) g(x) in Q[x] f(x) g(x) in Z[x]. Lemma f(x), g(x) Z[x], f(x) primitive polynomial, f(x) g(x) in Z[x] f(x) g(x) in Q[x]. Proof. f(x) g(x) in Z[x] h(x) Z[x] g(x) = h(x) f(x). h(x) Z[x] h(x) Q[x], f(x) g(x) in Q[x]. ( f(x) primitive.), f(x) g(x) in Q[x], l(x) Q[x] g(x) = l(x) f(x). l(x) Z[x]. Lemma l(x) l(x) = c(l) l (x), l (x) primitive polynomials. g(x) = c(l) (l (x) f(x)). f(x) l (x) primitive polynomials, Lemma l (x) f(x) primitive polynomial. Lemma c(g) = c(l). c(g) N, c(l) N, l (x) Z[x], l(x) = c(l) l (x) l(x) Z[x]., Q[x] Z[x]. f(x) Z[x] f(x) Q[x] f(x) f(x) = g(x) h(x), g(x), h(x) Q[x] deg(g(x)) deg(h(x)) deg(f(x)). f(x) Z[x] f(x) = m(x) n(x), m(x), n(x) Z[x]. Lemma. Lemma f(x) Z[x] f(x) = g(x) h(x) g(x), h(x) Q[x], m(x), n(x) Z[x] f(x) = m(x) n(x) deg(m(x)) = deg(g(x)) deg(n(x)) = deg(h(x)).

17 Rings Proof. Lemma g(x) = c(g) g (x) h(x) = c(h) h (x) g (x), h (x) Z[x] primitive polynomial. Proposition c(g) c(h) = c(g h) = c(f), f(x) Z[x], c(g) c(h) = c(f) N. m(x) = ( c(g) c(h) ) g (x) Z[x] n(x) = h (x) Z[x], f(x) = g(x) h(x) = ( c(g) g (x) ) (c(h) h (x) ) = ( c(g) c(h) ) g (x) h (x) = m(x) n(x). deg(m(x)) = deg(g (x)) = deg(g(x)) deg(n(x)) = deg(h (x)) = deg(h(x)). f(x) Z[x] f(x) = m(x) n(x), m(x), n(x) Z[x], m(x), n(x) Z[x] unit. m(x), n(x) Q[x] f(x) Q[x]., m(x), n(x) Z[x] unit, Q[x] unit. 2x + 2 Q[x] irreducible Z[x] 2x + 2 = 2 (x + 1), 2 x + 1 Z[x] unit ( 2 Q[x] unit), 2x + 2 Z[x] irreducible. Z[x] irreducible element Q[x] irreducible element. irreducible element divisor unit unit. Z[x] unit 1 1. Definition p(x) Z[x] (1) p(x) Z[x] divisor ±1 ±p(x), p(x) Z[x] irreducible element. (2) p(x) f(x) g(x) f(x), g(x) Z[x] p(x) f(x) p(x) g(x) p(x) Z[x] prime element. Lemma. Lemma p(x) Z[x] deg(p(x)) > 0. (1) p(x) irreducible element, p(x) primitive polynomial. (2) p(x) prime element, p(x) primitive polynomial. Proof. (1) p(x) irreducible. p(x) = c(p) p (x), c(p) N Z[x] p (x) Z[x], c(p) p(x) divisor. p(x) irreducible deg(p (x)) = deg(p(x)) > 0 c(p) = 1, p(x) primitive.

18 7.3. Polynomials over the Integers 135 (2) p(x) prime. p(x) = c(p) p (x), p(x) c(p) p (x). p(x) prime, p(x) c(p) p(x) p (x). deg(p(x)) > 0 p(x) c(p). p(x) p (x). λ(x) Z[x] p (x) = λ(x) p(x). p (x) = ( λ(x) c(p) ) p (x). Z[x] integral domain p (x) 0 λ(x) c(p) = 1. λ(x) c(p) Z[x] unit. c(p), λ(x) = c(p) = 1. p(x) primitive., Z[x] irreducible element prime element. Z[x] ideal principle ideal,. Q[x] irreducible element, Z[x] irreducible element Q[x] irreducible element. Lemma p(x) Z[x], deg(p(x)) > 0 p(x) primitive polynomial, p(x) Q[x] irreducible element p(x) Z[x] irreducible element. Proof. p(x) Z[x] irreducible element. p(x) Q[x] irreducible element, g(x), h(x) Q[x] 0 < deg(g(x)) < deg(p(x)), 0 < deg(h(x)) < deg(p(x)) p(x) = g(x) h(x). Lemma m(x), n(x) Z[x] deg(m(x)) = deg(g(x)), deg(n(x)) = deg(h(x)) p(x) = m(x) n(x). m(x) p(x) divisor. 0 < deg(m(x)) < deg(p(x)), m(x) ±1 m(x) ±p(x). p(x) Z[x] irreducible element. p(x) Q[x] irreducible element., p(x) Q[x] irreducible element. p(x) = m(x) n(x), m(x), n(x) Z[x]. p(x) Q[x] irreducible m(x) n(x) Q[x] unit, : m(x) = d! m(x) Z[x] d Z. p(x) = d n(x) d p(x). p(x) primitive, d = ±1. p(x) divisor ±1 ±p(x), p(x) Z[x] irreducible. Q field, F [x] Q[x]. Q[x] irreducible prime, Z[x] irreducible prime. Proposition p(x) Z[x]. p(x) Z[x] irreducible element, p(x) Z[x] prime element., p(x) Z[x] prime element, p(x) Z[x] irreducible element.

19 Rings Proof., deg(p(x)) = 0 p(x) Z. Z irreducible prime (Proposition 7.1.7), deg(p(x)) > 0. p(x) Z[x] irreducible element. Lemma primitive, Lemma p(x) Q[x] irreducible element. Proposition p(x) Q[x] prime element. f(x), g(x) Z[x] p(x) f(x) g(x) in Z[x], Lemma p(x) f(x) g(x) in Q[x]. p(x) Q[x] prime p(x) f(x) p(x) g(x) in Q[x]. Lemma p(x) f(x) p(x) g(x) in Z[x]. p(x) Z[x] prime element., p(x) Z[x] prime element. p(x) = m(x) n(x) m(x), n(x) Z[x]. p(x) m(x) n(x), p(x) n(x) p(x) m(x). p(x) n(x), λ(x) Z[x] n(x) = λ(x) p(x). n(x) = λ(x) (n(x) m(x) ) = ( λ(x) m(x) ) n(x). n(x) 0 Z[x] integral domain, λ(x) m(x) = 1. m(x) Z[x] unit, m(x) = ±1., p(x) m(x) n(x) = ±1. p(x) divisor ±1 ±p(x), p(x) irreducible element. Z[x], ideal principle ideal, Proposition irreducible element prime., f(x) f(x), polynomial. Theorem f(x) Z[x] 0, 1, 1 polynomial, p 1 (x),..., p r (x) Z[x], p i (x) Z[x] irreducible elements, f(x) = p 1 (x) n1 p r (x) n r, n i N, i {1,..., r}. f(x) f(x) = q 1 (x) m1 q s (x) ms, q i (x) Z[x] irreducible elements, r = s p i (x) = q i (x), n i = m i, i {1,..., r}. Proof., f(x) Z[x] irreducible elements. ( degree). deg(f(x)) = 0, f(x) N unit, Z (Theorem 7.1.8) f(x) irreducible elements. degree n polynomial. deg(f(x)) = n, f(x) irreducible,. f(x) irreducible., Z[x] polynomial irreducible degree polynomials ( 2x + 2). f(x) f(x) = c(f) f (x), f (x) Z[x]

20 7.3. Polynomials over the Integers 137 primitive polynomial. c(f) N, Theorem c(f) = 1 irreducible polynomials. f (x) irreducible elements. f (x) irreducible,. f (x) irreducible, Lemma f (x) Q[x] irreducible, f (x) = g(x) h(x) g(x), h(x) Q[x] 0 < deg(g(x)) < deg(f(x)) 0 < deg(h(x)) < deg(f(x)). Lemma m(x), n(x) Z[x] deg(m(x)) = deg(g(x)) deg(n(x)) = deg(h(x)) f (x) = m(x) n(x). deg(m(x)) < deg(f(x)) = n deg(n(x)) < n, m(x) n(x) irreducible elements. f (x) irreducible elements, f(x) = c(f)f (x) irreducible elements.. deg(f(x)) = 0, f(x) N, Theorem degree n polynomial. deg(f(x)) = n, f(x) = p 1 (x) n1 p r (x) nr = q 1 (x) m1 q s (x) ms, p i (x), q j (x), p i (x), q j (x) Z[x] irreducible elements. deg(f(x)) > 0, p i (x) polynomial degree 0, p 1 (x). Proposition p 1 (x) Z[x] prime element, p 1 (x) f(x), q j (x) polynomial p 1 (x), q 1 (x). p 1 (x) q 1 (x). q 1 (x) irreducible, divisor ±1 ±q 1 (x). deg(p 1 (x)) > 0 p 1 (x) q 1 (x), p 1 (x) = q 1 (x). f(x) f(x) = p 1 (x) n1 p 2 (x) n2 p r (x) n r = p 1 (x) m1 q 2 (x) m2 q s (x) m s. p 1 (x), p 1 (x) (p 1 (x) n 1 1 p 2 (x) n2 p r (x) n r p 1 (x) m 1 1 q 2 (x) m2 q s (x) m ) s = 0. p 1 (x) 0 Z[x] integral domain, p 1 (x) n 1 1 p 2 (x) n2 p r (x) nr p 1 (x) m 1 1 q 2 (x) m2 q s (x) ms = 0. g(x) = p 1 (x) n 1 1 p 2 (x) n2 p r (x) n r. p 1 (x) deg(p 1 (x)) > 0, deg(g(x)) = deg(f(x)) deg(p 1 (x)) < deg(f(x)) = n g(x) = p 1 (x) n 1 1 p 2 (x) n2 p r (x) nr = p 1 (x) m 1 1 q 2 (x) m2 q s (x) ms g(x), r = s p 1 (x) = q 1 (x),..., p r (x) = q r (x) n 1 = m 1, n 2 = m 2,..., n r = m r,.

21 Rings Theorem Z[x] irreducible elements Z. Lemma Z[x] irreducible element Q[x] irreducible. Z[x] irreducible elements. f(x) Z[x] irreducible. polynomial irreducible. Proposition (Eisenstein Criterion). f(x) = x n + a n 1 x n a 1 x + a 0 Z[x], n > 0. p N p a 0, p a 1,..., p a n 1 p 2 a 0, f(x) Z[x] irreducible element. Proof. c(f) = 1 f(x) primitive polynomial. f(x) irreducible in Z[x] f(x) degree n polynomials.. f(x) = g(x) h(x) g(x) = c r x r + + c 1 x + c 0 Z[x], 0 < r < n h(x) = d s x s + + d 1 x + d 0 Z[x], 0 < s < n. g(x) h(x) c 0 d 0 = a 0. p a 0 = c 0 d 0, p c 0 p d 0. p 2 c 0 d 0, c 0 d 0 p. c 0! p c 0 p d 0. g(x) h(x) c 0 d 1 + c 1 d 0 = a 1. p a 1 p c 0 p c 1 d 0. p d 0 p c 1. p c r. p c 0, p c 1,..., p c r 1, p c r. g(x) h(x) x r c 0 d r + c 1 d r c r 1 d 1 + c r d 0 = a r. ( s < r, d s+1 = = d r = 0) 0 < r < n p a r, p c 0,..., p c r 1, p c r d 0. p d 0, p c r. g(x) h(x) ( f(x) x n ) c r d s = 1. p c r c r d s = 1., f(x) Z[x] irreducible element., Lemma ( Lemma ) Proposition polynomials Q[x] irreducible.

22 7.4. Quotient Field of an Integral Domain Quotient Field of an Integral Domain Z integral domain Q field. Q Z field. Z Q integral domain D. integral domain D, S = {(a, b) a, b D, b 0}. S equivalence relation. S (a, b), (c, d) S, (a, b) (c, d) a d = c b. relation, Q a/b c/d, a, b, c, d Z b 0, d 0, a d = c b. relation equivalence relation: (equiv1): (a, b) S, D integral domain commutative, a b = b a. (a, b) (a, b). (equiv2): (a, b) (c, d), (c, d) (a, b). (a, b) (c, d) a d = c b. (c, d) (a, b) c b = a d,, (c, d) (a, b). (equiv3): (a, b) (c, d) (c, d) (e, f), (a, b) (e, f). a d = c b (7.1) c f = e d (7.2) (7.1) (7.2) a f = e b? (7.1) f, (a d) f = (c b) f = (c f) b. (7.2) (a d) f = (e d) b, d (a f e b) = 0. d 0, D zero divisor ( D integral domain), a f = e b., S equivalence relation, S. (a, b) S, [a, b] S (a, b). S S. S [a, b], a, b D b 0, (a, b) (c, d), S [a, b] = [c, d]. S. [a, b] S [c, d] S, : [a, b] + [c, d] = [a d + c b, b d] [a, b] [c, d] = [a c, b d].. well-defined. [a, b] + [c, d] [a, b] [c, d] S, b d 0. b 0 d 0 D integral domain, b d 0. [a, b] = [a, b ] [c, d] = [c, d ], [a, b] + [c, d] = [a, b ] + [c, d ] [a, b] [c, d] = [a, b ] [c, d ]. [a, b] + [c, d] = [a, b ] + [c, d ]

23 Rings (a d + c b) (b d ) = (a d + c b ) (b d). a b = a b c d = c d (a d + c b) (b d ) = (a b ) (d d ) + (c d ) (b b) = (a b) (d d ) + (c d) (b b) = (a d + c b ) (b d)., [a, b] [c, d] = [a, b ] [c, d ] (a c) (b d ) = (a c ) (b d). a b = a b c d = c d (a c) (b d ) = (a b ) (c d ) = (a b) (c d) = (a c ) (b d). S, S ring, (R1) (R8).,. S commutative ring with 1. S 0 [0, 1] 1 [1, 1]. [a, b] S [a, b] + [0, 1] = [a, b] [a, b] [1, 1] = [a, b]. S commutative D integral domain D commutative [a, b] [c, d] = [a c, b d] = [c, d] [a, b]. S field, [a, b] S [a, b] [0, 1] [c, d] S [a, b] [c, d] = [1, 1]. [a, b] [0, 1] a 0, [b, a] S. [a, b] [b, a] = [a b, a b] = [1, 1]., S 0 unit, S field, D quotient field fraction field. D quotient field S, D field.. isomorphic. S D, S subring D isomorphic. S D field F field subring D isomorphic, F subring S isomorphic. D quotient field. Proposition D integral domain, S D quotient field, D S injective ( ) ring homomorphism. Proof. φ : D S a D, φ(a) = [a, 1]. a, b D φ(a + b) = [a + b, 1] = [a, 1] + [b, 1] = φ(a) + φ(b) φ(a b) = [a b, 1] = [a, 1] [b, 1] = φ(a) φ(b). φ D S ring homomorphism. φ, ker(φ) = {0}. φ(0) = [0, 1] 0 ker(φ). a ker(φ),

24 7.4. Quotient Field of an Integral Domain 141 φ(a) = [a, 1] = [0, 1]., [a, 1] = [0, 1] a 1 = 0 1, a = 0. ker(φ) = {0}. Theorem D/ ker(φ) im(φ) Proposition ker(φ) = {0} D im(φ). im(φ) S subring (Lemma 6.3.3), D D quotient field S subring isomorphic. D quotient field field. Proposition D integral domain, S D quotient field. F field subring D isomorphic, F subring S isomorphic. Proof. ring homomorphism φ : D F. φ ring homomorphism ψ : S F. [a, b] S, ψ([a, b]) = φ(a) φ(b) 1. ψ well-defined. ψ([a, b]) F. [a, b] S, b 0, φ φ(b) F 0. F field φ(b) 1 F. ψ([a, b]) = φ(a) φ(b) 1 F. [a, b] = [c, d] ψ([a, b]) = ψ([c, d]). ( :,,.) a d = c b, φ(a) φ(b) 1 = φ(c) φ(d) 1. φ ring homomorphism φ(a d) = φ(a) φ(d) φ(c b) = φ(c) φ(b). a d = c b φ(a d) = φ(c b) φ(a) φ(d) = φ(c) φ(b). φ(d) 1 φ(b) 1 ( φ(b) φ(d) 0) φ(a) φ(b) 1 = φ(c) φ(d) 1. ψ well-defined. ψ ring homomorphism. [a, b], [c, d] S, ψ ψ([a, b] + [c, d]) = ψ([a d + c b, b d]) = φ(a d + c b) φ(b d) 1 ψ([a, b]) + ψ([c, d]) = φ(a) φ(b) 1 + φ(c) φ(d) 1. φ ring homomorphism, φ(b d) = φ(b) φ(d) φ(a d + c b) φ(b d) 1 = φ(a) φ(b) 1 + φ(c) φ(d) 1. ψ([a, b] + [c, d]) = ψ([a, b]) + ψ([c, d]).

25 Rings ψ([a, b] [c, d]) = ψ([a c, b d]) = φ(a c) φ(b d) 1 = ψ([a, b]) ψ([c, d]), ψ ring homomorphism. ψ, ker(ψ) = {[0, 1]}. [a, b] ker(ψ), ψ([a, b]) = φ(a) φ(b) 1 = 0. φ(b) φ(a) = 0. φ, a ker(φ) = {0}, a = 0. [a, b] = [0, 1]. ψ., S D quotient field, D S, [a, 1] a. [a, b] S a/b.