1 1.1 1.1 R R (1) R = 1 2 Z = 2 n Z (2) R 1.2 R C Z R 1.3 Z 2 = {(a, b) a Z, b Z Z 2 a, b, c, d Z (a, b) + (c, d) = (a + c, b + d), (a, b)(c, d) = (ac, bd) (1) Z 2 (2) Z 2? (3) Z 2 1.4 C Q[ 1] = {a + bi a, b Q (i = 1 1.5 C Z[ 1] = {a + bi a, b Z (i = 1 2 a + bi 2 = a 2 + b 2 1.6 R = 2 n Z, k N {0 k (1) R Q Z R Q (2) R 1.7 R = m n Z, m (1) R Q Z R Q (2) R 1
AI 2 1.2 1.8 n Z f n : Z Z f n (k) = nk ( k Z) f n n 1.9 R R f : R R f 1 1 {a R f(a) = 0 R = {0 R 1.10 R R f : R R f 1 : R R 1.11 Z Z id Z 1.12 Q Q id Q 1.13 Z Z 2 Z 2 1.3 1.14 f : Z 2 Z (1) a, b (m, n) Z 2 f((m, n)) = am + bn f((1, 0)) f((0, 1)) (2) f a, b (1, 0)(1, 0) = (1, 0), (0, 1)(0, 1) = (0, 1) 1.15 f : C C a R f(a) = a i = 1 (1) f(i) = i f(i) = i i 2 + 1 = 0 (2) f f 1.3 1.16 Z[x] f = x 4 +2x 3 3x 2 +4x 5 g = x 2 x+1 1.17 f Q[x] x 1 6, f x 2 + x + 1 x + 2 f x 3 1
AI 3 1.18 K a, b K f = f(x) K[x] K[x] x a x b 0 f (x a)(x b) 1.4 1.19 R I I R (1) R = Z, I = N {0 = {0, 1, 2, 3,... (2) R = Z, I = {n Z n (3) R = Z[x], I = {f Z[x] f = 0 f (4) R = Z[x], I = {f(x) Z[x] f(1) 2Z (5) R = Q[x], I = {f(x) Q[x] f( 2) = 0 1.20 I J R I + J I J R 1.21 R R R I R I R R 1.22 R R f : R R (1) I R f 1 (I ) R (2) f I R f(i) R (3) R I f(i) R R, I, f, R 1.23 (1) K {0 K 2 (2) R 0 1 R {0 R 2 R 1.24 K R f : K R f 1.25 I Z 2 (1) (a, b) I (a, 0) I (0, b) I (2) I 1 = {a Z (a, 0) I I 2 = {a Z (0, a) I Z (3) I = I 1 I 2 = {(a, b) a I 1, b I 2
AI 4 1.5 1.26 Z/7Z Z/7Z 1.27 R = Z/8Z (1) R (2) R (3) R 1.28 R = Z/12Z (1) R (2) R (3) R 1.29 R = Z/2Z S := R[x]/R[x](x 2 + x + 1) 1 1 Z Z/2Z x 2 + x + 1 = 1x 2 + 1x + 1 R[x] (1) S? S (2) S (3) S (2) 1.30 R C i = 1 ρ : R[x] C f = f(x) R[x] ρ(f) = f(i) (1) ρ (2) f(x) R[x] x 2 + 1 ax + b (a, b R) f(i) = ai + b (3) Ker ρ = R[x](x 2 + 1) (4) R[x]/Ker ρ C 1.31 R 1.32 Z 2
AI 5 1.6 1.33 Z n Z nz = Zn n (1) 24Z + 15Z (2) 2014Z + 2520Z (3) 15Z + 36Z + 16Z 1.34 Q[x] (1) Q[x](x 2 1) + Q[x](x 3 + 1) (2) Q[x](x 4 + x 2 + 1) + Q[x](x 4 + 2x 3 + x 2 1) 1.35 3978a + 3465b = 9 a, b 1.36 Q[x] a(x)(x 3 + 1) + b(x)(x 2 + 1) = 1 a(x), b(x) Q[x] 1.37 n 2 k k Z/nZ k (1) k Z/nZ kl + qn = 1 l, q (2) k Z/nZ k n (3) Z/16Z (4) Z/30Z 1.38 R a, b R 0 d a b a = a d, b = b d a, b R e = a b d (1) Ra Rb Re (2) ua + vb = 1 u, v (3) c R Ra Rb c = c(ua + vb ) c e (4) Ra Rb = Re 1.39 36Z 54Z Z 1.40 Z[x] I = 2Z[x] + xz[x] = {2f + xg f, g Z[x]
AI 6 1.41 Q R = 2 n Z, k N {0 k (1) I R m I = Rm R (I R {0 I ) (2) R {0 Rm (m ) 1.42 Q R = m n Z, m (1) I R m I = Rm R (2) R {0 R2 k (k = 0, 1, 2,... ) 1.7 1.43 R {0 R R 1.44 R R R (1) I R I R R (2) I R I R R 1.45 p (1) pz Z = {(np, m) n, m Z Z 2 (2) pz pz = {(np, mp) n, m Z Z 2 (3) {0 Z = {(0, m) m Z Z 2 (4) {(0, 0) Z 2 1.46 Z 2 1.47 c Z ρ c : Z[x] Z ρ c (f) = f(c) (f = f(x) Z[x]) (1) Ker ρ c = Z[x](x c) (2) Z[x](x c)
AI 7 1.48 c Z p J = Z[x](x c) + Z[x]p ρ c : Z[x] Z (1) f Z[x] f(c) p r f J r = 0 (2) π : Z Z/pZ (π(k) = k) Ker (π ρ c ) = J (3) J Z[x] 1.49 Q R = 2 n Z, k N {0 k (I R I Z ) 1.50 Q R = m n Z, m 1.51 R I 0 R π : R R/I 0 a R π(a) = a a R/I 0 (1) J R/I 0 I := π 1 (J) R I I 0 (2) I R I I 0 J := π(i) R/I 0 (3) I := {I I R I I 0, J := {J J R/I 0 Φ : I J Φ(I) = π(i) (I I) Ψ : J I Ψ(J) = π 1 (J) (J J ) Φ Ψ (4) I I I R π(i) R/I 0 (3) (5) I I (R/I 0 )/π(i) R/I R R/I 0 (R/I 0 )/π(i) (6) I I I R π(i) R/I 0 1.52 (1) Z/32Z? (2) Z/36Z?
AI 8 1.8 1.53 Z 96577 1.54 Q[x] x 4 + x 2 + 1 C[x] x 4 + x 2 + 1 1.55 R := Z[ 5] (1) 1 + 5 1 5 R 2 (2) 3, 1 + 5, 1 5 R 2 3 = (1 + 5)(1 5) = 6 (3) R R 2 (4) 2 3 R 2 (5) 1 + 5 1 5 R 1.56 (1) n f = x 2 n Q[x] f Q[x] n (2) I := Q[x]f Q[x] (3) Q[ n] = {a + b n a, b Q ρ : Q[x] Q[ n] g Q[x] ρ(g) = g( n) ρ (4) Q[x]/I Q[ n] Q[ n] 1.57 f = x 3 + x + 1 F 2 [x] (1 F 2 1 (1) f F 2 [x] (2) I := F 2 [x]f F 2 [x] (3) L := F 2 [x]/i 1.58 R = 2 n Z, k N {0 k (1) m m R m Z (2) R ( ) 1.59 R = m n Z, m (1) m m R (2) R ( )
AI 9 1.9 1.60 well-defined ( 1.61 K K K 1.62 Z[x] Q(x) 1.63 K K[x] x 1 x 3 + 3x 2 + 4 R (1) R K[x] (2) R K(x) 2 2.1 2.1 M R N 1 N 2 M R (1) N 1 + N 2 := {u 1 + u 2 u 1 N 1, u 2 N 2 M R (2) N 1 N 2 M R (3) u 1,... u m M Ru 1 + + Ru m = {a 1 u 1 + a m u m a 1,..., a m R M R 2.2 M N R f : M N R Ker f M R Im f N R 2.3 M N R f : M N R (1) f Ker f = {0 (2) f R f 1 : N M R 2.4 L, M, N R f : L M g : M N R g f : L N R 2.5 Z Z n f n : Z Z f n (a) = na ( a Z) (1) f n Z
AI 10 (2) Im f n Ker f n (3) f n Z n (4) f : Z Z Z n f = f n (5) f n n 2.6 M, N R M N R Hom R (M, N) f, g Hom R (M, N) a R M N f + g af (f + g)(u) = f(g) + g(u), (af)(u) = af(u) ( u M) (1) f + g af R (2) R Hom R (M, N) R (3) R R Hom R (R, N) N R 2.7 n 2 M := Z/nZ Z m f m : M M f m (x) = mx (x Z) (1) f m Z (2) m, l f m = f l m l n (3) f : M M Z 0 m n 1 f = f m m 2.8 n m Z/nZ Z/mZ Z 0 2.2 2.9 (1) M 3 := {(u 1, u 2, u 3 ) Z 3 u 1 + u 2 + u 3 = 0 Z 3 (2) f : M 3 M 3 f((u 1, u 2, u 3 )) = (u 2, u 3, u 1 ) f Z (1)
AI 11 2.10 V C n T : V V C V v 1,..., v n α T v 1 = αv 1, T v j = αv j + v j 1 (2 j n) v 1,..., v n V T 2.3 ( ) a b 2.11 a, b, c A = 0 c M 2 (Z) a, b, c A A 1 M 2 (Z) 2.12 n M := Z[ n] Z (1) M 1 n (2) a, b Z f : M M α M f(α) = (a + b n)α f Z (3) f 1, n (4) f Z a, b (5) n = 1 f Z a, b 2.13 M = {f Z[x] deg f 2 M 1, x, x 2 Z (1) n P : M M f(x) M P (f(x)) = (x + 1)f (x) nf(x) f (x) f(x) P Z (2) P 1, x, x 2 (3) P (4) P n Ker P
AI 12 2.5 2.14 ( ) ( ) ( ) (1) 9 6 20 (2) (3) 4 0 0 6 3 5 0 3 (4) ( ) 36 24 18 9 2.15 ( ) 2 6 4 2 6 4 6 0 8 (1) (2) 2 8 0 (3) 2 9 0 2 3 4 0 4 2 0 4 2 2.16 Q[x] ( ) ( ) x 1 1 x 1 1 (1) (2) (3) 0 x 2 0 x 1 ( ) x 1 0 0 x 2 2.17 Q[x] x 1 1 0 x 1 1 0 (1) 0 x 1 0 (2) 0 x 1 1 0 0 x 2 0 0 x 1 (3) x 1 0 0 0 x 2 0 0 0 x 3 2.6 2.18 2.4 f R 2.19 V = R 3 3 t (1, 1, 2) W V/W 2.20 2.15 (1) (4) Z? 2.21 2.16 (1) (3) Z? 2.22 2.17 (1) (3) Q[x]? 2.23 2.18 (1) (3) Q[x]?
AI 13 2.7 2.24 R M R Ann R M = {a R au = 0 ( u M) (1) Ann R M R (2) R = Z M = (Z/8Z) (Z/36Z) Ann Z M 2.25 R M 1 M 2 R M = M 1 M 2 k = 1, 2 p k : M M k p k ((u 1, u 2 )) = u k (u 1 M 1, u 2 M 2 ) (1) p k R (k = 1 (2) Ker p k (3) M 1 M 1 {0 = {(u 1, 0) u 1 M M 1 M 1 M 2 R M/M 1 M 2 R (1) (2) 2.26 240 2.27 M R 2.7 (1) M T M = {0 (2) M M R 2.28 M R M {0 M M (1) M M (2) M u M M = Ru Ann R u := {a R au = 0 R R R/Ann R u M
AI 14 2.8 Jordan 2.29 V K T : V V K W V W K[x] V T K[x] T (W ) W 2.30 Jordan ( ) 1 3 1 2 0 0 4 1 1 2 3 (1) (2) 0 1 0 (3) 6 0 4 (4) 8 2 2 0 2 0 0 1 3 1 4 6 1 2 2.31 A Jordan P 1 AP Jordan P 2.32 A M(n, C) I A = {f(x) C[x] f(a) = O A φ A (x) I A C[x] I A = C[x]φ(x) 2.33 A M(n, C) (1) A Jordan 1 A 1 (2) m A m = I n A : x m 1 A