Báo cáo nghiên cứu khoa học: " Vành cấu xạ và QF – vành." R R− M A ⊆ M A ⊆ e M A ⊆ ⊕ M A M A M A M R− M (C 1 ) M M M M (C 2 ) A B M A M B M (C 3 ) A B M A ∩ B = 0 A ⊕ B M (1 − C 1 ) M M M CS− (1 − C 1 ) M (C 1 ) (1 − C 1 ) (C 1 ) (C 2 ) (C 1 ) (C 3 ) (C 2 ) =⇒ (C 3 ) =⇒ =⇒ =⇒ =⇒ =⇒ (1 − C 1 ) M Σ− Σ− Σ − (1 − C 1 )) M (I) M (N) ) (1 − C 1 ) I N R R R R R R R a R b R a = aba R e R e 2 = e. R R A R l (A) = {b ∈ R | ba = 0, ∀a ∈ A} A r(A) = {b ∈ R | ab = 0, ∀a ∈ A} A A = {a} l(a), r(a) a l (A) r(A) R R R P − a ∈ R lr(a) = Ra R P − R R (C 2 ) R ⊕ R P − R P − a R b R aR = r(b) bR = r(a) Ra = l(b) Rb = l(a) R R a R b c R aR = r(b) r(a) = cR Ra = l(b) l (a) = cR R R R a R e R b c R aR = r(b) bR = r(a) Ra = l(c) Rc = l(a) a aR = R = r(0) r(a) = 0 = 0R b = 0 aR = r(b) bR = r(a) c = 0 Ra = l(c) Rc = l(a) a eR = r(1 − e) r(e) = (1 − e)R Re = l(1 − e) l(e) = R(1 − e) x ∈ eR x = ex (1 − e)x = (1 − e)ex = (e − e 2 )x = 0 x ∈ r(1 − e) x ∈ r(1 − e) (1 − e)x = x − ex = 0 x = ex ∈ eR eR = r(1 − e) x ∈ r(e) ex = 0 x = (1 −e)x ∈ (1−e)R x ∈ (1 −e)R x = (1 − e)x ex = e(1 − e)x = (e − e 2 )x = 0 x ∈ r(e) r(e) = (1 − e)R Re = l(1 − e) R(1 − e) = l(e) e R R R R R R R 0 a ∈ R a aR = r(1 − e) r(a) = (1 − e)R e R a = 0 0R = 0 = r(1 − 0) r(0) = R = (1 − 0)R a = 0 R aR R R aR = eR e R aR = r(1 − e) r(a) = (1 − e)R R R R R = Π i∈I R i R i i ∈ I a = (a i ) i∈I ∈ R a i ∈ R i R a i R i i ∈ I R R i i ∈ I a = (a i ) i∈I ∈ R R b = (b i ) i∈I ∈ R aR = r(b) bR = r(a) a i R i i ∈ I x i ∈ a i R i (i ∈ I) x = (x i ) i∈I ∈ aR = r(b) bx = (b i x i ) i∈I = (0) b i x i = 0, ∀i ∈ I x i ∈ r(b i ) a i R i ⊆ r(b i ), ∀i ∈ I x i ∈ r(b i )(i ∈ I) b i x i = 0, ∀i ∈ I x = (x i ) i∈I bx = (0) x ∈ r(b) = aR x = (x i ) i∈I = (a i ) i∈I (r i ) i∈I = (a i r i ) i∈I x i = a i r i , ∀i ∈ I x i ∈ a i R i , ∀i ∈ I r(b i ) ⊆ a i R i , ∀i ∈ I a i R i = r(b i ), ∀i ∈ I b i R i = r(a i ), ∀i ∈ I a i R i i ∈ I a i R i i ∈ I b i ∈ R i (i ∈ I) a i R i = r(b i ) b i R i = r(a i ) i ∈ I a = (a i ) i∈I ∈ R R b = (b i ) i∈I aR = r(b) bR = r(a) x = (x i ) i∈I ∈ aR x i ∈ a i R i = r(b i ), ∀i ∈ I b i x i = 0, ∀i ∈ I bx = (b i ) i∈I (x i ) i∈I = (b i x i ) i∈I = (0) x ∈ r(b) aR ⊆ r(b) x = (x i ) i∈I ∈ r(b) bx = (b i ) i∈I (x i ) i∈I = (b i x i ) = (0) b i x i = 0 x i ∈ r(b i ) = a i R i , ∀i ∈ I x ∈ aR r(b) ⊆ aR aR = r(b) bR = r(a) a R R {e i } n 1 a 1 ∈ e 1 Re 1 a 1 e 1 Re 1 a 1 + a 2 + + a n R a i e i Re i i ∈ { 2, , n} a 1 + a 2 + +a n R a i e i Re i i ∈ { 2, , n} a 1 + e 2 + + e n R a 1 + a 2 + + a n R a i e i Re i i ∈ { 2, , n} (a) ⇒ (b) a 1 , a 2 , , a n e 1 Re 1 , e 2 Re 2 , e n Re n (a 1 , a 2 , , a n ) e 1 Re 1 × e 2 Re 2 × × e n Re n a 1 + a 2 + + a n R (b) ⇒ (c) (c) ⇒ (d) e i e i Re i (d) ⇒ (e) a i = e i (e) ⇒ (a) a 2 , a n a i e i Re i i ∈ {2, , n} a 1 + a 2 + + a n R a 1 e 1 Re 1 b ∈ e 1 Re 1 r e 1 Re 1 (a 1 ) = b(e 1 Re 1 ) r e 1 Re 1 (b) = a 1 (e 1 Re 1 ) b ∈ R r(a 1 + a 2 + + a n ) = bR r(b) = (a 1 + a 2 + + a n )R 0 = b(a 1 + a 2 + + a n ) = ba 1 + ba 2 + + ba n ∈ e 1 R ⊕ e 2 R ⊕ ⊕ e n R ba i = 0 i ∈ { 2, , n} a i e i Re i be i = ba i a −1 i = 0 i ∈ {2, , n} be 1 = b(1 − e 2 − − e n ) = b e 1 b = b b = e 1 be 1 ∈ e 1 Re 1 x = e 1 xe 1 ∈ e 1 Re 1 a 1 x = 0 (a 1 + a 2 + + a n )x = (a 1 + a 2 + + a n )e 1 x 1 e 1 = 0 x ∈ r(a 1 +a 2 + + a n ) = bR x ∈ e 1 Re 1 ∩bR = be 1 Re 1 r e 1 Re 1 (a 1 ) ⊆ b (e 1 Re 1 ) x ∈ be 1 Re 1 x ∈ bR = r(a 1 +a 2 + +a n ) 0 = (a 1 + a 2 + + a n )x = a 1 x x ∈ r e 1 Re 1 (a 1 ) b(eRe) ⊆ r e 1 Re 1 (a 1 ) r e 1 Re 1 (a 1 ) = b (e 1 Re 1 ) a 1 (e 1 Re 1 ) = (a 1 +a 2 + +a n )(e 1 Re 1 ) ⊆ (a 1 +a 2 + +a n )R = r(b) a 1 (e 1 Re 1 ) ⊆ r e 1 Re 1 (b) x ∈ r e 1 Re 1 (b) x ∈ r(b) = (a 1 + a 2 + +a n )R x = e 1 xe 1 ∈ e 1 Re 1 x ∈ a 1 (e 1 Re 1 ) r e 1 Re 1 (b) ⊆ a 1 (e 1 Re 1 ) r e 1 Re 1 (b) = a 1 (e 1 Re 1 ) a i e 1 Re 1 R a u R au ua R R a ∈ R b ∈ R aR = r(b) r(a) = Rb ab = ba = 0 (ua)R = r(bu −1 ) r(ua) = (bu −1 )R (au)R = r(u −1 b) r(au) = (u −1 b)R x ∈ (ua)R x = uax 1 bu −1 x = bu −1 uax 1 = bax 1 = 0 x ∈ r(bu −1 ) (ua)R ⊆ r(bu −1 ) x ∈ r(bu −1 ) bu −1 x = 0 u −1 x ∈ r(b) = aR u −1 x = ax 1 x = (ua)x 1 ∈ (ua)R r(bu −1 ) ⊆ (ua)R (ua)R = r(bu −1 ) x ∈ (bu −1 R) x = (bu −1 )x 1 (ua)x = (ua)(bu −1 )x 1 = 0 x ∈ r(ua) (bu −1 R) ⊆ r(ua) x ∈ r(ua) uax = 0 ax = 0 x ∈ r(a) = bR x = bx 1 x = (bu −1 )ux 1 ∈ (bu −1 R) r(ua) ⊆ (bu −1 )R r(ua) = (bu −1 )R ua (au)R = r(u −1 b) r(au) = (u −1 b)R au a ∈ R a R b aba = a (ab) 2 = abab = ab ab R (ab)R = r(1 − ab) r(ab) = (1 − ab)R ab (ab)b −1 = a R R R R R (a) ⇒ (b ) (b) ⇒ (c) (b) ⇒ (d) (c) ⇒ (a) R R a ∈ R b ∈ R aR = r(b) r(a) = bR ab = ba = 0 R x ∈ R a = axa u = xax + b aua = a(xax + b)a = axaxa + aba = axa = a a(1 − xa) = 0 1 − xa ∈ r(a) = bR 1 − xa = by y ∈ R v = a + (1 − ax)y uv = vu = 1 uv = u(a+(1−ax)y) = ua+uy −uaxy = (xax+b)a+(xax+b)y −(xax+ b)axy = xaxa + ba + xaxy + by − xaxaxy − baxy = xa + xaxy + 1 − xa − xaxy = 1 uv = 1 uvu = u u(1 − vu) = 0 1 − vu ∈ r(u) r(u) = 0 t ∈ r(u) 0 = ut = (xax + b)t = xaxt + bt (∗) 0 = a(ut) = a(xaxt + bt) = (axa)xt + abt = axt (∗) bt = 0 t ∈ r(b) = aR t = at 1 t 1 ∈ R 0 = axt = axat 1 = (axa)t 1 = at 1 = t r(u) = 0 1 − vu =∈ r(u) vu = 1 a = aua u R (a) (d) ⇒ (a) M k k R = End(M k ) R R R R P − R R (C 2 ) R R Soc(R R ) = Soc( R R) R R R R R R (R⊕R) R R Soc(R R ) = Soc( R R) R R R R R R R P − R⊕R P − (R⊕R) R (C 2 ) (R ⊕ R) R CS (R ⊕ R) R R R R C 2 2 R = ZC 2 R = Π i∈I R i R i i ∈ I a = (a i ) i∈I ∈ R a i ∈ R i R a i R i i ∈ I R R i i ∈ I a = (a i ) i∈I R b = (b i ) i∈I , c = (c i ) i∈I aR = r(b) cR = r(a) a i R i i ∈ I x i ∈ a i R i (i ∈ I) x = (x i ) i∈I ∈ aR = r(b) bx = (b i x i ) i∈I = (0) b i x i = 0, ∀i ∈ I x i ∈ r(b i ) a i R i ⊆ r(b i ), ∀i ∈ I x i ∈ r(b i )(i ∈ I) b i x i = 0, ∀i ∈ I x = (x i ) i∈I bx = (0) x ∈ r(b) = aR x = (x i ) i∈I = (a i ) i∈I (r i ) i∈I = (a i r i ) i∈I x i = a i r i , ∀i ∈ I x i ∈ a i R i , ∀i ∈ I r(b i ) ⊆ a i R i , ∀i ∈ I a i R i = r(b i ), ∀i ∈ I c i R i = r(a i ), ∀i ∈ I a i R i i ∈ I a i R i i ∈ I b i , c i ∈ R i (i ∈ I) a i R i = r(b i ) c i R i = r(a i ) i ∈ I a = (a i ) i∈I ∈ R R b = (b i ) i∈I c = ( c i ) i∈I aR = r(b) cR = r(a) x = (x i ) i∈I ∈ aR x i ∈ a i R i = r(b i ), ∀i ∈ I b i x i = 0, ∀i ∈ I bx = (b i ) i∈I (x i ) i∈I = (b i x i ) i∈I = (0) x ∈ r(b) aR ⊆ r(b) x = (x i ) i∈I ∈ r(b) bx = (b i ) i∈I (x i ) i∈I = (b i x i ) = (0) b i x i = 0 x i ∈ r(b i ) = a i R i , ∀i ∈ I x ∈ aR r(b) ⊆ aR aR = r(b) cR = r(a) a R R R a ∈ R a R b ∈ R a = aba aR = r(ab − 1) r(a) = (1 − ba)R x ∈ aR x = ax 1 (ab−1)x = (ab−1)ax 1 = ( aba−a)x 1 = 0 x ∈ r(ab − 1) aR ⊆ r(ab − 1) x ∈ r(ab − 1) (ab − 1)x = abx − x = 0 x = abx ∈ aR r(ab − 1) ⊆ aR aR = r(ab − 1) x ∈ (1 − ba)R x = (1 − ba)x 1 ax = a(1 − ba)x 1 = (a − aba)x 1 = 0 x ∈ r(a) (1 − ba)R ⊆ r(a) x ∈ r(a) ax = 0 bax = 0 x = (1 − ba)x x ∈ (1 − ba)R r(a) ⊆ (1 − ba)R r(a) = (1 − ba)R a a R M k k R = End(M k ) R R M = ⊕ i∈I U i U i M F I A ⊕ (⊕ i∈F U i ) ⊆ e M. R R R (C 2 ) R QF − 2 R R Σ − (1 − C 1 ) R R Σ− R R (a) =⇒ (b) : R R R (C 2 ) R = ⊕ i∈I R i R i R CS− A R F ⊆ I A ⊕ (⊕ i∈F R i ) ⊆ e R R . V 1 = ⊕ i∈F R i V 2 = ⊕ i∈K R i K = I\F p 1 p 2 R R V 1 V 2 p 2 | A h = p 1 (p 2 | A ) −1 p 2 (A) −→ V 1 A = { x+h(x) | x ∈ p 2 (A)} h V 2 g : B −→ V 1 p 2 (A) ⊆ B ⊆ V 2 h V 2 C = {x + g(x) | x ∈ B} A ⊕ V 1 ⊆ e R R p 2 (A) = p 2 (A ⊕ V 1 ) ⊆ e p 2 (R R ) = V 2 p 2 (R R ) = V 2 p 2 (A) ⊆ e B ⊆ V 2 A ⊆ e C A R A = C p 2 (A) = B g = h k ∈ K X k = R k ∩ p 2 (A) X k = 0, ∀k ∈ K X k A k = {x + h(x) | x ∈ X k } X k ∼ = A k A k A A k ⊆ e P ⊆ R k ⊕ V 1 A k ∩ V 1 = 0 P ∩ V 1 = 0 p 2 | P h k = h | p 2 (A k ) h h k λ k = p 1 (p 2 | P ) −1 : p 2 (R R ) −→ V 1 λ k h k p 2 (P ) = p 2 (A k ) p 2 | P A k ⊆ e P A k = P A k R R R R (1 − C 1 )− A k ⊆ ⊕ R R X k ∼ = A k R R (C 2 ) X k ⊆ ⊕ R R X k ⊆ e R k ⊆ ⊕ R R X k = R k , ∀k ∈ F p 2 (A) = V 2 A ∼ = V 2 = ⊕ i∈K R i A ⊆ ⊕ R R R CS− R R Σ− (b) =⇒ (a) (b) ⇐⇒ (c) : R (R ⊕ R) R R R R R (C 2 ) (R ⊕ R) R R R R C 2 2 R = Z 4 C 2 . Báo cáo nghiên cứu khoa học: " Vành cấu xạ và QF – vành. " R R− M A ⊆ M A ⊆ e M A ⊆ ⊕ M A M A M A M R− M (C 1 ). (b) ⇐⇒ (c) : R (R ⊕ R) R R R R R (C 2 ) (R ⊕ R) R R R R C 2 2 R = Z 4 C 2