Báo cáo nghiên cứu khoa học: "Một số điều kiện để môđun có tính chất chuyển đổi là trơn" ppt

a ∈ R a = e + u
u R R
M End(M
R
)
M
A A = M

⊕ N = ⊕
i∈I
A
i
M

∼
=
M A

i
⊆ A
i
A

i
⊆
⊕
A
i
A = M

⊕ ( ⊕

i∈I
A

i
) M
R
R
R
R M
R
End(M
R
)
M R
C
1
M M
C
2
A B M A
M B
C
3
A B M A ∩ B = 0 A ⊕ B
M
M M
C
1
C
1

M
C
1
M C
1
C
2
M
M C
1
C
3
M
⇒ ⇒ ⇒ ⇒ ⇒ C
1
A ⊆ M, A ⊆
⊕
M End(M
R
) A
M M
M M
M M
u − dim(M) < ∞ M
A ⊆ M
f ∈ End(M
R
) f(A) ⊆ A
e R eRe (1 − e)R(1 − e)
R

M = M
1
⊕ M
2
⊕ ⊕ M
n
M
i
M
M
R
= M
1
⊕ M
2
⊕ ⊕ M
n
{e
1,
e
2
, , e
n
}
End(M
R
) e
1
+ e
2

+ + e
n
= 1
M
M
i
= Me
i
∀i = 1, , n e
i
End(M
R
)e
i
∼
=
End(Me
i
).
M
i
End(Me
i
) e
i
End(M
R
)e
i
∀i = 1, , n

End(M
R
) M
R
R
R
/
J(R)
R J(R)
R
M
R
M
R
X = M ⊕ M
X
A
1
= M ⊕ 0 , A
2
= 0 ⊕ M.
K = {(x, x), x ∈ M}.
f, g ∈ End(M
R
) f + g = 1
M
M

=


(f(x), −g(x)), x ∈ M

M

∼
=
M
X = M

⊕ K (x, y) ∈ X
(x, y) =

f(x − y), −g(x − y)

+

f(y) + g(x), f(y) + g(x)

∈ M

+ K.
X ⊆ M

+ K M

∩ K = 0 X = M

⊕ K
A


i
⊆ A
i
X = M

⊕ A

1
⊕ A

2
x ∈ M
(x, x) =

f(y), −g(y)

+ (x
1
, 0) + (0, x
2
).
f

, g

∈ End(M) f

(x) = x
2
g


(x) = x
1

ff

(x), ff

(x)

=

ff

(x), −gf

(x)

+ (0, 0) +

0, f

(x)

.

g.g

(x), g.g


(x)

=

− fg

(x), gg

(x)

+

g

(x), 0

+ (0, 0).

ff

(x), ff

(x)

=

f(y
1
), −g(y
1

)

+

g

ff

(x), 0

+

0, f

ff

(x)

.
f

ff

(x) = f

(x)
f

ff


= f

g

gg

= g

ff

, gg

S = End(M
R
)
x = f

(x) − g(y) x = f(y) + g

(x) y = (f

− g

)(x)
x = f(y)+g

(x) = (ff

+gg


)(x) ff

+gg

= 1
M
f ∈ End(M)
e
2
= e = ff

∈ fS 1 − e ∈ (1 − f)S
S = End(M
R
)
f, e ∈ End(M ) e
2
= e f(M) = Imf ⊆
⊕
M
Im(1−f ) = Kerf ⊆
⊕
M e

f(M)

⊆ f(M ) e

(1−f )(M)


⊆ (1−f)(M)
(ef)(M) ⊆ Ker(1−f) Imf = Ker(1−f) (1−f )

ef(M)

= 0
fef = ef (1 − f)e(1 − f) = e(1 − f)
e − ef = e(1 − f) = (1 − f)e(1 − f) = e − ef − fe + fef
= e − ef − fe + ef = e − fe.
ef = fe e S
f

f = f

(ff

)f = f

f(ff

) = (f

f)ff

= f(f

f)f

= f(f


ff

) = ff

.
g

g = gg

u = (f

− g

)
f − (1 − e) S f = (1 − e) + u S
M
R
M
R
u − dimM < ∞
M
R
M u−dimM < ∞ M =
n
⊕
i=1
U
i
U
i

U
i
⊆ M M
U
i
M End(U
i
)
End(U
i
) α, β ∈ End(U
i
)
(α − β) End(U
i
)
N = N
1
⊕ N
2
N
1
= (U
i
, 0)
∼
=
U
i
N

2
= (0, U
i
)
∼
=
U
i
f := (α, β) U
i
N f(x) = (α(x), β(x)) R
g ∈ End(U
i
) g(x
1
, x
2
) = (α − β)
−1
(x
1
− x
2
) gf = id
U
i
f
N = Imf ⊕ K = N
1
⊕ N

2
K ⊆
⊕
N Imf
∼
=
U
i
N

1
⊆
⊕
N
1
, N

2
⊆
⊕
N
2
N = Im f ⊕ N

1
⊕ N

2
N
1

= N

1
⊕ N

1
, N
2
= N

2
⊕ N

2
Imf
∼
=
N
/
N

1
⊕ N

2
∼
=
N

1

⊕ N

2
Imf
∼
=
U
i
N

1
= 0 N

2
= 0
N

1
= 0 N

2
= 0 N = Imf ⊕ N
1
⊕ N

2
N =
N
1
⊕N

2
N

2
 N
2
N
2
N

2
= 0 N = Imf ⊕N
1
π
2
: N → N
2
N N
2
π
2
|
Imf
: Imf → N
2
β = π
2
f : U
i
→ N

2
∼
=
U
i

β End(U
i
)

End(U
i
)
End(U
i
) U
i
M =
n
⊕
i=1
U
i
(1−C
1
) u−dim(M) < ∞
M (1 − C
1
) u − dim(M) < ∞
M

u−dim(
M
/
SocM
) < ∞
M = M
1
⊕ M
2
M
1
u − dim(M
2
) < ∞ M
2
M