Proof. Take the standard basis h
1
= e
11
−e
22
, h
2
= e
22
−e
33
, e
ij
(1 6 i 6= j 6 3), If I6= 0 is an ideal of s-
l(3,F), adh
1
(h
1
, h
2
, e
12
, e
13
, e
21
, e
23
, e
31
, e
32
) = (h
1
, h
2
, e
12
, e
13
, e
21
, e
23
, e
31
, e
32
)diag(0, 0, 2, 1, −2, −1, −1, 1),
adh
2
((h
1
, h
2
, e
12
, e
13
, e
21
, e
23
, e
31
, e
32
)) = (h
1
, h
2
, e
12
, e
13
, e
21
, e
23
, e
31
, e
32
)diag(0, 0, −1, 1, 1, 2, −1, −2),
If h
i
∈ I, then[h
i
e
ij
] = λe
ij
∈ I; if e
ij
(i 6= j) ∈ I, [e
ij
e
ij
] = e
ii
− e
jj
= h
i
∈ I, it means that I=L
unless I is vacant. If char F=3, then the centre of sl(3,F) is s(3,F), what’s more, s(3,F) is an ideal of
sl(3,F).
7.Prove that t(n, F ) and δ(n, F ), are self-normalizing subalgebra of gl(n, F ), whereas n(n, F ) has
normalizert(n, F ).
Proof. We firstly write down the normalizers N
gl(n,F )
(t(n, F )) = {x ∈ gl(n, F )|[x, t(n, F )] ⊂ t(n, F )},
N
gl(n,F )
(δ(n, F )) = {x ∈ gl(n, F )|[x, δ(n, F )] ⊂ δ(n, F )}, N
gl(gl(n,F ))
(n(n, F )) = {x ∈ gl(n, F )|[x, n(n, F )] ⊂
n(n, F )}, take the standard basis of t(n,F)e
ij
i < j, e
kl
∈ N
gl(n,F )
(t(n, F )), 1 ≤ k, l ≤ n, [e
kl
e
ij
] =
e
kl
e
ij
− e
ij
e
kl
= δ
li
e
kj
− δ
jk
e
il
∈ t(n, F )whenk > j, e
kj
= 0andi > l, e
il
= 0, hencek > j > i >
l, e
kl
= 0. Similarly, take the standard basis of δ(n, F ) e
ii
, 1 ≤ i ≤ n, e
kl
∈ N
gl(n,F )
(δ(n, F )), 1 ≤
k, l ≤ n, [e
kl
e
ii
] = e
kl
e
ii
− e
ii
e
kl
= δ
li
e
kl
− δ
ik
e
il
, when k 6= i, e
ki
= 0andi 6= le
il
= 0, only when
k=i=l, δ
li
e
kl
− δ
ik
e
il
∈ δ(n, F ), hence e
kl
∈ δ(n, F ) ,we notice that t(n, F ) = δ(n, F ) + n(n, F ),
[x + y, n(n, F )] = [x, n(n, F )] + [y, n(n, F )] ⊂ n(n, F ), where x ∈ δ(n, F ), y ∈ n(n, F ), hence
t(n, F ) ⊂ N
gl(n,F )
(n(n, F )), to the contrary, if y ∈ N
gl(n,F )
(n(n, F )), to prove y ∈ t(n, F ), set
y = e
kl
, for any e
ij
∈ n(n, F ) i < j, [e
kl
e
ij
] = e
kl
e
ij
− e
ij
e
kl
= δ
li
e
kj
− δ
jk
e
il
, when kleqj, e
kj
=
0, i ≤ l, e
il
= 0, when k ≤ j ≤ i ≤ l[e
lk
e
ij
] = 0, so e
kl
= 0
8.Prove that in each classical linear Lie algebra, the set of diagonal matrices is a self-normalizing subal-
gebra, when char F=0.
Proof. If D is the set of diagonal matrices of A
l
, {h
i
= e
ii
− e
i+1i+1
}, 1 ≤ i ≤ l are the standard
basis of D, N
A
l
(D) = {x ∈ A
l
|[x, D] ⊂ D}, if e
kl
∈ N
A
l
(D), [e
klh
i
] = [e
kl
e
ii
] − [e
kl
e
i+1 i+1
] =
δ
li
e
ki
− δ
ik
e
il
− δ
l i+1
e
k i+1
+ δ
i+1 k
e
i+1 l
lies in D, so k=l=i, or k=l=i+1, that e
kl
∈ D.
9.Prove Proposition 2.2
Proof. (i) Define a map ψ : L/Kerφ → Imφ by x + Kerφ 7→ φ(x), since ψ([x + Kerφ, y + Kerφ]) =
ψ([xy] + Kerφ) = φ([xy]) = [φ(x)φ(y)], and set ψ(x + Kerφ) = φ(x), ψ(x
0
+ Kerφ) = φ(x
0
),
ifφ(x) = φ(x
0
), then φ(x) − φ(x
0
) = φ(x − x
0
) = 0, it induces x − x
0
∈ Kerφ, x + Kerφ =
x
0
+ Kerφ, now we prove there exists a unique ψ : L/I → L
0
, making the following diagram commute
L
π
φ
//
L
0
ψ
L/I
define ψ : L/I → L
0
by x + I 7→ φ(x), if φ( x
1
) = φ(x
2
), it will induce that x
1
− x
2
∈ I, so ψ is
well-defined. if there exists another φ
0
such that ψ ◦ π = φ
0
, then (φ − φ
0
)(x) = (ψ ◦ π − ψ ◦ π)(x) = 0;
(ii) it’s easy to prove that J/I is an ideal of L/I, and define epic homomorphism π:L/I → L/J by
x+I 7→ x+J, x ∈ L, if and only if x ∈ J, π(x+I) = 0, Kerπ = J/I, from the standard homomorphism
theorem(i), we have L/I/J/I
∼
=
L/J, what’s more π is homomorphic because of x, y ∈ L, π([x +
I, y + I]) = π([xy] + I) = [xy] + J = [π(x + I), π(y + I)]; (iii) it’s easy to prove that I ∩ J is
an ideal of I, let π be the canonical map L → L/J, f be the restriction to π on I, i.e. f = π|
I
I →
I/J, f(I)
∼
=
I/Kerf, π(I + J)
∼
=
(I + J)/J, f(I) = π(J + I) = π(J) + π(I) = 0 + π|
I
(I = f(I),
5
m0_743127722022-11-08大佬博主好，这个文件是第一到第八节的答案，请问还有后面几节的答案吗
