索引
- Hw11：验证商模是模。
- Hw12： R R R是一个含幺环，证明
- 1） K e r ( f ) Ker\left( f \right) Ker(f)是 R − R- R−模，是 M M M的子模；
- 2） Im ( f ) \operatorname{Im}\left( f \right) Im(f)是 R − R- R−模，是 M ′ M' M′的子模；
- Hw13：陈述模态射基本定理。
- Hw14：证明 F r a c ( R ) Frac\left( R \right) Frac(R)是一个域，且 R R R是 F r a c ( R ) Frac\left( R \right) Frac(R)的子环。
- Hw15：当含幺交换环 R R R是整环时有 deg ( f g ) = deg ( f ) + deg ( g ) \deg \left( fg \right)=\deg \left( f \right)+\deg \left( g \right) deg(fg)=deg(f)+deg(g)
- Hw16：存在 { η 1 , . . . , η n } ⊆ R [ x 1 , x 2 , . . . , x n ] S n \left\{ {{\eta }_{1}},...,{{\eta }_{n}} \right\}\subseteq R{{\left[ {{x}_{1}},{{x}_{2}},...,{{x}_{n}} \right]}^{{{S}_{n}}}} {η1,...,ηn}⊆R[x1,x2,...,xn]Sn 满足 η i ≠ r σ j {{\eta }_{i}}\ne r{{\sigma }_{j}} ηi=rσj, ∀ r ∈ R , ∀ i , j ∈ { 1 , 2 , . . . , n } \forall r\in R,\text{ }\forall i,j\in \left\{ 1,2,...,n \right\} ∀r∈R, ∀i,j∈{1,2,...,n}，且 R [ η 1 , . . . , η n ] = R [ x 1 , . . . , x n ] S n R\left[ {{\eta }_{1}},...,{{\eta }_{n}} \right]=R{{\left[ {{x}_{1}},...,{{x}_{n}} \right]}^{{{S}_{n}}}} R[η1,...,ηn]=R[x1,...,xn]Sn.
- Hw17：证明 Z \mathbb{Z} Z是 P I D PID PID。
- Hw18：找一个 Z \mathbb{Z} Z的素理想但同时不是极大理想。
- Hw19：证明
7
∈
Z
[
−
5
]
7\in \mathbb{Z}\left[ \sqrt[{}]{-5} \right]
7∈Z[−5
]是不可约元素。 - Hw20：证明若存在 f : K 1 ↦ K 2 f:{{K}_{1}}\mapsto {{K}_{2}} f:K1↦K2是一个域态射，则 c h a r ( K 1 ) = c h a r ( K 2 ) char\left( {{K}_{1}} \right)=char\left( {{K}_{2}} \right) char(K1)=char(K2)。
Hw11：验证商模是模。
证明：
Proof:
Let
(
M
,
+
)
\left( M,+ \right)
(M,+) be one commutative group,
(
M
,
φ
)
\left( M,\varphi \right)
(M,φ) is an R-module,
N
<
M
N<M
N<M &
(
N
,
φ
)
\left( N,\varphi \right)
(N,φ) is a submodule of
(
M
,
φ
)
\left( M,\varphi \right)
(M,φ), in which
φ
:
R
⊙
M
→
M
r
⊙
m
1
→
m
2
\varphi :\begin{matrix} R & \odot & M & \to & M \\ r & \odot & {{m}_{1}} & \to & {{m}_{2}} \\ \end{matrix}
φ:Rr⊙⊙Mm1→→Mm2. Consider
(
M
/
N
,
ψ
)
\left( M/N,\psi \right)
(M/N,ψ), in which
M
/
N
=
{
m
‾
=
m
+
N
∣
m
∈
N
}
M/N=\left\{ \left. \overline{m}=m+N \right|m\in N \right\}
M/N={m=m+N∣m∈N} and
ψ
≅
φ
:
R
∗
M
/
N
→
M
/
N
r
∗
m
1
‾
→
m
2
‾
\psi \cong \varphi :\begin{matrix} R & * & M/N & \to & M/N \\ r & * & \overline{{{m}_{1}}} & \to & \overline{{{m}_{2}}} \\ \end{matrix}
ψ≅φ:Rr∗∗M/Nm1→→M/Nm2.
First we prove that
(
M
/
N
,
+
)
\left( M/N,+ \right)
(M/N,+) is a commutative group.
Commutativity law:
∀
m
1
,
m
2
∈
M
\forall {{m}_{1}},{{m}_{2}}\in M
∀m1,m2∈M,
{
(
m
1
+
N
)
+
(
m
2
+
N
)
=
(
m
1
+
m
2
)
+
(
N
+
N
)
=
(
m
1
+
m
2
)
+
N
(
m
2
+
N
)
+
(
m
1
+
N
)
=
(
m
2
+
m
1
)
+
(
N
+
N
)
=
(
m
2
+
m
1
)
+
N
m
1
+
m
2
=
m
2
+
m
1
\left\{ \begin{aligned} & \left( {{m}_{1}}+N \right)+\left( {{m}_{2}}+N \right)=\left( {{m}_{1}}+{{m}_{2}} \right)+\left( N+N \right)=\left( {{m}_{1}}+{{m}_{2}} \right)+N \\ & \left( {{m}_{2}}+N \right)+\left( {{m}_{1}}+N \right)=\left( {{m}_{2}}+{{m}_{1}} \right)+\left( N+N \right)=\left( {{m}_{2}}+{{m}_{1}} \right)+N \\ & {{m}_{1}}+{{m}_{2}}={{m}_{2}}+{{m}_{1}} \\ \end{aligned} \right.
⎩⎪⎨⎪⎧(m1+N)+(m2+N)=(m1+m2)+(N+N)=(m1+m2)+N(m2+N)+(m1+N)=(m2+m1)+(N+N)=(m2+m1)+Nm1+m2=m2+m1
⇒
m
1
‾
+
m
2
‾
=
m
2
‾
+
m
1
‾
\Rightarrow \overline{{{m}_{1}}}+\overline{{{m}_{2}}}=\overline{{{m}_{2}}}+\overline{{{m}_{1}}}
⇒m1+m2=m2+m1.
Association law:
∀
m
1
,
m
2
,
m
3
∈
M
\forall {{m}_{1}},{{m}_{2}},{{m}_{3}}\in M
∀m1,m2,m3∈M,
(
m
1
‾
+
m
2
‾
)
+
m
3
‾
=
m
1
‾
+
(
m
2
‾
+
m
3
‾
)
=
(
m
1
+
m
2
+
m
3
)
+
N
\left( \overline{{{m}_{1}}}+\overline{{{m}_{2}}} \right)+\overline{{{m}_{3}}}=\overline{{{m}_{1}}}+\left( \overline{{{m}_{2}}}+\overline{{{m}_{3}}} \right)=\left( {{m}_{1}}+{{m}_{2}}+{{m}_{3}} \right)+N
(m1+m2)+m3=m1+(m2+m3)=(m1+m2+m3)+N
Unit:：
0
‾
=
0
+
N
=
N
\overline{0}=0+N=N
0=0+N=N.
Reversibility:
∀
m
∈
M
\forall m\in M
∀m∈M,
∃
(
−
m
)
∈
M
\exists \left( -m \right)\in M
∃(−m)∈M，
m
‾
+
(
−
m
)
‾
=
(
m
+
N
)
+
(
−
m
+
N
)
=
(
m
−
m
)
+
N
=
N
=
0
‾
\overline{m}+\overline{\left( -m \right)}=\left( m+N \right)+\left( -m+N \right)=\left( m-m \right)+N=N=\overline{0}
m+(−m)=(m+N)+(−m+N)=(m−m)+N=N=0
And since
ψ
≅
φ
\psi \cong \varphi
ψ≅φ, it is obvious that
(
M
/
N
,
ψ
)
\left( M/N,\psi \right)
(M/N,ψ) is an R-module.
Hw12： R R R是一个含幺环，证明
1） K e r ( f ) Ker\left( f \right) Ker(f)是 R − R- R−模，是 M M M的子模；
证明：
需要证明两点——
①
K
e
r
(
f
)
<
M
Ker\left( f \right)<M
Ker(f)<M。
首先有
K
e
r
(
f
)
⊆
M
Ker\left( f \right)\subseteq M
Ker(f)⊆M
封闭性：
∀
x
1
,
x
2
∈
K
e
r
(
f
)
,
f
(
x
1
+
x
2
)
=
f
(
x
1
)
+
f
(
x
2
)
=
0
+
0
=
0
⇒
x
1
+
x
2
∈
K
e
r
(
f
)
\forall {{x}_{1}},{{x}_{2}}\in Ker\left( f \right),\text{ }f\left( {{x}_{1}}+{{x}_{2}} \right)=f\left( {{x}_{1}} \right)+f\left( {{x}_{2}} \right)=0+0=0\text{ }\Rightarrow \text{ }{{x}_{1}}+{{x}_{2}}\in Ker\left( f \right)
∀x1,x2∈Ker(f), f(x1+x2)=f(x1)+f(x2)=0+0=0 ⇒ x1+x2∈Ker(f)
结合律：继承
M
M
M的加法结合律。
单位元：由于
f
(
0
)
=
0
f\left( 0 \right)=0
f(0)=0有
0
∈
K
e
r
(
f
)
0\in Ker\left( f \right)
0∈Ker(f)，作为
M
M
M的单位元，
0
0
0也是
K
e
r
(
f
)
Ker\left( f \right)
Ker(f)的单位元。
可逆性：
∀
x
∈
K
e
r
(
f
)
\forall x\in Ker\left( f \right)
∀x∈Ker(f)，有
f
(
0
)
=
f
(
−
x
+
x
)
=
f
(
−
x
)
+
f
(
x
)
=
f
(
−
x
)
+
0
=
0
⇒
f
(
−
x
)
=
0
⇒
−
x
∈
K
e
r
(
f
)
f\left( 0 \right)=f\left( -x+x \right)=f\left( -x \right)+f\left( x \right)=f\left( -x \right)+0=0\Rightarrow f\left( -x \right)=0\Rightarrow -x\in Ker\left( f \right)
f(0)=f(−x+x)=f(−x)+f(x)=f(−x)+0=0⇒f(−x)=0⇒−x∈Ker(f).
②
R
[
K
e
r
(
f
)
]
⊆
K
e
r
(
f
)
R\left[ Ker\left( f \right) \right]\subseteq Ker\left( f \right)
R[Ker(f)]⊆Ker(f)
∀
x
∈
K
e
r
(
f
)
\forall x\in Ker\left( f \right)
∀x∈Ker(f)，
∀
r
∈
R
\forall r\in R
∀r∈R，考虑
f
(
r
x
)
=
r
f
(
x
)
=
r
×
0
=
0
⇒
r
x
∈
K
e
r
(
f
)
⇒
R
[
K
e
r
(
f
)
]
⊆
K
e
r
(
f
)
f\left( rx \right)=rf\left( x \right)=r\times 0=0\text{ }\Rightarrow \text{ }rx\in Ker\left( f \right)\text{ }\Rightarrow \text{ }R\left[ Ker\left( f \right) \right]\subseteq Ker\left( f \right)
f(rx)=rf(x)=r×0=0 ⇒ rx∈Ker(f) ⇒ R[Ker(f)]⊆Ker(f)
2） Im ( f ) \operatorname{Im}\left( f \right) Im(f)是 R − R- R−模，是 M ′ M' M′的子模；
证明：
需要证明两点——
①
Im
(
f
)
<
M
′
\operatorname{Im}\left( f \right)<M'
Im(f)<M′。
首先有
Im
(
f
)
⊆
M
′
\operatorname{Im}\left( f \right)\subseteq M'
Im(f)⊆M′.
封闭性：
∀
y
1
,
y
2
∈
Im
(
f
)
\forall {{y}_{1}},{{y}_{2}}\in \operatorname{Im}\left( f \right)
∀y1,y2∈Im(f)，
∃
x
1
,
x
2
∈
M
\exists {{x}_{1}},{{x}_{2}}\in M
∃x1,x2∈M，使得
f
(
x
1
)
=
y
1
&
f
(
x
2
)
=
y
2
⇒
f
(
x
1
+
x
2
)
=
f
(
x
1
)
+
f
(
x
2
)
=
y
1
+
y
2
⇒
y
1
+
y
2
∈
Im
(
f
)
f\left( {{x}_{1}} \right)={{y}_{1}}\text{ }\And \text{ }f\left( {{x}_{2}} \right)={{y}_{2}}\text{ }\Rightarrow \text{ }f\left( {{x}_{1}}+{{x}_{2}} \right)=f\left( {{x}_{1}} \right)+f\left( {{x}_{2}} \right)={{y}_{1}}+{{y}_{2}}\text{ }\Rightarrow \text{ }{{y}_{1}}+{{y}_{2}}\in \operatorname{Im}\left( f \right)
f(x1)=y1 & f(x2)=y2 ⇒ f(x1+x2)=f(x1)+f(x2)=y1+y2 ⇒ y1+y2∈Im(f)
结合律：继承
M
′
M'
M′的结合律。
单位元：
f
(
0
)
=
0
f\left( 0 \right)=0
f(0)=0，
0
∈
Im
(
f
)
0\in \operatorname{Im}\left( f \right)
0∈Im(f)也是
Im
(
f
)
\operatorname{Im}\left( f \right)
Im(f)的单位元。
可逆性：
∀
y
∈
Im
(
f
)
,
∃
x
∈
M
,
s
.
t
.
f
(
x
)
=
y
\forall y\in \operatorname{Im}\left( f \right),\text{ }\exists x\in M,\text{ }s.t.\text{ }f\left( x \right)=y
∀y∈Im(f), ∃x∈M, s.t. f(x)=y。考虑
−
x
∈
M
-x\in M
−x∈M和
f
(
−
x
)
f\left( -x \right)
f(−x)：
f
(
0
)
=
f
(
x
−
x
)
=
f
(
x
)
+
f
(
−
x
)
=
y
+
f
(
−
x
)
=
0
f\left( 0 \right)=f\left( x-x \right)=f\left( x \right)+f\left( -x \right)=y+f\left( -x \right)=0
f(0)=f(x−x)=f(x)+f(−x)=y+f(−x)=0
⇒
f
(
−
x
)
=
−
y
∈
Im
(
f
)
\Rightarrow f\left( -x \right)=-y\in \operatorname{Im}\left( f \right)
⇒f(−x)=−y∈Im(f).
②
R
[
Im
(
f
)
]
⊆
Im
(
f
)
R\left[ \operatorname{Im}\left( f \right) \right]\subseteq \operatorname{Im}\left( f \right)
R[Im(f)]⊆Im(f)。
∀
y
∈
Im
(
f
)
,
∃
x
∈
M
,
s
.
t
.
f
(
x
)
=
y
\forall y\in \operatorname{Im}\left( f \right),\text{ }\exists x\in M,\text{ }s.t.\text{ }f\left( x \right)=y
∀y∈Im(f), ∃x∈M, s.t. f(x)=y，
∀
r
∈
R
\forall r\in R
∀r∈R，有
r
y
=
r
f
(
x
)
=
f
(
r
x
)
,
r
x
∈
M
ry=rf\left( x \right)=f\left( rx \right),\text{ }rx\in M
ry=rf(x)=f(rx), rx∈M，所以有
r
y
∈
Im
(
f
)
ry\in \operatorname{Im}\left( f \right)
ry∈Im(f)，所以
R
[
Im
(
f
)
]
⊆
Im
(
f
)
R\left[ \operatorname{Im}\left( f \right) \right]\subseteq \operatorname{Im}\left( f \right)
R[Im(f)]⊆Im(f)。
Hw13：陈述模态射基本定理。
-
f : M → M ′ f:M\to M' f:M→M′是一个模态射，则 Ker ( f ) ◃ M \text{Ker}\left( f \right)\triangleleft M Ker(f)◃M，且 ∃ f ‾ : M / Ker ( f ) → M ′ \exists \overline{f}:M/\text{Ker}\left( f \right)\to M' ∃f:M/Ker(f)→M′，使得以下的图交换：
注：
其中映射 P P P可取 P : M → M / K e r ( f ) m → m K e r ( f ) P:\begin{matrix} M & \to & M/Ker\left( f \right) \\ m & \to & mKer\left( f \right) \\ \end{matrix} P:Mm→→M/Ker(f)mKer(f)， f ‾ \overline{f} f 可取 f ‾ : M / K e r ( f ) → M ′ m K e r ( f ) → f ( m ) \overline{f}:\begin{matrix} M/Ker\left( f \right) & \to & M' \\ mKer\left( f \right) & \to & f\left( m \right) \\ \end{matrix} f:M/Ker(f)mKer(f)→→M′f(m)
由于 K e r ( f ) ◃ M Ker\left( f \right)\triangleleft M Ker(f)◃M，所以 M / K e r ( f ) M/Ker\left( f \right) M/Ker(f)是一个群，且在博客《群态射，环态射，域态射》中的“群态射例子”部分已证明 P P P也是一个群态射。
此外 f ‾ \overline{f} f也是一个群态射，还存在群同构 M / K e r ( f ) ≅ Im ( f ) ⊆ M ′ M/Ker\left( f \right)\cong \operatorname{Im}\left( f \right) \subseteq M' M/Ker(f)≅Im(f)⊆M′。 -
N ◃ M , S < M N\triangleleft M,\text{ }S<M N◃M, S<M，则 S N < M , S ⋂ N ◃ S SN<M,\text{ }S\bigcap N\triangleleft S SN<M, S⋂N◃S，且： S N / N ≅ S / ( S ⋂ N ) SN/N\cong S/\left( S\bigcap N \right) SN/N≅S/(S⋂N)（这里的 ≅ \cong ≅是存在群同构（双射群态射）的意思，下同）；
-
N ◃ M N\triangleleft M N◃M，则 M / N M/N M/N的所有（正规？）子群均形如 K / N K/N K/N ( N ◃ K < M ) \left( N\triangleleft K<M \right) (N◃K<M)，且
K / N ◃ M / N ⇔ K ◃ M K/N\triangleleft M/N\text{ }\Leftrightarrow \text{ }K\triangleleft M K/N◃M/N ⇔ K◃M，此时我们有： ( M / N ) / ( K / N ) ≅ M / K \left( M/N \right)/\left( K/N \right)\cong M/K (M/N)/(K/N)≅M/K。
Hw14：证明 F r a c ( R ) Frac\left( R \right) Frac(R)是一个域，且 R R R是 F r a c ( R ) Frac\left( R \right) Frac(R)的子环。
证明：
给定
R
R
R是一个交换整环，考虑结构
(
F
r
a
c
(
R
)
,
+
,
×
)
\left( Frac\left( R \right),+,\times \right)
(Frac(R),+,×)，其中
+,
×
\text{+,}\times
+,×分别定义如下
r
1
s
1
+
r
2
s
2
=
r
1
s
2
+
r
2
s
1
s
1
s
2
\frac{{{r}_{1}}}{{{s}_{1}}}+\frac{{{r}_{2}}}{{{s}_{2}}}=\frac{{{r}_{1}}{{s}_{2}}+{{r}_{2}}{{s}_{1}}}{{{s}_{1}}{{s}_{2}}}
s1r1+s2r2=s1s2r1s2+r2s1
r
1
s
1
×
r
2
s
2
=
r
1
r
2
s
1
s
2
\frac{{{r}_{1}}}{{{s}_{1}}}\times \frac{{{r}_{2}}}{{{s}_{2}}}=\frac{{{r}_{1}}{{r}_{2}}}{{{s}_{1}}{{s}_{2}}}
s1r1×s2r2=s1s2r1r2
其中
r
1
,
r
2
∈
R
,
s
1
,
s
2
∈
R
\
{
0
}
{{r}_{1}},{{r}_{2}}\in R,\text{ }{{s}_{1}},{{s}_{2}}\in R\backslash \left\{ 0 \right\}
r1,r2∈R, s1,s2∈R\{0}。
-
(
F
r
a
c
(
R
)
,
+
)
\left( Frac\left( R \right),+ \right)
(Frac(R),+)是交换群。
加法封闭性： r 1 s 2 + r 2 s 1 ∈ R {{r}_{1}}{{s}_{2}}+{{r}_{2}}{{s}_{1}}\in R r1s2+r2s1∈R，且由于 R R R是整环， s 1 , s 2 ≠ 0 {{s}_{1}},{{s}_{2}}\ne 0 s1,s2=0，有 s 1 s 2 ≠ 0 {{s}_{1}}{{s}_{2}}\ne 0 s1s2=0。于是
r 1 s 1 + r 2 s 2 = r 1 s 2 + r 2 s 1 s 1 s 2 ∈ F r a c ( R ) \frac{{{r}_{1}}}{{{s}_{1}}}+\frac{{{r}_{2}}}{{{s}_{2}}}=\frac{{{r}_{1}}{{s}_{2}}+{{r}_{2}}{{s}_{1}}}{{{s}_{1}}{{s}_{2}}}\in Frac\left( R \right) s1r1+s2r2=s1s2r1s2+r2s1∈Frac(R)
加法结合律：
( r 1 s 1 + r 2 s 2 ) + r 3 s 3 = r 1 s 1 + ( r 2 s 2 + r 3 s 3 ) = r 1 s 2 s 3 + r 2 s 1 s 3 + r 3 s 1 s 2 s 1 s 2 s 3 \left( \frac{{{r}_{1}}}{{{s}_{1}}}+\frac{{{r}_{2}}}{{{s}_{2}}} \right)+\frac{{{r}_{3}}}{{{s}_{3}}}=\frac{{{r}_{1}}}{{{s}_{1}}}+\left( \frac{{{r}_{2}}}{{{s}_{2}}}+\frac{{{r}_{3}}}{{{s}_{3}}} \right)=\frac{{{r}_{1}}{{s}_{2}}{{s}_{3}}+{{r}_{2}}{{s}_{1}}{{s}_{3}}+{{r}_{3}}{{s}_{1}}{{s}_{2}}}{{{s}_{1}}{{s}_{2}}{{s}_{3}}} (s1r1+s2r2)+s3r3=s1r1+(s2r2+s3r3)=s1s2s3r1s2s3+r2s1s3+r3s1s2
加法交换律：
r 1 s 1 + r 2 s 2 = r 2 s 2 + r 1 s 1 = r 1 s 2 + r 2 s 1 s 1 s 2 \frac{{{r}_{1}}}{{{s}_{1}}}+\frac{{{r}_{2}}}{{{s}_{2}}}=\frac{{{r}_{2}}}{{{s}_{2}}}+\frac{{{r}_{1}}}{{{s}_{1}}}=\frac{{{r}_{1}}{{s}_{2}}+{{r}_{2}}{{s}_{1}}}{{{s}_{1}}{{s}_{2}}} s1r1+s2r2=s2r2+s1r1=s1s2r1s2+r2s1
加法单位元： 0 = 0 s , ∀ s ∈ R \ { 0 } 0=\frac{0}{s},\text{ }\forall s\in R\backslash \left\{ 0 \right\} 0=s0, ∀s∈R\{0}
加法可逆性： ∀ r 1 s 1 ∈ F r a c ( R ) \forall \frac{{{r}_{1}}}{{{s}_{1}}}\in Frac\left( R \right) ∀s1r1∈Frac(R)，其加法逆元为 − r 1 s 1 \frac{-{{r}_{1}}}{{{s}_{1}}} s1−r1 -
(
F
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\left( Frac\left( R \right),\times \right)
(Frac(R),×)是半群，且
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\left( Frac\left( R \right)\backslash \left\{ 0 \right\},\times \right)
(Frac(R)\{0},×)是交换群。
运算封闭性： ∀ r 1 , r 2 ∈ R , s 1 , s 2 ∈ R \ { 0 } \forall {{r}_{1}},{{r}_{2}}\in R,\text{ }{{s}_{1}},{{s}_{2}}\in R\backslash \left\{ 0 \right\} ∀r1,r2∈R, s1,s2∈R\{0}， r 1 r 2 ∈ R , s 1 s 2 ∈ R \ { 0 } {{r}_{1}}{{r}_{2}}\in R,\text{ }{{s}_{1}}{{s}_{2}}\in R\backslash \left\{ 0 \right\} r1r2∈R, s1s2∈R\{0}， r 1 s 1 × r 2 s 2 = r 1 r 2 s 1 s 2 ∈ F r a c ( R ) \frac{{{r}_{1}}}{{{s}_{1}}}\times \frac{{{r}_{2}}}{{{s}_{2}}}=\frac{{{r}_{1}}{{r}_{2}}}{{{s}_{1}}{{s}_{2}}}\in Frac\left( R \right) s1r1×s2r2=s1s2r1r2∈Frac(R)且当 r 1 ≠ 0 , r 2 ≠ 0 {{r}_{1}}\ne 0,\text{ }{{r}_{2}}\ne 0 r1=0, r2=0时有 r 1 r 2 s 1 s 2 ≠ 0 \frac{{{r}_{1}}{{r}_{2}}}{{{s}_{1}}{{s}_{2}}}\ne 0 s1s2r1r2=0
结合律和交换律：继承 R R R的结合律和交换律。
乘法单位元： 1 = s s , ∀ s ∈ R \ { 0 } 1=\frac{s}{s},\text{ }\forall s\in R\backslash \left\{ 0 \right\} 1=ss, ∀s∈R\{0}
可逆性： ∀ r 1 s 1 ∈ F r a c ( R ) , r 1 , s 1 ≠ 0 \forall \frac{{{r}_{1}}}{{{s}_{1}}}\in Frac\left( R \right),\text{ }{{r}_{1}},{{s}_{1}}\ne 0 ∀s1r1∈Frac(R), r1,s1=0，其乘法逆元为 s 1 r 1 \frac{{{s}_{1}}}{{{r}_{1}}} r1s1 - 分配律显然也满足。
综上，
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r
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Frac\left( R \right)
Frac(R)是域。
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R\subseteq Frac\left( R \right)
R⊆Frac(R)是子环。
Hw15：当含幺交换环 R R R是整环时有 deg ( f g ) = deg ( f ) + deg ( g ) \deg \left( fg \right)=\deg \left( f \right)+\deg \left( g \right) deg(fg)=deg(f)+deg(g)
证明：
给定一个含幺交换环
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{{G}_{1}}=\left\{ \left. x \right|{{a}_{x}}\ne 0 \right\},\text{ }{{G}_{2}}=\left\{ \left. y \right|{{a}_{y}}\ne 0 \right\}.
G1={x∣ax=0}, G2={y∣ay=0}.
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m=max(G1)=deg(f), n=max(G2)=deg(g)，此时
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f = ∑ i = 0 m a i x i , g = ∑ j = 0 n b j x j . f=\sum\limits_{i=0}^{m}{{{a}_{i}}{{x}^{i}}},\text{ }g=\sum\limits_{j=0}^{n}{{{b}_{j}}{{x}^{j}}}. f=i=0∑maixi, g=j=0∑nbjxj.
⇒ f g = ∑ i = 0 m ∑ j = 0 n a i b j x i + j = ∑ k = 0 m + n ( ∑ i + j = k a i b j ) x k \Rightarrow fg=\sum\limits_{i=0}^{m}{\sum\limits_{j=0}^{n}{{{a}_{i}}{{b}_{j}}{{x}^{i+j}}}}=\sum\limits_{k=0}^{m+n}{\left( \sum\limits_{i+j=k}^{{}}{{{a}_{i}}{{b}_{j}}} \right){{x}^{k}}} ⇒fg=i=0∑mj=0∑naibjxi+j=k=0∑m+n(i+j=k∑aibj)xk
R R R是整环， a m , b n ∈ R , a m , b n ≠ 0 {{a}_{m}},{{b}_{n}}\in R,\text{ }{{a}_{m}},{{b}_{n}}\ne 0 am,bn∈R, am,bn=0 ⇒ a m b n ≠ 0 \Rightarrow {{a}_{m}}{{b}_{n}}\ne 0 ⇒ambn=0 ⇒ deg ( f g ) = m + n \Rightarrow \deg \left( fg \right)=m+n ⇒deg(fg)=m+n - 当
G
1
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∅
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{{G}_{1}}=\varnothing \text{ }or\text{ }{{G}_{2}}=\varnothing
G1=∅ or G2=∅时，
f g = 0 ⇒ deg ( f g ) = − ∞ = 0 + ( − ∞ ) = − ∞ + 0 = − ∞ + ( − ∞ ) = deg ( f ) + deg ( g ) fg=0\Rightarrow \deg \left( fg \right)=-\infty =0+\left( -\infty \right)=-\infty +0=-\infty +\left( -\infty \right)=\deg \left( f \right)+\deg \left( g \right) fg=0⇒deg(fg)=−∞=0+(−∞)=−∞+0=−∞+(−∞)=deg(f)+deg(g).
综上，结论得证。
Hw16：存在 { η 1 , . . . , η n } ⊆ R [ x 1 , x 2 , . . . , x n ] S n \left\{ {{\eta }_{1}},...,{{\eta }_{n}} \right\}\subseteq R{{\left[ {{x}_{1}},{{x}_{2}},...,{{x}_{n}} \right]}^{{{S}_{n}}}} {η1,...,ηn}⊆R[x1,x2,...,xn]Sn 满足 η i ≠ r σ j {{\eta }_{i}}\ne r{{\sigma }_{j}} ηi=rσj, ∀ r ∈ R , ∀ i , j ∈ { 1 , 2 , . . . , n } \forall r\in R,\text{ }\forall i,j\in \left\{ 1,2,...,n \right\} ∀r∈R, ∀i,j∈{1,2,...,n}，且 R [ η 1 , . . . , η n ] = R [ x 1 , . . . , x n ] S n R\left[ {{\eta }_{1}},...,{{\eta }_{n}} \right]=R{{\left[ {{x}_{1}},...,{{x}_{n}} \right]}^{{{S}_{n}}}} R[η1,...,ηn]=R[x1,...,xn]Sn.
-
（构造方法一：适用于 ∀ n ∈ Z > 0 \forall n\in {{\mathbb{Z}}_{>0}} ∀n∈Z>0）令 η i = σ i + r i j , ∀ r i j ∈ R \ { 0 } {{\eta }_{i}}={{\sigma }_{i}}+{{r}_{ij}},\text{ }\forall {{r}_{ij}}\in R\backslash \left\{ 0 \right\} ηi=σi+rij, ∀rij∈R\{0}。
证明：
∀ n ∈ Z > 0 \forall n\in {{\mathbb{Z}}_{>0}} ∀n∈Z>0，令 η i = σ i + r i j , ∀ r i j ∈ R \ { 0 } {{\eta }_{i}}={{\sigma }_{i}}+{{r}_{ij}},\text{ }\forall {{r}_{ij}}\in R\backslash \left\{ 0 \right\} ηi=σi+rij, ∀rij∈R\{0}，则有
σ i = η i − r i j {{\sigma }_{i}}={{\eta }_{i}}-{{r}_{ij}} σi=ηi−rij
⇒ R [ x 1 , . . . , x n ] S n = R [ σ 1 , . . . , σ n ] ⊆ R [ η 1 , . . . , η n ] \Rightarrow R{{\left[ {{x}_{1}},...,{{x}_{n}} \right]}^{{{S}_{n}}}}=R\left[ {{\sigma }_{1}},...,{{\sigma }_{n}} \right]\subseteq R\left[ {{\eta }_{1}},...,{{\eta }_{n}} \right] ⇒R[x1,...,xn]Sn=R[σ1,...,σn]⊆R[η1,...,ηn]
∀ g ( x 1 , . . . , x n ) = ∑ k = 0 n g k η k ∈ R [ η 1 , . . . , η n ] \forall g\left( {{x}_{1}},...,{{x}_{n}} \right)=\sum\limits_{k=0}^{n}{{{g}_{k}}{{\eta }_{k}}}\in R\left[ {{\eta }_{1}},...,{{\eta }_{n}} \right] ∀g(x1,...,xn)=k=0∑ngkηk∈R[η1,...,ηn]，其中 g k ∈ R {{g}_{k}}\in R gk∈R， ∀ σ ∈ S n \forall \sigma \in {{S}_{n}} ∀σ∈Sn，有
σ ( g ) = σ ( ∑ k = 0 n g k η k ) = ∑ k = 0 n g k σ ( η k ) = ∑ k = 0 n g k η k = g ⇒ g ∈ R [ x 1 , . . . , x n ] S n \sigma \left( g \right)=\sigma \left( \sum\limits_{k=0}^{n}{{{g}_{k}}{{\eta }_{k}}} \right)=\sum\limits_{k=0}^{n}{{{g}_{k}}\sigma \left( {{\eta }_{k}} \right)}=\sum\limits_{k=0}^{n}{{{g}_{k}}{{\eta }_{k}}}=g\text{ }\Rightarrow \text{ }g\in R{{\left[ {{x}_{1}},...,{{x}_{n}} \right]}^{{{S}_{n}}}} σ(g)=σ(k=0∑ngkηk)=k=0∑ngkσ(ηk)=k=0∑ngkηk=g ⇒ g∈R[x1,...,xn]Sn
从而有
R [ η 1 , . . . , η n ] ⊆ R [ x 1 , . . . , x n ] S n R\left[ {{\eta }_{1}},...,{{\eta }_{n}} \right]\subseteq R{{\left[ {{x}_{1}},...,{{x}_{n}} \right]}^{{{S}_{n}}}} R[η1,...,ηn]⊆R[x1,...,xn]Sn
所以有
R [ η 1 , . . . , η n ] = R [ x 1 , . . . , x n ] S n R\left[ {{\eta }_{1}},...,{{\eta }_{n}} \right]=R{{\left[ {{x}_{1}},...,{{x}_{n}} \right]}^{{{S}_{n}}}} R[η1,...,ηn]=R[x1,...,xn]Sn
对 η i = σ i + r i j {{\eta }_{i}}={{\sigma }_{i}}+{{r}_{ij}} ηi=σi+rij假设 ∃ σ k \exists {{\sigma }_{k}} ∃σk和 ∃ t ∈ R \exists t\in R ∃t∈R，使得 η i = t σ k {{\eta }_{i}}=t{{\sigma }_{k}} ηi=tσk，则有
deg ( η i ) = deg ( t σ k ) = deg ( σ k ) ⇒ k = i \deg \left( {{\eta }_{i}} \right)=\deg \left( t{{\sigma }_{k}} \right)=\deg \left( {{\sigma }_{k}} \right)\text{ }\Rightarrow \text{ }k=i deg(ηi)=deg(tσk)=deg(σk) ⇒ k=i
而考虑式子 σ i + r i j = t σ i {{\sigma }_{i}}+{{r}_{ij}}=t{{\sigma }_{i}} σi+rij=tσi即 r i j = ( t − 1 ) σ i {{r}_{ij}}=\left( t-1 \right){{\sigma }_{i}} rij=(t−1)σi，有
deg [ ( t − 1 ) σ i ] = deg ( r i j ) = 0 ( r i j ≠ 0 ) , deg ( σ i ) = i ≥ 1 ⇒ t = 1 \begin{matrix} \deg \left[ \left( t-1 \right){{\sigma }_{i}} \right]=\deg \left( {{r}_{ij}} \right)=0\text{ }\left( {{r}_{ij}}\ne 0 \right),\text{ }\deg \left( {{\sigma }_{i}} \right)=i\ge 1 \\ \Rightarrow t=1 \\ \end{matrix} deg[(t−1)σi]=deg(rij)=0 (rij=0), deg(σi)=i≥1⇒t=1
而 σ i + r i j = σ i {{\sigma }_{i}}+{{r}_{ij}}={{\sigma }_{i}} σi+rij=σi对 r i j ≠ 0 {{r}_{ij}}\ne 0 rij=0显然是不成立的，因此证得 ∀ t ∈ R , ∀ i , j ∈ { 1 , 2 , . . . , n } , η i ≠ t σ j \forall t\in R,\text{ }\forall i,j\in \left\{ 1,2,...,n \right\},{{\eta }_{i}}\ne t{{\sigma }_{j}} ∀t∈R, ∀i,j∈{1,2,...,n},ηi=tσj. -
（构造方法二）对 n = 2 n=2 n=2，有 σ 1 = x 1 + x 2 , σ 2 = x 1 x 2 {{\sigma }_{1}}={{x}_{1}}+{{x}_{2}},\text{ }{{\sigma }_{2}}={{x}_{1}}{{x}_{2}} σ1=x1+x2, σ2=x1x2，可令 η 1 = σ 1 + σ 2 , η 2 = σ 1 − σ 2 {{\eta }_{1}}={{\sigma }_{1}}+{{\sigma }_{2}},\text{ }{{\eta }_{2}}={{\sigma }_{1}}-{{\sigma }_{2}} η1=σ1+σ2, η2=σ1−σ2。
对 ∀ n ∈ Z ≥ 3 \forall n\in {{\mathbb{Z}}_{\ge 3}} ∀n∈Z≥3，令 η i = ∑ k ≠ i σ k {{\eta }_{i}}=\sum\limits_{k\ne i}^{{}}{{{\sigma }_{k}}} ηi=k=i∑σk，并证明如下。
首先我们证明 ∀ r ∈ R , ∀ i , j ∈ { 1 , 2 , . . . , n } \forall r\in R,\text{ }\forall i,j\in \left\{ 1,2,...,n \right\} ∀r∈R, ∀i,j∈{1,2,...,n}, η i ≠ r σ j {{\eta }_{i}}\ne r{{\sigma }_{j}} ηi=rσj。
假设 η i = r σ j {{\eta }_{i}}=r{{\sigma }_{j}} ηi=rσj，则有
∑ k ≠ i σ k = r σ j ⇒ ∑ k ≠ i σ k − r σ j = 0 ⇒ deg ( ∑ k ≠ i σ k − r σ j ) = deg ( 0 ) = − ∞ \begin{matrix} \sum\limits_{k\ne i}^{{}}{{{\sigma }_{k}}}=r{{\sigma }_{j}} \\ \Rightarrow \sum\limits_{k\ne i}^{{}}{{{\sigma }_{k}}}-r{{\sigma }_{j}}=0 \\ \Rightarrow \deg \left( \sum\limits_{k\ne i}^{{}}{{{\sigma }_{k}}}-r{{\sigma }_{j}} \right)=\deg \left( 0 \right)=-\infty \\ \end{matrix} k=i∑σk=rσj⇒k=i∑σk−rσj=0⇒deg(k=i∑σk−rσj)=deg(0)=−∞
而由于 n ≥ 3 n\ge 3 n≥3，必有 deg ( ∑ k ≠ i σ k − r σ j ) ≥ 1 \deg \left( \sum\limits_{k\ne i}^{{}}{{{\sigma }_{k}}}-r{{\sigma }_{j}} \right)\ge 1 deg(k=i∑σk−rσj)≥1，因此产生矛盾。
然后我们要证明 # { η i } = n \#\left\{ {{\eta }_{i}} \right\}=n #{ηi}=n 和 R [ η 1 , . . . , η n ] = R [ x 1 , . . . , x n ] S n R\left[ {{\eta }_{1}},...,{{\eta }_{n}} \right]=R{{\left[ {{x}_{1}},...,{{x}_{n}} \right]}^{{{S}_{n}}}} R[η1,...,ηn]=R[x1,...,xn]Sn。
考虑等式
t 1 η 1 + t 2 η 2 + . . . + t n η n = 0 − − − ( 16.1 ) ∑ k = 1 n [ ( ∑ j ≠ k ( t j ) ) σ k ] = 0 ⇒ ∑ j ≠ 1 t j = 0 , ∑ j ≠ 2 t j = 0 , . . . , ∑ j ≠ n t j = 0 − − − ( 16.2 ) \begin{matrix} {{t}_{1}}{{\eta }_{1}}+{{t}_{2}}{{\eta }_{2}}+...+{{t}_{n}}{{\eta }_{n}}=0\text{ }---\left( 16.1 \right) \\ \sum\limits_{k=1}^{n}{\left[ \left( \sum\limits_{j\ne k}^{{}}{\left( {{t}_{j}} \right)} \right){{\sigma }_{k}} \right]}=0 \\ \Rightarrow \sum\limits_{j\ne 1}^{{}}{{{t}_{j}}}=0,\text{ }\sum\limits_{j\ne 2}^{{}}{{{t}_{j}}}=0,\text{ }...,\text{ }\sum\limits_{j\ne n}^{{}}{{{t}_{j}}}=0\text{ }---\left( 16.2 \right) \\ \end{matrix} t1η1+t2η2+...+tnηn=0 −−−(16.1)k=1∑n[(j=k∑(tj))σk]=0⇒j=1∑tj=0, j=2∑tj=0, ..., j=n∑tj=0 −−−(16.2)
令 T = ( t 1 t 2 . . . t n ) T T={{\left( \begin{matrix} {{t}_{1}} & {{t}_{2}} & ... & {{t}_{n}} \\ \end{matrix} \right)}^{T}} T=(t1t2...tn)T, η = ( η 1 η 2 . . . η n ) T \eta ={{\left( \begin{matrix} {{\eta }_{1}} & {{\eta }_{2}} & ... & {{\eta }_{n}} \\ \end{matrix} \right)}^{T}} η=(η1η2...ηn)T, σ = ( σ 1 σ 2 . . . σ n ) T \sigma ={{\left( \begin{matrix} {{\sigma }_{1}} & {{\sigma }_{2}} & ... & {{\sigma }_{n}} \\ \end{matrix} \right)}^{T}} σ=(σ1σ2...σn)T, A = ( 0 1 1 . . . 1 1 0 1 ⋱ 1 1 1 0 ⋱ 1 ⋮ ⋮ ⋮ ⋱ 1 1 1 1 . . . 0 ) A=\left( \begin{matrix} 0 & 1 & 1 & ... & 1 \\ 1 & 0 & 1 & \ddots & 1 \\ 1 & 1 & 0 & \ddots & 1 \\ \vdots & \vdots & \vdots & \ddots & 1 \\ 1 & 1 & 1 & ... & 0 \\ \end{matrix} \right) A=⎝⎜⎜⎜⎜⎜⎛011⋮1101⋮1110⋮1...⋱⋱⋱...11110⎠⎟⎟⎟⎟⎟⎞, 有
{ A σ = η ( from η i = ∑ k ≠ i σ k ) T T η = 0 ( from ( 16.1 ) ) A T = 0 ( from ( 16.2 ) ) . \left\{ \begin{aligned} & A\sigma =\eta \text{ }\left( \text{from }{{\eta }_{i}}=\sum\limits_{k\ne i}^{{}}{{{\sigma }_{k}}} \right) \\ & \\ & {{T}^{T}}\eta =0\left( \text{from }\left( \text{16}\text{.1} \right) \right) \\ & AT=0\left( \text{from }\left( \text{16}\text{.2} \right) \right) \\ \end{aligned} \right.. ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎧Aσ=η ⎝⎛from ηi=k=i∑σk⎠⎞TTη=0(from (16.1))AT=0(from (16.2)).
A = ( 0 1 1 . . . 1 1 0 1 ⋱ 1 1 1 0 ⋱ 1 ⋮ ⋮ ⋮ ⋱ 1 1 1 1 . . . 0 ) − r 1 + r i → r i → ( 0 1 1 . . . 1 1 − 1 0 . . . 0 1 0 − 1 . . . 0 ⋮ ⋮ ⋮ ⋱ 0 1 0 0 . . . − 1 ) r i + r 1 → r 1 → ( n − 1 0 0 . . . 0 1 − 1 0 . . . 0 1 0 − 1 . . . 0 ⋮ ⋮ ⋮ ⋱ 0 1 0 0 . . . − 1 ) A=\left( \begin{matrix} 0 & 1 & 1 & ... & 1 \\ 1 & 0 & 1 & \ddots & 1 \\ 1 & 1 & 0 & \ddots & 1 \\ \vdots & \vdots & \vdots & \ddots & 1 \\ 1 & 1 & 1 & ... & 0 \\ \end{matrix} \right)\underrightarrow{-{{r}_{1}}+{{r}_{i}}\to {{r}_{i}}}\left( \begin{matrix} 0 & 1 & 1 & ... & 1 \\ 1 & -1 & 0 & ... & 0 \\ 1 & 0 & -1 & ... & 0 \\ \vdots & \vdots & \vdots & \ddots & 0 \\ 1 & 0 & 0 & ... & -1 \\ \end{matrix} \right)\underrightarrow{{{r}_{i}}+{{r}_{1}}\to {{r}_{1}}}\left( \begin{matrix} n-1 & 0 & 0 & ... & 0 \\ 1 & -1 & 0 & ... & 0 \\ 1 & 0 & -1 & ... & 0 \\ \vdots & \vdots & \vdots & \ddots & 0 \\ 1 & 0 & 0 & ... & -1 \\ \end{matrix} \right) A=⎝⎜⎜⎜⎜⎜⎛011⋮1101⋮1110⋮1...⋱⋱⋱...11110⎠⎟⎟⎟⎟⎟⎞−r1+ri→ri⎝⎜⎜⎜⎜⎜⎛011⋮11−10⋮010−1⋮0.........⋱...1000−1⎠⎟⎟⎟⎟⎟⎞ ri+r1→r1⎝⎜⎜⎜⎜⎜⎛n−111⋮10−10⋮000−1⋮0.........⋱...0000−1⎠⎟⎟⎟⎟⎟⎞
⇒ det ( A ) = ( n − 1 ) det ( − I n − 1 ) = ( n − 1 ) ( − 1 ) ( n − 1 ) + ( n − 2 ) + . . . + 1 det ( I 1 ) = ( n − 1 ) ( − 1 ) n ( n − 1 ) 2 ≠ 0 \begin{aligned} & \Rightarrow \det \left( A \right)=\left( n-1 \right)\det \left( -{{I}_{n-1}} \right) \\ & =\left( n-1 \right){{\left( -1 \right)}^{\left( n-1 \right)+\left( n-2 \right)+...+1}}\det \left( {{I}_{1}} \right) \\ & =\left( n-1 \right){{\left( -1 \right)}^{\frac{n\left( n-1 \right)}{2}}} \\ & \ne 0 \\ \end{aligned} ⇒det(A)=(n−1)det(−In−1)=(n−1)(−1)(n−1)+(n−2)+...+1det(I1)=(n−1)(−1)2n(n−1)=0
⇒ \Rightarrow ⇒ A A A是可逆矩阵。
所以齐次线性方程组 A T = 0 AT=0 AT=0的解为 T = ( 0 , 0 , . . . , 0 ) T T={{\left( 0,0,...,0 \right)}^{T}} T=(0,0,...,0)T，显然也就有
T = ( 0 , 0 , . . . 0 ) T ⇔ T T η = 0 T={{\left( 0,0,...0 \right)}^{T}}\Leftrightarrow {{T}^{T}}\eta =0 T=(0,0,...0)T⇔TTη=0
所以 { η 1 , η 2 , . . . , η n } \left\{ {{\eta }_{1}},{{\eta }_{2}},...,{{\eta }_{n}} \right\} {η1,η2,...,ηn}是线性无关向量组, # { η i } = n \#\left\{ {{\eta }_{i}} \right\}=n #{ηi}=n.
A σ = η σ = A − 1 η R [ x 1 , x 2 , . . . , x n ] S n = R [ σ 1 , . . . , σ n ] ⊆ R [ η 1 , . . . , η n ] \begin{matrix} A\sigma =\eta \\ \sigma ={{A}^{-1}}\eta \\ R{{\left[ {{x}_{1}},{{x}_{2}},...,{{x}_{n}} \right]}^{{{S}_{n}}}}=R\left[ {{\sigma }_{1}},...,{{\sigma }_{n}} \right]\subseteq R\left[ {{\eta }_{1}},...,{{\eta }_{n}} \right] \\ \end{matrix} Aσ=ησ=A−1ηR[x1,x2,...,xn]Sn=R[σ1,...,σn]⊆R[η1,...,ηn]
另一方面，显然有 ∀ g ∈ R [ η 1 , . . . , η n ] , g ∈ R [ x 1 , . . . , x n ] S n ⇒ R [ η 1 , . . . , η n ] ⊆ R [ x 1 , . . . , x n ] S n \forall g\in R\left[ {{\eta }_{1}},...,{{\eta }_{n}} \right],\text{ }g\in R{{\left[ {{x}_{1}},...,{{x}_{n}} \right]}^{{{S}_{n}}}}\Rightarrow R\left[ {{\eta }_{1}},...,{{\eta }_{n}} \right]\subseteq R{{\left[ {{x}_{1}},...,{{x}_{n}} \right]}^{{{S}_{n}}}} ∀g∈R[η1,...,ηn], g∈R[x1,...,xn]Sn⇒R[η1,...,ηn]⊆R[x1,...,xn]Sn。
因此可得到 R [ η 1 , . . . , η n ] = R [ x 1 , . . . , x n ] S n R\left[ {{\eta }_{1}},...,{{\eta }_{n}} \right]=R{{\left[ {{x}_{1}},...,{{x}_{n}} \right]}^{{{S}_{n}}}} R[η1,...,ηn]=R[x1,...,xn]Sn。
Hw17：证明 Z \mathbb{Z} Z是 P I D PID PID。
Proof:
First of all obviously
Z
\mathbb{Z}
Z is a unital commutative ring. So
Z
\mathbb{Z}
Z is a PID equals to proposition: for any ideal
I
⊆
Z
I\subseteq \mathbb{Z}
I⊆Z,
∃
d
∈
Z
,
s
.
t
.
I
=
d
Z
\exists d\in \mathbb{Z},\text{ }s.t.\text{ }I=d\mathbb{Z}
∃d∈Z, s.t. I=dZ.
Let
I
⊆
Z
I\subseteq \mathbb{Z}
I⊆Z be one ideal of
Z
\mathbb{Z}
Z.
If
I
=
{
0
}
I=\left\{ 0 \right\}
I={0}, then we have
I
=
0
Z
I=0\mathbb{Z}
I=0Z.
If
I
≠
{
0
}
I\ne \left\{ 0 \right\}
I={0}, then
∃
a
∈
I
,
a
≠
0
\exists a\in I,\text{ }a\ne 0
∃a∈I, a=0. Because
±
1
∈
Z
\pm 1\in \mathbb{Z}
±1∈Z,
I
Z
⊆
I
I\mathbb{Z}\subseteq I
IZ⊆I, we have
−
a
∈
I
-a\in I
−a∈I,
thus there must be at least one positive element
b
>
0
,
b
∈
I
b>0,\text{ }b\in I
b>0, b∈I. Denote the minimum positive of
I
I
I by
c
c
c, that is
∀
t
∈
I
>
0
,
t
≥
c
\forall t\in {{I}_{>0}},\text{ }t\ge c
∀t∈I>0, t≥c
Then obviously we have
c
Z
=
{
n
c
=
(
n
)
⋅
(
c
+
c
+
.
.
.
+
c
⏞
∣
n
∣
)
∣
n
=
±
1
,
±
2
,
±
3
,
.
.
.
}
⊆
I
c\mathbb{Z}=\left\{ \left. nc= \left( n \right)\centerdot \left( \overbrace{c+c+...+c}^{\left| n \right|} \right) \right|n=\pm 1,\pm 2,\pm 3,... \right\}\subseteq I
cZ=⎩⎨⎧nc=(n)⋅⎝⎛c+c+...+c
Assume that
∃
d
∈
I
,
∀
n
∈
Z
,
d
≠
n
c
\exists d\in I,\text{ }\forall n\in \mathbb{Z},\text{ }d\ne nc
∃d∈I, ∀n∈Z, d=nc, then
∃
n
1
∈
Z
\exists {{n}_{1}}\in \mathbb{Z}
∃n1∈Z, s.t.
∣
n
1
∣
c
<
d
<
∣
n
1
+
1
∣
c
,
\left| {{n}_{1}} \right|c<d<\left| {{n}_{1}}+1 \right|c,
∣n1∣c<d<∣n1+1∣c,
thus we have
∣
n
1
+
1
∣
c
−
d
∈
I
&
0
<
(
n
1
+
1
)
c
−
d
<
c
,
\left| {{n}_{1}}+1 \right|c-d\in I\text{ }\And \text{ }0<\left( {{n}_{1}}+1 \right)c-d<c,
∣n1+1∣c−d∈I & 0<(n1+1)c−d<c,
which contradicts to the condition “c is the minimum positive element in
I
I
I”. Thus we have
∀
t
∈
I
,
∃
n
t
∈
Z
,
s
.
t
.
t
=
n
t
c
,
\forall t\in I,\text{ }\exists {{n}_{t}}\in \mathbb{Z},\text{ }s.t.\text{ }t={{n}_{t}}c,
∀t∈I, ∃nt∈Z, s.t. t=ntc,
which means
I
⊆
c
Z
I\subseteq c\mathbb{Z}
I⊆cZ.
Thus we have
I
=
c
Z
I=c\mathbb{Z}
I=cZ, which means all ideals of
Z
\mathbb{Z}
Z are principal ideals. Thus
Z
\mathbb{Z}
Z is PID.
Hw18：找一个 Z \mathbb{Z} Z的素理想但同时不是极大理想。
解：
{
0
}
\left\{ 0 \right\}
{0}满足题目要求。
一方面，由于
Z
\mathbb{Z}
Z是整环，有
∀
a
,
b
∈
Z
,
a
b
=
0
⇒
a
=
0
or
b
=
0
\forall a,b\in \mathbb{Z},\text{ }ab=0\text{ }\Rightarrow \text{ }a=0\text{ or}\text{ }b=0
∀a,b∈Z, ab=0 ⇒ a=0 or b=0，即有
a
b
∈
{
0
}
⇒
a
∈
{
0
}
or
b
∈
{
0
}
ab\in \left\{ 0 \right\}\text{ }\Rightarrow \text{ }a\in \left\{ 0 \right\}\text{ or}\text{ }b\in \left\{ 0 \right\}
ab∈{0} ⇒ a∈{0} or b∈{0}，所以平凡理想
{
0
}
\left\{ 0 \right\}
{0}是素理想。
另一方面，存在
Z
\mathbb{Z}
Z的其他理想
d
Z
,
∀
d
∈
Z
\
{
0
}
d\mathbb{Z},\text{ }\forall d\in \mathbb{Z}\backslash \left\{ 0 \right\}
dZ, ∀d∈Z\{0}，
{
0
}
⊆
d
Z
\left\{ 0 \right\}\subseteq d\mathbb{Z}
{0}⊆dZ，所以
{
0
}
\left\{ 0 \right\}
{0}不是极大理想。
Hw19：证明
7
∈
Z
[
−
5
]
7\in \mathbb{Z}\left[ \sqrt[{}]{-5} \right]
7∈Z[−5
证明：
设
7
=
(
a
+
b
−
5
)
(
c
+
d
−
5
)
7=\left( a+b\sqrt[{}]{-5} \right)\left( c+d\sqrt[{}]{-5} \right)
7=(a+b−5
⇒
49
=
(
a
+
b
−
5
)
(
a
−
b
−
5
)
(
c
+
d
−
5
)
(
c
−
d
−
5
)
=
(
a
2
+
5
b
2
)
(
c
2
+
5
d
2
)
⇒
a
2
+
5
b
2
=
1
,
7
,
49
\begin{aligned} & \Rightarrow 49=\left( a+b\sqrt[{}]{-5} \right)\left( a-b\sqrt[{}]{-5} \right)\left( c+d\sqrt[{}]{-5} \right)\left( c-d\sqrt[{}]{-5} \right)=\left( {{a}^{2}}+5{{b}^{2}} \right)\left( {{c}^{2}}+5{{d}^{2}} \right) \\ & \Rightarrow {{a}^{2}}+5{{b}^{2}}=1,7,49 \\ \end{aligned}
⇒49=(a+b−5
但是
a
+
b
−
5
&
c
+
d
−
5
a+b\sqrt[{}]{-5}\text{ }\And \text{ }c+d\sqrt[{}]{-5}
a+b−5
所以有
a
2
+
5
b
2
=
7
{{a}^{2}}+5{{b}^{2}}=7
a2+5b2=7，此方程在
Z
\mathbb{Z}
Z中无解。
故7没有（不可分解）不可逆因子，故7不可约。
Hw20：证明若存在 f : K 1 ↦ K 2 f:{{K}_{1}}\mapsto {{K}_{2}} f:K1↦K2是一个域态射，则 c h a r ( K 1 ) = c h a r ( K 2 ) char\left( {{K}_{1}} \right)=char\left( {{K}_{2}} \right) char(K1)=char(K2)。
证明：
给定两个域
(
K
1
,
+
,
×
)
,
(
K
2
,
⊕
,
⊗
)
\left( {{K}_{1}},+,\times \right),\left( {{K}_{2}},\oplus ,\otimes \right)
(K1,+,×),(K2,⊕,⊗)和一个域态射
f
:
K
1
↦
K
2
f:{{K}_{1}}\mapsto {{K}_{2}}
f:K1↦K2。
首先域态射
f
f
f一定是单射，即
∀
k
11
,
k
12
∈
K
1
,
k
11
≠
k
12
\forall {{k}_{11}},{{k}_{12}}\in {{K}_{1}},\text{ }{{k}_{11}}\ne {{k}_{12}}
∀k11,k12∈K1, k11=k12，有
f
(
k
11
)
≠
f
(
k
12
)
f\left( {{k}_{11}} \right)\ne f\left( {{k}_{12}} \right)
f(k11)=f(k12)。
设
K
1
{{K}_{1}}
K1的
+
,
×
+,\times
+,×单位元分别为
0
+
,
1
×
{{0}^{+}},{{1}^{\times }}
0+,1×，
K
2
{{K}_{2}}
K2的
⊕
,
⊗
\oplus ,\otimes
⊕,⊗单位元分别为
0
⊕
,
1
⊗
{{0}^{\oplus }},{{1}^{\otimes }}
0⊕,1⊗，则一定有
f
(
0
+
)
=
0
⊕
,
f
(
1
×
)
=
1
⊗
f\left( {{0}^{+}} \right)={{0}^{\oplus }},\text{ }f\left( {{1}^{\times }} \right)={{1}^{\otimes }}
f(0+)=0⊕, f(1×)=1⊗.
定义
r
‾
:
=
1
×
+
1
×
+
.
.
.
+
1
×
⏟
r
,
r
‾
∗
:
=
1
⊗
+
1
⊗
+
.
.
.
+
1
⊗
⏟
r
,
\overline{r}:=\underbrace{{{1}^{\times }}+{{1}^{\times }}+...+{{1}^{\times }}}_{r},\ \ {{\overline{r}}^{*}}:=\underbrace{{{1}^{\otimes }}+{{1}^{\otimes }}+...+{{1}^{\otimes }}}_{r},
r:=r
C
K
1
=
{
r
‾
∣
r
=
1
,
2
,
.
.
.
}
C
K
2
=
{
r
‾
∗
∣
r
=
1
,
2
,
.
.
.
}
{{C}_{{{K}_{1}}}}=\left\{ \left. \overline{r} \right|r=1,2,... \right\}\text{ }{{C}_{{{K}_{2}}}}=\left\{ \left. {{\overline{r}}^{*}} \right|r=1,2,... \right\}
CK1={r∣r=1,2,...} CK2={r∗∣r=1,2,...}
首先有
f
(
C
K
1
)
⊆
C
K
2
f\left( {{C}_{{{K}_{1}}}} \right)\subseteq {{C}_{{{K}_{2}}}}
f(CK1)⊆CK2，即有
f
:
C
K
1
→
C
K
2
f:{{C}_{{{K}_{1}}}}\to {{C}_{{{K}_{2}}}}
f:CK1→CK2。
∀
r
‾
∈
C
K
1
\forall \overline{r}\in {{C}_{{{K}_{1}}}}
∀r∈CK1，有
f
(
r
‾
)
=
f
(
1
×
+
1
×
+
.
.
.
+
1
×
⏟
r
)
=
f
(
1
×
)
⊕
f
(
1
×
)
⊕
.
.
.
⊕
f
(
1
×
)
⏟
r
=
1
⊗
⊕
1
⊗
⊕
.
.
.
⊕
1
⊗
⏟
r
=
r
‾
∗
∈
C
K
2
\begin{aligned} & f\left( \overline{r} \right)=f\left( \underbrace{{{1}^{\times }}+{{1}^{\times }}+...+{{1}^{\times }}}_{r} \right) \\ & =\underbrace{f\left( {{1}^{\times }} \right)\oplus f\left( {{1}^{\times }} \right)\oplus ...\oplus f\left( {{1}^{\times }} \right)}_{r} \\ & =\underbrace{{{1}^{\otimes }}\oplus {{1}^{\otimes }}\oplus ...\oplus {{1}^{\otimes }}}_{r} \\ & ={{\overline{r}}^{*}}\in {{C}_{{{K}_{2}}}} \\ \end{aligned}
f(r)=f(r
其次有
f
:
C
K
1
→
C
K
2
f:{{C}_{{{K}_{1}}}}\to {{C}_{{{K}_{2}}}}
f:CK1→CK2是双射。
∀
r
‾
∗
∈
C
K
2
,
\forall {{\overline{r}}^{*}}\in {{C}_{{{K}_{2}}}},
∀r∗∈CK2,有
r
‾
∗
=
1
⊗
⊕
1
⊗
⊕
.
.
.
⊕
1
⊗
⏟
r
=
f
(
1
×
)
⊕
f
(
1
×
)
⊕
.
.
.
⊕
f
(
1
×
)
⏟
r
=
f
(
1
×
+
1
×
+
.
.
.
+
1
×
⏟
r
)
=
f
(
r
‾
)
\begin{aligned} & {{\overline{r}}^{*}}=\underbrace{{{1}^{\otimes }}\oplus {{1}^{\otimes }}\oplus ...\oplus {{1}^{\otimes }}}_{r} \\ & =\underbrace{f\left( {{1}^{\times }} \right)\oplus f\left( {{1}^{\times }} \right)\oplus ...\oplus f\left( {{1}^{\times }} \right)}_{r} \\ & =f\left( \underbrace{{{1}^{\times }}+{{1}^{\times }}+...+{{1}^{\times }}}_{r} \right) \\ & =f\left( \overline{r} \right) \\ \end{aligned}
r∗=r
⇒
∃
r
‾
∈
C
K
1
,
s
.
t
.
f
(
r
‾
)
=
r
‾
∗
\Rightarrow \exists \overline{r}\in {{C}_{{{K}_{1}}}},\text{ }s.t.\text{ }f\left( \overline{r} \right)={{\overline{r}}^{*}}
⇒∃r∈CK1, s.t. f(r)=r∗，所以
f
:
C
K
1
→
C
K
2
f:{{C}_{{{K}_{1}}}}\to {{C}_{{{K}_{2}}}}
f:CK1→CK2是满射。又
f
:
C
K
1
→
C
K
2
f:{{C}_{{{K}_{1}}}}\to {{C}_{{{K}_{2}}}}
f:CK1→CK2是单射，所以
f
:
C
K
1
→
C
K
2
f:{{C}_{{{K}_{1}}}}\to {{C}_{{{K}_{2}}}}
f:CK1→CK2是双射。
所以有
#
C
K
1
=
#
C
K
2
\text{ }\!\!\#\!\!\text{ }{{C}_{{{K}_{1}}}}=\#{{C}_{{{K}_{2}}}}
# CK1=#CK2，即
c
h
a
r
(
K
1
)
=
c
h
a
r
(
K
2
)
char\left( {{K}_{1}} \right)=char\left( {{K}_{2}} \right)
char(K1)=char(K2)。
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