参考文献:Ideals, Varieties, and Algorithms (4th ed.) [Cox, Little & O’Shea 2015-06-14]
前置文章:仿射簇 和 Groebner基、消元 和 扩展
Nullstellensatz
给定仿射簇
V
⊆
k
n
V \subseteq k^n
V⊆kn,那么就可以写出一个理想:
I
(
V
)
=
{
f
∈
k
[
x
1
,
⋯
,
x
n
]
:
f
(
a
)
=
0
,
∀
a
∈
V
}
\pmb I(V) = \{ f \in k[x_1,\cdots,x_n]: f(a)=0 ,\forall a \in V \}
III(V)={f∈k[x1,⋯,xn]:f(a)=0,∀a∈V}
给定理想
I
∈
k
[
x
1
,
⋯
,
x
n
]
I \in k[x_1,\cdots,x_n]
I∈k[x1,⋯,xn],那么就可以写出一个仿射簇:
V
(
I
)
=
{
a
∈
k
n
:
f
(
a
)
=
0
,
∀
f
∈
I
}
\pmb V(I) = \{ a \in k^n: f(a) = 0, \forall f \in I \}
VVV(I)={a∈kn:f(a)=0,∀f∈I}
然而,上述的对应关系不是一一的。例如,对于任意域
k
k
k,
<
x
>
<x>
<x>和
<
x
2
>
<x^2>
<x2>是
k
[
x
]
k[x]
k[x]上的不同的理想,但是它们的簇相同:
V
(
x
)
=
V
(
x
2
)
=
{
0
}
\pmb V(x) = \pmb V(x^2) = \{0\}
VVV(x)=VVV(x2)={0};又或者不是代数闭域的
R
R
R,那么
1
,
1
+
x
2
,
1
+
x
2
+
x
4
1,1+x^2,1+x^2+x^4
1,1+x2,1+x2+x4都没有实数根,从而
V
(
1
)
=
V
(
1
+
x
2
)
=
V
(
1
+
x
2
+
x
4
)
=
∅
\pmb V(1) = \pmb V(1+x^2) = \pmb V(1+x^2+x^4) = \empty
VVV(1)=VVV(1+x2)=VVV(1+x2+x4)=∅
弱零点定理(The Weak Nullstellensatz):令 k k k是代数封闭域(algebraically closed field), I ∈ k [ x 1 , ⋯ , x n ] I \in k[x_1,\cdots,x_n] I∈k[x1,⋯,xn]是理想,它满足 V ( I ) = ∅ V(I)=\empty V(I)=∅,那么 I = k [ x 1 , ⋯ , x n ] I=k[x_1,\cdots,x_n] I=k[x1,⋯,xn]
The original German name Nullstellensatz: a word formed, in typical German fashion, from three simpler words: Null (= Zero), Stellen (= Places), Satz (= Theorem).
也就是说,对于代数闭域,空簇和全空间一一对应。如果令
k
=
C
k = C
k=C,那么就是Fundamental Theorem of Algebra for multivariable polynomials
:如果一组多项式生成了
C
[
x
1
,
⋯
,
x
n
]
C[x_1,\cdots,x_n]
C[x1,⋯,xn]中的真理想,那么它们一定在
C
n
C^n
Cn内有公共零点。
consistency algorithm:
- 给定代数闭域 k k k上的多项式组 f 1 , ⋯ , f s ∈ k [ x 1 , ⋯ , x n ] f_1,\cdots,f_s \in k[x_1,\cdots,x_n] f1,⋯,fs∈k[x1,⋯,xn],判断是否有解
- 首先计算理想 I = < f 1 , ⋯ , f s > I=<f_1,\cdots,f_s> I=<f1,⋯,fs>的约化Groebner基,字典序任意
- 如果这组基是 { 1 } \{1\} {1},那么方程组没有公共零点
- 如果这组基不是 { 1 } \{1\} {1},那么方程组一定有至少一个公共零点
Hilbert零点定理(Hilbert’s Nullstellensatz):令
k
k
k是代数封闭域,对于
f
,
f
1
,
⋯
,
f
s
∈
k
[
x
1
,
⋯
,
x
n
]
f,f_1,\cdots,f_s \in k[x_1,\cdots,x_n]
f,f1,⋯,fs∈k[x1,⋯,xn],那么
f
∈
I
(
V
(
f
1
,
⋯
,
f
s
)
)
f \in \pmb I(\pmb V(f_1,\cdots,f_s))
f∈III(VVV(f1,⋯,fs))
当仅当存在
m
∈
N
+
m \in N^+
m∈N+,使得
f
m
∈
I
:
=
<
f
1
,
⋯
,
f
s
>
⊆
k
[
x
1
,
⋯
,
x
n
]
f^m \in I := <f_1,\cdots,f_s> \subseteq k[x_1,\cdots,x_n]
fm∈I:=<f1,⋯,fs>⊆k[x1,⋯,xn]
proof:
⇐ \Leftarrow ⇐,由于 f m ∈ I f^m \in I fm∈I,那么 ( f m ) ( V ( I ) ) = 0 (f^m)(\pmb V(I))=0 (fm)(VVV(I))=0,这仅当 f ( V ( I ) ) = 0 f(\pmb V(I))=0 f(VVV(I))=0,于是 f ∈ I ( V ( I ) ) f \in \pmb I(\pmb V(I)) f∈III(VVV(I))
⇒
\Rightarrow
⇒,我们巧妙地构造如下理想,
I
~
=
<
f
1
,
⋯
,
f
s
,
1
−
y
f
>
⊆
k
[
x
1
,
⋯
,
x
n
,
y
]
\tilde I = <f_1,\cdots,f_s,1-yf> \subseteq k[x_1,\cdots,x_n,y]
I~=<f1,⋯,fs,1−yf>⊆k[x1,⋯,xn,y]
对于点
(
a
1
,
⋯
,
a
n
,
a
n
+
1
)
∈
k
n
+
1
(a_1,\cdots,a_n,a_{n+1}) \in k^{n+1}
(a1,⋯,an,an+1)∈kn+1,由于
f
∈
I
(
V
(
f
1
,
⋯
,
f
s
)
)
f \in \pmb I(\pmb V(f_1,\cdots,f_s))
f∈III(VVV(f1,⋯,fs))
- 如果 ( a 1 , ⋯ , a n ) ∈ V ( f 1 , ⋯ , f s ) (a_1,\cdots,a_n) \in \pmb V(f_1,\cdots,f_s) (a1,⋯,an)∈VVV(f1,⋯,fs),那么 f ( a 1 , ⋯ , a n ) = 0 f(a_1,\cdots,a_n)=0 f(a1,⋯,an)=0,从而 1 − a n + 1 f ( a 1 , ⋯ , a n ) = 1 ≠ 0 1-a_{n+1}f(a_1,\cdots,a_n)=1 \neq 0 1−an+1f(a1,⋯,an)=1=0,因此 ( a 1 , ⋯ , a n , a n + 1 ) ∉ V ( I ~ ) (a_1,\cdots,a_n,a_{n+1}) \notin \pmb V(\tilde I) (a1,⋯,an,an+1)∈/VVV(I~)
- 如果 ( a 1 , ⋯ , a n ) ∉ V ( f 1 , ⋯ , f s ) (a_1,\cdots,a_n) \notin \pmb V(f_1,\cdots,f_s) (a1,⋯,an)∈/VVV(f1,⋯,fs),那么存在 1 ≤ i ≤ s 1 \le i \le s 1≤i≤s,使得 f i ( a 1 , ⋯ , a n ) ≠ 0 f_i(a_1,\cdots,a_n) \neq 0 fi(a1,⋯,an)=0,于是 ( a 1 , ⋯ , a n , a n + 1 ) ∉ V ( I ~ ) (a_1,\cdots,a_n,a_{n+1}) \notin \pmb V(\tilde I) (a1,⋯,an,an+1)∈/VVV(I~)
所以,
V
(
I
~
)
=
∅
\pmb V(\tilde I) = \empty
VVV(I~)=∅,根据 the Weak Nullstellensatz,
1
∈
I
~
1 \in \tilde I
1∈I~,
1
=
∑
i
=
1
s
p
i
(
x
1
,
⋯
,
x
n
,
y
)
f
i
+
q
(
x
1
,
⋯
,
x
n
,
y
)
(
1
−
y
f
)
1 = \sum_{i=1}^s p_i(x_1,\cdots,x_n,y)f_i + q(x_1,\cdots,x_n,y)(1-yf)
1=i=1∑spi(x1,⋯,xn,y)fi+q(x1,⋯,xn,y)(1−yf)
令
y
=
1
/
f
y = 1/f
y=1/f,那么
1
=
∑
i
=
1
s
p
i
(
x
1
,
⋯
,
x
n
,
1
/
f
)
f
i
1 = \sum_{i=1}^s p_i(x_1,\cdots,x_n,1/f)f_i
1=i=1∑spi(x1,⋯,xn,1/f)fi
两边同乘
f
m
f^m
fm,
m
∈
N
+
m \in N^+
m∈N+足够大,可以消去分母,那么就有
f
m
=
∑
i
=
1
s
(
p
i
(
x
1
,
⋯
,
x
n
,
1
f
)
⋅
f
m
)
f
i
f^m = \sum_{i=1}^s \left( p_i\left(x_1,\cdots,x_n,\frac{1}{f}\right) \cdot f^m \right)f_i
fm=i=1∑s(pi(x1,⋯,xn,f1)⋅fm)fi
因此,
f
m
∈
I
f^m \in I
fm∈I,证毕。
Ideal - Variety Correspondence
令 V V V是任意的仿射簇,如果对于某正整数 m m m,有 f m ∈ I ( V ) f^m \in \pmb I(V) fm∈III(V),那么 f ∈ I ( V ) f \in \pmb I(V) f∈III(V)。因为任意点 a ∈ V a \in V a∈V,有 ( f ( a ) ) m = 0 (f(a))^m = 0 (f(a))m=0,如果 f ( a ) ≠ 0 f(a) \neq 0 f(a)=0,那么矛盾。
根理想:理想 I I I是根的(radical),它满足如果 ∃ m ≥ 1 , f m ∈ I \exist m\ge 1,f^m \in I ∃m≥1,fm∈I则导致 f ∈ I f \in I f∈I。一个簇 V V V, I ( V ) I(V) I(V)是根理想。
理想的根:令
I
⊆
k
[
x
1
,
⋯
,
x
n
]
I \subseteq k[x_1,\cdots,x_n]
I⊆k[x1,⋯,xn]是理想,定义它的根(radical of
I
I
I):
I
:
=
{
f
∣
f
m
∈
I
,
∃
m
≥
1
}
\sqrt I := \{f | f^m \in I,\exist m \ge 1\}
I
可以证明,
I
⊆
I
⊆
k
[
x
1
,
⋯
,
x
n
]
I \subseteq \sqrt I \subseteq k[x_1,\cdots,x_n]
I⊆I
强零点定理(The Strong Nullstellensatz):令
k
k
k是代数封闭域,
I
∈
k
[
x
1
,
⋯
,
x
n
]
I \in k[x_1,\cdots,x_n]
I∈k[x1,⋯,xn]是理想,那么
I
(
V
(
I
)
)
=
I
\pmb I(\pmb V(I))=\sqrt I
III(VVV(I))=I
Ideal - Variety Correspondence:令
k
k
k是任意域,
-
映射
I : a f f i n e v a r i e t i e s → i d e a l s \pmb I: affine\,\,varieties \to ideals III:affinevarieties→idealsV : i d e a l s → a f f i n e v a r i e t i e s \pmb V: ideals \to affine\,\,varieties VVV:ideals→affinevarieties
是反包含的(inclusion-reversing)
- 如果 I 1 ⊆ I 2 I_1 \subseteq I_2 I1⊆I2,那么 V ( I 1 ) ⊇ V ( I 2 ) \pmb V(I_1) \supseteq \pmb V(I_2) VVV(I1)⊇VVV(I2)
- 如果 V 1 ⊆ V 2 V_1 \subseteq V_2 V1⊆V2,那么 I ( V 1 ) ⊇ I ( V 2 ) \pmb I(V_1) \supseteq \pmb I(V_2) III(V1)⊇III(V2)
-
对于任意的簇 V V V,都有
V ( I ( V ) ) = V \pmb V(\pmb I(V)) = V VVV(III(V))=V
即 I \pmb I III总是双射。 -
对于任意的理想 I I I,都有
V ( I ) = V ( I ) \pmb V(\sqrt I) = \pmb V(I) VVV(I)=VVV(I) 因此 V \pmb V VVV不是单射。
-
如果 k k k是代数闭域,那么限制到根理想上的映射
I : a f f i n e v a r i e t i e s → r a d i c a l i d e a l s \pmb I: affine\,\,varieties \to radical\,\,ideals III:affinevarieties→radicalidealsV : r a d i c a l i d e a l s → a f f i n e v a r i e t i e s \pmb V: radical\,\,ideals \to affine\,\,varieties VVV:radicalideals→affinevarieties
是反包含的双射(inclusion-reversing bijections),两者互为逆映射。
因此,只要在代数闭域上工作,那么:
- 关于仿射簇的任意问题,都可以转化为关于根理想的代数(algebra)问题
- 关于根理想的任意问题,都可以转化为关于仿射簇的几何(geometry)问题
已知理想 I = < f 1 , ⋯ , f s > I = <f_1,\cdots,f_s> I=<f1,⋯,fs>,有如下三个问题:
- Radical Generators:如何计算
I
\sqrt I
I
的一组基? - Radical Ideal:如何判断 I I I是否是根理想?
- Radical Membership:给定
f
∈
k
[
x
1
,
⋯
,
x
n
]
f \in k[x_1,\cdots,x_n]
f∈k[x1,⋯,xn],如何判断
f
∈
I
f \in \sqrt I
f∈I
是否成立?
Radical Membership
令
k
k
k是任意域,理想
I
=
<
f
1
,
⋯
,
f
s
>
⊆
k
[
x
1
,
⋯
,
x
n
]
I = <f_1,\cdots,f_s> \subseteq k[x_1,\cdots,x_n]
I=<f1,⋯,fs>⊆k[x1,⋯,xn]
给定某多项式
f
∈
k
[
x
1
,
⋯
,
x
n
]
f \in k[x_1,\cdots,x_n]
f∈k[x1,⋯,xn],构造理想
I
~
=
<
f
1
,
⋯
,
f
s
,
1
−
y
f
>
⊆
k
[
y
,
x
1
,
⋯
,
x
n
]
\tilde I = <f_1,\cdots,f_s,1-yf> \subseteq k[y,x_1,\cdots,x_n]
I~=<f1,⋯,fs,1−yf>⊆k[y,x1,⋯,xn]
利用 Hilbert’s Nullstellensatz 的证明过程,有
f
∈
I
⟺
∃
m
∈
N
+
,
f
m
∈
I
⟺
1
∈
I
~
f \in \sqrt I \iff \exists m \in N^+,f^m \in I \iff 1 \in \tilde I
f∈I
radical membership algorithm:
- 给定任意域
k
k
k上的理想
I
=
<
f
1
,
⋯
,
f
s
>
I=<f_1,\cdots,f_s>
I=<f1,⋯,fs>,判断某多项式
f
f
f是否属于
I
\sqrt I
I
- 首先计算理想 I ~ = < f 1 , ⋯ , f s , 1 − y f > \tilde I = <f_1,\cdots,f_s,1-yf> I~=<f1,⋯,fs,1−yf>的约化Groebner基,字典序任意
- 如果这组基是
{
1
}
\{1\}
{1},那么
f
∈
I
f \in \sqrt I
f∈I
- 如果这组基不是
{
1
}
\{1\}
{1},那么那么
f
∉
I
f \notin \sqrt I
f∈/I
Radical Generators
令
f
∈
k
[
x
1
,
⋯
,
x
n
]
f \in k[x_1,\cdots,x_n]
f∈k[x1,⋯,xn],令
I
=
<
f
>
I=<f>
I=<f>是主理想(principal ideal),假设
f
=
c
f
1
e
1
f
2
e
2
⋯
f
r
e
r
f = c f_1^{e_1} f_2^{e_2} \cdots f_r^{e_r}
f=cf1e1f2e2⋯frer是唯一素分解,那么
I
=
<
f
>
=
<
f
1
f
2
⋯
f
r
>
\sqrt{I} = \sqrt{<f>} = <f_1f_2\cdots f_r>
I
它的根理想也是主理想,其生成元为
f
r
e
d
=
f
1
f
2
⋯
f
r
f_{red} = f_1f_2\cdots f_r
fred=f1f2⋯fr,叫做
f
f
f的约化(reduction)。如果
f
=
f
r
e
d
f = f_{red}
f=fred,那我们称
f
f
f是 reduced 或者 square-free 的多项式。
如果域
k
⊇
Q
k \supseteq \mathbb Q
k⊇Q,那么主理想
I
=
<
f
>
I=<f>
I=<f>的根为
I
=
<
f
r
e
d
>
\sqrt{I} = <f_{red}>
I
f
r
e
d
=
f
G
C
D
(
f
,
∂
f
∂
x
1
,
⋯
,
∂
f
∂
x
n
)
f_{red} = \dfrac{f}{GCD(f, \frac{\partial f}{\partial x_1}, \cdots, \frac{\partial f}{\partial x_n})}
fred=GCD(f,∂x1∂f,⋯,∂xn∂f)f
可以利用后面介绍的理想的交来计算GCD,从而直接计算出
f
r
e
d
f_{red}
fred,不必做困难的素分解。
Sums, Products, Intersections
理想的和
令
I
,
J
I,J
I,J是环
k
[
x
1
,
⋯
,
x
n
]
k[x_1,\cdots,x_n]
k[x1,⋯,xn]上的理想,那么它们的和(sum)
I
+
J
=
{
f
+
g
:
f
∈
I
,
g
∈
J
}
I+J = \{ f+g: f \in I,g \in J \}
I+J={f+g:f∈I,g∈J}
-
I + J I+J I+J是包含 I , J I,J I,J的最小理想。
-
如果 I = < f 1 , ⋯ , f r > I=<f_1,\cdots,f_r> I=<f1,⋯,fr>, J = < g 1 , ⋯ , g s > J=<g_1,\cdots,g_s> J=<g1,⋯,gs>,那么
I + J = < f 1 , ⋯ , f r , g 1 , ⋯ , g s > ⊆ k [ x 1 , ⋯ , x n ] I+J = <f_1,\cdots,f_r,g_1,\cdots,g_s> \subseteq k[x_1,\cdots,x_n] I+J=<f1,⋯,fr,g1,⋯,gs>⊆k[x1,⋯,xn] -
根据上述性质,可以推出:
< f 1 , ⋯ , f r > = < f 1 > + ⋯ + < f r > <f_1,\cdots,f_r> = <f_1> + \cdots + <f_r> <f1,⋯,fr>=<f1>+⋯+<fr>即环 k [ x 1 , ⋯ , x n ] k[x_1,\cdots,x_n] k[x1,⋯,xn]中任意的理想(根据 Hilbert Basis Theorem 它们都是有限生成的),都可以分解为有限个主理想的和。
-
令 I , J I,J I,J是环 k [ x 1 , ⋯ , x n ] k[x_1,\cdots,x_n] k[x1,⋯,xn]上的理想,那么
V ( I + J ) = V ( I ) ∩ V ( J ) \pmb V(I+J) = \pmb V(I) \cap \pmb V(J) VVV(I+J)=VVV(I)∩VVV(J)
任意的 a ∈ V ( I + J ) a \in \pmb V(I+J) a∈VVV(I+J),由于 I ⊆ I + J I \subseteq I+J I⊆I+J,从而 a ∈ V ( I ) a \in \pmb V(I) a∈VVV(I),同理有 a ∈ V ( J ) a \in \pmb V(J) a∈VVV(J),于是 V ( I + J ) ⊆ V ( I ) ∩ V ( J ) \pmb V(I+J) \subseteq \pmb V(I) \cap \pmb V(J) VVV(I+J)⊆VVV(I)∩VVV(J)任意的 a ∈ V ( I ) ∩ V ( J ) a \in \pmb V(I) \cap \pmb V(J) a∈VVV(I)∩VVV(J),对于 h ∈ I + J h \in I+J h∈I+J,有 h = f + g , f ∈ I , g ∈ J h = f + g,f \in I,g \in J h=f+g,f∈I,g∈J,那么 h ( a ) = f ( a ) + g ( a ) = 0 + 0 = 0 h(a) = f(a)+g(a) = 0+0=0 h(a)=f(a)+g(a)=0+0=0,因此 a ∈ ( I + J ) a \in \pmb(I+J) a∈(((I+J),于是 V ( I + J ) ⊇ V ( I ) ∩ V ( J ) \pmb V(I+J) \supseteq \pmb V(I) \cap \pmb V(J) VVV(I+J)⊇VVV(I)∩VVV(J)
理想的积
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I\cdot J = \{ f_1g_1 + \cdots + f_rg_r: f_i \in I,g_j \in J \}
I⋅J={f1g1+⋯+frgr:fi∈I,gj∈J}
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I ⋅ J I \cdot J I⋅J是理想, I ⋅ J ⊆ I I\cdot J \subseteq I I⋅J⊆I 且 I ⋅ J ⊆ J I\cdot J \subseteq J I⋅J⊆J
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如果 I = < f 1 , ⋯ , f r > I=<f_1,\cdots,f_r> I=<f1,⋯,fr>, J = < g 1 , ⋯ , g s > J=<g_1,\cdots,g_s> J=<g1,⋯,gs>,那么
I ⋅ J = < f i g j : 1 ≤ i ≤ r , 1 ≤ j ≤ s > I \cdot J = <f_ig_j: 1 \le i \le r,1 \le j \le s> I⋅J=<figj:1≤i≤r,1≤j≤s> -
令 I , J I,J I,J是环 k [ x 1 , ⋯ , x n ] k[x_1,\cdots,x_n] k[x1,⋯,xn]上的理想,那么
V ( I ⋅ J ) = V ( I ) ∪ V ( J ) \pmb V(I\cdot J) = \pmb V(I) \cup \pmb V(J) VVV(I⋅J)=VVV(I)∪VVV(J)
任意的 a ∈ V ( I ⋅ J ) a \in \pmb V(I \cdot J) a∈VVV(I⋅J),它使得 f ( a ) g ( a ) = 0 , ∀ f ∈ I , ∀ g ∈ J f(a)g(a)=0,\forall f \in I,\forall g \in J f(a)g(a)=0,∀f∈I,∀g∈J,如果 ∃ g ∈ J , g ( a ) ≠ 0 \exists g \in J,g(a) \neq 0 ∃g∈J,g(a)=0,那么一定有 f ( a ) = 0 , ∀ f ∈ I f(a)=0,\forall f \in I f(a)=0,∀f∈I,从而 a ∈ V ( I ) a \in \pmb V(I) a∈VVV(I),反之亦然,于是 V ( I ⋅ J ) ⊆ V ( I ) ∪ V ( J ) \pmb V(I\cdot J) \subseteq \pmb V(I) \cup \pmb V(J) VVV(I⋅J)⊆VVV(I)∪VVV(J)任意的 a ∈ V ( I ) ∪ V ( J ) a \in \pmb V(I) \cup \pmb V(J) a∈VVV(I)∪VVV(J),要么 f ( a ) = 0 , ∀ f ∈ I f(a)=0,\forall f \in I f(a)=0,∀f∈I,要么 g ( a ) = 0 , ∀ g ∈ J g(a)=0,\forall g \in J g(a)=0,∀g∈J,所以 f ( a ) g ( a ) = 0 , ∀ f ∈ I , ∀ g ∈ J f(a)g(a)=0,\forall f \in I,\forall g \in J f(a)g(a)=0,∀f∈I,∀g∈J,因此任意的 h ∈ I ⋅ J h \in I \cdot J h∈I⋅J,都有 h ( a ) = ∑ 0 = 0 h(a) = \sum 0 = 0 h(a)=∑0=0,从而 a ∈ V ( I ⋅ J ) a \in \pmb V(I \cdot J) a∈VVV(I⋅J),于是 V ( I ⋅ J ) ⊇ V ( I ) ∪ V ( J ) \pmb V(I\cdot J) \supseteq \pmb V(I) \cup \pmb V(J) VVV(I⋅J)⊇VVV(I)∪VVV(J)
理想的交
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I\cap J = \{ f: f \in I, f \in J \}
I∩J={f:f∈I,f∈J}
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I ∩ J I\cap J I∩J是理想,且 I ⋅ J ⊆ I ∩ J I \cdot J \subseteq I \cap J I⋅J⊆I∩J
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令 I , J I,J I,J是环 k [ x 1 , ⋯ , x n ] k[x_1,\cdots,x_n] k[x1,⋯,xn]上的理想,那么
V ( I ∩ J ) = V ( I ) ∪ V ( J ) = V ( I ⋅ J ) \pmb V(I \cap J) = \pmb V(I) \cup \pmb V(J) = \pmb V(I \cdot J) VVV(I∩J)=VVV(I)∪VVV(J)=VVV(I⋅J)
由于 I ⋅ J ⊆ I ∩ J I \cdot J \subseteq I \cap J I⋅J⊆I∩J,因此 V ( I ∩ J ) ⊆ V ( I ⋅ J ) = V ( I ) ∪ V ( J ) \pmb V(I \cap J) \subseteq \pmb V(I \cdot J) = \pmb V(I) \cup \pmb V(J) VVV(I∩J)⊆VVV(I⋅J)=VVV(I)∪VVV(J)任意的 a ∈ V ( I ) ∪ V ( J ) a \in \pmb V(I) \cup \pmb V(J) a∈VVV(I)∪VVV(J),要么 f ( a ) = 0 , ∀ f ∈ I f(a)=0,\forall f \in I f(a)=0,∀f∈I,要么 g ( a ) = 0 , ∀ g ∈ J g(a)=0,\forall g \in J g(a)=0,∀g∈J,自然地对于任意的 h ∈ I ∩ J h \in I \cap J h∈I∩J都有 h ( a ) = 0 h(a)=0 h(a)=0,从而 a ∈ V ( I ∩ J ) a \in \pmb V(I \cap J) a∈VVV(I∩J),于是 V ( I ∩ J ) ⊇ V ( I ) ∪ V ( J ) \pmb V(I \cap J) \supseteq \pmb V(I) \cup \pmb V(J) VVV(I∩J)⊇VVV(I)∪VVV(J)
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令 I , J I,J I,J是环 k [ x 1 , ⋯ , x n ] k[x_1,\cdots,x_n] k[x1,⋯,xn]上的任意理想,那么
I ∩ J = I ∩ J \sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J} I∩J=I ∩J
如果 f ∈ I ∩ J f \in \sqrt{I \cap J} f∈I∩J,那么 f m ∈ I ∩ J f^m \in I \cap J fm∈I∩J,于是 f m ∈ I f^m \in I fm∈I,即 f ∈ I f \in \sqrt{I} f∈I ,同理可证 f ∈ J f \in \sqrt{J} f∈J ,于是 I ∩ J ⊆ I ∩ J \sqrt{I \cap J} \subseteq \sqrt{I} \cap \sqrt{J} I∩J ⊆I ∩J 如果 f ∈ I ∩ J f \in \sqrt{I} \cap \sqrt{J} f∈I
∩J ,那么 f m ∈ I f^m \in I fm∈I且 f p ∈ J f^p \in J fp∈J,那么 f m + p ∈ I ⋅ J ⊆ I ∩ J f^{m+p} \in I\cdot J \subseteq I \cap J fm+p∈I⋅J⊆I∩J,因此 f ∈ I ∩ J f \in \sqrt{I \cap J} f∈I∩J ,于是 I ∩ J ⊇ I ∩ J \sqrt{I \cap J} \supseteq \sqrt{I} \cap \sqrt{J} I∩J ⊇I ∩J
给定环
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f(t)I := <f(t)h(x): h(x) \in I>
f(t)I:=<f(t)h(x):h(x)∈I>
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对于 I = < f 1 , ⋯ , f s > I=<f_1,\cdots,f_s> I=<f1,⋯,fs>,那么
f ( t ) I = < f ( t ) f 1 ( x ) , ⋯ , f ( t ) f s ( x ) > f(t)I = <f(t)f_1(x),\cdots,f(t)f_s(x)> f(t)I=<f(t)f1(x),⋯,f(t)fs(x)> -
如果 g ( x , t ) ∈ f ( t ) I g(x,t) \in f(t)I g(x,t)∈f(t)I, a ∈ k a \in k a∈k是任意的域元素,那么
g ( x , a ) ∈ I g(x,a) \in I g(x,a)∈I
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I \cap J = (tI + (1-t)J) \cap k[x_1,\cdots,x_n]
I∩J=(tI+(1−t)J)∩k[x1,⋯,xn]
algorithm for computing intersections of ideals:
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给定 I = < f 1 , ⋯ , f r > I=<f_1,\cdots,f_r> I=<f1,⋯,fr>, J = < g 1 , ⋯ . g s > J = <g_1,\cdots.g_s> J=<g1,⋯.gs>,构造
t I + ( 1 − t ) J = < t f 1 , ⋯ , t f r , ( 1 − t ) g 1 , ⋯ , ( 1 − t ) g s > tI + (1-t)J = <tf_1,\cdots,tf_r,(1-t)g_1,\cdots,(1-t)g_s> tI+(1−t)J=<tf1,⋯,tfr,(1−t)g1,⋯,(1−t)gs> -
字典序 t > x 1 > ⋯ > x n t>x_1>\cdots>x_n t>x1>⋯>xn,计算它的Groebner基 G G G
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集合 G ′ = G ∩ k [ x 1 , ⋯ , x n ] G' = G \cap k[x_1,\cdots,x_n] G′=G∩k[x1,⋯,xn]是理想 I ∩ J I \cap J I∩J的一组Groebner基
特别地,如果
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I=<f>,J=<g>⊆k[x1,⋯,xn]是主理想,那么
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I∩J也是主理想,且它的生成元为:
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h = LCM(f,g)
h=LCM(f,g)
algorithm for computing the least common multiple:
- 为了计算 f , g ∈ k [ x 1 , ⋯ , x n ] f,g \in k[x_1,\cdots,x_n] f,g∈k[x1,⋯,xn]的最小公倍数,构造理想 I = < f > , J = < g > I=<f>,J=<g> I=<f>,J=<g>
- 利用algorithm for computing intersections of ideals,计算 I ∩ J I \cap J I∩J的任意一组Groebner基 G G G
- 由于 I ∩ J I \cap J I∩J仍是主理想,那么任取 h ∈ G h \in G h∈G,都有 h = L C M ( f , g ) h = LCM(f,g) h=LCM(f,g)
algorithm for computing the greatest common divisor:
- 为了计算 f , g ∈ k [ x 1 , ⋯ , x n ] f,g \in k[x_1,\cdots,x_n] f,g∈k[x1,⋯,xn]的最大公因子,
- 利用algorithm for computing the least common multiple,计算出 h = L C M ( f , g ) h = LCM(f,g) h=LCM(f,g)
- 输出 G C D ( f , g ) = f ⋅ g h GCD(f,g) = \dfrac{f \cdot g}{h} GCD(f,g)=hf⋅g
注意,欧几里得算法对于多变量多项式的GCD会失效!当然,上述算法的效率也不算高,实践中使用更高效的1993年 DAVENPORT, SIRET and TOURNIER 的算法。
更多推荐
代数几何:仿射簇-理想的对应关系,理想的和、积、交
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