Algebraic Linguistics Linguistic Focus 3 Hilbert Zero Point Theorem

Linguistic Focus

3

Hilbert Zero Point Theorem  

　　  TANAKA Akio

<Hilbert zero point theorem 1>

k is algebraically closed field.

Maximum ideal of polynomial in n variables over k,  k [ X₁, …, X_n ], is given by ( X₁-a₁, …, X_n – a_n ) being consisted from kⁿ’s arbitrary elements ( a₁, …, a_n ).

[Proof]

Maximum ideal k [ X₁, …, X_n ]     m

Extension field      R = k [ X₁, …, X_n ] / m

R’s transcendence degree over k     r

From Noetherian normalizing theorem*, (1) there exist algebraically independent elements of y₁,…, yn∈ R over k and (2) R is integral over k [y₁, …, y_r].

Prime ideal of k [y₁, …, y_r]     p

Prime ideal of R     p^-

From rising theorem**     p^- ∩k [y₁, …, y_r] = p

From R is field, p^- = ( 0 ) therefore p^- = ( 0 ).

Therefore k [y₁, …, y_r] is field.

Therefore r = 0

R is k’s algebraic extension.

K is algebraically closed field.  Therefore k = R.

k [ X₁, …, X_n ] = k + m

X_i = a_i + m_i

X_i - a_i ∈ m

( X₁-a₁, …, X_n – a_n ) ⊂ m

( X₁-a₁, …, X_n – a_n ) is maximum ideal.

Therefore ( X₁-a₁, …, X_n – a_n ) = m

[Notes]

Noetherian normalizing theorem*

k is field.

R is finitely generated integral domain over k.

R’s transcendence degree over k is n.

There exist arbitrary elements x₁, …, x_n in R, and R becomes integral over k [x₁, …, x_n ].

Rising theorem**,

R is ring.

S is subring in R.

R is integral over S.

Prime ideal of S is p.

Prime ideal of R is p^-.

There exists p^- ∩ S = p.

<Hilbert zero point theorem 2>

k is algebraically closed field.

When a is ideal of k [ X₁, …, X_n ] and g∈k [ X₁, …, X_n ] and all the points of Z ( a ) is 0,

g∈√a

[Proof]

k [ X₁, …, X_n ] is Noetherian ring.

There exist finite elements f₁, …, f_m ∈a.

A = (f₁, …, f_m )

From lemma*, there exists a certain natural number l.

g^l ∈ a

g ∈ √a

[Note]

Lemma*

k [ X₁, …, X_n ]  ∋ g, f₁, …, f_m

Arbitrary point of Aⁿ    P = ( a₁, …, a_n )

Given condition

P ∈ Z ({ f₁, …, f_m }) → g ( a₁, …, a_n ) = 0

Conclusion is next.

There exist natural number l and elements h₁, …, h_m in k [ X₁, …, X_n ]. 

g^l  = ∑^m _i=1 h_if_i

Tokyo October 1, 2007

Sekinan Research Field of Language

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