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Hilbert Zero Point Theorem
<Hilbert zero point theorem 1>
k is algebraically closed field.
Maximum ideal of polynomial in n variables over k, k [ X1, …, Xn ], is given by ( X1-a1, …, Xn – an ) being consisted from kn’s arbitrary elements ( a1, …, an ).
[Proof]
Maximum ideal k [ X1, …, Xn ] m
Extension field R = k [ X1, …, Xn ] / 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, …, yr].
Prime ideal of k [y1, …, yr] p
Prime ideal of R p-
From rising theorem** p- ∩k [y1, …, yr] = p
From R is field, p- = ( 0 ) therefore p- = ( 0 ).
Therefore k [y1, …, yr] is field.
Therefore r = 0
R is k’s algebraic extension.
K is algebraically closed field. Therefore k = R.
k [ X1, …, Xn ] = k + m
Xi = ai + mi
Xi - ai ∈ m
( X1-a1, …, Xn – an ) ⊂ m
( X1-a1, …, Xn – an ) is maximum ideal.
Therefore ( X1-a1, …, Xn – an ) = 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, …, xn in R, and R becomes integral over k [x1, …, xn ].
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, …, Xn ] and g∈k [ X1, …, Xn ] and all the points of Z ( a ) is 0,
g∈√a
[Proof]
k [ X1, …, Xn ] is Noetherian ring.
There exist finite elements f1, …, fm ∈a.
A = (f1, …, fm )
From lemma*, there exists a certain natural number l.
gl ∈ a
g ∈ √a
[Note]
Lemma*
k [ X1, …, Xn ] ∋ g, f1, …, fm
Arbitrary point of An P = ( a1, …, an )
Given condition
P ∈ Z ({ f1, …, fm }) → g ( a1, …, an ) = 0
Conclusion is next.
There exist natural number l and elements h1, …, hm in k [ X1, …, Xn ].
gl = ∑m i=1 hifi
Tokyo October 1, 2007
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