< 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 ].
[Proof]
R is integral over k.
Elements of R exist. y1, …, ym
R [y1, …, ym ]
Variable Y1, …, Ym
Surjective homomorphism φ: k [Y1, …, Ym ] → k [ y1, …, ym ] Yi ↦ yi
Ker φ = p
From homomorphism theorem*
Isomorphism k [Y1, …, Ym ] / p ≅ k [ y1, …, ym ]
R is integral domain.
p is prime ideal.
From definition of transcendence degree**
m ≥ n
When m = n, y1, …, ym is algebraically independent.
x1 = y1 , …, xm = ym
When m ≻ n , conditions are below.
Subring S is in R, S is generated by elements m-1 over k. And integral over S in R exists.***
[Notes]
<Homomorphism theorem*>
Ring R1, R2
Homomorphism of ring φ: R1 → R2
φ leads injective homomorphism φ- from quotient ring R1 / Ker φ to R2.
When φ is surjective, φ- is isomorphism.
<Definition of transcendence degree**>
Number of transcendental basis in field extension K ⊃ k.
Transcendental basis is subset S ⊂ K. S satisfies the next.
(1) S is algebraically independent over k.
(2) K ⊃ k ( S ) is algebraic extension.
<Subring S is in R, S is generated by elements m-1 over k. And integral over S in R exists.***>
[Proof]
Transcendental degree in R over k n
n < m
y1, …, ym is not algebraically independent over k.
0 ≠polynomial f (Y1, …, Ym ) ∈ k [Y1, …, Ym ]
f (y1, …, ym ) = 0
Arbitrary natural numbers r2, …, rm
Z2 = Y2 –Y1r2, …, Zm = Ym – Y1rm
z2 = y2 –y1r2, …, zm = ym –y1rm
f ( y1, z2 + y1r2, …, zm + y1rm ) = 0
S = [z2, …, zm ]
Monomial in f (y1, …, ym ) aY1l1…Ymlm
aY1l1…Ymlm = aY1l1 ( Z2 + Y1r2 ) l1… (Zm + Y1 rm) l1
Maximum dimensional term
aY1l1+r2l2+…+rmlm
Tokyo October 2, 2007
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