Noether normalization lemma

This is a standard result in commutative algebra (cf. for example, Zariski-Samuel, vol. 2, Ch. 7, 7). However there is a fairly straightforward and geometric proof, using only the Noether normalization lemma and based on the Norm, so I think it is worthwhile proving the algebraic version too. This proof is due to J. Tate. 

Let d i 0 by Noether s normalization lemma there exist bv. ..,Dd algebraically independent in b such that B is integral over Klbv bdJ. it follows easily that there exists c c A such that b1. .., bd c Bc and Bt is integral over Aclb,.. in particular Be is an Ac[bvJ ]-module of finite type. Therefore It is possible to find an exact sequence of Ajb. j-modules... 

Noether s normalization lemma, 2-6 Non-degenerate curve, 11 -5 Non-special curve, 11 -6... 

Noether s Normalization Lemma. Let R be an integral domain, finitely generated over a field k. If R has transcendence degree n over k, then there exist elements x, ..., xn E R, algebraically independent over k, such that R is integrally dependent on the subring k[xi,..., xn] generated by the x s. 

Geometric form of Noether s normalization lemma. Let X be an affine variety of dimension n. Then there exists a finite surjective morphism n ... 

Proof (Based on suggestions of G. Stolzenberg). Since the theorem is a local statement (in the Zariski topology), we can suppose that X is affine. By Noether s normalization lemma (geometric form), there exists a finite surjective morphism... 

We also need a new form of Noether s Normalization lemma ... 

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