Bourbaki

Corollary 1

Tag 0CRL, permanent. Cite it as [Bourbaki, Tag 0CRL].

Commutative Algebra, chapter VIII, § 6. Dimension des anneaux gradués

Suppose that H0H_0 is a field and that HH is a finitely generated H0H_0-algebra. Put n=dim⁡(H)n = \dim(H). There exist homogeneous elements a1,...,ana_1, ..., a_n of HH, all of the same degree >0> 0, such that the H0H_0-homomorphism φ:H0[X1,...,Xn]→H\varphi : H_0[X_1, ..., X_n] \to H defined by φ(Xi)=ai,i=1,...,n\varphi(X_i) = a_i, i = 1, ..., n, is injective and makes HH into a finite H0[X1,...,Xn]H_0[X_1, ..., X_n]-algebra. If HH is generated by H1H_1 as an H0H_0-algebra and if H0H_0 is infinite, the aia_i may be supposed to be of degree 1.

By Prop. 6 there exists an H0H_0-homomorphism φ\varphi of the indicated form which makes HH into a finite H0[X1,...,Xn]H_0[X_1, ..., X_n]-algebra. By Theorem 1 of § 2, No. 3, one then has

dim⁡(H0[X1,...,Xn]/(Ker φ))=dim⁡(H); \dim(H_0[X_1, ..., X_n]/(\mathrm{Ker}\ \varphi)) = \dim(H);
since one has
dim⁡(H)=n=dim⁡(H0[X1,...,Xn]), \dim(H) = n = \dim(H_0[X_1, ..., X_n]),
and since H0[X1,...,Xn]H_0[X_1, ..., X_n] is an integral domain, this implies Ker φ={0}\mathrm{Ker}\ \varphi = \{0\}.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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