Bourbaki

Exercise 5, § 6

Commutative Algebra, VIII, p. 100

Let ρ:A→B\rho : A \to B be an injective homomorphism of noetherian rings making BB a finitely generated AA-algebra. We call degree of transcendence of BB over AA, and denote it by dA(B)d_A(B), the largest integer dd such that there exists an injective homomorphism of AA-algebras b:A[X1,...,Xd]→Bb : A[X_1, ..., X_d] \to B. (a) Prove that if σ:B→C\sigma : B \to C satisfies the same hypotheses as ρ\rho, one has the inequality

dA(C)⩾dA(B)+dB(C). d_A(C) \geqslant d_A(B) + d_B(C) .
(b) Prove that if AA and BB are integral, dA(B)d_A(B) is equal to the degree of transcendence of the field of fractions of BB over that of AA. Moreover if in aa one supposes CC integral, one has the equality dA(C)=dA(B)+dB(C)d_A(C) = d_A(B) + d_B(C). (c) Prove that if BB is a subalgebra of A[v,v−1]A[v, v^{-1}] containing A[v]A[v], where vv is an indeterminate, one has dA(B)=1d_A(B) = 1.

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