Bourbaki

Corollary 3

Tag 04RB, permanent. Cite it as [Bourbaki, Tag 04RB].

Commutative Algebra, chapter V, § 3. Finitely generated algebras over a field

Let A be an integral domain, B a finitely generated A-algebra and b an element of B such that zbn≠0zb^n \neq 0 for all z≠0z \neq 0 in A and every integer n>0n > 0. Let φ:A→B\varphi : A \to B be the canonical homomorphism; there exists a≠0a \neq 0 in A such that, for every homomorphism f from A to an algebraically closed field L such that f(a)≠0f(a) \neq 0, there exists a homomorphism g from B to L such that g(b)≠0g(b) \neq 0 and f=g∘φf = g \circ \varphi.

The hypothesis on b implies that, if h is the canonical homomorphism x↦x/1x \mapsto x/1 of B to B[b−1]B[b^{-1}], the homomorphism h∘φh \circ \varphi of A to B[b−1]B[b^{-1}] is injective. By Corollary 1 there therefore exist an element a≠0a \neq 0 of A and a subring B′B' of B[b−1]B[b^{-1}] such that B[b−1,a−1]B[b^{-1}, a^{-1}] is integral over B′[a−1]B'[a^{-1}] and B′B' is isomorphic to a polynomial algebra A[Y1,…,Yn]A[Y_1, \ldots, Y_n]. Let f be a homomorphism from A to an algebraically closed field L such that f(a)≠0f(a) \neq 0; there exists a homomorphism from A[Y1,…,Yn]A[Y_1, \ldots, Y_n] to L extending f and hence there exists a homomorphism f′f' from B′B' to L extending f. As f′(a)≠0f'(a) \neq 0 in L, there exists a homomorphism f′′f'' from B′[a−1]B'[a^{-1}] to L such that

f′′(x/an)=f′(x)⋅(f(a))−n f''(x/a^n) = f'(x) \cdot (f(a))^{-n}

for all x∈B′x \in B' and all n>0n > 0 (Chapter II, §2, no. 1, Proposition 1). Finally, as B[b−1,a−1]B[b^{-1}, a^{-1}] is integral over B′[a−1]B'[a^{-1}], there exists a homomorphism f′′′f''' from B[b−1,a−1]B[b^{-1}, a^{-1}] to L extending f′′f'' (§2, no. 1, Corollary 4 to Theorem 1). If j:x↦x/1j : x \mapsto x/1 is the canonical homomorphism from B to B[b−1,a−1]B[b^{-1}, a^{-1}], g=f′′′∘jg = f''' \circ j solves the problem for j(b)j(b) is invertible in B[b−1,a−1]B[b^{-1}, a^{-1}] and hence f′′′(j(b))≠0f'''(j(b)) \neq 0 in L.

Note that, if B is assumed to be an integral domain and A⊂BA \subset B in Corollary 3, the hypothesis on b is equivalent to b≠0b \neq 0.

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

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