Bourbaki

Theorem 2 (Principle of extension of algebraic identities)

Tag 06B6, permanent. Cite it as [Bourbaki, Tag 06B6].

Algebra, chapter IV, § 2. Zeros of polynomials

— Suppose that A is an infinite integral domain. Let g1,…,gm,fg_1, \ldots, g_m, f be elements of A[(Xi)i∈I]A[(X_i)_{i \in I}] and assume the following hypotheses:

a) g1≠0,…,gm≠0g_1 \neq 0, \ldots, g_m \neq 0; b) for all x∈A1x \in \mathbf{A}^1 such that g1(x)≠0,…,gm(x)≠0g_1(x) \neq 0, \ldots, g_m(x) \neq 0, we have f(x)=0f(x) = 0. Then f=0f = 0.

For if f≠0f \neq 0, we have fg1…gm≠0fg_1 \ldots g_m \neq 0 (IV, p. 9, Prop. 8), hence there exists x∈A1x \in \mathbf{A}^1 such that f(x)g1(x)…gm(x)≠0f(x)g_1(x) \ldots g_m(x) \neq 0 (IV, p. 18, Cor. 2), which contradicts the hypothesis.

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

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Transcribed from the printed edition, 2003, Springer. Not translated.