Bourbaki

Lemma 2

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

Algebra, chapter IV, § 5. Symmetric tensors and polynomial mappings

— Let N be an A-module, n an integer ≥0\geq 0 and

f=m0+m1X+⋯+mnXn∈N[X]. f = m_0 + m_1 X + \cdots + m_n X^n \in N[X].

Suppose that there exist α0,α1,…,αr∈A\alpha_0, \alpha_1, \ldots, \alpha_r \in A such that f(α0)=⋯=f(αr)=0f(\alpha_0) = \cdots = f(\alpha_r) = 0, and such that for i≠ji \neq j the homothety of ratio αi−αj\alpha_i - \alpha_j in N is injective, then f=0f = 0.

(This lemma generalizes the Cor. of IV, p. 16.)

The lemma clearly holds for n=0n = 0; we shall prove it by induction on nn. We have

f(X)=f(X)−f(α0)=∑i=1nmi(Xi−α0i)=(X−α0)g(X) f(X) = f(X) - f(\alpha_0) = \sum_{i=1}^n m_i (X^i - \alpha_0^i) = (X - \alpha_0) g(X)

where g is an element of N[X] of the form m0′+m1′X+…+mn−1′Xn−1m_0' + m_1' X + \ldots + m_{n-1}' X^{n-1}. The hypotheses of the lemma imply that g(α1)=⋯=g(αr)=0g(\alpha_1) = \cdots = g(\alpha_r) = 0, hence g=0g = 0 by the induction hypothesis, and so f=0f = 0.

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.