Bourbaki

Proposition 3

Tag 06C2, permanent. Cite it as [Bourbaki, Tag 06C2].

Algebra, chapter IV, § 4. Formal power series

Let u=∑aνXν∈A[[I]]u = \sum a_\nu X^\nu \in \mathbf{A}[[\mathbf{I}]] and mm an integer >0> 0. For every n∈Nn \in \mathbf{N} let (αν,n)ν∈N(I)(\alpha_{\nu, n})_{\nu \in \mathbf{N}^{(I)}} be the family of coefficients of unu^n. If α0m=0\alpha_0^m = 0, then αν,n=0\alpha_{\nu, n} = 0 for n≥∣ν∣+mn \geq |\nu| + m.

Let ν∈N(I)\nu \in \mathbf{N}^{(I)} and n∈Nn \in \mathbf{N}. We have

αν,n=∑ν(1)+⋯+ν(n)=ναν(1)⋯αν(n). \alpha_{\nu,n} = \sum_{\nu(1) + \cdots + \nu(n) = \nu} \alpha_{\nu(1)} \cdots \alpha_{\nu(n)} .

If n≥∣ν∣+mn \geq |\nu| + m and ν(1)+⋯+ν(n)=v\nu(1) + \cdots + \nu(n) = v, we have ∣ν(1)∣+⋯+∣ν(n)∣≤n−m|\nu(1)| + \cdots + |\nu(n)| \leq n - m. We thus have v(r)=0v(r) = 0 and so αv(r)=ar\alpha_{v(r)} = a_r, for at least mm distinct values of rr; it follows that αν(1)⋯αν(n)=0\alpha_{\nu(1)} \cdots \alpha_{\nu(n)} = 0, whence the result.

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

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Cites

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