Bourbaki

Lemma 2

Tag 0A6R, permanent. Cite it as [Bourbaki, Tag 0A6R].

Lie Groups and Lie Algebras, chapter II, § 8. Convergence of the Hausdorff series (ultrametric case)

v(n)⩽v(n!)⩽(n−1)θv(n) \leqslant v(n!) \leqslant (n - 1)\theta and v(n)⩽(log⁡n)/(log⁡p)v(n) \leqslant (\log n)/(\log p) for every integer n⩾1n \geqslant 1. v(n!)=vp(n!)=(n−S(n))θ⩽(n−1)θv(n!) = v_p(n!) = (n - S(n))\theta \leqslant (n - 1)\theta by Lemma 1. On the other hand, n⩾pv(n)n \geqslant p^{v(n)}, whence v(n)⩽(log⁡n)/(log⁡p)v(n) \leqslant (\log n)/(\log p). Let I={U,V}I = \{ U, V \} be a set of two elements and let

H=∑r,s⩾0Hr,s(U,V)∈L^Q(I) \mathbf{H} = \sum_{r,s \geqslant 0} H_{r,s}(U, V) \in \hat{\mathbf{L}}_{\mathbf{Q}}(I)
be the Hausdorff series (§ 6, no. 4, Definition 1). Let Z(p)\mathbf{Z}_{(p)} be the local ring of Z\mathbf{Z} relative to the prime ideal (p)(p) and (eb)b∈B(e_b)_{b \in B} a basis of LZ(p)(I)\mathbf{L}_{\mathbf{Z}_{(p)}}(I) over Z\mathbf{Z} (§ 2, no. 11, Theorem 1). It is also a basis of LQ(I)\mathbf{L}_{\mathbf{Q}}(I) over Q\mathbf{Q}.

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 volume. Not translated.