Bourbaki

Example 4

Tag 0CU3, permanent. Cite it as [Bourbaki, Tag 0CU3].

Commutative Algebra, chapter IX, § 1. Vecteurs de Witt

We have

F0=X0p+pX1 F_0 = X_0^p + p X_1
F1=X1p+pX2−∑i=0p−1(pi)pp−i−1X0piX1p−i. F_1 = X_1^p + p X_2 - \sum_{i=0}^{p-1} \binom{p}{i} p^{p-i-1} X_0^{pi} X_1^{p-i}.

Since we have Φn(F0,...,Fn)≡Φn(X0p,...,Xnp)mod  .pn+1\Phi_n(F_0, ..., F_n) \equiv \Phi_n(X_0^p, ..., X_n^p) \mod. p^{n+1}. A for every n∈Nn \in \mathbf{N} (formulas (2) and (15)), it follows from prop. 1, b) that we have Fn≡Xnpmod  .pF_n \equiv X_n^p \mod. p. A for every n∈Nn \in \mathbf{N}.

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

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