Proposition 5

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

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

Let $A$ be a ring of characteristic $p$ (A, V, p. 2). Whatever the elements $a$ and $b$ of $W(A)$ , and the positive integers $m, n$ , one has, if $a = (a_n)_{n \in \mathbf{N}}$ ,

(51) \quad F(a) = (a_n^p)_{n \in \mathbf{N}}

(52) \quad p.a = VF(a) = FV(a) = (0, a_0^p, a_1^p, ...)

(53) \quad V^m(a) \times V^n(b) = V^{m+n}(F^n(a) \times F^m(b)) .

Formula (51) results from Example 4 of No. 3. One immediately deduces from this the equality

VF(a) = FV(a) = (0, a_0^p, a_1^p, ...) ,

and the equality $p.a = FV(a)$ has been proved (No. 5, prop. 3), whence (52).

Let us prove (53). According to formula (37) (where one substitutes $V^n(b)$ for $b$ ), one has

(54) \quad V^m(a) \times V^n(b) = V^m(a \times F^m(V^n(b))) .

From formula (37), one also deduces

(55) \quad V^n(F^m(b)) \times a = V^n(F^m(b) \times F^n(a)) .

Formula (53) then results from (54) and (55) and from the relation $F^m \circ V^n = V^n \circ F^m$ , itself a consequence of (51).

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

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