Bourbaki

Exercise 22, § 1

Commutative Algebra, IX, p. 47

Let AA be a Z(p)\mathbf{Z}_{(p)}-algebra. We denote by DA\mathscr{D}_A the algebra over W(A)W(A) generated by two indeterminates FF and VV subject to the relations

Fa=(Fa).Ffor a∈W(A)aV=V(Fa)for a∈W(A)FV=pVaF=Vafor a∈W(A), \begin{align*} Fa &= (Fa).F & \text{for } a \in W(A) \\ aV &= V(Fa) & \text{for } a \in W(A) \\ FV &= p \\ VaF &= Va & \text{for } a \in W(A), \end{align*}
where FF and VV denote respectively the Frobenius and shift homomorphisms of W(A)W(A). a) Let BB be an AA-algebra. Making FF and VV act on W(B)W(B) by the Frobenius and shift homomorphisms of W(B)W(B), we endow W(B)W(B) with a DA\mathscr{D}_A-module structure. Analogously, for each integer n⩾1n \geqslant 1, Wn(B)W_n(B) is a DA\mathscr{D}_A-module. b) For every element xx of DA\mathscr{D}_A, there exists a finite-support family (ai)i∈Z(a_i)_{i \in \mathbf{Z}} of elements of W(A)W(A), characterized by the equality
x=∑n⩾1a−nFn+a0+∑n⩾1Vnan. x = \sum_{n \geqslant 1} a_{-n} F^n + a_0 + \sum_{n \geqslant 1} V^n a_n.
c) Suppose W(A)W(A) is an integral domain. Let x∈DAx \in \mathscr{D}_A. We denote by δ(x)\delta(x) the greatest integer nn such that ana_n is nonzero. If xx and yy are two nonzero elements of DA\mathscr{D}_A, one has δ(xy)=δ(x)+δ(y)\delta(xy) = \delta(x) + \delta(y). Deduce that DA\mathscr{D}_A is an integral domain. d) If AA is a perfect ring of characteristic pp, one can replace the condition VaF=VaVaF = Va for a∈W(A)a \in W(A) by the condition VF=pVF = p. The left ideal DAV\mathscr{D}_A V of DA\mathscr{D}_A is two-sided. e) If kk is a perfect field of characteristic pp, Dk\mathscr{D}_k is noetherian. (Consider Dk\mathscr{D}_k as quotient of the ring generated over W(k)W(k) by two indeterminates XX and YY subject to the relations XY=YXXY = YX and Xa=(Fa)X,aY=Y(Fa)Xa = (Fa)X, aY = Y(Fa) for a∈W(k)a \in W(k). Then apply III, § 2, No. 8, corollary 2 to Theorem 1 and Exercise 10.)

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