Bourbaki

Exercise 3, § 1

Algebra, V

Let AA be a commutative ring. For every polynomial f∈A[X]f \in A[X] we write in the polynomial ring A[X,Y]A[X, Y] of polynomials in two indeterminates over AA,

f(X+Y)=∑m=0∞Δmf(X)Ym, f(X + Y) = \sum_{m=0}^{\infty} \Delta_m f(X) Y^m,

where Δ\Delta is thus an AA-linear mapping of A[X]A[X] into itself. a) If ZZ is an indeterminate and τZ\tau_Z the AA-linear mapping of A[X]A[X] into A[X,Z]A[X, Z] such that τZf=f(X+Z)\tau_Z f = f(X + Z), show that τZ∘Δ=Δ∘τZ\tau_Z \circ \Delta = \Delta \circ \tau_Z (where on the right Δ\Delta is the mapping defined above in the ring B[X]B[X], where B=A[Z]B = A[Z]). Deduce that in A[X]A[X] we have

ΔmΔn=ΔnΔm=(m+nm)Δm+n \Delta_m \Delta_n = \Delta_n \Delta_m = \binom{m+n}{m} \Delta_{m+n}

for integers m,n≥0m, n \geq 0 (cf. IV, p. 8). b) Suppose that AA is a ring of characteristic p>0p > 0 and for each integer k≥0k \geq 0 put Dk=ΔpkD_k = \Delta_{p^k}. Show that if f∈A[X]f \in A[X] and g∈A[Xpk]g \in A[X^{p^k}] then Dk(fg)=g⋅Dkf+f⋅DkgD_k(fg) = g \cdot D_k f + f \cdot D_k g. Every polynomial f∈A[X]f \in A[X] can be uniquely written in the form f(X)=∑m=0∞Xmpkgm(X)f(X) = \sum_{m=0}^{\infty} X^{mp^k} g_m(X) where deg⁡(gm)<pk\deg(g_m) < p^k; show that

Dkf(X)=∑m=0∞mX(m−1)pkgm(X) D_k f(X) = \sum_{m=0}^{\infty} m X^{(m-1)p^k} g_m(X)

and deduce that Dkp=0D_k^p = 0.

c) For every integer m>0m > 0 let m=α0+α1p+…+αkpkm = \alpha_0 + \alpha_1 p + \ldots + \alpha_k p^k be the expansion of mm in base pp (Set Theory, III, p. 177), where thus 0≤a,<p0 \leq a, < p for 0≤j≤k0 \leq j \leq k; suppose that aj≠0a_j \neq 0. Show that

Δm=(1α0!D0α0)(1α1!D1α1)…(1αk!Dkαk). \Delta_m = \left( \frac{1}{\alpha_0!} D_0^{\alpha_0} \right) \left( \frac{1}{\alpha_1!} D_1^{\alpha_1} \right) \ldots \left( \frac{1}{\alpha_k!} D_k^{\alpha_k} \right).

(If Δm′\Delta'_m is the right-hand side of this relation, note that if we introduce the expansion f(X+Y)=∑q=0∞fq(Y)Xqf(X + Y) = \sum_{q=0}^{\infty} f_q(Y) X^q, we have Δm′f(X+Y)=∑q=0∞fq(Y)Δm′(Xq)\Delta'_m f(X + Y) = \sum_{q=0}^{\infty} f_q(Y) \Delta'_m(X^q) and deduce that this may be written Δm′f(X+Y)=fm(X)+Yg(X,Y)\Delta'_m f(X + Y) = f_m(X) + Yg(X, Y), where gg is a polynomial.)

Solution

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