If $F:V\to W$ is a smooth at $a\in V$ function between finite-dimensional vector spaces over $\mathbb{R}$, then we have $$ F(x) = \sum_{k=0}^N\frac{1}{k!}(D^kF)(a)[(x-a)^{\otimes k}]+\text{remainder}, $$ where $(D^kF)(a):\mathrm{Sym}^k(V)\to W$ are $\mathbb{R}$-linear maps. Algebraically, this is a statement about writing tuples of homogeneous polynomials of same degree as a multilinear map, whose coefficients are the respective mixed partial derivatives (at $a$).

Now, let $R$ be a commutative ring, in which $k!$ are invertible. I would like to be able to reference the analogous statement about the formal Taylor expansion at $0$ of tuples of formal power series $F:=(F_1,\dots,F_n)\in R[[X_1,\dots,X_m]]^n$ (or equivalently, tuples of polynomials). This seems like a completely standard fact, but I have not been able to pinpoint it in the literature.

I am also curious, is there a workaround statement in the case that $1/k!$ don't exist in $R$ (e.g. if $R$ has positive characteristic)?

  • $\begingroup$ Regarding your last question, you should treat $(1/k!)D^k$ as a symbol. The operator it defines often makes sense even if $1/k!$ doesn't. $\endgroup$ – Donu Arapura Dec 11 '17 at 13:09
  • $\begingroup$ @DonuArapura: I see, so you mean simply compute with $D^k$ as if $\mathrm{char}(R)=0$ and then formally cancel out the factorials (for example instead of getting $0$ in positive characteristic) like a physicist? :-) After much search, it seems the right keyword is Hasse derivatives: math.fontein.de/2009/10/02/… Apparently they work for any commutative $R$, and a Taylor-like expansion holds. $\endgroup$ – M.G. Dec 11 '17 at 17:09

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