# Derivation of formal power series

The basic idea of this question is to see if there is any other derivations than 'formal derivations'.

Let $$\mathbb{K}$$ be a field. Given a commutative $$\mathbb{K}$$-algebra $$A$$, a derivation of $$A$$ is a $$\mathbb{K}$$-linear map $$D:A\rightarrow A$$ satisfying $$D(ab)=D(a)b+ aD(b)$$. Consider the case when $$\mathbb{K}=\mathbb{R}$$ and the algebra $$A$$ is the formal power series $$\mathbb{R}[[x]]$$. One derivation of $$\mathbb{R}[[x]]$$ is $$p(x)\frac{\partial}{\partial x}$$, defined formally as $$p(x) \frac{\partial}{\partial x} \left(\sum_{n=0}^\infty c_n x^n \right)=p(x)\cdot \left(\sum_{n=0}^\infty n\cdot c_n x^{n-1}\right)$$ for any $$p(x)\in \mathbb{R}[[x]]$$. My question is that, are there any derivations that is not of the form $$p(x)\frac{\partial}{\partial x}$$ as above? If yes, is there any reference/proof/example for existence of such derivations?

This question arose when I was thinking of vector fields of a $$\mathbb{Z}$$-graded manifold as a derivation of smooth functions. With my notion of graded manifolds, smooth functions are formal power series of virtual coordinates.

When I was thinking about it, since $$\mathbb{R}[[x]]$$ is not generated by $$\{x\}$$ as an algebra, there should be some other derivations but I could not make it clear. Any help to figure it out would be greatly appreciated.

• When $K$ has char. zero, there are nonzero derivations of $K((x))$ that vanish on $K[x]$. But I don't know it it can be arranged to map $K[[x]]$ into itself.
– YCor
Commented Feb 3, 2021 at 8:27

Every $$\mathbb{K}$$-derivation $$D$$ of $$\mathbb{K}[[x_1,\dots,x_n]]$$ has the standard form $$D(f)=\sum_{i=1}^n p_i \frac{\partial f}{\partial x_i},$$ where of course $$p_i=D(x_i)$$.
Indeed, let $$\mathfrak{m}$$ be the maximal ideal of $$\mathbb{K}[[x_1,\dots,x_n]]$$: then clearly $$D(\mathfrak{m}^{N+1})\subset \mathfrak{m}^N$$ for all $$N\geq0$$, so $$D$$ is continuous for the $$\mathfrak{m}$$-adic topology. The claim follows because the formula holds when $$f$$ is a polynomial, $$\mathbb{K}[x_1,\dots,x_n]$$ is dense in $$\mathbb{K}[[x_1,\dots,x_n]]$$, and $$\mathbb{K}[[x_1,\dots,x_n]]$$ is Hausdorff.
• I wonder what's the answer for derivations of the ring $R$ of germs of smooth functions near the origin in $\mathbb{R}^n$. This maps surjectively to $\mathbb{R}[\![x_1,\dotsc,x_n]\!]$ with kernel $\mathfrak{m}^\infty\neq 0$. Commented Feb 3, 2021 at 9:43
• Thanks for your answer. I have been thinking about it but I was not sure why it $D(\mathfrak{m}^{N+1}) \subset \mathfrak{m}^N$ has to be satisfied. With this answer, my question is, in some sense, equivalent to understand why it HAS TO satisfy that continuity condition.
• @sock $\mathfrak m^{N+1}$ is generated by elements of the form $\prod_{i=1}^{N+1} f_i$ with $f_i \in \mathfrak m$. Now apply $D(ab)=a D(b)+ D(a) b$ $N$ times.... Commented Feb 5, 2021 at 2:56
• @sock More generally, the same argument shows that for any ring $R$, ideal $I$, and $R$-module $M$, every derivation $D:R\to M$ is $I$-adically continuous (more precisely, $D(I^{N+1})\subset I^N M$). This does not help very much if $M$ is not $I$-adically separated: for instance, as YCor observed, there are nonzero derivations on $\mathbb{K}[[x]]$ with values in $\mathbb{K}(\!x)\!)$, that vanish on $\mathbb{K}[x]$. Commented Feb 5, 2021 at 10:02