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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.

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  • $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Feb 3, 2021 at 8:27

1 Answer 1

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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.

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    $\begingroup$ 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 $\mathfrak{m}^\infty\neq 0$. $\endgroup$ Commented Feb 3, 2021 at 9:43
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    $\begingroup$ 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. $\endgroup$
    – sock
    Commented Feb 5, 2021 at 2:49
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    $\begingroup$ @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.... $\endgroup$
    – Will Sawin
    Commented Feb 5, 2021 at 2:56
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    $\begingroup$ @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 $\mathbb{K}[x]$. $\endgroup$ Commented Feb 5, 2021 at 10:02
  • $\begingroup$ A more detailed proof can be found under theorem 1.5.2 in page 12 of www-users.mat.umk.pl/~anow/ps-dvi/pol-der.pdf. $\endgroup$
    – exfret
    Commented Aug 5, 2021 at 13:06

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