Derivations annihilated by powers of the augmentation ideal Consider an augmented commutative ring $R$, with augmentation ideal $\varpi$. Let $\delta$ be a derivation of $R$. The example I have in mind is $R=\mathbb F_p[x]/(x^{p^i})$ and $\delta=d/dx$, though I would like statements as general as possible.
I would like to know whether $\varpi^m f=0$ implies $\varpi^{m+1}\delta(f)=0$.
My intuition is that $\delta$ maps $\varpi^m$ to $\varpi^{m-1}$, so somehow "divides by $\varpi$". I have found neither proof of the above statement nor counter-example.
Bonus questions: is the property true for local rings ($\varpi$ is the unique maximal ideal)? graded local rings?
Many thanks in advance!
 A: For the sake of completeness, here is the proof I suggested in the comments,
in some more detail.
Lemma 1. Let $\mathbf{k}$ be a commutative ring. Let $A$ be a $\mathbf{k}
$-algebra. Let $I$ be a two-sided ideal of $A$. Let $f:A\rightarrow A$ be a
derivation. Then, $f\left(  I^{n+1}\right)  \subseteq I^{n}$ for every
$n\in\mathbb{N}$.
Lemma 1 is Proposition 1.21 in my Collected trivialities on algebra
derivations, where I prove it by straightforward induction on $n$.
Now, your claim follows from the following fact:
Corollary 2. Let $\mathbf{k}$ be a commutative ring. Let $A$ be a
$\mathbf{k}$-algebra. Let $I$ be a two-sided ideal of $A$. Let $f:A\rightarrow
A$ be a derivation. Let $a\in A$ and $n\in\mathbb{N}$ be such that $I^{n}a=0$.
Then, $I^{n+1}f\left(  a\right)  =0$.
Proof of Corollary 2. We must prove that $I^{n+1}f\left(  a\right)  =0$. In
other words, we must prove that $gf\left(  a\right)  =0$ for every $g\in
I^{n+1}$. So let us fix $g\in I^{n+1}$. We have $f\left(  \underbrace{g}_{\in
I^{n+1}}\right)  \in f\left(  I^{n+1}\right)  \subseteq I^{n}$ (by Lemma 1),
and thus $f\left(  g\right)  a\in I^{n}a=0$. In other words, $f\left(
g\right)  a=0$.
But $\underbrace{g}_{\in I^{n+1}=II^{n}}a\in I\underbrace{I^{n}a}_{=0}=0$, so
that $ga=0$.
Since $f$ is a derivation, we have $f\left(  ga\right)  =gf\left(  a\right)
+\underbrace{f\left(  g\right)  a}_{=0}=gf\left(  a\right)  $. Hence,
$gf\left(  a\right)  =f\left(  \underbrace{ga}_{=0}\right)  =f\left(
0\right)  =0$. This is exactly what we wanted to prove. Thus, Corollary 2 holds.
