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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!

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  • $\begingroup$ To the non-bonus question: The Leibniz identity (generalized to multiple factors) yields $\delta\left(i_1 i_2 \cdots i_{m+1} f\right) = \sum_{k=1}^{m+1} i_1 i_2 \cdots i_{k-1} \delta\left(i_k\right) i_{k+1} i_{k+2} \cdots i_{m+1} f + i_1 i_2 \cdots i_{m+1} \delta\left(f\right)$ for all $i_1, i_2, \ldots, i_{m+1} \in \varpi$. Argue that both the left hand side and each addend of the sum on the right hand side are zero. Conclude that the second addend of the right hand side is zero, and thus $\varpi^{m+1} \delta\left(f\right) = 0$. $\endgroup$
    – darij grinberg
    Commented Nov 4, 2015 at 21:27
  • $\begingroup$ (Notice that $R$ needs not be commutative here, and $\varpi$ can be any two-sided ideal, not necessarily the augmentation ideal.) $\endgroup$
    – darij grinberg
    Commented Nov 4, 2015 at 21:28
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    $\begingroup$ Actually there is an even simpler argument: First prove $\delta\left(I^{m+1}\right) \subseteq I^m$ (for instance, by induction over $m$, or by the generalized Leibniz identity as above); thus, $\delta\left(I^{m+1}\right) f = 0$. Then, argue using $\delta\left(if\right) = \delta\left(i\right) f + i \delta\left(f\right)$ for $i \in I^{m+1}$. $\endgroup$
    – darij grinberg
    Commented Nov 4, 2015 at 21:30
  • $\begingroup$ Thanks a lot! Would you mind typing your answer as an answer, so you would get appropriate credit? $\endgroup$
    – grok
    Commented Nov 4, 2015 at 22:41

1 Answer 1

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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.

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