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I'm very certain that this is nothing new under the sun but this is something I wondered yesterday and couldn't find resources online. Let $R$ be a (commutative, unital) ring and let $I$ be an ideal in $R$. I say that a function $f:R\rightarrow R$ is $I$-differentiable if for all $x\in R$ we have $f(x+I)-f(x)\subset I$. The set $D(I)$ of all such functions is easily seen to be a subring of the ring $\mathrm{Fun}(R,R)$ of all $R$-valued functions on $R$ that contains all polynomial functions, all $I$-invariant functions and is stable und composition. On the other hand, in many cases such as $R=\mathbb{Z}$ and $I=n\mathbb{Z}$, $D(I)$ is easily seen to be not the whole ring (e.g. no nontrivial function with finite support is $n\mathbb{Z}$-differentiable). So my question is: What precisely are all the $n\mathbb{Z}$-differentiable $\mathbb{Z}$-valued functions on $\mathbb{Z}$? I'm suspecting $D(n\mathbb{Z})$ to actually be the smallest subring that contains all polynomial functions, all $I$-invariant functions and is stable under composition, but would be delighted if it was actually larger.

Thank you so much for your help and for providing further literature!

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    $\begingroup$ The $I$-differentiable functions are exactly the ones which induce a well-defined function $R/I \to R/I$ $\endgroup$
    – math54321
    Sep 6, 2023 at 20:08
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    $\begingroup$ So e.g. for $I = n \mathbb{Z}$, you can think of an $n \mathbb{Z}$-differentiable function just as a collection of $n$ endofunctions of $\mathbb{Z}$ $\endgroup$
    – math54321
    Sep 6, 2023 at 20:22
  • $\begingroup$ Ah, okay, that makes sense; thank you! $\endgroup$
    – j grk
    Sep 7, 2023 at 7:40

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