Fix a commutative ring $A$, a derivation $\partial_A$ on $A$, and a multiplicative set $S$ in $A$. Then let $B = S^{-1}A$, $f : A \to B$ be the natural map, and $\partial_B$ be the natural extension of $\partial_A$ to $B$. We of course always have $$f(\ker \partial_A) \subseteq \ker \partial_B.$$ Broadly speaking, under what situations can we guarantee that this an equality?
Here are some observations.
- If $A$ is a field, there's of course nothing to do.
- If $A$ is of positive characteristic $p$, there will likely be problems (even if $A$ is a domain). For example, if we take $A = \mathbb{F}_p[t]$ with $\partial_A$ being differentiation with respect to $t$, then $\ker \partial_A = \mathbb{F}_p[t^p]$. If we let $B = \mathrm{Frac}(A) = \mathbb{F}_p(t)$, then $\ker \partial_B = \mathbb{F}_p(t^p)$, which is of course strictly larger.
- If $A$ is not a domain, there will likely be problems (even in characteristic 0). For example, let $A$ be the ring of smooth functions $\mathbb{R} \to \mathbb{R}$ with $\partial_A$ being usual differentiation. Then $\ker \partial_A = \mathbb{R}$. Moreover, we know that there exist $f, g \in A$ such that the Wronskian $W(f,g) = fg' - f'g = 0$ despite the fact that $f$ and $g$ are $\mathbb{R}$-linearly independent. (An explicit example of such a pair can be found in Bôcher's note from 1900 "On linear dependence of functions of one variable.") So, if we let $S$ be the multiplicative subset generated by $g$, note that $$\partial_B(f/g) = -W(f,g)/g^2 = 0,$$ so $f/g \in \ker \partial_B$ but it is not in $f(\ker \partial_A) = \mathbb{R}$ since $f$ and $g$ are $\mathbb{R}$-linearly independent.
- The subring of constants in a differential field is itself necessarily a field. So, if $A$ is a domain and $B = \mathrm{Frac}(A)$, we know that $\ker \partial_B$ must be a field, so there will be problems if $\ker \partial_A$ is not already a field. For example, let $A = \mathbb{Q}[x,y]$ and $\partial_A$ is differentiation with respect to $y$, so that $\ker \partial_A = \mathbb{Q}[x]$. Taking $B = \mathrm{Frac}(A) = \mathbb{Q}(x,y)$, we get $\ker \partial_B = \mathbb{Q}(x)$.
So... putting together these observations...
- Do we have equality when $A$ is a domain such that $\ker \partial_A$ is a field of characteristic 0? (When $A$ is a domain, we may as well assume that $B = \mathrm{Frac}(A)$.)
- If the above conditions aren't sufficient, are there any further (non-tautological) conditions we can impose on $A$ and $\partial_A$ to guarantee equality?
In addition to these specific questions, I'd love to hear anything people have to say about the broad question above also (whether that be in the form of general statements or fun counterexamples).
Proposition 3.1 in https://arxiv.org/pdf/1304.7663v2.pdf shows that the desired equality is true when $A$ is a differentially simple $\mathbb{Q}$-algebra. This doesn't fully answer the question I posted in full generality, but I think this is enough for me.