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Consider an abelian algebra, $R$, over the field $K$ with the properties that every residue field of $R$ is (canonically) isomorphic to $K$ (I'm not sure but I think this is necessary, otherwise we could be talking about $\mathbb{R}[x]$ as a $\mathbb{Q}$-algebra) and that for every maximal ideal $m$, $R/m^2\cong K\oplus m/m^2$ as additive groups (and let $\phi:R\rightarrow R/m^2$ be the natural quotient map). For every $s \in R$ call the $K$ factor of $\phi(s)$ the 'value of $s$ at $m$' and call the $m/m^2$ factor 'the derivative of $s$ at $m$'. Under what conditions does 'the derivative of $s$ vanishes at each $m$' imply that 'the value of $s$ is the same at every $m$'? What are the conditions if we talk about jet spaces instead of tangent spaces (i.e. replace every instance of $m^2$ with $m^k$ for a specific or arbitrarily large $k$)?

There are obvious examples and counterexamples. The ring of polynomials over a field of characteristic 0 satisfies this condition, as does the ring of smooth functions on a connected manifold (and maybe even $p$ times differentiable functions, even though the Zariski tangent space isn't the normal tangent space). Any ring of functions on a disconnected space obviously doesn't satisfy this, but there are also connected counterexamples, such as the ring of polynomials over a field of positive characteristic, although in that case talking about arbitrarily large jet spaces gives you an analogous statement (i.e. if every jet of a polynomial is constant, then the polynomial is constant).

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    $\begingroup$ Does not the answer depend on the choice of an isomorphism $R/m^2\cong R/m\oplus m/m^2$? I don't see any canonical map which one might require to be an isomorphism. There are maps $m/m^2\to R/m^2\to R/m$, one may require this to be a split short exact sequence, but a choice of the splitting still has to be made, and I think the answer will depend on it. $\endgroup$ Dec 9, 2015 at 20:33
  • $\begingroup$ Yes that's true. I'll edit the question to make it more specific. $\endgroup$ Dec 9, 2015 at 21:07
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    $\begingroup$ @მამუკაჯიბლაძე There is a canonical choice of splitting map $R/m\to R/m^2$, which identifies $R/m$ with $K$ then includes into $R/m^2$ by the structure map (the map making $R/m^2$ into a $K$-algebra). $\endgroup$ Dec 9, 2015 at 21:46
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    $\begingroup$ There was no $K$ in the question when მამუკა ჯიბლაძე responded. $\endgroup$ Dec 9, 2015 at 22:03

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Let me propose a more natural version of your question that can be asked about more algebras:

What can we say about an element $r \in R$ of a commutative algebra $R$ over a field $k$ given that $D(r)$ vanishes for every $k$-linear derivation $D : R \to M$ ($M$ an $R$-module)? In particular, must we have $r \in k$?

This condition should imply but in general should be stronger than your condition. It's equivalent to the condition that the universal $k$-linear derivation $d : R \to \Omega_{R/k}$ into Kähler differentials vanishes on $r$. I don't know what to say at this level of generality, unfortunately.

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  • $\begingroup$ Does this work in positive characteristic? My impression was that the Kähler differentials on a polynomial ring behave exactly like ordinary formal derivatives so in $\mathbb{F}_p[x]$ you would have $d(x^p)=0$ for every derivation $d$. $\endgroup$ Dec 10, 2015 at 8:28
  • $\begingroup$ @Exomnium: I'm not sure what you mean by "does this work" because I haven't made a claim about what the answer is. The OP already acknowledged that the case of positive characteristic is troublesome: it's in fact the case that every $p^{th}$ power lies in the kernel of every derivation. $\endgroup$ Dec 10, 2015 at 17:15
  • $\begingroup$ Sorry, previous (now deleted) remark was dumb. If $R = k[t]/t^{100}$, then $t^2$ has deriviative zero at the only point of $\mathrm{Spec}(R)$, but $d t^2$ is not zero. If $R$ is a finitely generated domain, they should be the same. $\endgroup$ Dec 10, 2015 at 21:20

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