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MO question 19282 is about integral epimorphisms of commutative rings, and a counterexample is given to surjectivity. What about the case of the Frobenius endomorphism of a commutative, characteristic $p$ ring $R$: if it is epimorphic, is it surjective ?

This comes up naturally when one considers the Frobenius twist functor on $R$-algebras, its right adjoint, and especially the question whether the unit and counit of this adjunction are isomorphisms.

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  • $\begingroup$ I take it "epimorphism" is meant in categorical sense? Some authors use it for surjective ring homomorphisms. $\endgroup$
    – Wojowu
    Commented Jul 31, 2021 at 20:24
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    $\begingroup$ Interpreting derivations $R\to M$ as homomorphisms $R\to R\oplus M\cdot\varepsilon$, the condition implies that $\Omega^1_R=0$. I don’t know if this observation is useful. $\endgroup$
    – Piotr Achinger
    Commented Jul 31, 2021 at 20:33
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    $\begingroup$ P.S. I remember that in a normal ring $R$, the kernel of $d:R\to \Omega^1_R$ equals $R^p$, so in this case the answer to your question should be yes. $\endgroup$
    – Piotr Achinger
    Commented Jul 31, 2021 at 20:44
  • $\begingroup$ @Wojowu yes, I mean an epimorphism of unitary commutative rings. $\endgroup$
    – Matthieu Romagny
    Commented Aug 3, 2021 at 9:18
  • $\begingroup$ @Piotr thank you! In fact I think there are many specific cases where one can prove that F is surjective. The general case still eludes me (and I'm not even sure if surjectivity holds). $\endgroup$
    – Matthieu Romagny
    Commented Aug 3, 2021 at 9:21

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