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If $R$ is a ring of characteristic $p\gt 0$, what does $R^{1/p}$ mean?

I am not sure how to search for it, since I don't know a name for it. From the notation, it seems to be a ring consisting of the p-th roots of all elements of $R$, but a rigorous definition would be nice and perhaps even a name.

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Where did you see this notation in the first place? –  Álvaro Lozano-Robledo Aug 24 '11 at 15:31
    
@Alvaro: In several papers that I am reading. None of them give a name or reference it or define it. –  B M Aug 24 '11 at 15:56
    
@B M: I've added your question to the body of the message, rather than just relying on the title. It's better to make the post self-contained, and not rely on the subject line for content. –  Arturo Magidin Aug 24 '11 at 17:19

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up vote 8 down vote accepted

If $R$ is a domain, it has a fraction field $K$ which in turn has an algebraic closure $\bar K=\Omega$.
This latter field has a well-known Frobenius automorphism $Frob:\Omega \to \Omega: x\mapsto x^p$.
The ring you are looking after is the image of $R$ under its inverse automorphism, namely the ring $$R^{1/p}= Frob^{-1}(R)$$ You can iterate this process and get rings $R^{1/p},R^{1/p^2},R^{1/p^3} ,\ldots \subset \Omega\;$ whose union is symbolically denoted $R^{1/p^\infty}$ .

If $R$ is not a domain I think you should be very wary and I definitely don't want to say anything about that case.

An example The simplest non trivial example might be the polynomial ring $R=\mathbb F_p[X]$, for which we have $R^{1/p}=\mathbb F_p[X^{1/p}]$.

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Thanks, that helps a lot. –  B M Aug 24 '11 at 16:02

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