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Let $R=\mathbb{Z}_p[[X]]$ where $\mathbb{Z}_p$ denotes the $p$-adic integers and $p$ is a prime. Then what is $R_{(p)}$ $(R$ localised at the ideal $pR)$ $?$

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    $\begingroup$ I'm probably missing something obvious here, but I don't see how $R$ localized at $pR$, which is a prime ideal of $R$, can have characteristic $p$. Possibly you mean to take the residue field of the localization? (Also, please fix the formatting on $R_{(p)}$, which is currently being typeset as $R_(p)$; you missed the braces around the $(p)$.) $\endgroup$ Jan 26, 2014 at 15:45
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    $\begingroup$ The p-adic completion of $R_{(p)}$ is Fontaine's ring $\mathcal{O}_\mathcal{E} = \lbrace \sum_{n \in \mathbb{Z}} a_n X^n : a_n \in \mathbb{Z}_p, lim_{n \to - \infty} a_n = 0\rbrace$. This in turn is a Cohen ring for its residue field $\mathbb{F}_p ((X))$. (From which also $R_{(p)}/p \simeq \mathbb{F}_p ((X))$.) I do not know an explicit description of $R_{(p)}$ inside its completion. What kind of description do you need? $\endgroup$ Jan 26, 2014 at 15:57
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    $\begingroup$ I see, the identification of the relevant elements of the two rings is not what one might expect. For instance, $(p+X)^{-1}$ is $p^{-1}-p^{-2}X+p^{-3}X^2-\cdots$ in $\mathbb Q_p((X))$, but $X^{-1}-X^{-2}p+X^{-3}p^2-\cdots$ in $\mathcal{O_E}$. $\endgroup$ Jan 26, 2014 at 17:13
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    $\begingroup$ @Robert: Yes, it’s a discrete valuation ring. $\endgroup$ Jan 26, 2014 at 17:20
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    $\begingroup$ @EmilJeřábek: Yes, that's a trap which I once learned the hard way (in a seminar talk). $R_{(p)}$ can be embedded in both $\mathcal{O}_\mathcal{E}$ and $\mathbb{Q}_p((X))$, but as your example shows, these two are not contained (in a naive way) in a common overring, so one cannot "intersect" them. (And I do not know an explicit description of $R_{(p)}$ in $\mathbb{Q}_p((X))$ either.) $\endgroup$ Jan 26, 2014 at 17:57

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$R$ is a UFD and its primes up to associates are: $p$, and the irreducible polynomials in $\mathbb{Z}_p[X]$ that are "distinguished," or "Weierstrass," that is, that are congruent mod $p$ to $X^n$, where $n$ is the degree of the polynomial. See Theorem 1.4 of http://arxiv.org/pdf/1107.4860v4.pdf

Thus, $R_{(p)}$ is gotten by inverting all of these except $p$. Specifically, elements of $R$ are of the form $f/g$, where $f$ is an element of $R$ and $g$ is a distinguished polynomial in $\mathbb{Z}_p[X]$ (which is a product of irreducible distinguished polynomials). Thus $g = X^n + ph$, where $h$ is a polynomial in $\mathbb{Z}_p[X]$ of degree less than $n$.

This all can be arrived at from the $p$-adic Weierstrass preparation theorem.

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  • $\begingroup$ $X$ itself is a distinguished polynomial. $\endgroup$ Jan 28, 2014 at 6:45
  • $\begingroup$ Good point. I will edit my answer. $\endgroup$ Jan 28, 2014 at 6:55

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