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Let $\mathcal{O}$ be a complete discrete valuation ring with algebraically closed residue field $k$ of characteristic $p>0$. Let $\pi\in \mathcal{O}$ generate the maximal ideal and suppose $(p)=(\pi^e)$.

With this information, can one determine the highest $p$-th power root of unity that lies in $\mathcal{O}$? My gut feeling is that if $n\in\mathcal{O}$ is a $p^a$th root of unity, then $a\leq e$. Is this right?

Thanks for any help.

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One can bound the order of $p$-power roots of unity in $K$ in terms of the ramification index. Let $K$ be the fraction field of $\mathscr{O}$. Since $\mathscr{O}$ has mixed characteristic (as $(p)=(\pi^e)\neq 0$), there is a unique continuous embedding $\mathbf{Z}_p\hookrightarrow\mathscr{O}$ which extends uniquely to a ring map $\mathbf{Q}_p\hookrightarrow K$. Suppose that $\zeta_n$ is a $p^n$-th root of unity in $K$. We then have $\mathbf{Q}_p(\zeta_n)\subseteq K$. Since ramification indices are multiplicative, we then get $e=e(K/\mathbf{Q}_p)=e(K/\mathbf{Q}_p(\zeta_n))e(\mathbf{Q}_p(\zeta_n)/\mathbf{Q}_p)\geq p^{n-1}(p-1)$, so $n$ is constrained by the requirement that $p^{n-1}(p-1)\leq e$.

In response to the question about the embedding $\mathbf{Z}_p\hookrightarrow\mathscr{O}$, it follows from completeness and the assumption of mixed characteristic. We have $p\in(\pi)$, so for each $n\geq 1$, $p^n\in(\pi^n)$. Thus the composite $\mathbf{Z}\hookrightarrow\mathscr{O}\to\mathscr{O}/(\pi^n)$ descends to a homomorphism $\mathbf{Z}/p^n\mathbf{Z}\to\mathscr{O}/(\pi^n)$. These maps are compatible as $n$ varies, so on taking inverse limits, one gets a homomorphism $\mathbf{Z}_p=\varprojlim_n\mathbf{Z}/p^n\mathbf{Z}\to\mathscr{O}=\varprojlim\mathscr{O}/(\pi^n)$ which is continuous for the $p$-adic topology on the source and the $\pi$-adic topology on the target, by construction. It sends $p$ to a non-zero element of the maximal ideal of $\mathscr{O}$ (since $\mathscr{O}$ has characteristic zero), and therefore must be injective, since every element of $\mathbf{Z}_p$ is a power of $p$ times a unit.

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  • $\begingroup$ Thanks- this is exactly what I was after. Do you have a reference for the continuous embedding you mention? $\endgroup$
    – Andy
    Apr 12, 2014 at 17:30

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