This may be a completely trivial question, but I haven’t seen it stated in any of the references I checked. Is the characteristic of a ring $R$ equal to that of its completions? This is true for the rationals, for example, by Ostrowski‘s theorem but I don’t see how to show it holds in general.
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$\begingroup$ The link about completions refers to many kind of completions, and it's not obvious whether there is one given precise meaning of completion to make the question clear-cut. Do you have a single example of "completion" of a ring that is not injective? $\endgroup$– YCorCommented Nov 30, 2020 at 11:42
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$\begingroup$ Very well. By a ring completion here I mean the projective limit of the quotient rings $\lim_{\leftarrow n}R/I^n$ for any proper ideal. I do not have a non-injective example and my question was whether one ever exists. $\endgroup$– MOnewbieCommented Nov 30, 2020 at 11:53
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$\begingroup$ Yes, it exists. Take $R=R_1\times R_2$ where $R_1$ has characteristic $0$ and $R_2$ has characteristic $n>0$, and $I=R_1\times\{0\}$. (Note that this definition of "completion" doesn't encompass the rationals, the example you mentioned.) $\endgroup$– YCorCommented Nov 30, 2020 at 11:55
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$\begingroup$ It can also happen that $I=I^2$, in which case the completion is $R/I$. $\endgroup$– Laurent Moret-BaillyCommented Nov 30, 2020 at 13:50
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1$\begingroup$ However, it is true that the $m$-adic completion $\hat{R}$ of a Noetherian local ring $R$ with maximal ideal $m$ has the same characteristic as $R$. It shares a lot of other properties with $R$ too. Indeed, the map $R \rightarrow \hat{R}$ is faithfully flat with trivial closed fiber. $\endgroup$– Neil EpsteinCommented Dec 1, 2020 at 16:19
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