April 20th
Today I learned a little about the relationship between localizing a global field and localizing a ring. For concreteness, we fix $K$ a number field and $\mf p$ a (finite) prime. In order to differentiate notation, we let $\mathcal O_\mf p$ be the $\mf p$-adic integers, and we let\[\mathcal O_{\mathcal O\setminus\mf p}=\left\{\frac ab\in K:b\notin\mf p\right\}\]be the localization of the ring.
To begin with, we remark that these are definitely not equivalent objects. For example, $\mathcal O_{\mathcal O\setminus\mf p}\subseteq K$ is countable, but thinking about\[\mathcal O_\mf p=\varprojlim\mathcal O_K/\mf p^\bullet\mathcal O_K\]as coherent sequences, this set bijects into $(\mathcal O_K/\mf p)^\NN$—each next term in the coherent sequence has $|\mathcal O_K/\mf p|$ possibilities because $(\mathcal O_K/\mf p^{n+1})/(\mathcal O_K/\mf p^n)\cong\mathcal O_K/\mf p).$ And we know that\[|\RR|=\left|\{0,1\}^\NN\right|=\left|(\ZZ/2\ZZ)^\NN\right|\le\left|(\mathcal O_K/\mf p)^\NN\right|\le\left|(\ZZ/2^{\op N(\mf p)}\ZZ)^\NN\right|=|\RR|.\]In particular, $\mathcal O_\mf p$ is uncountable.
However, we do have an embedding $\mathcal O_{\mathcal O\setminus\mf p}\to\mathcal O_\mf p$ which inherits a lot of its properties from the embedding $K\to K_\mf p.$
Proposition. We have a dense embedding $\mathcal O_{\mathcal O\setminus\mf p}\to\mathcal O_\mf p.$
The idea is to take the dense embedding $K\to K_\mf p$ and restrict the input to $\mathcal O_{\mathcal O\setminus\mf p}.$ In particular, for any $\frac ab\in\mathcal O_{\mathcal O\setminus\mf p}$ with $b\notin\mf p,$ we have that\[\left|\frac ab\right|_\mf p=\frac{|a|_\mf p}{|b|_\mf p}\le\frac11=1,\]so this restriction does go into $\mathcal O_\mf p\subseteq K_\mf p.$ So we have an embedding $\mathcal O_{\mathcal O\setminus\mf p}\to\mathcal O_\mf p.$
However, we know stronger. In particular, if we pick up $k\in K\cap\mathcal O_\mf p,$ then we claim $k\in\mathcal O_{\mathcal O\setminus\mf p}.$ We may write $k=\frac ab$ with $a,b\in\mathcal O_K,$ and because $a,b\in\mathcal O_\mf p$ (via the above embedding), we can write $a=a_0\varpi^\alpha$ and $b=b_0\varpi^\beta,$ where $\varpi\in\mf p^2\setminus\mf p$ and $a_0,b_0\notin\mf p.$ In particular, $a/b\in\mathcal O_\mf p$ requires\[0\le\nu_\mf p(a/b)=\nu_\mf p(a_0)-\nu_\mf p(b_0)+\alpha-\beta=\alpha-\beta.\]This lets us write\[\frac ab=\frac{a_0\varpi^\alpha}{b_0\varpi^\beta}=\frac{a_0\varpi^{\alpha-\beta}}{b_0}\in\mathcal O_{\mathcal O\setminus\mf p}.\]Thus, the elements that go into $\mathcal O_\mf p$ in the embedding are ones that start in $\mathcal O_{\mathcal O\setminus\mf p}.$
This tells us our embedding is dense pretty quickly. Fixing any open ball $B(a,r)\subseteq\mathcal O_\mf p\subseteq K_\mf p,$ we know that $K\subseteq K_\mf p$ is dense, so we can find $k\in K\cap B(a,r).$ But then $k\in\mathcal K\cap O_\mf p,$ so the above argument requires $k\in\mathcal O_{\mathcal O\setminus\mf p}.$ So $k$ provides our element of $\mathcal O_{\mathcal O\setminus\mf p}\cap B(a,r).$ $\blacksquare$
From the point of view of number theory, the above lemma basically is telling me to think about $\mathcal O_{\mathcal O\setminus\mf p}$ as the $\mf p$-adic integers in $K.$ I don't personally use them much, but they do have some of the usual properties associated with the $\mf p$-adic integers. For example, we have the following.
Proposition. We have that $\mathcal O_{\mathcal O\setminus\mf p}$ is a discrete valuation ring with $\mf p\mathcal O_{\mathcal O\setminus\mf p}$ its only maximal ideal.
We classify the ideals of $\mathcal O_{\mathcal O\setminus\mf p}.$ Let $I$ be any ideal of $\mathcal O_{\mathcal O\setminus\mf p}.$ All elements of $\mathcal O_{\mathcal O\setminus\mf p}$ have $\nu_\mf p$ bounded below by $0$ because the denominator cannot have elements divisible by $\mf p.$ So we can find $a\in I$ such that\[\nu_\mf p(\alpha)=\inf_{\beta\in I}\nu_\mf p(\beta)\]by the well-order of $\NN.$ Now we claim that $I=\mf p^{\nu_\mf p(\alpha)}\mathcal O_{\mathcal O\setminus\mf p}.$ Indeed, $I\subseteq\mf p^{\nu_\mf p(\alpha)}\mathcal O_{\mathcal O\setminus\mf p}$ because the numerator of any element of $I$ must have at least $\nu_\mf p(\alpha)$ powers of $\mf p.$
On the other hand, we can write any element of $\mf p^{\nu_\mf p(\alpha)}\mathcal O_{\mathcal O\setminus\mf p},$ as\[\underbrace{p_1\cdot\ldots\cdot p_{\nu_\mf p(\alpha)}}_{\in\mf p}\cdot\frac xy=\alpha\cdot\underbrace{\frac{p_1}{\alpha}}_{\in\mathcal O_{\mathcal O\setminus\mf p}}\cdot p_2\cdot\ldots\cdot p_{\nu_\mf p(\alpha)}\cdot\frac xy\in\alpha\mathcal O_{\mathcal O\setminus\mf p}.\]In particular, this element lives in $I,$ which completes the proof. $\blacksquare$