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Let $K = \mathbb{Q}_5(\sqrt{5})$. As I learned from my last post, it is possible to write $K$ as the completion of $(k,v)$ where $k$ is a number field (i.e. a finite extension of $\mathbb{Q}$) and a valuation $v$ on $k$ defined by some prime ideal $\mathfrak{p}$ in $\mathcal{O}_k$. While I am still digesting the answer of mentioned post, I think it makes sense to choose $k = \mathbb{Q}(\sqrt{5})$. Now the question remains on how to choose the prime ideal $\mathfrak{p}$ (which is I think not shown in the answer).

I know that $\mathbb{Q}_5$ is the completion of $\mathbb{Q}$ wrt. the $5$-adic value which comes from the prime ideal $(5) \subseteq \mathcal{O}_\mathbb{Q} = \mathbb{Z}$ (cf. also this post of mine). Now I think that the $\mathfrak{p}$ we are looking for has something to do with $(5)$ but I was not able to advance further from this point. Question: How to explicitly describe $\mathfrak{p}$ (say, by finitely many explicit generators)?

ANT Learner
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    Well how does $(5)$ decompose in your $K$, i.e. what primes $\mathfrak p$ lie above it? I would try one of those and hope that either there is only one (i.e. $(5) = \mathfrak{p}^e$ for some $e$, presumably $e=2$), or if there are more, it doesn't matter which one I choose. – Torsten Schoeneberg Jan 15 '21 at 01:16
  • @TorstenSchoeneberg: Ah, I just learned about factorization of ideals of the ring of integers into prime ideals recently. I may try your suggested approach. General question: You said that it can happen that there are more "prime factors" but it does not matter here which one I choose to describe my extension in this example. Can you always do that for arbitrary finite extensions of $\mathbb{Q}_p$? – ANT Learner Jan 15 '21 at 08:37

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$$\Bbb{Z}_5[\sqrt5]= \varprojlim_{n\to \infty} \Bbb{Z}[\sqrt5]/( \sqrt5^n)= \varprojlim_{n\to \infty} \Bbb{Z}[\sqrt5]/(5^n)$$ Well $O_{\Bbb{Q}(\sqrt5)}=\Bbb{Z}[\frac{1+\sqrt5}2]$ but we don't really care here.

The valuation extended to $\Bbb{Q}(\sqrt5)$ is $$v(a)= n\text{ if } \ a\in ( \sqrt5^n),\not \in ( \sqrt5^{n+1}), \qquad v(a/b)=v(a)-v(b), v(0)=\infty$$

$\Bbb{Q}_5(\sqrt5)=\Bbb{Z}_5[\sqrt5][5^{-1}]$ is the completion of $\Bbb{Q}(\sqrt5)$ for $v$.

reuns
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  • Would you mind to explain what the first half of your answer has exactly to do with the original question? And in your case, $\mathfrak{p} = (\sqrt{5})$, right? If so, how can I see that it is indeed a prime ideal in $\mathcal{O}_K$? – ANT Learner Jan 15 '21 at 08:31
  • What do you get when trying to show that $(\sqrt 5)$ is a maximal ideal of $\Bbb{Z}[\sqrt5]$ – reuns Jan 15 '21 at 17:25
  • If $(\sqrt{5})$ is a maximal ideal of $\mathbb{Z}[\sqrt{5}]$, then the quotient $\mathbb{Z}[\sqrt{5}]/(\sqrt{5})$ is a field. Is that what you mean? I still don't see where we are heading. – ANT Learner Jan 15 '21 at 17:52
  • Why don't you state clearly what you don't understand ? If it is unclear to you then clearly you need to work out $\mathbb{Z}[\sqrt{5}]/(\sqrt{5})$ making it simpler (the ideal contains $5$) and finding that it is a field. – reuns Jan 15 '21 at 18:20
  • So I understand why $\Bbb{Z}5[\sqrt5]=\varprojlim{n\to \infty} \Bbb{Z}[\sqrt5]/(5^n)$. I do not understand is why you are bringing up $O_{\Bbb{Q}(\sqrt5)}=\Bbb{Z}[\frac{1+\sqrt5}2]$, if you say yourself that it does not matter to our answer. But the most crucial part I do not understand is why $\Bbb{Q}_5(\sqrt5)=\Bbb{Z}_5[\sqrt5][5^{-1}]$ is the completion of $\Bbb{Q}(\sqrt5)$ for $v$. For me, a completion is the quotient of all Cauchy sequences (here in $\Bbb{Q}(\sqrt{5})$) mod the ideal of zero sequences. It looks like you use a result I am not aware of. Would you mind explaining that? – ANT Learner Jan 16 '21 at 18:43
  • That's what I said in your previous question, $\varprojlim \Bbb{Z}/(5^n)$ and $\Bbb{Z}_5$ are the same ring of limits of sequences of integers that converge modulo every $5^n$, just constructed differently. It works the same way for the finite extensions thereof. – reuns Jan 16 '21 at 19:59
  • Sorry, but I feel like I am stuck here. I am already stuck with why $\Bbb{Q}_5(\sqrt{5}) = \Bbb{Z}_5[\sqrt{5}][5^{-1}]$, not to mention what the completion (which is an analytical definition to me) has to do with inverse limits (which is more an algebraic concept). – ANT Learner Jan 17 '21 at 02:32
  • Just try and see what you get. $\Bbb{Q}_5= \Bbb{Z}_5[5^{-1}]$ gives that $\Bbb{Q}_5[\sqrt 5]=\Bbb{Z}_5[\sqrt 5][5^{-1}]$ – reuns Jan 17 '21 at 02:37
  • The completion and inverse limit aren't just abstract definitions, they are very concrete things, and they are the same ring, in the same way that the decimal expansion defines a real number and is in the completion of $\Bbb{Q}$. – reuns Jan 17 '21 at 02:39