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This an exercise in Algebraic Number Theory written by Jürgen Neukirch. It is in chapter $2$, section $5$. The question is as follows:

For a $\mathfrak{p}$-adic number field $K$, every subgroup of finite index in $K^*$ is both open and closed.

Here is what I thought:

I think that we may use proposition $(5.7)$ on page $140$, if we denote the subgroup as $G$, then there is a subgroup corresponding to $G$ in $$\mathbb{Z}\oplus\mathbb{Z}/(q-1)\mathbb{Z}\oplus\mathbb{Z}/p^\alpha\mathbb{Z}\oplus\mathbb{Z}_p^d$$ with finite index too, as $Z$, $Z/(q-1)Z$, and $Z/p^\alpha Z$ have discrete topology, and the only subgroups with finite index of $Z_p$ are of the form $p^n Z_p$ for some integer n, which is both open and closed in $Z_p$, so I think we may use these facts to solve this question. I am not sure if this is right, so I hope we can discuss it and I could learn more. Thank you for seeing and answering it.

Here is the proposition $(5.7)$ on page $140$ in Algebraic Number Theory written by Jürgen Neukirch:

$(5.7)$ Proposition: Let $K$ be a local field and $q = p^f$ the number of elements in the residue class field. Then the following hold.

(i) If $K$ has characteristic $0$, then one has (both algebraically and topologically) $$K^*\cong\mathbb{Z}\oplus\mathbb{Z}/(q-1)\mathbb{Z}\oplus\mathbb{Z}/p^\alpha\mathbb{Z}\oplus\mathbb{Z}_p^d$$ $\qquad$ where $a>=0$ and $d$=[$K$:$Q_p$].

(ii) If K has characteristic p, then one has (both algebraically and topologically) $$ K^*\cong\mathbb{Z}\oplus\mathbb{Z}/(q-1)\mathbb{Z}\oplus\mathbb{Z}_p^N $$

Actually my basic idea is to analyze the same topological structure problem on a homeomorphism but easier algebraic group as $\mathbb{Z}\oplus\mathbb{Z}/(q-1)\mathbb{Z}\oplus\mathbb{Z}/p^\alpha\mathbb{Z}\oplus\mathbb{Z}_p^d$ or $\mathbb{Z}\oplus\mathbb{Z}/(q-1)\mathbb{Z}\oplus\mathbb{Z}_p^N $. And we know $Z$, $Z/(q-1)Z$ and $Z/p^\alpha Z$ all have discrete topology and their subgroups are open and closed, since the quotient group is discrete. To $Z_p$, its only subgroup of finite index is of the form $p^nZ_p$ for some $n$, and this subgroup is both open and closed in $Z_p$, so every subgroup with finite index of $\mathbb{Z}\oplus\mathbb{Z}/(q-1)\mathbb{Z}\oplus\mathbb{Z}/p^\alpha\mathbb{Z}\oplus\mathbb{Z}_p^d$ or $\mathbb{Z}\oplus\mathbb{Z}/(q-1)\mathbb{Z}\oplus\mathbb{Z}_p^N $ is both open and closed. So every subgroup with finite index of $K^*$ is both open and closed.

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Here is a closely related question: math.stackexchange.com/questions/17745/…. It seems to me that both answers given there (one by Matt E, one by me) can be carried over to answer your question: note in particular that $K^{\times} \cong \mathbb{Z} \times \mathcal{O}_K^{\times}$, so the question quickly reduces to the structure of the unit group of the valuation ring. Note also that this is pretty similar to Lalit Jain's answer below, so if you're working on fleshing that out, keep going! –  Pete L. Clark May 25 '11 at 18:21
    
@Peter L.Clark:Thank you very much! But could you please have a look at my idea,I want to know if it is right,thanks. –  curiosity May 25 '11 at 18:27
    
@curiosity: it's hard to easily follow what you write without having Neukirch's book in hand (which I don't at the moment). If you made it more self-contained -- e.g., by explaining all your notation -- more people would probably take a look at it. –  Pete L. Clark May 25 '11 at 18:48
    
@Peter L.Clark: thank you for your advise!....I am fresh and still have some trouble in using this...and I have just seen your answer about "nth powers in the p-adics",it is very helpful. –  curiosity May 25 '11 at 18:51
    
My issue with your argument is that you could have a subgroup that is not a product of subgroups...unless I am misunderstanding your point. Either way, I do think that the kind of argument outlined in Pete L.Clark's post or mine below is fairly common and a good tool to know. For example, it is very useful in Class Field Theory. –  Lalit Jain May 25 '11 at 21:30
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1 Answer

up vote 2 down vote accepted

Here is a couple hints. Since the group is finite index it must contain some $K^{*n}$ for some $n$ so it suffices to show that $K^{*n}$ is open for all $n.$ To do this we just need to show that $K^{*n}$ contains some neighbourhood of the identity or equivalently that it contains $U^{(m)} = 1+\mathfrak{p}^m$ for some $m.$ My final hint on how to do this is to use log and exp. There are other ways but this is a great application of these functions.

If you still have trouble I can post a more detailed answer.

EDITED:

Here is some more detail. Note that I wrote this up a while ago for an assignment so I apologize for mistakes.

For sufficiently large $m$ we note that $U^{(m)} \cong \mathfrak{p}^m$ by the mutually inverse homeomorphisms (and isomorphisms), $\exp $ and $\log.$ In particular, choose $m$ large enough so that this isomorphism is true for $U^{(m - v_p(n))}.$ Then if $\alpha \in U^{(m)},$ $\log \alpha \in \mathfrak{p}^{m}$ and $\frac{\log \alpha}{n} \in \mathfrak{p}^{m - v_p(n)}.$ But then $\exp(\frac{\log \alpha}{n}) \in U^{(m - v_p(n))}$ and in addition $\exp(\frac{\log \alpha}{n})^n = \exp(\log(\alpha)) = \alpha.$ Thus $\exp(\frac{\log \alpha}{n})$ is an $n$-th root of $\alpha$ so $\alpha \in K^{\ast n}.$ Since $\alpha$ was arbrirtary this shows that $U^{(m)}\subset K^{\ast n}$ so we are done.

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thank you, in fact, I have thought about the method you just show me....thank you! And I do have problems in the details,so if you can show me the details,that would be so nice! And thank you again for your answer. –  curiosity May 25 '11 at 18:19
    
It would be very kind of you if you can write down the other way you have mentioned,which require a great application of these functions....thank you! –  curiosity May 25 '11 at 18:56
    
Thank you very much,this is helpful. –  curiosity May 25 '11 at 19:51
    
If possible, could you please take a look at my idea ?thanks. –  curiosity May 25 '11 at 19:53
    
Does it mean that $K^{*n}$ form a basis of the element $1$? Thank you. –  curiosity May 25 '11 at 20:04
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