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There is a problem in Neukirch's Algebraic Number Theory,which is in Charpter $2$ section $5$ .

The problem is :

If $K$ is a $p-adic$ number field, then the groups $K^{*n}$, for $n$ belongs to $N$, form a basis of neighbourhoods of $1$ in $K^*$.

I think it is very clear that $K^{*n}$ is an open subgroup of $K^*$ for each integer $n$,but how to check they form a basis of neighbourhood of $1$ in $K^*$? This really puzzled me these days,so please offer me some help and show how to solve it,thank you very much!

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1 Answer 1

up vote 5 down vote accepted

First of all, a proof that the sets $(K^{\times})^n$ are open can be found here.

As for the fact that these sets form a n.h. basis of $1$, this is not actually true if you give $K^{\times}$ its usual topology, but it is true if you replace $K^{\times}$ by $\mathcal O_K^{\times}$ (where $\mathcal O_K$ is the ring of integers of $K$).

Verifying this is (in my view) easier than proving the openness.

Hint: Consider what happens when $n$ is divisible by a huge power of $p$.

Extended hint, in response to comment from OP below: If you are not familiar with these ideas, then it would be best to start with the case $K = \mathbb Q_p$. Consider $(\mathbb Z_p^{\times})^{p-1}$. Show that this consists precisely of the set $1 + p \mathbb Z_p$.

Now if you take an integer $x$ that is $\equiv 1$ mod $p$, what can you say about $x^{p^n}$? It is certainly congruent to $1$ mod $p$, but can you do any better?

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Thank you very much! But could you please show your answer with details,I am not quite familiar with those topological concepts. Thanks! –  curiosity May 26 '11 at 18:15
    
@curiosity: Dear curiosity, I have extended my hint. If it still doesn't make sense, then perhaps you should ask another, more basic question, say about the nature of the topology on $\mathbb Q_p$. You won't be able to solve Neukirch's question unless you have a solid understanding of the topology on $\mathbb Q_p$ and $\mathbb Q_p^{\times}$. Regards, –  Matt E May 26 '11 at 18:25
    
Dear,MattE.Thank you for your extended hint and advise, but I can understand what happens when we consider this question in $Q_p$,but I still can not make it that clear when the field is a $p−adic$ number field....could you please excuse my foolishness and show me how to prove $O_k ^{*n}$ form a basis of neighbourhood of $1$? Thank you! –  curiosity Jun 1 '11 at 13:32
    
@curiosity: Dear curiosity, The multiplicative group of the residue field of $\mathcal O_K$ has finite order, so raising to that order, we land in subgrop of $\mathcal O_K^{\times}$ that is trivial mod the uniformizer. Now this subgroup is pro-$p$, and so raising to higher powers of $p$, we can put ourselves inside any open n.h of $1$. (In short, it is exactly the same as the $\mathbb Q_p$-case, just with $p-1$ replaced by the order of the multiplicative group of the residue field.) Regards, –  Matt E Jun 3 '11 at 1:48

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