Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to learn how to apply local class field theory and I thought about trying to enumerate some low degree abelian extensions of $\mathbb{Q}_p$. The easiest case is the quadratic extensions i.e. the cyclic extensions of degree 2. But here no local class field theory is needed as quadratics are easy to classify by just working directly with the polynomials and finally counting the index of the subgroup of squares.

My question is how to do the cyclic of degree 3 case? I know that local class field theory gives me an isomorphism

$\mathbb{Q}_p^\times/\textrm{Nm}_{K/\mathbb{Q}_p}(K^\times)\to \textrm{Gal}(K/\mathbb{Q}_3)\simeq C_3.$

Hence, to somehow list the abelian extensions of degree 3, I would need to be able to somehow classify the open subgroups of index 3 in $\mathbb{Q}_p^\times$. Is there any simple way to do that or is there an easier approach for solving the original problem? I guess it might be difficult to actually write down each cyclic degree 3 extension, but just knowing the number of them would already help.

Is there any simple method that extends to cyclic or dihedral extensions of degree 4?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

You are looking at this in the right way, by local class field theory to find the number of extensions of a given degree it suffices to figure out what the subgroups of index three of $\mathbb{Q}_p^{ ast}$ there are.

Let's consider a specific example, say $p=5.$ In this case $\mathbb{Q}_5^{ \ast} = \mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}\times U_1$ where $U_1 = 1+\mathfrak{p}.$ Finding a subgroup of index 3 is the same as finding surjective homomorphisms to $\mathbb{Z}/3\mathbb{Z}.$ Clearly $\mathbb{Z}/4\mathbb{Z}$ must be in the kernel of such a map. In addition $U_1^3 = U_1$ so $U_1$ must also be in the kernel (you can use power series or log/exp to see this). There are exactly two surjective group homomorphisms from $\mathbb{Z}\rightarrow \mathbb{Z}/3\mathbb{Z}$ given by sending 1 to 1 or 2, so we see that there must be two such extensions.

In general similar tricks should work, but I think you have to work out specific cases. $p=2$ is a little problematic also, but it still works out.

share|improve this answer
    
Thanks for the help. That decomposition of $\mathbb{Q}_p$ was new to me even though it's actually really obvious. Are there any other useful decompositions of $p$-adics that I should know of. :) –  dstt Apr 25 '11 at 1:11
    
BTW, I worked on an extension of this and noticed that your argument is slightly flawed. A subgroup of index 3 only determines a surjective homomorphism into $\mathbb{Z}/3\mathbb{Z}$ up to an automorphism of $\mathbb{Z}/3\mathbb{Z}$. It follows that in this case one subgroup gives you two distinct maps. Hence, there is actually only one cyclic extension of $\mathbb{Q}_5$. –  dstt Apr 26 '11 at 3:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.