Can Kummer Extensions of p-adics Prove a Contradiction?

[yonat83]

Hi everyone,

This is my first post here, so first of all, Congrats on this website, it seems very instructive and usefull!

I m currently studying p-adic number fields for my master degree and have a problem I can't seem to resolve.

Consider the p-adic local field F=Q_3 and its 2 extensions k1=Q_3(z_13) and k2=Q_3(z_3), when z_n denotes a primitive n-th root of unity.
Since (3, 13) = 1, k1 is unramified over F and k2 is obviously totally ramified over F.
Moreover, [k1:F] = 3 is the order of 3 in the multiplicative group Z_13* since 3^3=27=13*2+1.
on the other hand [k2:F] = 3-1 = 2.
Both are cyclic sub-extensions of k1k2 / F

Now, consider the extension k1k2 / k2. This is a cyclic extension of degree 3 and by ramification theory, its unramified.
Now by Kummer theory, since z_3 is in k2 its generated by some root a^(1/3) for some a in k2. We can assume a is integral.
But now, reducing mod P, when P is the maximal ideal of k2 over (3), the equation x^3 = a (mod P) has solutions in the residue field of k2 since the power-3 map is just the frobenius automorphism on the residue field of k2. Hence the residue field extension corresponding to k1k2 / k2 is trivial and then k1k2 is totally ramifie over k2! That's a contradiction...

So there is abviously a mistake somewhere in my "proof" but I can't find out where it is...
I could really use some help,

thanks a lot

Yonathan

 

[Hurkyl]

So there is abviously a mistake somewhere in my "proof" but I can't find out where it is...

You seemed to have argued the generator of k2 lies in the residue class field of k2, but haven't said anything about the generator of k1.

 

[yonat83]

I don't care much about k1: the only thing I need to know about it is that it is cyclic unramified of degree 3... I know for sure there is such a field over F from class field theory for exemple

 

[Hurkyl]

But you don't have appeared to use k1 at all.


Oh, I think you were assumed that any degree three extension could be defined by a polynomial of the form

f(x) = x³ - a​

? That is definitely false -- no polynomial of that form is ever irreducible in the residue class field of Q₃ or its finite extensions, and so any algebraic extension of Q₃ whose residue class field is a degree 3 extension of the residue class field of Q₃ cannot be defined by a polynomial of this form.

 

[yonat83]

Ok, but k2 contains a primitive 3 root of unity, doesn't Kummer theory exactly says that 3-cyclic extensions of k2 are ALL generated by such a polynomial?

 

[Hurkyl]

Sigh, this question is going to bother me all day.

 

[yonat83]

Ok, so I think the problems comes from the fact that reducing the equation x^p-a modulo P does NOT give a defining equation for the coresponding residue field extension.

Then the question arising is how can one determine such a defining equation on residue fields, given the local extension? More generaly, how can one compute ramification groups of Kummer p-extensions?

 

[Hurkyl]

Ok, so I think the problems comes from the fact that reducing the equation x^p-a modulo P does NOT give a defining equation for the coresponding residue field extension.

Right. I think it in terms of rings of integers -- for example, the ring of integers in the number field \mathbb{Q}(\sqrt{5}) is not R_1 = \mathbb{Z}[\sqrt{5}]. The residue class field at 2 is the finite field with 4 elements, but the latter ring only generates the subfield of 2 elements.

Of course, if you took the full ring of integers R_2 = \mathbb{Z}[(1 + \sqrt{5})/2], then reducing modulo 2 would, in fact, give the finite field of 4 elements.

The index of R₁ in R₂ is 2, so 2 is the only prime over which there is a problem. (Of course, this could also be seen directly)

 

[Hurkyl]

IIRC, some of the prime factors of the discriminant of a polynomial tell you where the bad behavior described above can occur, and the remaining factors tell you where ramification occurs.

 

[Hurkyl]

So, Artin-Schreier theory tells us about the degree-p extensions of a finite field of characteristic p.

I'd like to say the following statement is true, but my memory of padic fields is fuzzy so I'm not sure:

There is a one-to-one correspondence between finite, unramified extensions of F and finite extensions of its residue field​

For the record, I know nothing about ramification groups (at least by that name), so I can't help you there.

 

[yonat83]

Indeed the correspondence you're talking about is true.

Now, forget about ramification groups: In a first step, I only need to find a simple (or maybe more complicated) criterion for deciding whether the local field defined by x^p - a is ramified or not.

Artin-Shreier theory, in my opinion, is for no use here since I m starting with a polynomial over a 0 characteristic field.

Do you remember exactly how does the discriminant plays in this problem?

 

[yonat83]

Ok, I found the solution in a book:

You have to look at the equation X^p - a = 0 mod P^(p*e/(p-1))
when P stands for the prime ideal of your field, p the characteristic and e the absolute ramification index:
It has a solution iff the extension is unramified.

In the case of a more general Kummer extension of degree p^n, n>1 the solution is much more complex and you can look at

http://www.springerlink.com/content/l54k280u0255l652/

for a solution.

Thanks for your help!

 

[Hurkyl]

Artin-Shreier theory, in my opinion, is for no use here since I m starting with a polynomial over a 0 characteristic field.

The point was to use Artin-Schreier on the residue field. But that's mainly because I know finite fields a lot better than padic fields.

Thanks for your help!

np! I wish I could have done more.