Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If $K_n$ is the field $\mathbb{Q}_p(\mu_{p^n})$, then it's easy to see that the relative different $\mathcal{D}(K_n / K_{n-1})$ is $(p)$ for all $n \ge 2$.

What happens if I take an arbitrary, probably totally ramified, finite extension $L/\mathbb{Q}_p$ and look at the tower $L_n = LK_n$? It's clear that $\mathcal{D}(L_n / L_{n-1})$ divides $(p)$, and one can show (using a general result of Tate on $\mathbb{Z}_p$-extensions) that its valuation tends to 1 as $n \to \infty$; but is it true that it's equal to $(p)$ for all sufficiently large $n$?

share|cite|improve this question
I assume you mean to take $L/\mathbb{Q}_p$ finite? (If $L = K_\infty$ the result is obviously false.) –  David Loeffler Feb 3 '11 at 16:57
Maybe Coates-Greenberg's "deeply ramified" extensions will help. –  Chris Wuthrich Feb 3 '11 at 18:01
Definitely not true. It's midnight now, I'll try to write up an explanation tomorrow. –  Lubin Feb 4 '11 at 8:01

3 Answers 3

up vote 6 down vote accepted

Take $K=Q_p$ and $L/K$ finite. It is known that the sequence $\{ p^n v_p( \mathcal{D}(L_n/K_n) ) \}_n$ is eventually constant (it's basically the valuation of the different of the extension $E_L/E_K$ which you get from the theory of the field of norms). This and the transitivity of the different should allow you to answer your question.

As for the proof of my claim, see proposition 4.5 of

share|cite|improve this answer

Here's a relatively easy counterexample. Take $p=2$, $L=\mathbb{Q}_2(2^{1/3})$. I did some direct computation and saw that $v_2(\mathfrak{D}^{L_2}_L)=2/3$, $v_2(\mathfrak{D}^{L_3}_{L_2})=5/6$. Looks like there's a pattern. But there's a better way.

We're in a situation where not only $L/\mathbb{Q}_2$ is tamely ramified of degree $3$, but also every $L_n/K_n$. Now we use the functoriality of the Hasse-Herbrand transition function: if $k\subset F\subset K$, then $\varphi^K_k=\varphi^F_k\circ \varphi^K_F$. Use the relation $\varphi^{L_n}_{\mathbb{Q}_2}=\varphi^{K_n}_{\mathbb{Q}_2}\circ \varphi^{L_n}_{K_n}= \varphi^L_{\mathbb{Q}_2}\circ \varphi^{L_n}_ L$ and the fact that a tamely ramified extension has all the transition-function action at the origin. That is, the function is $y=x$ for $x\le 0$, but $y=x/e$ for $x\ge 0$, where $e$ is the ramification index. So as real functions, $\varphi^L_{\mathbb{Q}_2}=\varphi^{L_n}_{K_n}$, namely this is just $y=x/3$. Consequently, the transition function of $L_n/L$ is gotten by conjugating that of $K_n/{\mathbb{Q}_2}$ with the tame transition function. The effect is to multiply all coordinates of vertex points by $3$.

But we also know the transition function of $K_n$ over ${\mathbb{Q}_2}$: its vertices are at all $(2^{i-1}-1,i-1)$ for $2\le i\le n$.The new vertices are at $(3,3)$, $(9,6)$, $(21,9)$, etc. This means that the lower breaks of $L_n/L$ are at $3(2^{i-1}-1)$ for $2\le i\le n$, and in particular the unique break of $L_n/L_{n-1}$ is at $3(2^{n-1}-1)$.

Now use the formula \begin{align}{ v_F(\mathfrak{D}^F_k)=\sum_{j\ge 0}\bigl(|G_j|-1\bigr) }\end{align} where the $G$'s are the lower ramification groups, and where in this case all the numbers being added up are $1$ or $0$, to see that $v_{L_n}\bigl(\mathfrak{D}^{L_n} _ {L_{n-1}}\bigr)=3(2^{n-1}-1)+1=3\cdot2^{n-1}-2$. Divide by the ramification number of $L_n$ over $\mathbb{Q}_2$ to get $1-1/(3\cdot 2^{n-2})$, agreeing with my computations for $n=2$ and $n=3$.

It's not an issue of tame versus wild ramification in the extension $L/\mathbb{Q}_2$, either. I used $L=\mathbb{Q}_2(2^{1/4})$ to find that the numbers are $1-3/2^m$; the argument is similar but a bit more delicate, since you have no a priori idea of what the transition function of $L_n/K_n$ might be.

share|cite|improve this answer

The answer is 'no'. The different is usualy distinct from $(p)$; if you have a tower for whose upper level the different equals $(p)$, then the same is true for all other levels (possibly, one needs $p>2$ here). See Proposition 1.3 in

share|cite|improve this answer

Your Answer


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.