0
$\begingroup$

I'm looking for an example of degree $n$ ramified but not totally ramified example over $\Bbb Q_p$.

I can find degree $n$ ramified extension, for example, $\Bbb Q_p({p^{1/n}})/\Bbb Q_p$. $p^{1/n}$'s minimal polynomial over $\Bbb Q_p$ is $x^n-p$ ($p$-eisenstein polynomial) and this is degree $n$, so the extension degree is exactly $n$. And norm of ${p^{1/n}}$ is $-p$, this is prime element of $\Bbb Q_p$, so the extension is totally ramified.

But this kind of construction only generates totally ramified cases.. But I'm looking for a degree $n$ and ramified but not totally ramified.

Could you give me an example of degree $n$ ramified but not totally ramified example over $\Bbb Q_p$?

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

If $n$ is prime, then any extension will be either unramified or totally ramified, because $n=ef$. Suppose that $n$ is not prime and $n=ml$ with $m,l>1$. Take a degree $m$ totally ramified extension, say $\Bbb Q_p(p^{1/m})$ and a degree $l$ unramified extension $\Bbb Q_p(\zeta_{w})$ where $w=p^l-1$. Then we can take the composite field $\Bbb Q_p(p^{1/m},\zeta_{w})$ this has degree $ml=n$. Indeed $x^m-p$ is is still Eisenstein over the field $\Bbb Q_p(\zeta_w)$ because $p$ is a prime element in the ring of integers (as $\Bbb Q_p(\zeta_w)/\Bbb Q_p$ is unramified). Because the ramification index and inertia degree are multiplicative, this extension has ramification index $m$ and inertia degree $l$, so it's ramified, but not totally ramified.

$\endgroup$
12
  • 1
    $\begingroup$ @dandelion when I said "p-adic field", I meant any finite extension of $\Bbb Q_p$ (so any local field of char 0). It's true that the ring of integers of $\Bbb Q_p(\zeta_w)$ is a DVR, too. $\endgroup$
    – Lukas Heger
    Commented Sep 22, 2021 at 16:39
  • 1
    $\begingroup$ @dandelion I think that should be in every book covering local fields, e.g. Neukirch's Algebraic Number Theory. The basic idea is that for a finite extension $L/\Bbb Q_p$, there's a unique extension of the p-adic valuation to $L$ and the ring of integers of $L$ is the valuation ring associated to that valuation. But the valuation remains discrete, so the valuation ring is a DVR, hence a PID, hence a UFD. $\endgroup$
    – Lukas Heger
    Commented Sep 22, 2021 at 18:08
  • 1
    $\begingroup$ Oh, that is clear view. I was troubled because I took DVR as local PID that is not a field, but by seeing characterization by valuation , it was clear. Thank you. $\endgroup$
    – Poitou-Tate
    Commented Sep 22, 2021 at 18:15
  • 1
    $\begingroup$ @dandelion discrete means that the value group is isomorphic to $\Bbb Z$, if $L$ is a finite extension of $\Bbb Q_p$ with ramification index $e$, then the extended valuation will have value group $\frac{1}{e}\Bbb Z$, but that's still isomorphic to $\Bbb Z$. $\endgroup$
    – Lukas Heger
    Commented Sep 23, 2021 at 5:21
  • 2
    $\begingroup$ @dandelion yes. If you want a discrete valuation with exactly Z as a valuation group, you can just multiply the extended valuation by the ramification index. This won't change the valuation ring. $\endgroup$
    – Lukas Heger
    Commented Sep 23, 2021 at 15:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .