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What can be said about extensions à la $\mathbb{Q}_p(\sqrt[n]{a})/\mathbb{Q}_p$? Ramification behaviour, valuation ring, ...?

I find it hard to say anything general - for example, as a function of the $p$-adic valuation of $n$ and/or $a$. Of course some special cases are rather easy to handle, and I understand what happens when $v_p(a) = 0$. This might be a hard question, or a question for which there is a standard reference - I didn't find one - or something rather easy, in which case I'm just missing something.

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    $\begingroup$ Honestly, I don't think myself that it is such an ill-posed question, and I really don't see how to make it much better. I could ask under what conditions we get unramified/tamely ramified/wildly ramified/... extensions, but honestly, I don't know what type of conditions I should be looking for, so it is difficult to make the question more focused. If anyone knows how to make this a better question, please edit or do some suggestions! $\endgroup$
    – Wanderer
    Jan 21, 2010 at 0:20
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    $\begingroup$ As for motivation, I could say that I started thinking about this sort of things when I was a bit confused (during a seminar) that Serre stated that in his modularity conjecture paper that $\mathbb{Q}_2(\sqrt{5})/\mathbb{Q}_2$ is unramified (which is indeed not hard to see). And it helps me also to understand my number theory course better - it is in no way related to homework, by the way. It is quite a natural question to ask. But I don't think this type of motivation would make the question much better! $\endgroup$
    – Wanderer
    Jan 21, 2010 at 0:28
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    $\begingroup$ Out of interest, if the OP really is still learning this stuff, I would press them to give a rigorous definition of "Q_p(a^{1/n})". Already that's a slightly interesting exercise! $\endgroup$ Jan 21, 2010 at 7:53
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    $\begingroup$ To give an example of what I mean in the "make this rigorous" comment: what is Q_7(2^{1/4})? Q_7 already contains one 4th root of 2, but it doesn't contain all four of them. So is Q_7(2^{1/4}) equal to Q_7 or to the unramified quadratic extension? $\endgroup$ Jan 21, 2010 at 15:15
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    $\begingroup$ Dear Anweshi, Unless $a = 1$, there is no notion of a primitive root of $a$. $\endgroup$
    – Emerton
    Jan 22, 2010 at 15:21

2 Answers 2

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If $n$ is prime to $p$, then ${\mathbb Q}\_p(a^{1/n})$ is unramified if $n | v_p(a)$, and is tamely ramified otherwise. To see this, we note that we may first of all divide $a$ by powers of $p^n$, and so assume that $0 \leq v_p(a) < n.$

If in fact $v_p(a)=0$, i.e. $a$ is a unit, then the extension is unramified, and the ring of integers is equal to ${\mathbb Z}\_p[a^{1/n}]$ (by Hensel's lemma, since $x^n - a$ is then a separable equation mod $p$).

Otherwise, if $0 < v_p(a) < n,$ we get a tamely ramified extension (essentially by the definition of tamely ramified).

If $p | n$ then the situation is a little more complicated. For example, if $n = p$ and $0 < v_p(a) < p,$ then the extension is wildy ramified.

If $a$ is a unit, then we may write $a = \zeta u,$ where $\zeta$ is a $(p-1)$st root of 1 and $u \equiv 1 \bmod p,$ and since $\zeta^p = \zeta,$ we see that ${\mathbb Q}\_p(a^{1/p}) = {\mathbb Q}\_p(u^{1/p}).$ Now (supposing that $p$ is odd, for simplicity) if $u \equiv 1 \bmod p^2,$ then $u$ is in fact a $p$th power in ${\mathbb Q}\_p,$ and so the extension is trivial. On the other hand, if $u \equiv 1 \bmod p,$ but not mod $p^2$, then the extension is wildy ramified of degree $p$, with ring of integers equal to ${\mathbb Z}\_p[u^{1/p}].$

To see this last claim, note that if $X^p - u = 0,$ and we write $Y = X - 1$, then $(Y + 1)^p - u = 0,$ i.e. $Y^p + pY^{p-1} + \cdots + p Y + (u-1) = 0,$ and so $Y$ satisfies an Eisenstein polynomial of degree $p$. This implies that the extension is wildly ramified of degree $p$, that $Y$ is a uniformizer in the extension, and that the ring of integers is equal to ${\mathbb Z}\_p[Y] = {\mathbb Z}\_p[u^{1/p}].$

Added in response to Keith Conrad's comments below: As Keith points out, the extension ${\mathbb Q}_p(a^{1/n})$ is not really well-defined unless ${\mathbb Q}_p$ contains the $n$th roots of $1$, or equivalently, if $n$ divides $p-1$ (or 2 if $p = 2$).

But note e.g. if $p$ does not divide $n$, then adding the $n$th roots of unity gives an unramified extension of ${\mathbb Q}_p(a^{1/n})$, and so the ramification behaviour is independent of the choice of $n$th root, while in the case when $n = p$ also treated above, adjoining the $p$th roots of unity is a tamely ramified extension of ${\mathbb Q}_p$, so the claims regarding wild ramification are independent of the choice of $p$th root.

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    $\begingroup$ The notation a^(1/n) is inherently ambiguous (which n-th root of a?) and it is not necessarily the case that an n-th root generates an extension of degree n. For example, consider f(x) = x^3 - 10 over Q_3. Since |f(4)|_3 < |f'(4)|_3^2, by Hensel's lemma 10 is a cube in Z_3, but that does not mean the extension Q_3(10^(1/3))/Q_3 is trivial because 10^(1/3) doesn't stand for any definite cube root. One root of x^3 - 10 is in Q_3 and the other two are not (because Q_3 does not contain a nontrivial cube root of 1). Certain cube roots of 10 over Q_3 generate a quadratic extension. $\endgroup$
    – KConrad
    Jan 21, 2010 at 16:26
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    $\begingroup$ Serre pointed out to me once a paper of Chevalley in which C. made an error by not taking into account that a^(1/n) can have different behavior over a base field for different choices of n-th root. $\endgroup$
    – KConrad
    Jan 21, 2010 at 16:28
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    $\begingroup$ Here is the paper I mentioned above: Deux théorèmes d'arithmétique. J. Math. Soc. Japan 3 (1951), 36--44. There is a reference in the Math Reviews review that a result in this paper is used by Weil in a paper which is reviewed "above". That was back in the days when Math Reviews was in hard copy with reviews on paper, one after another. The paper by Weil came right before Chevalley's in the same journal: Sur la théorie du corps de classes. (French) J. Math. Soc. Japan 3, (1951). 1--35 $\endgroup$
    – KConrad
    Jan 21, 2010 at 16:34
  • $\begingroup$ @Emerton, I couldn't understand the 5th and 6th paragraph. In 2nd paragraph you said that the extension is unramified if $a$ is unit. While in the 5th and 6th paragraph, you said that the extension can be widely totally ramified assuming $a$ unit and $u \equiv 1$ mod $(p)$ not mod ($p^2$). Can you please explain? $\endgroup$
    – MAS
    Jan 12 at 13:42
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It's possible to delve more deeply into the ramification structure in the more complicated wild case referred to in Emerton's answer. If $n=p^m$, then Coleman calculated the conductor of the Kummer extension ${\mathbb Q}_p(\zeta_n,\sqrt[n]{a})/{\mathbb Q}_p(\zeta_n)$ using his reciprocity law and one should be able to use this to determine the ramification filtration of ${\mathbb Q}_p(\sqrt[n]{a})/\mathbb{Q}_p$.

Romyar Sharifi does just that here to determine the ramification groups of the maximal (nonabelian) Kummer extension ${\mathbb Q}_p\left(\sqrt[p^{\infty}]{{\mathbb Q}_p^\times}\right)/\mathbb{Q}_p$.

In particular, one finds that the (upper) ramification jumps are of the form $i$, $i+\frac{1}{p-1}$ and $i+\frac{1}{p(p-1)}$ for integers $i$.

In a similar vein and also by computing conductors, Viviani finds the ramification groups of the extension ${\mathbb Q}_p(\zeta_{p^m},\sqrt[p^m]{a})/{\mathbb Q}_p$. as long as $p^2$ doesn't divide $a$.

One trick he uses is to notice that for instance if $p$ exactly divides $a$ then $$\frac{(1-\zeta_p)}{\sqrt[p]{a}\;\sqrt[p^2]{a}\cdots\sqrt[p^m]{a}}$$ is a uniformizer of ${\mathbb Q}_p(\zeta_p,\sqrt[p^m]{a})$ and observes that the proofs could be simplified and generalized if one was able to write down uniformizers in further extensions.

So does anyone know how to do this ? Can one explicitly write down a uniformizer for the field ${\mathbb Q}_p(\zeta_{p^m},\sqrt[p^m]{a})$ in a similar way ?

At a stretch, this might have some application in integral $p$-adic Hodge theory. The arithmetically profinite extension $K(\zeta_{p^\infty},\sqrt[p^\infty,]{\pi})/K$ where $\pi$ is a prime in $K$ makes an appearance in Tong Liu's extension of Breuil and Kisin's work and the more one knows about this extension and it's field of norms the better for calculations/proofs.

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