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It is known that a finite extension $K/\mathbb{Q}_{p}$ is totally ramified if and only if $K = \mathbb{Q}_{p}(\alpha)$ where $\alpha$ is a root of Eisenstein polynomial. Is there any totally ramified extension that is not of the form $\mathbb{Q}_{p}(\sqrt[n]{pu})$ for some $u\in \mathbb{Z}_{p}^{\times}$? Every degree 2 totally ramified extensions has this form, but I don't know whether this is also true for degree 3 or higher. Thanks in advance.

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    $\begingroup$ Lately I’ve been convincing myself of things that were not true, but I have convinced myself just now that $\Bbb Q_2(\sqrt3\,)$ is not of the form you claim for quadratic extensions. Could you check? $\endgroup$ – Lubin Apr 21 at 2:42
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    $\begingroup$ Yes, this is a phenomenon of wild ramification, when the residue characteristic divides the degree of the extension. For instance, I believe that $X^3+3X+3$ over $\Bbb Q_3$ gives another example. It’s a matter of the Herbrand transition function, or if you like, the location of the “breaks” in the ramification filtration. $\endgroup$ – Lubin Apr 21 at 5:10
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    $\begingroup$ @Lubin If you find the time to flesh that out in an answer, I'm sure I won't be the only user who will appreciate it :-) $\endgroup$ – Jyrki Lahtonen Apr 21 at 7:12
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    $\begingroup$ I’ll see what I can do, @JyrkiLahtonen. Got lots on my plate the next 48 hours. $\endgroup$ – Lubin Apr 21 at 18:10
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    $\begingroup$ Thanks, @Lubin. We are not in a hurry here. $\endgroup$ – Jyrki Lahtonen Apr 21 at 18:11
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In response to Jyrki Lahtonen’s request, I’ll make an attempt to describe what’s going on here.

The Hasse-Herbrand transition function is a concave polygonal real-valued function on $\Bbb R$ that encapsulates much (but not all) of the information that comes out of the study of the higher ramification of a separable extension of local fields. You can read all about the subject in Chapter IV of Serre’s Corps Locaux (translated as Local Fields). What you see below will look nothing like Serre’s treatment, however. The least of the differences is that the traditional coordinatization of the plane, as in Serre, places the vertex describing the tame part of a totally ramified extension at the origin. My coordinatization puts this vertex at $(1,1)$.

Part I is to describe the Newton Copolygon. I won’t relate it to the more familiar Polygon, but you will see the connection. Let $f(X)=\sum_na_nX^n\in\mathfrak o[X]$, where for specificity’s sake I will suppose that $\mathfrak o$ is the ring of integers in a finite extension $k$ of $\Bbb Q_p$, and that we are using the (additive) valuation $v$ on $k$ normalized so that $v(p)=1$. For each nonzero monomial $a_nX^n$, draw the half-plane $\Pi_n$ described in $\Bbb R^2$ as all points $(\xi,\eta)$ satisfying $\eta\le n\xi+v(a_n)$. Then form the convex set $\bigcap_n\Pi_n$. This is the copolygon, but I hope I won’t confuse matters too badly by calling the “copolygon function” the function $v_f$ whose graph is the boundary of the convex set just described. You see, for instance, that if $f(X)=pX+pX^2+X^3$ the copolygon’s boundary has only one vertex, at $(\frac12,\frac32)$, with slope $3$ to the left and slope $1$ to the right. You see without difficulty that as long as $g$ has no constant term, $v_{f\circ g}=v_f\circ v_g$.

Part II. Without saying what the “lower breaks” and “upper breaks” of the ramification filtration of the Galois group of a Galois extension $K\supset k\supset\Bbb Q_p$ are, I simply proclaim that the Herbrand function is the polygonal real-valued function $\psi^K_k$ whose only vertices are at each break point $(\ell_i,u_i)$. The lovely fact about the transition functions is that if $L\supset K\supset k$, then $\psi^L_k=\psi^K_k\circ\psi^L_K$. The transition function $\psi^K_k$ is an invariant of the extension, not depending on any choices.

Part III is to relate these two polygonal functions, though this is not the place to explain why they are connected. Though the traditional description of the transition function, as in Serre, always starts from a Galois group, you will notice that there is no mention of groups below. For simplicity, I’ll describe only $\psi^k_{\Bbb Q_p}$ for $k$ totally ramified over $\Bbb Q_p$, since that’s enough to answer Saewoo Lee’s question.

Let $\mathfrak o$ be the ring of integers of $k$, and $\pi$ a prime element (generator of the maximal ideal), and let $F(X)$ be the minimal $\Bbb Q_p$-polynomial for $\pi$. Form the polynomial $f(X)=F(X+\pi)$, so that $f$ has no constant term. Now take the copolygon function $v_f$ of this $f$, and stretch it horizontally by a factor of $e^k_{\Bbb Q_p}=[k:\Bbb Q_p]$, to get $\psi^k_{\Bbb Q_p}$. That is, $\psi^k_{\Bbb Q_p}(\xi)=v_f(\xi\,/\,e)$.

Let’s work out three examples, namely $\Bbb Q_2(\sqrt{2u}\,)$, $\Bbb Q_2(\sqrt3\,)$, and $\Bbb Q_3(\rho)$ where the minimal polynomial for $\rho$ is $X^3-3X-3$.

First, over $\Bbb Q_2$, a prime is $\pi=\sqrt{2u}$, minimal polynomial $F(X)=X^2-2u$, giving $f(X)=X^2+2\pi X$. The copolygon has unique vertex at $(\frac32,3)$, and the transition function has unique vertex at $(3,3)$. (The initial segment of $\psi^K_k$ will always have slope $1$.)

Second, over $\Bbb Q_2$, a choice for a prime of $\Bbb Z_2[\sqrt3\,]$ is $\sqrt3-1$, with minimal polynomial $F(X)=X^2+2X-2$, so that $f(X)=X^2+2\pi X+2X=X^2+2(1+\pi)X$. The polygon has its one vertex at $(1,2)$, so that $\psi$ has its vertex at $(2,2)$, enough to show that $\Bbb Q_2(\sqrt3\,)$ is not of form $\Bbb Q_2(\sqrt{2u}\,)$.

Third, over $\Bbb Q_3$ with $F(X)=X^3-3X-3$, we get $f(X)=X^3+3\rho X^2+3\rho^2X-3X$, in which only the monomials $X^3$ and $3(\rho-1)X$ count, so that the copolygon has its vertex at $(\frac12,\frac32)$, and the transition function’s vertex is at $(\frac32,\frac32)$.

I’ll leave it to you to show that the vertex of the transition function for $\Bbb Q_3(\sqrt[3]{3u}\,)$ is at $(\frac52,\frac52)$. (Don’t be surprised that these vertices don’t have integral coordinates. That’s guaranteed only for normal, abelian extensions, by Hasse-Arf, and the cubic extensions here are neither.)

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There is a general theorem that every tamely totally ramified extension of $\mathbf Q_p$ with degree $n$ has the form $\mathbf Q_p(\sqrt[n]{\pi})$ for some prime $\pi$ in $\mathbf Z_p$, so $\pi = pu$ for a unit $u$ in $\mathbf Z_p$. (There is a similar theorem over other local fields.) So if you want a totally ramified extension not of that form you need $n$ to be divisible by $p$.

Let's try $n=p$. Something we can say about extensions $\mathbf Q_p(\sqrt[p]{pu})$ for $p>2$ is that they are not Galois over $\mathbf Q_p$: a field $K$ containing a full set of roots of $x^p - pu$ must contain the nontrivial $p$th roots of unity, and those have degree $p-1$ over $\mathbf Q_p$ so $[K:\mathbf Q_p]$ is divisible by $p-1$. Therefore $[K:\mathbf Q_p] \not= p$ when $p>2$. Thus a Galois totally ramified extension of $\mathbf Q_p$ having degree $p$ can't have the form $\mathbf Q_p(\sqrt[p]{pu})$.

Every totally ramified abelian Galois extension of $\mathbf Q_p$ with degree divisible by $p$ contains a subextension with degree $p$ since the Galois group has a subgroup of index $p$: in an abelian group of order $n$ there is a subgroup of each order dividing $n$ and thus also a subgroup of each index dividing $n$ by using a subgroup of order equal to the complementary factor in $n$ of the desired index. Subextensions of totally ramified extensions are totally ramified and subextensions of abelian Galois extensions are abelian Galois extensions. Thus all we need to do now is find a totally ramified abelian Galois extension of $\mathbf Q_p$ with degree divisible by $p$ and inside of it there are extensions of degree $p$, all of which are examples of the kind being sought (not having the form $\mathbf Q_p(\sqrt[n]{pu})$).

The easiest choice is a cyclotomic extension: $\mathbf Q_p(\zeta_{p^2})$ where $\zeta_{p^2}$ is a root of unity of order $p^2$. This field has degree $p^2-p$ over $\mathbf Q_p$, with cyclic Galois group $(\mathbf Z/p^2\mathbf Z)^\times$, so the field contains a unique subextension with degree $p$ over $\mathbf Q_p$, namely the field fixed by the unique subgroup of the Galois group with order $(p^2-p)/p = p-1$. That subgroup is the solutions to $a^{p-1} \equiv 1 \bmod p^2$, and a generator of this extension over $\mathbf Q_p$ is $\sum_{a^{p-1} = 1} \zeta_{p^2}^a$ where the sum runs over all solutions of $a^{p-1} \equiv 1 \bmod p^2$.

Example when $p=3$: $a^2 \equiv 1 \bmod 9$ has solutions $\pm 1 \bmod 9$ and $\zeta_{9} + \zeta_9^{-1}$ has minimal polynomial $f(x) = x^3 - 3x + 1$. (Then $f(x+1) = x^3 + 3x^2 - 3$ and $f(x-1)$ is not Eisenstein at $3$.) I did my calculation of the minimal polynomial in $\mathbf C$, which is okay since a primitive $p$th-power root of unity has the same degree over $\mathbf Q_p$ as it does over $\mathbf Q$, so the structure of the intermediate fields in a $p$th-power cyclotomic extension over $\mathbf Q_p$ and over $\mathbf Q$ are the same.

Example when $p=5$: solutions to $a^4 \equiv 1 \bmod 25$ are $1, 7, 18$, and $24$, and $\zeta_{25} + \zeta_{25}^7 + \zeta_{25}^{18} + \zeta_{25}^{24}$ has minimal polynomial over $\mathbf Q_5$ equal to $g(x) = x^5 - 10x^3 + 5x^2 + 10x + 1$. (Note $g(x-1) = x^5 - 5x^4 + 25x^2 - 25x + 5$; the polynomial $g(x+1)$ is not Eisenstein at $5$.)

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  • $\begingroup$ Thanks! So if I understood correctly, the two examples are Galois extension of degree $p$ which can't be of the form, and they are Galois because they are subextensions of cyclotomic extension (which is abelian). Is this right? $\endgroup$ – Seewoo Lee Apr 23 at 20:13
  • $\begingroup$ That is correct. $\endgroup$ – KCd Apr 23 at 20:28

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