MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Over the p-adics, every Galois group is solvable. Does this imply that the quintic (and higher-order polynomials for that matter) can be solved by radicals over $\mathbb{Q}_p$?

EDIT: The original place I learned that the p-adic galois groups were solvable was in Milne's Algebraic Number Theory text (Chapter 7, Cor 7.59).

As was pointed out the comments, I should clarify that I meant to ask 2 questions. Namely, whether the general quintic can be solved by radicals in this context (still no) and whether any given one can be (which I now believe is yes).

share|cite|improve this question
@Jon: You may want to add the reference to the fact that the Galois groups of finite extensions of $\mathbb{Q}_p$ are solvable. It follows from the fact that the absolute Galois group of $\mathbb{Q}_p$ is pro-$p$ which, I think, was proved by Koch or Yakovlev in the 60s, and from the recent result of Segal and Nikolov about finite index subgroups of pro-$p$-groups. Is there a simpler proof? – Mark Sapir Apr 9 '12 at 13:18
Yes: the Hilbert ramification groups provide a family of normal subgroups of the decomposition group of a global extension. – Franz Lemmermeyer Apr 9 '12 at 13:33
@Franz: Thanks! – Mark Sapir Apr 9 '12 at 14:01
«The quintic can be solved by radicals» can mean either «the general quntic has a solution by radicals» or «the roots of every specific quintic can be constructed using radicals from the field of coefficients». It would probably be best if the question were explicit about which of the two meanins it is about. (The wording suggests to me the second meaning is intended...) – Mariano Suárez-Alvarez Apr 9 '12 at 16:41
OK, I am beginning to understand. Analogously: over the reals every Galois group is solvable, since any polynomial can be factored into factors of degrees 1 and 2. That doesn't mean there is a formula for the zeros in terms of the coefficients of the original polynomial. – Gerald Edgar Apr 9 '12 at 17:56

Even though every extension of $\mathbb{Q}_p$ is solvable, I don't think one can write down a formula for the solution to a general quintic in terms of radicals; if one could, then the $S_5$-extension $\mathbb{Q}_p(r_1,...,r_5)/\mathbb{Q}_p(e_1,...,e_5)$ would be solvable, where the $e_i$ are the elementary symmetric polynomials in the $r_i$.

share|cite|improve this answer
What I meant was that there doesn't exist a formula for the roots of a quintic in terms of the coefficients. Here, the $e_i$ are of course the coefficients. It is weird, though -- the roots can all be expressed using radical operations, and they only depend on the coefficients, but there isn't a general formula in terms of radicals of the coefficients to spit them out! – Raju Apr 9 '12 at 14:56
@RK: Yes, it is a nice illustration of the fact that Abel's and Galois' formulations are different. – Mark Sapir Apr 9 '12 at 15:10
This answer and the comments are enlightening, thanks! – Martin Brandenburg Apr 9 '12 at 20:27
Nevertheless, Lazard and Dummit give a formula analogous to Cardano's for solving any solvable quintic; see my second answer below. – Chandan Singh Dalawat May 26 '12 at 7:50

If you want to solve the quintic over the $p$-adics, two cases naturally arise : $p\neq5$ and $p=5$.

Suppose first that $p\neq5$ and let $f\in\mathbf{Q}_p[T]$ be an irreducible polynomial of degree $5$. Then the extension $K$ obtained by adjoining a root $\alpha$ of $f$ to $\mathbf{Q}_p$ is always contained in $F(\root5\of{F^\times})$, where $F=\mathbf{Q}_p(\root5\of1)$. So you see immediately that $\alpha$ can be expressed by radicals.

In fact, $f$ can always be taken to be of the form $f=T^5-x$ in the generic case when $K$ is (totally) ramified over $\mathbf{Q}_p$, so if you wish I can claim to have solved the quintic by radicals by just saying that $\alpha=\root5\of x$.

Now let $f\in\mathbf{Q}_5[T]$ be an irreducible polynomial of degree $5$. Then the extension $K$ obtained by adjoining a root $\alpha$ of $f$ to $\mathbf{Q}_5$ is always contained in $F(\root5\of{F^\times})$, where $F=\mathbf{Q}_5(\root4\of{\mathbf{Q}_5^\times})$. Here again you see that $\alpha$ can be expressed by radicals.

There is nothing special about the prime $5$ or the base field $\mathbf{Q}_p$. You can replace $5$ by any prime $l$, and $\mathbf{Q}_p$ by any finite extension thereof, and you will get similar results depending on whether $l\neq p$ or $l=p$.

You can even replace $\mathbf{Q}_p$ by a finite extension of $\mathbf{F}_p((\pi))$, provided you replace the "radical" $\root p\of{x}$ (which denotes a root of the binomial $T^p-x$) by its characteristic-$p$ cousin $\wp^{-1}(x)$ (which denotes a root of the trinomial $T^p-T-x$).

share|cite|improve this answer

The fact is true for every field $K$ of characteristic 0: every finite algebraic extension of $K$ with solvable Galois group is inside an extension obtained by adding radicals (i.e. solutions of equations $x^n=a$). The field is separable since its characteristic is 0. Hilbert's Theorem 90 (see Google) holds over every field of characteristic 0. Therefore if a finite extension $K'$ of $K$ contains primitive roots of 1 of degree $n$, then any extension $E$ of $K'$ with cyclic Galois group of order $n$ is obtained by adding a root of the equation $x^n=a$ for some $a\in K'$. Now you can apply the fundamental theorem of Galois theory.

share|cite|improve this answer
So what is the solution of the general quinting over $Q_p?$ – Igor Rivin Apr 9 '12 at 14:03
@Kevin: Thanks! Is it only pro-solvable then? I thought Franz agreed above that it is pro-$p$. – Mark Sapir Apr 9 '12 at 15:23
I deleted my comments: these were wrong (because of Kevin's remark). – Mark Sapir Apr 9 '12 at 15:25
The simplest way to see that the absolute Galois group $G$ of $k=\mathbf{Q}_p$ is not a pro-$p$-group is to remark that $k$ has quadratic extensions (for example $k(\sqrt{p})$) for $p$ odd, and cubic extensions (for example $k(\root3\of2)$) for $p=2$. – Chandan Singh Dalawat Apr 9 '12 at 18:38
At the same time, it is true that the most interesting part of $G$ is a pro-$p$-group : there is a canonical filtration $V\subset T\subset G$ by closed subgroups in which the quotients $G/T$ and $T/V$ are fully understood, and the wild inertia subgroup $V$ is a pro-$p$-group. – Chandan Singh Dalawat Apr 9 '12 at 18:43

Dave Dummit's paper "Solving Solvable Quintics" constructs a sextic out of the coefficients of the quintic which has a rational root if and only if the quintic has a solvable Galois Group. Although this is stated over $\mathbb{Q}$ I believe that it applies over any field of characteristic 0 or > 5.

share|cite|improve this answer

Try Lazard (Daniel), Solving quintics by radicals, in The legacy of Niels Henrik Abel, 207–225, Springer, Berlin, 2004.

MR2077574 (2005g:12002) says : Let $F$ be a field of characteristic different from 2 and 5. Let $f$ be a univariate irreducible polynomial of degree 5 over $F$. The polynomial $f$ is said to be solvable by radicals if the Galois group over $F$ of the field generated by all the roots of $f$ is solvable. In the paper the author gives a formula for solving by radicals any polynomial $f$ of degree 5 which is solvable by radicals. The field extension which is generated by the radicals which appear in the result is always minimal, when only one root is produced, as well as when all roots are given. This formula has been implemented in Maple.

Reviewed by Jerzy Urbanowicz.

share|cite|improve this answer
This paper as well as its Maple implementation can be found on Lazard's home page: – Martin Brandenburg May 26 '12 at 10:31
Great ! I didn't succeed in locating his homepage, so many thanks for having done so. – Chandan Singh Dalawat May 26 '12 at 10:52

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.