6
$\begingroup$

Set $h(x) = x^5+x^4+x^3+x^2+x-1$, let $L$ be the splitting field of $h$ over $\mathbb{Q}$, and let $p$ be a prime of $L$ lying over $2$.

What is the isomorphism class of the inertia group $I_p$, and how do I find it?

I would be more happy with having some procedure for finding inertia groups, rather than only the isomorphism class of this specific group.

I would also like to find the field $O_L/p$, and the decomposition group $D_p$ if possible.

Some useful information is: $h$ is irreducible over the rational numbers, and $h(x)(x-1) = x^6 - 2x + 1$ which allows for an easy irreducible factorization mod $2$: $h(x) = (x + 1)(x^2 + x + 1)^2$.

Improve this question
$\endgroup$
3
  • $\begingroup$ As a first step, have you been able to compute the Galois group of the splitting field of $h$ over $\mathbf Q_2$? I worked out that it is $D_4$ (group of symmetries of a square). $\endgroup$
    – KConrad
    Commented Aug 16, 2015 at 4:10
  • $\begingroup$ @KConrad Sadly enough no, I have just used a computer to see that it factors as a linear factor times a quartic factor in $\mathbb{Q}_2$. How do I see that it is $D_4$? $\endgroup$
    – Pablo
    Commented Aug 16, 2015 at 4:20
  • $\begingroup$ You can use the local fields database of Jones and Roberts, or look it up in their paper (see 3.5, and 4.1 for identifying quartic types). Knowing how to compute the automorphism group helps, which is by Panayi root-finding as explained there. math.la.asu.edu/~jj/localfields hobbes.la.asu.edu/localfields/database.pdf $\endgroup$
    – ABCDveve
    Commented Aug 16, 2015 at 9:54

2 Answers 2

Reset to default
3
$\begingroup$

This answer produces $I$ somewhat indirectly. So it actually does not what the OP asked for.

The decomposition group $D$, which is the Galois group of $h(x)$ over $\mathbb Q_2$, can be computed as follows: Using resultants, one sees that the minimal polynomial over $\mathbb Q$ of the difference of two distinct roots of $h(x)$ is \begin{equation} H(y)=y^{20}+6y^{18}+21y^{16}+46y^{14}-116y^{12}+694y^{10}+1837y^8-1810y^6-1527y^4+8560y^2+9584. \end{equation} From the factorization of $h(x)$ over $\mathbb Q_2$ we know that $D$ is a transitive subgroup of $S_4$. Over $\mathbb Q_2$ the polynomial $H(y)$ factors into irreducibles of degrees $4,4,4,8$. The only transitive subgroup of $S_4$ which has these orbit lengths on the $20$ pairs of distinct elements of $\{1,2,3,4,5\}$ is the dihedral group of order $8$.

As Will Sawin already remarked, the inertia group $I$ is a subgroup of $S_2\times S_2$. On the other hand, $D/I$, as a Galois group of an extension of a finite field, is cyclic. This forces $|I|\ge4$, because a dihedral group of order $8$ modulo a normal subgroup of order $\le2$ isn't cyclic. Thus $I=S_2\times S_2$.

There is a paper by Sybilla Beckmann which describes a method to compute inertia groups under certain restrictions. Her theorem does not apply here. The paper finishes with some remarks about how to possibly extend the methods.

Improve this answer
$\endgroup$
4
  • 2
    $\begingroup$ Knowing how $h(x) \bmod 2$ factors does not by itself tell you the residue field degree for the splitting field of $h(x)$ over $\mathbf Q_2$. Consider the polynomials $F(x) = x^2+1$ and $G(x) = x^2 + 3$. Both factor mod 2 as $(x+1)^2$, but their respective splitting fields over $\mathbf Q_2$ are $\mathbf Q_2(i)$ and $\mathbf Q_2(\sqrt{-3}) = \mathbf Q_2(\zeta_3)$. As extensions of $\mathbf Q_2$ the first of these is ramified and the second is unramified. $\endgroup$
    – KConrad
    Commented Aug 16, 2015 at 14:33
  • 1
    $\begingroup$ @KConrad: Thanks for pointing out the mistake. So I used an alternative argument which unfortunately is even more indirect. $\endgroup$
    – Peter Mueller
    Commented Aug 16, 2015 at 15:14
  • $\begingroup$ @PeterMueller Just being curious: If you were given this polynomial $H$ over $\mathbb{Q}$ without any connection to this problem, could you prove directly that it is irreducible? $\endgroup$
    – Pablo
    Commented Aug 20, 2015 at 9:21
  • 1
    $\begingroup$ @Pablo: The only thing I see: Modulo $5$ $H$ factors into degree $5$ irreducibles, and modulo $19$ it factors into degree $4$ irreducibles, so $H$ need to be irreducible. $\endgroup$
    – Peter Mueller
    Commented Aug 20, 2015 at 9:43
2
$\begingroup$

The inertia group acts on the roots and preserves the reduction modulo $2$. So it is a subgroup of the group of permutations of the roots preserving the reduction mod $2$, which is $S_2 \times S_2 = \mathbb Z/2 \times \mathbb Z/2$.

To compute which one it is, you need to use Hensel lifting to actually factor the polynomial into one linear and two quadratic factors over $W(\mathbb F_4)$ and see what you need to adjoin the square roots of. It's easy to see if two square roots correspond to nontrivial extensions and whether they correspond to the same extension.

The point is that you have a factorization mod $2$, and you can use it to get a factorization mod $4$, and mod $8$, etc., until the roots separate, and then you can compute it. Probably you will not have to go very far.

Improve this answer
$\endgroup$
7
  • 1
    $\begingroup$ I don't believe this leads to a method that works in general. To illustrate it is not enough to look at congruences deep enough to "separate roots", note that $(x-ip)(x+ip) = x^2$ in $(\mathbf{Z}/(p^2))[x]$ for every $0 < i < p$. The reason "separating" roots is not sufficient is that there is no unique factorization over $\mathbf{Z}/(p^n)$ for $n > 1$, so one needs more information (such as how ramified the splitting field really is) to know that some congruential factors are actually reductions of factors over the (perhaps quite ramified) splitting field. $\endgroup$
    – grghxy
    Commented Aug 16, 2015 at 0:32
  • $\begingroup$ @WillSawin I have just used Magma to see that $h$ factors as a linear factor times a quartic factor in $\mathbb{Q}_2$. What can we conclude from this? $\endgroup$
    – Pablo
    Commented Aug 16, 2015 at 4:30
  • 1
    $\begingroup$ @grghxy Yes, what I want to say is that you can keep looking at different congruences until you can compute $b^2-4ac$ modulo the group of squares in $W(\mathbb F_4)$, which includes the elements $1$ mod $8$. So we must go until the discriminant is nonzero and then go two extra powers of $2$. $\endgroup$
    – Will Sawin
    Commented Aug 16, 2015 at 14:02
  • 1
    $\begingroup$ @Pablo Peter Mueller's answer seems to give you what you want. Can Magma factor it over the degree 2 unramified extension of $\mathbb Q_2$. You should then get two quadratic factors, and whether the ratio of the discriminants of the two factors is a perfect square tells you whether the inertia group is $\mathbb Z/2$ or $\mathbb Z/2 \times \mathbb Z/2$. $\endgroup$
    – Will Sawin
    Commented Aug 16, 2015 at 14:04
  • 1
    $\begingroup$ @Will Sawin: Doesn't the fact that $h(x)$ has all its roots in $\mathbb F_4$ already show that $|D/I|\le 2$? Alternatively, $I=\mathbb Z/2$ isn't possible, because then $D/I$ weren't cyclic. $\endgroup$
    – Peter Mueller
    Commented Aug 16, 2015 at 14:28

You must log in to answer this question.

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