Timeline for Finding the inertia group
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 16, 2015 at 14:42 | comment | added | Pablo | @WillSawin Magma factors the polynomial over this field and gives all the information. | |
| Aug 16, 2015 at 14:41 | comment | added | Pablo | @PeterMueller you are right! The inertia group is of order $4$ and Magma shows both this, and the Galois group $D_8$. I will accept you answer (it would be even better if you include your last argument there). | |
| Aug 16, 2015 at 14:28 | comment | added | Peter Mueller | @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. | |
| Aug 16, 2015 at 14:04 | comment | added | Will Sawin | @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 $\mathbb Z/2 \times \mathbb Z/2$. | |
| Aug 16, 2015 at 14:02 | comment | added | Will Sawin | @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 $2$. | |
| Aug 16, 2015 at 4:30 | comment | added | Pablo | @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? | |
| Aug 16, 2015 at 0:32 | comment | added | grghxy | 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. | |
| Aug 15, 2015 at 23:44 | history | answered | Will Sawin | CC BY-SA 3.0 |