Timeline for Decomposition and inertia groups, and reduction modulo primes
Current License: CC BY-SA 3.0
11 events
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| Jun 29, 2017 at 6:53 | comment | added | Lior Bary-Soroker | I am a bit confused. In Dedekind result, shouldn't the condition that $\bar{f}$ is separable be added? If it is true as stated, can one provide a reference? | |
| Feb 22, 2017 at 13:33 | comment | added | 352506 | Done. I also added the necessary hypothesis that $p$ does not divide any $e_i$. | |
| Feb 22, 2017 at 13:33 | history | edited | 352506 | CC BY-SA 3.0 |
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| Feb 22, 2017 at 13:26 | history | edited | 352506 | CC BY-SA 3.0 |
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| Feb 22, 2017 at 13:07 | comment | added | KConrad | Please fix the statement of Beckmann's theorem in your original question so it includes the hypothesis that you left out. | |
| Feb 22, 2017 at 12:31 | comment | added | 352506 | Unlike Dedekind's theorem, Beckmann's result has a chance of being generalized almost verbatim to the case $R=\mathbb Q[T]$, since the inertia groups would still be cyclic. I was wondering if this has been worked out somewhere. | |
| Feb 22, 2017 at 12:28 | comment | added | 352506 | @KConrad: I omitted an additional condition in Beckmann's result: writing $f(x)=g_1(x)^{e_1}\cdots g_m(x)^{e_m}+p\cdot h(x)$ in $\mathbb Z[x]$, the polynomial $\bar h(x)$ should not be divisible by any of the $\bar g_i(x)$. This condition is not met in your example because $h(x)=-12x^4 + x^2 - 75$, so $\bar h(x)=x^2$ is divisible by $\bar g_1(x)=x$. | |
| Feb 22, 2017 at 4:14 | comment | added | KConrad | permutation of the roots that is a 4-cycle on four roots and fixes the other two roots, which is an odd permutation of the roots. But the discriminant of the polynomial is a perfect square, so the action of ${\rm Gal}(N/\mathbf Q)$ on the roots is by even permutations only: the Galois group contains no 4-cycle on the roots. So something is wrong. I got this example from an answer I posted on MO a number of years ago: mathoverflow.net/questions/21247/… | |
| Feb 22, 2017 at 4:10 | comment | added | KConrad | I am suspicious about the way you describe Beckmann's result. Consider the polynomial $f(x) = x^6 - 35x^4 + 3x^2 - 225$. It is irreducible over $\mathbf Q$ and $f(x) \equiv x^4(x^2+1) \bmod 3$, so in your notation $g_1(x) = x$, $e_1 = 4$, $g_2(x) = x^2+1$, and $e_2 = 1$. You are saying that in the splitting field $N$ of $f(x)$ over $\mathbf Q$ that any prime lying over 3 has an inertia group generated by an element whose permutation action on the roots of $f(x)$ "includes" (do you mean equals?) a product of one 4-cycle and two 1-cycles. That means ${\rm Gal}(N/\mathbf Q)$ contains a [contd] | |
| Feb 22, 2017 at 4:05 | comment | added | KConrad | Dedekind's theorem relies on the residue fields of the Dedekind domain being finite, and thus having cyclic Galois groups. It extends to a base ring equal to the algebraic integers of any number field, whose residue fields are also finite fields. But with $\mathbf Q[T]$ as the base ring, its residue fields are number fields and decomposition groups need not be cyclic. What kind of substitute would you want? | |
| Feb 21, 2017 at 18:58 | history | asked | 352506 | CC BY-SA 3.0 |