Timeline for Galois Groups of a family of polynomials

8 events

when toggle format	what		by	license	comment
Nov 5, 2010 at 0:59	history	edited	David E Speyer	CC BY-SA 2.5	added 114 characters in body
Nov 4, 2010 at 21:03	comment	added	Dror Speiser		@Victor: 71 is 3 mod 4, which is in accordance with the sign pattern above.
Nov 4, 2010 at 20:53	comment	added	Victor Miller		Also, the $p^{p-3}$ part comes from looking at the Newton polygon.
Nov 4, 2010 at 20:51	comment	added	Victor Miller		@David, thanks for catching that. I'll redo the calculation, but magma reports that $f_{71}$ has galois group the full symmetric group so the discriminant would have to be negative.
Nov 4, 2010 at 20:46	comment	added	David E Speyer		"If q is prime then the only way that both can vanish over $F_{q^k}$ is if $x=1$". Or if $q$ divides $n$ and $x$ is zero. But in that case we also have $q\|n(n+1)$, so your final conclusion is correct.
Nov 4, 2010 at 20:43	comment	added	Victor Miller		Observe that $(x-1)^2 f_n(x) = x^{n+1} - x - n(x-1)$ so the derivative is $(n+1)(x^n-1)$. If $q$ is prime then the only way that both can vanish over $\mathbb{F}_{q^k}$ is if $x=1$. But $f_n(1) = n(n+1)/2$, so the primes dividing the discriminant must divide $n(n+1)$.
Nov 4, 2010 at 20:42	history	edited	David E Speyer	CC BY-SA 2.5	added 248 characters in body
Nov 4, 2010 at 20:36	history	answered	David E Speyer	CC BY-SA 2.5