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Jan
28
awarded  Nice Answer
Jan
11
comment Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$
Thanks for the updates, that does seem to be less hassle than I had thought (although there's still too many computations for my liking!).
Jan
11
comment Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$
Thanks for adding the details, this is more or less what I expected (i.e. some explicit computations along the lines of Rene's answer) but it's nice to know I wasn't missing something slicker.
Jan
11
comment Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$
@egreg Sorry to unaccept, I did like this proof, but in the end the answer I came up with seemed to use fewer computations and can be applied to other situations (including when your base field doesn't necessarily lie in $\mathbb{C}$).
Jan
11
accepted Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$
Jan
11
comment Prove a finite group has non-normal Sylow p-subgroup of order p if $n_{p}\neq 1\pmod{p^2}$
@JiayuanW You're welcome.
Jan
3
answered Galois Group of $X^p - 2$, p-an odd prime
Jan
3
comment Prove a finite group has non-normal Sylow p-subgroup of order p if $n_{p}\neq 1\pmod{p^2}$
@JiayuanW I've updated the answer to better explain why (all but one of)the orbits have size divisible by $p^2$. In general, the behavior of intersections of Sylow subgroups is complex, I don't know if it can be described easily but my guess would be that the behavior can be as complicated as you could imagine.
Jan
3
revised Prove a finite group has non-normal Sylow p-subgroup of order p if $n_{p}\neq 1\pmod{p^2}$
added 80 characters in body
Jan
2
awarded  Nice Question
Jan
2
answered Prove a finite group has non-normal Sylow p-subgroup of order p if $n_{p}\neq 1\pmod{p^2}$
Jan
2
reviewed Reviewed Prove a finite group has non-normal Sylow p-subgroup of order p if $n_{p}\neq 1\pmod{p^2}$
Jan
1
answered Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$
Dec
30
comment Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$
Ah, that's great, the direct proof works out, thanks a lot. It's possibly worth noting that you don't actually need to compute real and imaginary parts, once you know that $x_1$ and $x_2$ are conjugates, the argument shows that $x_1+x_2$ has degree $4$ over $\mathbb{Q}$ without writing it out explicitly.
Dec
30
comment Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$
Kevin, thanks for your answer but the details are exactly what I'm having trouble with. Perhaps you could help fill them in?
Dec
30
comment Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$
@ReneSchipperus To clarify, I understand the method, my point is that there seems to be more work to it than you seem to realise, which is why I'm not such a big fan. But anyway, thanks again.
Dec
30
comment Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$
@ReneSchipperus Obviously there's some miscommunication going on here, but I don't know any way to explain myself better, so I think it would be best to let the issue lie. Thanks for your time.
Dec
30
comment Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$
What would be the best way to show that $\alpha$ is degree $4$ over $\mathbb{Q}$? It seems "obvious ", but then things like $\sqrt{\sqrt{20}+6}$ end up being only degree $2$... The minimal polynomial seems not obviously irreducible, although you can explicitly describe the (irrational) roots, so you can just show that it doesn't split as a product of quadratics, which isn't so bad, but then the whole part becomes a lot of computation when put together...
Dec
30
comment Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$
@ReneSchipperus It's not impossible, it just implies that all of your coefficients are $0$. This gives you the four equations I mention. Perhaps you should square out the expression and then collect like terms to see what I mean.
Dec
30
comment Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$
@Rene My point is that where you say "and if we square this we get a linear relationship between..." there is still quite a lot of computation to do before you can conclude $a=b=c=d=0$, which is the result that you want to end up with.