Timeline for Which quartic fields contain the 4th roots of unity in their Galois closure?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 27, 2016 at 13:23 | vote | accept | Stanley Yao Xiao | ||
| S Oct 27, 2016 at 9:01 | history | suggested | Aurel | CC BY-SA 3.0 |
made title match the actual question
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| Oct 27, 2016 at 8:46 | review | Suggested edits | |||
| S Oct 27, 2016 at 9:01 | |||||
| Oct 26, 2016 at 22:21 | answer | added | R.P. | timeline score: 7 | |
| Oct 26, 2016 at 21:57 | comment | added | R.P. | Yes, I realized that my claim about $D_4$ was too hasty. I misremembered an algebra exercise from long ago (which is why I thought it was elementary). In the $D_4$ case, we must have a unique quadratic subextension $K/F/\mathbb{Q}$, whereas $K'/\mathbb{Q}$ acquires two more. I don't immediately see how to compute these, although there must be simple formulas in terms of generators of $K$ over $F$ and of $F$ over $\mathbb{Q}$... | |
| Oct 26, 2016 at 21:49 | comment | added | Gerry Myerson | @René, if $K$ is $Q(\root4\of2)$, then the Galois group of the normal closure $K'$ is the dihedral group, and $i$ is in $K'$ but not in $K$. | |
| Oct 26, 2016 at 21:41 | comment | added | Noam D. Elkies | $x^4 + x^3 + x^2 - 2$ has Galois group $S_4$ and discriminant $-46^2$. [Ask gp: for(a=0,2,for(b=-2,2,for(c=-2,2,for(d=-2,2,P=Pol([1,a,b,c,d]);if(polisirreducible(P),if(issquare(-poldisc(P)),if(polgalois(P)[1]==4!,print(P)))))))) ] | |
| Oct 26, 2016 at 21:26 | comment | added | R.P. | I think you must make a case distinction according to the Galois group. If the Galois group of $K'/\mathbb{Q}$ is $C_4$, $V_4$ or $D_4$, and $i \in K'$, then $i$ must be in $K$ already. (For $D_4$, this is an elementary field theory exercise.) If the group is $A_4$, then there are no index $2$ subgroups, so $i \in K'$ can't occur. If the group is $S_4$, then $i \in K'$ is equivalent to $\mathbb{Q}(i)$ being the unique quadratic subextension of $K'/\mathbb{Q}$, so this is equivalent to the discriminant of $K$ being $-1$ times a square. (I wonder if this last case actually occurs, though.) | |
| Oct 26, 2016 at 21:09 | history | edited | GH from MO |
edited tags
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| Oct 26, 2016 at 21:08 | history | asked | Stanley Yao Xiao | CC BY-SA 3.0 |