Timeline for Is Lehmer's polynomial solvable?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 2, 2015 at 19:18 | vote | accept | Stanley Yao Xiao | ||
| Dec 2, 2015 at 1:40 | comment | added | Noam D. Elkies | @YCor already answered by David Speyer, but not necessary for the original question once you know that solvable Galois $\Leftrightarrow$ solvable by radicals: clearly if $x$ satisfied this condition then so would $y = x + x^{-1}$. | |
| Dec 2, 2015 at 1:35 | comment | added | Robert Israel | If you insist on fully expanding them in all their glory, expressions such as the Salem root $x$ in the second paragraph are indeed rather fearsome, but they are not so bad if obtained in a sequence of steps, such as (I hope I got this right): $$ \eqalign{a_1 &= (748 + 12 i \sqrt{5871})^{1/3}\cr a_2 &= \sqrt{a_1 + 16 + 112/a_1}\cr a_3 &= \sqrt{12 \sqrt{6}/a_2 + 32 - a_1 - 112/a_1}\cr y &= (a_2 + a_3) \sqrt{6}/12\cr x &= \left(y + \sqrt{y^2-4}\right)/2\cr}$$ | |
| Dec 2, 2015 at 1:33 | comment | added | David E Speyer | @YCor If $M/L/K$ is a tower of fields, with $M$ and $L$ both Galois over $K$, then $Gal(M/K) \to Gal(L/K)$ is always surjective. Once proves this very early in a Galois theory course; there are a couple of different routes. | |
| Dec 2, 2015 at 1:17 | comment | added | YCor | Why is the homomorphism from the Galois group of the degree 10 Lehmer polynomial P(x) given by the OP to the Galois group of your degree 5 polynomial Q(y) is surjective? | |
| Dec 1, 2015 at 23:58 | comment | added | Noam D. Elkies | Oops, corrected now. I noticed this soon after posting (but by then I was e-incommunicado for an hour or so). I had already checked that the Galois group is not contained in $A_5$ (either because the discriminant is not square, or because the cycle structure $\{2,3\}$ is odd), and forgot to mention this argument that the Galois group cannot be $A_5$. Not that it matters for the present application because $A_5$ is also unsolvable . . . | |
| Dec 1, 2015 at 23:52 | history | edited | Noam D. Elkies | CC BY-SA 3.0 |
Corrected group theory argument
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| Dec 1, 2015 at 22:53 | comment | added | Jarek Kuben | "a subgroup of $S_5$ of order divisible by $30$, and the only such subgroup is $S_5$ itself".. I guess such things happen even to the best of us :) | |
| Dec 1, 2015 at 21:59 | history | answered | Noam D. Elkies | CC BY-SA 3.0 |