Timeline for Is there a $\mathbb{Q}$-linear map $T$ over $\mathbb{Q}[x]$ such that for all polynomials $Gal(T(f))$ $\simeq$ The commutator subgroup of $Gal(f)$?

Current License: CC BY-SA 3.0

12 events

when toggle format	what		by	license	comment
Apr 13, 2017 at 12:58	history	edited	CommunityBot		replaced http://mathoverflow.net/ with https://mathoverflow.net/
Apr 13, 2017 at 12:19	history	edited	CommunityBot		replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Sep 29, 2015 at 18:12	history	edited	Ali Taghavi	CC BY-SA 3.0	added 45 characters in body
Sep 28, 2015 at 11:58	history	edited	j.c.	CC BY-SA 3.0	improve title
Sep 28, 2015 at 11:12	history	edited	Ali Taghavi	CC BY-SA 3.0	added 135 characters in body
Sep 28, 2015 at 11:05	history	edited	Ali Taghavi	CC BY-SA 3.0	deleted 393 characters in body
Sep 28, 2015 at 8:54	vote	accept	Ali Taghavi
Sep 25, 2015 at 22:22	answer	added	Peter Mueller		timeline score: 17
Sep 24, 2015 at 4:37	answer	added	Anthony Quas		timeline score: 4
Sep 23, 2015 at 22:16	comment	added	Arturo Magidin		The problem mixes multiplicative and additive structures of $\mathbb{Q}[x]$, which likely makes this very hard. For example, if such a $T$ existed, there would exist two polynomials $q$ and $p$, both of which split over $\mathbb{Q}$, with the property that $p-aq$ splits over $\mathbb{Q}$ if and only if $a$ is a perfect cube ($a\in\mathbb{Q}$), and if $a$ is not a perfect cube, then $p-aq$ is a product of a polynomial that splits and the power of an irreducible cubic polynomial with cyclic Galois group of order $3$ ($p=T(x^3)$, $q=T(1)$); it seems hard to find $p,q$, or show no such exist.
Sep 23, 2015 at 21:22	comment	added	Johannes Hahn		Can you say something about the motivation for this question? Why on earth should would one even suspect that such a thing might exist?
Sep 23, 2015 at 19:03	history	asked	Ali Taghavi	CC BY-SA 3.0