3
$\begingroup$

In this question, the field of rational numbers is denoted by $\mathbb{Q}$. The Galois group of a polynomial $f$ is denoted by $Gal(f)$. The commutator subgroup of a group $G$ is denoted by $G'$.

Is there a $\mathbb{Q}$- linear map $T$ over $\mathbb{Q}[x]$ such that for all polynomials $f\in \mathbb{Q}[x]$ we have $$Gal(T(f))=(Gal(f))'$$

More generally we can consider the following situation;

Assume that $T$ is a a linear map over $\mathbb{Q}[x]$ and $\mathcal{F}$ is a functor over the category of groups. we say that $T$ is Galois-related to $\mathcal{F}$ if we have $$Gal(T(f))=\mathcal{F}(Gal(f))\;\;\;\text{For all polynomial } f \in \mathbb{Q}[x]$$

What are some non trivial examples of such situation?

$\endgroup$
  • 10
    $\begingroup$ I would prefer if the field of rational numbers were not denoted by $\mathbb{Q}[x]$. $\endgroup$ – guest Sep 16 '15 at 19:10
  • $\begingroup$ @guest lollllll! $\endgroup$ – 6005 Sep 16 '15 at 19:23
  • $\begingroup$ I’m not sure what a “$\Bbb Q$-linear map over $\Bbb Q[x]$” might be. Would that be a $\Bbb Q$-linear transformation of this infinite-dimensional $\Bbb Q$-space? $\endgroup$ – Lubin Sep 16 '15 at 21:13
  • $\begingroup$ @guest thanks for the comment. I revised that. $\endgroup$ – Ali Taghavi Sep 23 '15 at 13:01
  • $\begingroup$ @Lubin Yes a linear map over $\mathbb{Q}[x]$ where $\mathbb{Q}[x]$ is counted as a $\mathbb{Q}$-vector space. $\endgroup$ – Ali Taghavi Sep 23 '15 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.