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I thank J.C. and Arturo Magidin for interesting suggestions in the comments to this question

In this post the field of rational numbers is denoted by $\mathbb{Q}$. The space of polynomials with rational coefficients is denoted by $\mathbb{Q}[x]$. The commutator subgroup of a group $G$ is denoted by $G'$. The isomorphism between two groups is denoted by $\simeq$

We learn from the answer to the following question

Is there a $\mathbb{Q}$-linear map $T$ on $\mathbb{Q}[x]$ such that for all polynomials $\mathrm{Gal}(T(f))$ is the commutator subgroup of $\mathrm{Gal}(f)$?

that there does not exist a $\mathbb{Q}$-linear map $T$ on $\mathbb{Q}[x]$ such that for all $f\in \mathbb{Q}[x]$ we have $$Gal(T(f))\simeq(Gal(f))'$$

Motivated by the above fact we ask the following question:

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

What, if any, are some non trivial examples of such situation? (A non identity functor?Some interesting examples?)

Improve this question
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  • $\begingroup$ @j.c. $\mathcal{F}$ is a functor on category of groups, for example, in the linked question I considered $\mathcal{F}= commutator$ and according to the answer to that question I realized that this functor is Galoois related to no linear map $T$. $\endgroup$
    – Ali Taghavi
    Commented Sep 28, 2015 at 12:07
  • $\begingroup$ Thanks, I was confused! It might help to include a little more background from the other question to keep this one self-contained. Also it may help to use the term "endofunctor". $\endgroup$
    – j.c.
    Commented Sep 28, 2015 at 12:09
  • $\begingroup$ @DimaSustretov $T$ is $\mathbb{Q}$-linear on vector space $\mathbb{Q}[x]$. Please see the linked question. $\endgroup$
    – Ali Taghavi
    Commented Sep 28, 2015 at 12:11
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    $\begingroup$ I think the title is misleading. You are not seeking a "functorial approach to Galois theory". You are asking a question about functors and Galois theory, but you are not trying to approach Galois theory functorially. $\endgroup$
    – Arturo Magidin
    Commented Sep 28, 2015 at 16:27
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    $\begingroup$ I don't see any reason to expect that such things exist, or any use to which they might be put if they did. The operation sending a polynomial to its Galois group is very sensitive and I don't expect it to have basically any good behavior. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 28, 2015 at 21:11

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