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 $Gal(f)$? I  asked this  question at MSE but I did not receive an answer. So I ask it at MO:
We denote the field of rational numbers 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 $$Gal(T(f))=(Gal(f))'$$

For a related post see the following question:
Endofunctors on the category of groups which are Galois- related to a linear map on $\mathbb{Q}[x]$
 A: I haven't worked out the details, but I think the following shows that the answer must be no. If $T$ were such a map, it should decrease the degree of polynomials. Otherwise, it should be easy to come up with a polynomial $p$ whose image has bigger Galois group than $p$ does. Secondly, every quadratic polynomial has Abelian Galois group, so that its image must factorize over $\mathbb Q$. How hard could it be to show that $T(X^2)\in\text{lin}(1,X)$? Similarly for $T(1)$ and $T(X)$. Now $T$ has a kernel since it maps a 3-dimensional space into a 2-dimensional space, say $q=aX^2+bX+c$. Now for any polynomial $p$ that factors over $\mathbb Q$, so must $p+\lambda q$ for arbitrary $\lambda\in\mathbb Q$.
This cannot be!
A: The answer is no. First we see that $T$ is injective: Let $0\ne g\in\mathbb Q[X]$ with $T(g)=0$. Then $\textrm{Gal}(f-\alpha g)'=\textrm{Gal}(f)'$ for all $f\in\mathbb Q[X]$ and $\alpha\in\mathbb Q$. Pick $\beta\in\mathbb Q$ with $g(\beta)\ne0$, and $f\in\mathbb Q[X]$ of degree $n$ larger than $\deg g$ and $2$ such that $\textrm{Gal}(f)=S_n$. Then $\textrm{Gal}(f)'=S_n'=A_n$. But the polynomial $f(X)-\frac{f(\beta)}{g(\beta)}g(X)$ has the root $\beta$, so its Galois group is a subgroup of $S_{n-1}$. But no subgroup of $A_{n-1}=S_{n-1}'$ is isomorphic to $A_n$.
Thus we find $f\in\mathbb Q[X]$ of degree $2$ such that $F=T(f)$ has degree at least $2$. Since $\textrm{Gal}(F)=\textrm{Gal}(f)'=1$, $F$ splits into linear factors. Next pick $\gamma\in\mathbb Q$ such that $T(X+\gamma)=T(X)+\gamma T(1)$ has no common root with $F$. (Recall that $T(1)\ne0$.) Then $F$ and $G=T(X+\gamma)$ are relatively prime. As $f+\alpha(X+\gamma)$ has degree at most $2$ for all rational $\alpha$, we see that $F(X)+\alpha G(X)$ splits into linear factors for all rational $\alpha$. However, $F(X)+YG(X)$ is irreducible in $\mathbb Q[Y][X]$, so by Hilbert's irreducibility theorem there are infinitely many $\alpha\in\mathbb Q$ such that $F(X)+\alpha G(X)$ is irreducible. On the other hand, we saw that this polynomial factors into linear factors, so $\deg F\le 1$, contrary to $\deg F\ge2$.
