I asked this [question at MSE](https://math.stackexchange.com/questions/1438481/a-question-on-galois-theory) 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 $G'$.



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> Is there a $\mathbb{Q}$-linear map $T$ over $\mathbb{Q}[x]$ such that for all polynomials $f\in \mathbb{Q}[x]$ we have $$Gal(T(f))=(Gal(f))'$$
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For a related post see the following question:


[Endofunctors on the category of groups which are Galois- related to a linear map on $\mathbb{Q}[x]$](https://mathoverflow.net/questions/219455/endofunctors-on-the-category-of-groups-which-are-galois-related-to-a-linear-map)



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 $Gal(f)$?
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 $\deg F\ge2$.


