The Galois group of a specialized polynomial

Question

Let $P(t,x)\in\mathbb Q[t,x]$ be irreducible with Galois group $G$ over $\mathbb Q(t)$. It is known that if $t_0\in\mathbb Q$ is such that $P(t_0,x)$ is separable, then the Galois group of this specialized polynomial is isomorphic to a subgroup of $G$. Is this true also if $P(t_0,x)$ is not separable? I haven't been able to find a proof or a counterexample.

Answer 1

There's something a little misleading about the formulation of this question. Given such a polynomial $P(t,x)$, one has not only a Galois group $G$, but a natural action of $G$ on $d = \mathrm{deg}(P)$ points well defined up to conjugation. Given a separable specialization $P(t_0,x)$, it is true that the Galois group $H$ of this specialization is isomorphic (abstractly) to a subgroup of $G$, but it is also true that there is an inclusion $H \rightarrow G$ such that the action of $H$ on the $d$ roots of $P(t_0,x)$ is the restriction of the given permutation representation of $G$. When $P(t_0,x)$ is not separable, the Galois group $H$ no longer has any obvious action on $d$ points, and so one must settle for your weaker formulation that there merely exists an abstract inclusion from $H$ to $G$.

Still, the answer to this weaker formulation is still no. The easiest way to construct a counterexample is to consider isotrivial covers. Let $K/\mathbf{Q}$ be any field of degree $d$ with Galois group $G$. Let $\theta \in K$ be a primitive element. Let $\alpha \in K$ be any element at all. Now consider the minimal polynomial of $t \theta + \alpha$ over $\mathbf{Q}(t)$. This will define a degree $d$ polynomial $P(t,x)$. For a generic specialization $t = t_0 \in \mathbf{Q}$, the element $t_0 \theta + \alpha \in K$ will also be a primitive root, and so the Galois group will be $G$. But the specialization at $t = 0$ will have $\alpha$ as a root. In fact, the corresponding polynomial will be a power of the minimal polynomial of $\alpha$. So your claim is now the following: if $H$ is the Galois group of the splitting field of the minimal polynomial of $\alpha$, then $H$ is a subgroup of $G$. But this is absurd --- by Galois theory, $H$ is transparently a quotient of $G$. So it suffices to consider an example of a pair $(G,H)$ with $H$ a quotient of $G$ so that $H$ is not a subgroup of $G$. This doesn't happen for groups $G$ of extremely low order which perhaps indicate why you failed to find a counterexample. Perhaps the easiest example is $G = \mathrm{GL}_2(\mathbf{F}_3)$, which acts faithfully on $8$ points, and is a central extension of $\mathrm{PGL}_2(\mathbf{F}_3) = S_4$. If $S_4$ was a subgroup of $G$, it would have to be normal, but then $S_4$ has no automorphisms, so since $Z(S_4)$ is trivial this would force $G$ to be $S_4 \times \mathbf{Z}/2\mathbf{Z}$, which it is not. Thus the answer is no.

You can also easily write down an explicit example if you like. First write down a $G$-extension (say coming from the $y$-coordinate of the $3$-torsion of the elliptic curve $y^2 = x^3 + x + 1$)

$$K = \mathbf{Q}[\theta]/(961 - 558\theta^4 - 216 \theta^6 - 27 \theta^8),$$

which has Galois closure with Galois group $G = \mathrm{GL}_2(\mathbf{F}_3)$, and then let $\alpha = \theta^2$, whose Galois closure has Galois group $S_4$. Then let $P(t,x)$ be the minimal polynomial of $\alpha + t \theta$, which is:

$$923521 - 536238*t^4 - 207576*t^6 - 25947*t^8 - 2144952*t^2*x - 1245456*t^4*x - 207576*t^6*x - 1072476*x^2 - 1868184*t^2*x^2 - 518940*t^4*x^2 - 415152*x^3 - 415152*t^2*x^3 + 259470*x^4 + 120528*t^2*x^4 + 15066*t^4*x^4 + 241056*x^5 + 60264*t^2*x^5 + 76788*x^6 + 5832*t^2*x^6 + 11664*x^7 + 729*x^8$$

Then $P(t,x)$ has Galois group $G$ but $P(0,x) = (27x^4 + 216x^3 + 558x^2 - 961)^2$ has Galois group $S_4$.