I have been using the following result:

Given a polynomial $f(x,t)$ of degree $n$ in $\mathbb{Q}[x,t]$, if a rational specialization of $t$ results in a separable polynomial $g(x)$ of the same degree, then the Galois group of $g$ over $\mathbb{Q}$ is a subgroup of that of $f$ over $\mathbb{Q}(t)$.

However, I have been unable to prove this for myself, and cannot seem to find a proof of it anywhere. Is there an elementary proof? And if not, can anyone direct me to a source containing one, or at least explain the general principle?

My need to understand the result arose from considering the following:

If I specialize $t$ such that $g$ factorizes as $x^k.h(x)$, where h is an irreducible polynomial of degree $n-k$, is it legitimate to surmise that the Galois group of $h$ over $\mathbb{Q}$ is a subgroup of the original?

If nothing else, I'd be very grateful for an answer to this!

I have been using the following result:

Given a polynomial $f(x,t)$ of degree $n$ in $\mathbb{Q}[x,t]$, if a rational specialization of $t$ results in a separable polynomial $g(x)$ of the same degree, then the Galois group of $g$ over $\mathbb{Q}$ is a subgroup of that of $f$ over $\mathbb{Q}(t)$.

However, I have been unable to prove this for myself, and cannot seem to find a proof of it anywhere. Is there an elementary proof? And if not, can anyone direct me to a source containing one, or at least explain the general principle?

My need to understand the result arose from considering the following:

If I specialize $t$ such that $g$ factorizes as $x^k.h(x)$, where $h$ is an irreducible polynomial of degree $n-k$, is it legitimate to surmise that the Galois group of $h$ over $\mathbb{Q}$ is a subgroup of the original?

If nothing else, I'd be very grateful for an answer to this!