TITLE: Uniformly solvable families of polynomials
QUESTION [13 upvotes]: It is a famous theorem that there is no "quintic formula", i.e. there is no formula which expresses the roots of a quintic polynomial $x^5+a_4x^4+\cdots+a_0$ in terms of $a_4,\ldots,a_0$ and rational constants using the field operations and taking $n$th roots. But the standard proof of this actually proves something which on its face seems stronger: that there are particular quintics with roots that cannot be expressed in terms of rational numbers using these operations. The non-existence of a quintic formula merely requires that there be no "uniform" way of doing so. I'm wondering if this is actually a stronger statement.
To make this rigorous: Given $f(a_0,\ldots,a_k,x)\in \mathbb Q(a_0,\ldots,a_k)[x]$, we can define the family of polynomials associated to $f$ to be the set of polynomials $f(t_0,\ldots,t_k,x)\in \mathbb Q[x]$ where $t_0,\ldots,t_k\in \mathbb Q$ are such that this expression is defined. For example, the family of polynomials associated to $x^2+a_1x+a_0$ is the set of monic quadratic polynomials. If there an $f\in \mathbb Q(a_0,\ldots,a_k)[x]$ which does not admit a formula for its roots (in the variable $t$) in terms of elements of $\mathbb Q(a_0,\ldots,a_k)$, field operations and taking $n$th roots, yet every polynomial in the family of polynomials associated to $f$ does?
From the point of view of Galois theory, the question becomes whether there exist polynomials $f\in \mathbb Q(a_0,\ldots,a_k)[x]$ which are not solvable over $\mathbb Q(a_0,\ldots,a_k)$, but evaluating them at any point $(t_0,\ldots,t_k)\in\mathbb Q$ (for which the result is defined) gives a polynomial solvable over $\mathbb Q$?

REPLY [5 votes]: The answer is no. The following is a theorem of Hilbert:

Let $f(x_1, x_2, \ldots, x_n, t)$ be a polynomial with rational
  coefficients. Let $K_{gen}$ be the splitting field of $f$ over
  $\mathbb{Q}(x_1, \ldots, x_n)$, thought of as a polynomial in $t$. For
  $a_1$, ..., $a_n$ in $\mathbb{Q}$, let $K(a_1, \ldots, a_n)$ be the
  splitting field of $f(a_1, \ldots, a_n, t)$ over $\mathbb{Q}$. There
  are infinitely many $(a_1,\ldots, a_n)$ for which
  $\mathrm{Gal}(K_{gen}/\mathbb{Q}(x_1, \ldots, x_n)) \cong \mathrm{Gal}(K(a_1,
\ldots, a_n)/\mathbb{Q})$.

Infinitely many is a very weak statement; the right statement is that this is true for all but a "thin set" of $n$-tuples of rational numbers. There are many theorems saying that thin sets are small. See this question for some more discussion in the case that $G$ is $S_n$.
The theorem which gets the famous name is Hilbert's Irreducibility Theorem: If $f(x_1, \ldots, x_n, t)$ is irreducible as a polynomial in $\mathbb{Q}(x_1, \ldots, x_n)[t]$ then, for all most all $(a_1, \ldots, a_n) \in \mathbb{Q}^n$, the polynomial $f(a_1, \ldots, a_n)(t)$ is irreducible as a polynomial in $\mathbb{Q}[t]$. The deduction of the result I stated from this one is pretty straightforward (it is the first application of Hilbert's irreducibility theorem mentioned in Wikipedia) but I had trouble finding you a reference that writes it out carefully. 
If you are happy with the language of modern algebraic geometry, I recommend the first three chapters of these notes by Serre. If you are willing to read something lengthy, this Master's thesis looks like a beautiful winding tour through Hilbert's Irreducibility Theorem and its many applications