Trouble understanding the general polynomial of degree $n$ Herstein defines the general polynomial of degree $n$ over a field $F$ as the polynomial $p(x) = x^n + a_1x^{n-1} + \cdots + a_n \in F(a_1, \cdots, a_n)[x]$. From here, since it was previously shown that the group of automorphisms of this field is $S_n$, we show that polynomial is not solvable by radicals over $F$.
But this whole result doesn't make much sense to me. Before, we treated of $F(x_1, \cdots, x_n)$ where these "$x_i$" were just symbols, like the "$x$" in a polynomial $p(x)$. And now, we are taking the coefficients of $p(x)$, which can very well be inside the field $F$.
For example, the polynomial $x^5 - 7x + 2$ is in $\mathbb{Q}[x]$, and so $\mathbb{Q}(a_1, \cdots, a_5) = \mathbb{Q}$, and this seems to contradict the earlier result that the group was $S_5$.
I know I am making a huge mess, but I just can't seem to really grasp the concept, and would love some sort of enlightenment. Since I am self studying, I don't really have a professor to go to.
Thanks a lot in advance!
 A: My mistake above was to think I had calculated the splitting field of $p(x)$, when, in reality I had calculated the field over which we are working on. Let me elaborate.
We want to get to the splitting field $F(x_1, \cdots, x_n)$, where these are the roots of a polynomial. What we have to start with are, by the Newton-Girard identities, the elementary symmetric functions on these roots, namely, the coefficients of the polynomial $p(x)$, and their generated field, $F(a_1, \cdots, a_n)$.
Dietrich Burde was right in his comment - nothing was wrong, except my interpretation of the problem. When I said $\mathbb{Q}(a_1, \cdots, a_5) = \mathbb{Q}$ in the example, I was absolutely right. The thing is, this "collapsed" field (by which I mean, the result after plugging in the actual particular values) is the particular field of elementary symmetric functions over the roots of this polynomial, which remain unknown. So I didn't obtain $F(x_1, \cdots, x_n)$, as I thought I had, but rather $F(a_1, \cdots, a_n)$.
And the theorem I stated in the question says that $\operatorname{o}(\operatorname{Aut}_{\mathbb{Q}}\mathbb{Q}(x_1, \cdots, x_5)) = 5!$, whereas, by my original reasoning, I thought I had $\operatorname{o}(\operatorname{Aut}_{\mathbb{Q}}\mathbb{Q}) = 1 \neq 5!$ as it should be.
EDIT: I believe the same reasoning explains the apparent contradiction that there are indeed quintics solvable by radicals: when actually plugging in the values of the roots into the field $F(x_1, \cdots, x_n)$, some of them may have already been in the previous field, so the underlying result is much more complex than the theorem on rational functions predicted and the bound may end up being much smaller than $n!$
