I'm studying Galois theory at the moment. It seems to me that the fundamental motivation for the Galois group is the following.
We know that if a rational function of the roots is invariant under every permutation, then the value of that rational function is in the ground field. The Galois group is the set of permutations that makes the converse true.
To make this definition sensible, it needs to be the case that the converse is actually not true for the full symmetric group. Is this the case?
Specifically, the first thing my textbook (Edwards) proves about the Galois group (and also how Galois himself first introduced it), is:
Let $\psi$ be a polynomial in $n$ variables with coefficients in $K$. Then $\psi(r_1...r_n)\in K$ iff it's equal to $\psi(\sigma(r_1)...\sigma(r_n))$ for every permutation $\sigma$ in the Galois group.
The "if" is true even if we replace "Galois group" with "Symmetric group", so it seems to me that this theorem is only worth proving if the "only if" is only actually true for the Galois group. Or maybe I have it wrong - is it that the "only if" is actually true for the symmetric group, and the Galois group is something like the minimal group for which this is true? I suppose what I'm really asking is, to what degree and with what limitations does the above property characterize the Galois group?
You’ve left a lot of things unspecified in your question, and to answer it I have to make sure we’re on the same page on a couple of issues. When you say “the roots”, I’m assuming you’re talking about the set of all roots of a specific $K$-irreducible polynomial $P(X)$. Second, I’m usually discomfited by the expression “only if” when it’s used alone. Seems to me that “A only if B” means “if A, then B”, in other words “A $\Rightarrow$ B”. If I’m wrong on either of these, you can stop reading right now.
Your proposed statement for checking would thus be: “If $\psi(T_1,\cdots,T_4)\in K[T_1,\cdots,T_4]$ satisfies the condition that $\psi(r_1,\cdots,r_4)\in K$ then $\psi(r_1,\cdots,r_4)=\psi(r_{\sigma1},\cdots,r_{\sigma4})$ for all $\sigma\in\mathcal S_4$”. Here’s my counterexample: for my $K$, I’m going to take $\mathbb Q$, and for my $P(X)$ I’m going to take the fifth cyclotomic polynomial $X^4+X^3+X^2+X+1$, roots $r_j$ being $\zeta^j$ for $1\le j\le4$, and $\zeta$ being any primitive fifth root of unity. Your proposed statement is definitely not true: take $\psi=T_1T_4$. For, $\zeta\zeta^4=1$, but there are permutations of the four roots that destroy this relation.