Irreducible Polynomials, and Galois Groups I have a small question about some concepts.
I know that if I have a polynomial $p(x)$ over a field $K$ and an extension field $F$, that elements of $\text{Gal}(F/K)$ permute the roots of $p(x)$ which lie in in $F$. 
I know that in general not all permutations are allowed, but suppose I have an irreducible polynomial, not necessarily monic. Then is every permutation of the roots an element of $\text{Gal}(F/K)$? 
Thanks!
 A: No. The size of the Galois group is the same as the degree of the extension. Suppose $p$ has degree $n$. Then if $p$ splits completely when you adjoin a single root, like for example the minimal polynomial of $\zeta_n=e^{2\pi i/n}$ over $\Bbb{Q}$, $\Phi_n$, the degree of the extension will be the degree of $p$, or $n$. But then there are $n$ automorphisms of the field, but $n!$ permutations of the $n$ roots of $p$.
To see that $\Phi_n$ splits completely, note that $\zeta_n$ is a root of $x^n-1$, and the roots of $x^n-1$ are $\zeta_n^k$ for $k=0,\ldots,n-1$, so $x^n-1$ splits completely in $\Bbb{Q}[\zeta_n]$. Since $\Phi_n$ is the minimal poly for $\zeta_n$, we have $\Phi_n \mid x^n-1$, so it splits completely in $\Bbb{Q}[\zeta_n]$ as claimed.
In fact, in order to get every permutation of the roots, you can never add more than one root every time you extend the field to create the splitting field, except for the last extension, which will necessarily be quadratic, and one more extension will give the remaining two roots. 
