Consider the following family of polynomials in $K[x,y]$, where $K$ has characteristic zero:
$f_n(x,y)=(x+y)^n+(x-1)y^n,$
for $n\geq 3$. I can prove that $f_n(x,y)$ has an irreducible factor of degree $n-1$ in $x$. I also know that the galois group of $f_n$ over $K(y)$ is the symmetric group of degree $n-1$, but am having trouble proving this.
Here is an alternative form for $f_n$: make the substitution $x\rightarrow xy$ and divide by $y^n$. this gives:
$g_n(x,y)=(x+1)^n+yx-1.$
Substituting $-n$ for $y$ in $g_n(x,y)$ we get:
$g_n(x,-n)=x^2h(x),$
where $h(x)$ is separable (EDIT: $h(x)$ is the subject of this question).
I believe that, because there is one double root and the rest simple, we can somehow use Hensel's Lemma to show that the galois group of the original polynomial over $K(y)$ has a subgroup containing a transposition. But this is all a bit outside of my area of expertise, and I'm not sure about the details. Can anyone explain why this is true, or otherwise?
Alternatively, substituting $x\rightarrow x-1$ into $g_n(x,y)$ gives us a polynomial which factors as:
$(x-1)(x^{n-1}+x^{n-2}+\ldots +x^2+x+y+1)$
It seems as though it shouldn't be hard to show that some specialisation of $y$ into this gives a polynomial with galois group $S_{n-1}$ over $K$...but I'm well and truly stuck.
Any advice much appreciated!
