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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.

Consider the following family of polynomials in $K[x,y]$, where $K$ has characteristic zero:
$f_n(x,y)=(x+y)^n+(x-1)y,$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$...but I'm well and truly stuck. Any advice much appreciated! 2 added 146 characters in body Consider the following family of polynomials in$K[x,y]$, where$K$has characteristic zero:$f_n(x,y)=(x+y)^n+(x-1)y,$for$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.