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Let $K$ be the splitting field of a polynomial $f\in \mathbb Q[x]$, which is irreducible mod 3, with $G:=Gal(K|\mathbb Q)=S_n$ (symmetric group). Let $U$ be a subgroup of $G$ with fixed field $F:=fix(U)$ and write $F=\mathbb Q(\alpha)$. Now let $x,y \in \mathbb N$ with $gcd(x,3)=1$ and $gcd(y,3)=1$.

Why is the polynomial $F_\alpha(X):=X^3-3X-6\alpha\frac{x}{y} \in F[X]$ irreducible over $K$?

It looks like an Eisenstein polynomial.

Is here ramification theory needed?

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This is clearly false without further assumtions on $\alpha$. Take $U=G$, so that $F=\mathbb{Q}=\mathbb{Q}(\alpha)$ for any $\alpha\in \mathbb{Q}$. Take $x=y=1$, $\alpha=33$ for a concrete counterexample. –  Alex B. May 31 '12 at 14:41
    
Thank you for your answer. I forgot to mention that $gcd(\alpha,3)=1$. –  david75 May 31 '12 at 14:50
    
Your condition says that $3$ is inert in $K$. If $x$ is a root of $F_\alpha$ in $K$ then $3$ divides $x$ thus $3^2$ divides $6\alpha$, contradicting $\mathrm{gcd}(\alpha,3)=1$. –  François Brunault May 31 '12 at 16:30
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