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Let $K$ be the splitting field of a polynomial $f\in \mathbb Q[x]$, which is irreducible $\mod 3$, with $G:=\text{Gal}(K|\mathbb Q)=S_n$ (symmetric group). Let $U$ be a subgroup of $G$ with fixed field $F:=\operatorname{fix}(U)$ 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 $F_\alpha(X):=X^3-3X-\dfrac{6\alpha x}{y} \in F[X]$ irreducible over $K$?

It looks like an Eisenstein polynomial.

Is here ramification theory needed?

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  • $\begingroup$ 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. $\endgroup$
    – Alex B.
    Commented May 31, 2012 at 14:41
  • $\begingroup$ Thank you for your answer. I forgot to mention that $gcd(\alpha,3)=1$. $\endgroup$
    – david75
    Commented May 31, 2012 at 14:50
  • $\begingroup$ 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 $\mathrm{gcd}(\alpha,3)=1$. $\endgroup$
    – François Brunault
    Commented May 31, 2012 at 16:30

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