Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.