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Consider the polynomial $f(x)=x^4+6x^3+32x^2+17x-15$ and let $\alpha\in\mathbb{C}$ be a root of $f$. How can I show that $\mathbb{Q}(\alpha)$ has no subfield of degree 2 over $\mathbb{Q}$? I have an hint: consider the Galois group of $f$ over $\mathbb{F}_2$ and $\mathbb{F}_3$.

The Galois group over $\mathbb{F}_2$ is $\mathbb{Z}/4\mathbb{Z}$ and over $\mathbb{F}_3$ is $\mathbb{Z}/3\mathbb{Z}$, so $G:=Gal_\mathbb{Q}(f)$ has a 4-cycle and a 3-cycle (and of course $G<S_4$). I don't know how to continue, any help?

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1 Answer 1

up vote 3 down vote accepted

The Galois group $G$ is a subgroup of $S_4$ containing a four-cycle and a three-cycle. $G$ thus has order at least $4 \cdot 3=12$, but $G$ is not $A_4$ (since a four-cycle is odd), so $G$ is all of $S_4$.

A degree $2$ subextension of the splitting field $F$ of $f$ corresponds to an index $2$ subgroup of $G$. Thus there is exactly one degree $2$ subextension, $K$ say, corresponding to $A_4$. Suppose $K$ is contained in $\mathbb{Q}(a)$. Then (since $[\mathbb{Q}(a):\mathbb{Q}]=4$), $[\mathbb{Q}(a):K] =2$. Thus $\operatorname{Gal}(F/\mathbb{Q}(a))$ is an index $2$ subgroup of $\operatorname{Gal}(F/K)=A_4$. But $A_4$ has no index $2$ subgroups.

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