Let $f(x)$ be an irreducible polynomial of degree $4$ with rational coefficients, let $\alpha$ be a root of $f$ and set $L=\mathbb{Q}(\alpha)$ (say $\alpha \in \mathbb{C}$). Let $K$ be the splitting field of $f(x)$ over $L$ and assume that $L$ contains a quadratic subfield $\mathbb{Q} \subset M \subset L$ for $M$ a quadratic extension. Suppose $K$ properly contains $L$. Then, show that $K/\mathbb{Q}$ has as Galois group $D_{8}$.

The options for $\mathrm{Gal}(K/\mathbb{Q})$ (the extension is Galois) would be the transitive subgroups of $S_{4}$, namely the Klein 4-group, the cyclic group generated by $(1234)$, $D_{8}$, $A_{4}$ and $S_{4}$ (up to isomorphism).

Now, if $M$ is a quadratic extension of $\mathbb{Q}$, I have shown that $\alpha$ will satisfy a polynomial $g(x)$ of degree $2$ with coefficients in $M$ (in particular, this will be the minimal polynomial of $\alpha$ in $M$). Thus, it follows that $g|f$ in $M[x]$ since $\alpha$ is as well a root of $f$. One has:


However, I'm not sure on how to show that $[K:L]=2$ for the result to hold, and as well on how to use the fact that $g|f$ to get the result.

What would be the best way to proceed in this case? Thanks.

  • $\begingroup$ Did you mean $\Bbb{Q}\subset M\subset L$? Is $D_4$ or $D_8$ what you call the Sylow-2 subsgroup of $S_4$? You use both (and both are in use by various authors). $\endgroup$ – Jyrki Lahtonen Mar 28 '16 at 20:31
  • 2
    $\begingroup$ If my guess is correct, the point is that, by Galois correspondence, the intermediate field $M$ corresponds to a subgroup of index two of $Gal(K/\Bbb{Q})$. If you search our site, the group theory tag in particular, you will learn that $A_4$ does not have index two subgroups. And also (consequently) $A_4$ is the only such subgroup of $S_4$. These observations will after due argument severely restrict your choices for the Galois group. $\endgroup$ – Jyrki Lahtonen Mar 28 '16 at 20:37

One can show that since $[L : \mathbb{Q}] = 4$ and $L$ has a quadratic subextension, $L = \mathbb{Q}(\beta)$ for some $\beta \in L$ such that the minimal polynomial of $\beta$ over $\mathbb{Q}$ is of the form $g(x) = x^4 + ax^2 + b$. Applying the quadratic formula to $g(x)$ shows that the degree of the splitting field over $\mathbb{Q}$ of $g(x)$, which is $K$, is at most $8$. Since by hypothesis $[K : L] \neq 1$, we have must have $[K : \mathbb{Q}] = 8$.

The fact that $L/\mathbb{Q}$ is not normal implies by the Galois correspondence that $\textrm{Gal}(K/L)$ is not a normal subgroup, hence $\textrm{Gal}(K/\mathbb{Q})$ is not abelian. It also implies that $\textrm{Gal}(K/\mathbb{Q}) \cong D_8$, since $Q_8$ is the only other non-abelian order $8$ group and all of the subgroups of $Q_8$ are normal.


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