# Quartic extensions

Let $k$ be a perfect field and let $K/k$ be a quartic extension. Let $L/k$ be the normal (= Galois) closure of $K/k$. What structure can the Galois group $\text{Gal}(L/k)$ have? Of course it can be $\mathbb{Z}/4\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ if $K/k$ is Galois, but what are the other possibilities?

Edit: does the answer change if we assume that $L/k$ contains a quadratic extension of $k$?

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–  Brandon Carter Oct 20 '11 at 14:27
This paper by Keith Conrad may be helpful. –  SL2 Oct 20 '11 at 14:28
Many books on Galois theory will discuss this. The possible groups are transitive subgroups of S_4. Up to isomorphism, besides the two groups you mentioned the options are S_4, A_4, and D_4. –  KCd Oct 20 '11 at 14:30

Being finite and separable the extension $K/k$ is necessarily simple, so $K=k(\alpha)$ for some element $\alpha\in K$. Let $p(x)$ be the minimal polynomial of $\alpha$. The field $L$ is then the splitting field of $\alpha$, so over $L$ we have a factorization $$p(x)=(x-\alpha_1)(x-\alpha_2)(a-\alpha_3)(x-\alpha_4)$$ with $\alpha=\alpha_1$. The Galois group $G=Gal(L/k)$ can then be any transitive subgroup of the group of permutations of the set $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$. So $G\le S_4$, it must be transitive, and its order must be a multiple of 4. In addition to the ones you listed the dihedral group of 8 elements and the groups $A_4$ and $S_4$ are all possibilities.
Edit: To answer the added question. If $L/k$ contains a quadratic extension $F/k$, then Galois correspondence tells us that $G$ must have a subgroup $H$ of index two such that $F=Inv(H)$ (or $H=Gal(L/F)$). This rules out the possibility that $G\simeq A_4$. The other possibilities remain.
Translating the question about fields into the equivalent question about groups, we readily see that one can in principle obtain any group $G$ that has a subgroup of index 4 that doesn't contain a nontrivial normal subgroup of $G$ (since containing a normal subgroup would correspond to the quartic being contained in a smaller Galois extension). Equivalently, we want $G$ to have a core-free subgroup of index 4. Now, a group $G$ has a core-free subgroup of index $n$ if and only if it can be embedded as a transitive subgroup of the symmetric group on $n$ letters (exercise), so in principle, without further information about $k$, you can get precisely the transitive subgroups of $S_4$, which are $C_4$, $V_4$, $D_8$, $A_4$, and $S_4$. Of course, for any given $k$ not all these groups may be realisable as Galois group (e.g. if $k$ is finite, then you can only obtain $C_4$) but that's a different question.