Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $k$?

share|cite|improve this question
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
up vote 3 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.