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The groups whose subgroups are totally ordered by inclusion are easy to classify; they are subgroups of $\mathbb{Z}/p^{\infty} = \text{colim } \mathbb{Z}/p^k$ for some prime $p$, and thus $\mathbb{Z}/p^{\infty}$ or finite cyclic of prime power order. What about fields?

Is it possible to classify (or at least, give some properties and examples) the field extensions $E/F$, whose intermediate fields are totally ordered by inclusion?

If $E/F$ is a Galois extension, then we may rephrase the condition: The closed subgroups of $Gal(E/F)$ are totally ordered. It can be shown that then there is a prime $p$ such that $Gal(E/F)$ is pro-$p$-cyclic and thus isomorphic to $\mathbb{Z}/p^k$ for some $k \geq 0$ or to $\mathbb{Z}_p = \lim_k \mathbb{Z}/p^k$. Thus $E/F$ is built up out of cyclic Galois extensions $F_{i+1} / F_i$ degree $p$ (for example $E = \mathbb{F}_{q^{p^\infty}}, F = \mathbb{F}_q$). In characteristic $p$, cyclic Galois extensions of degree $p$ are characterized by a Theorem of Artin-Schreier. In characteristic $q \neq p$ ($q=0$ allowed), there is a characterization if $F_i$ contains a primitive $p$th root of unity. What can be said if this is not the case?

Now do not assume that $E/F$ is Galois. Here is a simple observation:

The intermediate fields of $E/F$ are totally ordered iff $E/F$ is algebraic and the finite intermediate fields of $E/F$ are totally ordered.

Proof: $\Rightarrow:$ If $t$ is a variable, then the intermediate fields of $F(t)/F$ are not totally ordered, consider $F(t^2)$ and $F(t^3)$. $\Leftarrow:$ Let $K,L$ be intermediate fields which are not compatible. Choose $a \in K - L, b \in L - K$. Then $F(a), F(b)$ are finite extensions, which are not compatible, contradiction.

An immediate consequence is, that we may first restrict to the finite case: Namely every $E/F$ as above is a directed union of finite subextensions, whose intermediate fields are totally ordered by inclusion; and vice versa.

Note that we cannot restrict the degree of $E/F$. Namely, $S_{n-1}$ is a maximal subgroup of $S_n$. Since $S_n=Gal(E/F)$ for some Galois extension $E/F$, if $K$ is the fixed field of $S_{n-1}$, the extension $K/F$ has degree $n$ and no nontrivial intermediate fields at all.

What about inseparable extensions? And what happens if we take the normal closure? Perhaps we can reduce everything to the Galois case?

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@Martin: I was confused for a little while by your terminology. I now think you mean not that the subgroups themselves can be totally ordered but that the lattice of subgroups is totally ordered: of course these are totally different concepts. Could you rewrite to clarify? –  Pete L. Clark Oct 6 '10 at 19:11
    
For the Galois case, every finite quotient must be cyclic. Choosing a compatible system of generators shows that the group must be topologically cyclic, hence isomorphic to $\mathbb{Z}_p$. –  Kevin Ventullo Oct 7 '10 at 4:47
    
@Pete: I've added "by inclusion", that should suffice (it does not make sense to consider total orders on the individual subfields; and besides then we sould speak of totally orderable or something like that). @Kevin: Thanks! I will edit my question. –  Martin Brandenburg Oct 7 '10 at 10:30
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1 Answer

In the separable case, as the question already notes, what you're really asking about is finite extensions and ultimately group theory. A finite, separable field extension is equivalent to transitive permutation group $G$ acting on a finite set $X$. This is in turn equivalent to a pair of groups $H \subseteq G$, with $X = G/H$, up to the equivalence of quotienting by a $G$-normal subgroup that lies in $H$. The invariant partitions of $X$ correspond to intermediate subgroups, which then in the Galois theory setting correspond to intermediate fields. If there are no invariant partitions, then $G$ is called primitive. Also, for every permutation group $G/H$, it is easy to find some field extension modeled by $G/H$, even though it is a famous open problem to always find such an extension of $\mathbb{Q}$.

So the question is when the $G$-invariant partitions of $X = G/H$ are totally ordered. The first remark is that if they are totally ordered, then actually the combinatorics of these partitions is completely determined by their relative indices. In other words, $X$ is divided into $n_1$ parts, then each part is divided into $n_2$ parts, and so on, and there is no structure other than this sequence of integers. In the Galois example where $H$ is trivial and $G = \mathbb{Z}/p^k$, each $n_i = p$.

So, $G$ is some group that acts on this tower of partitions. The second remark is that for a given tower of partitions, there is a maximal choice of $G$, namely all of the automorphisms of the tower of partitions. This is an iterated wreath product of the symmetric groups $S_{n_i}$. This is an example where there are no other invariant partitions, hence they are totally ordered. In fact, any iterated wreath product of primitive groups works just as well. These constructions correspond in Galois theory to starting with a base field $K$ and adjoining one root each of a sequence of primitive polynomials (primitive in the sense that the Galois group acts primitively on the roots), chosen "generically" so that the total Galois group is the wreath product.

If $\mathcal{P}$ is the tower of partitions, then the general example is a group $G \subseteq \text{Aut}(\mathcal{P})$ that happens not to have any other invariant partitions. I think that it's a complicated question to find all examples. From this viewpoint, it seems a bit accidental that $$\mathbb{Z}/p^k \subseteq S_p \wr S_p \wr \cdots \wr S_p$$ is still big enough that the invariant partitions are linearly ordered. So I don't think that there is any principle of reducing to the Galois case.

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