# Towers of perfect fields of mostly order $p^t$

This question is strongly related to this question. I'm curious about the following result. Fix a prime $p$ and let $L$ be a perfect field and $q_1(t),\dots,q_n(t) \in L[t]$ irreducible polynomials such that $\deg q_i = p^{a_i}$ with $\beta_i$ a root of $q_i(t)$. Set $L_i=L_{i-1}(\beta_i)$ where $L_0=L$ and assume that for $i<n$ we have $q_i(t)$ irreducible over $L_{i-1}$ and $q_n(t)$ has no roots in $L_{n-1}$. Is it the case that an irreducible factor of $q_n(t)$ in $L_{n-1}[t]$ has degree a power of $p$.

By induction the result only needs to be checked for two polynomials $f(t)$ and $q(t)$. I can't seem to make any headway in this case. I'd also be interested in a positive (or even better negative) result in the case $L=\mathbb Q$ and $f(t)$ and $q(t)$ satisfying the desired hypothesis. A positive result in this case is somewhat less interesting, because it doesn't give me everything I need. And of course feel free to assume $p=3$ if it makes you happy.

Update 1/29: I've worked through various examples in Sage seeing how degree $9$ polynomials factor over number fields of degree $9$, so far nothing has contradicted the hypothesis. In regards to proving it I would like to imagine there's some clever proof that goes as follows. Let $f(t),g(t)$ be two irreducible polynomials of degrees power of $p$, $\alpha$ a root of $f(t)$ and let $E$ be the splitting field of $f(t)$ and $g(t)$. Consider the action of the subgroup of $\mathrm{Gal}(E/L)$ that fixes $L(\alpha)$ on the roots of $g(t)$. Then I'd like to imagine that there's some clever way to show the orbit of one of the roots of $g(t)$ is a power of $p$ and deduce the theorem.

I've cross-posted this to mathoverflow here.

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Could you edit so that the polynomials are not called $p_i$? I get confused with degrees $p^m$ running around too. Reading more closely, my ever-weakening eyes are also confused by your use of $\alpha_i$ and $a_i$. –  Lubin Jan 27 '13 at 22:57
@Lubin I've changed the $p_i$ to $q_i$ and the $\alpha$ to $\beta$ I hope that helps. –  JSchlather Jan 28 '13 at 1:40

[Edited: I originally posted an invalid "proof" that the answer is yes. But actually the answer is no, as indicated in my answer to your mathoverflow crosspost. There I gave an explicit counterexample. I'm leaving this incorrect "proof" up for now, in case it might be of interest to anyone. The mistake is subtle: it's in the claim that distinct elements of $T$ are coprime. That isn't true, even though the elements of $T$ are distinct monic irreducibles. This is possible because the elements of $T$ aren't necessarily in $K[x]$, they're a collection of irreducible polynomials over different constant fields. Anyway, my original incorrect argument is below.]

Here's an alternative approach to what I gave in my comment to your mathoverflow question. Let $f(x)\in L[x]$ be a monic irreducible polynomial over a field $L$, and let $K/L$ be a finite separable extension. Let $J$ be the Galois closure of $K/L$, and let $G$ be the Galois group of $J/L$. Then the action of $G$ on $J$ extends naturally to an action on $J[x]$. Let $g(x)\in K[x]$ be a monic irreducible polynomial which divides $f(x)$. Let $T=\{\sigma(g):\sigma\in G\}$. Then every element of $T$ also divides $f(x)$, and the elements of $T$ are distinct monic irreducibles and hence are pairwise coprime, so $f(x)$ is divisible by the product $u(x):=\prod_{h\in T} h(x)$. But $u(x)$ is fixed by each element of $G$, so it is an element of $L[x]$. Hence $u(x)$ is a monic polynomial in $L[x]$ which divides the monic irreducible polynomial $f(x)$, so $u(x)=f(x)$. It follows that $\deg(g)$ divides $\deg(f)$: in fact, $\deg(g)=\deg(f)/\# T$.

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I've posted an example where this doesn't work on mathoverflow, but I'll recreate this here. Consider $L=\mathbb Q$ and $K=\mathbb Q(\sqrt[9]{2}$ then $x^{27}-2=(x^3 - \sqrt[9]{2})(x^6 + \sqrt[9]{2}x^3 + (\sqrt[9]{2})^2)(x^{18} + (\sqrt[9]{2})^3x^9 + (\sqrt[9]{2})^6)$. In particular $18$ and $6$ don't divide $27$. My original conjecture was much the same as what you're proving, but it turns out that each factor need not have prime power order. –  JSchlather Feb 5 '13 at 20:01
Oops, I see my mistake, not all the $h(x)$'s are in $K[x]$. Ok, I'll think about it further. –  Michael Zieve Feb 5 '13 at 20:08
I've accept this answer, but what I'm actually accepting is the counterexample presented here. –  JSchlather Feb 5 '13 at 22:46

A few thoughts.

For each $\alpha$, the stabilizer $Stab(\alpha)$ splits the roots of $g$, $\{ \beta_i\}$, in a few orbits. Denote by $O(\alpha)$ this decomposition and note $\mathcal O$ the image of $O$. The Galois group is also seen to act on $\mathcal O$ (if $\sigma(\alpha_i) = \alpha_j$, then $\sigma(O(\alpha_i)) = O(\alpha_j)$)

For any $\alpha$, $Stab(\alpha) \subseteq Stab(O(\alpha))$, and $\# O^{-1}(\{O(\alpha)\})$ is the index of this subgroup. Furthermore, since $G$ is transitive on the $\alpha_i$, it is transitive on $\mathcal O$, so that every orbit decomposition has the same number of preimages. Since the number of roots of $f$ is a power of $p$, the size of $\mathcal O$ is also a power of $p$.

So we can focus on those orbit decompositions rather than on the roots of $f$, and try to start from $\mathcal O$ instead.

Suppose we are given a set $\mathcal O$ of orbit decompositions of $\{ \beta_i\}$. This set induces a subgroup $G$ of compatible permutations of $\{ \beta_i\}$. We want $G$ to be transitive on $\mathcal O$ and $Stab(O)$ to be transitive on the orbits described in $O$ (for any $O$), so obviously the more permutations we put in $G$ the better, so the $G$ we pick only depends on $\mathcal O$, and the transitivity conditions only depends on $\mathcal O$ too.

If we can find such a suitable $\mathcal O$ whose cardinal is a power of $p$, then we have a candidate Galois Group for a counterexample.

Here are some simple ideas that don't work : Let $\mathcal O$ be all the partitions of $\{ \beta_i\}$ into sets of size $s_1+ \ldots + s_k$ where $\sum s_j = p^a$ and no $s_j$ is a power of $p$. Then $G$ is the full permutation group on $\{ \beta_i\}$, but $\# \mathcal O$ is not a power of $p$ : it is a product of integers like $(a+b)!/a!b!$ and $(kb)!/b!^k k!$, one of which has $p^a !$ in the numerator, which forces $a=1$ or $b=1$, hence an orbit of size $1$, which we didn't pick.

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Thanks, this was useful but I still couldn't make any progress. So I'm not sure if there's something more subtle going on here or it's just not true and the counterexample is at a high degree. I'll probably award you the bounty in a few hours so that it doesn't go to waste and post this on mathoverflow. –  JSchlather Feb 5 '13 at 14:23