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Suppose $K$ is an algebraic number field, and $a \in K$. Let $n$ be a positive integer. The polynomial $t^n - a \in K[t]$ splits as a product of irreducible factors of degrees $d_1, \dots, d_r$.

Is it necessarily the case that the smallest of the $d_i$ divides all the others?

More generally: is the set $\{d_1, \dots, d_r\}$ always totally ordered by divisibility?

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  • $\begingroup$ Please use a high-level tag like meta.mathoverflow.net/q/1075 $\endgroup$ Commented Dec 2 at 0:51
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    $\begingroup$ Here is some lazy analysis of question 1 I can't quite finish: Letting $L$ be the splitting field of $x^n - a$ over $K$ and considering the action of $\operatorname{Gal}(L/K)$ on the roots, we have that the $d_i$ are the sizes of the orbits. Thus the question becomes whether there is any subgroup $S$ of $Aff(\mathbb Z / (n)) = \{\begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} | a \in \mathbb Z / (n) ^\times, b \in \mathbb Z / (n)\}$ such that the size of the smallest orbit of the action of $S$ on $\mathbb Z / (n)$ given by $x \mapsto a x + b$ does not divide the sizes of the larger ones. $\endgroup$ Commented Dec 2 at 19:38
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    $\begingroup$ My temptation is to analyze this action prime by prime, and indeed when $n$ is an (odd?) prime power I think I can show the statement holds. However I get stuck when trying to glue together my understanding at each prime into some global understanding, the problem is that $S$ might be a good deal smaller than the direct product of its images modulo each prime dividing $n$. $\endgroup$ Commented Dec 2 at 19:45
  • $\begingroup$ I think I have successfully completed a proof along the lines suggested in my earlier comments that the answer to question 1 is yes. I will write it up sometime in the next two weeks if no one else does (remind me if I forget). The point is that if one proves that not only does the statement hold for $n = p^e$, but also that the smallest orbit is a power of $p$ in this case, then the smallest orbit modulo $p_1^{a_1} p_2^{a_2} ... p_k ^ {a _ k}$ is just the product of the smallest orbits modulo each $p_i^{a_i}$, which divides the lcm of any collection of orbits mod the $p_i^{a_i}$ one chooses. $\endgroup$ Commented 21 hours ago

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The answer to the second question is negative. For example, if $K=\mathbb{Q}$ and $a=1$, then the $d_i$'s are the $\varphi(m)$'s for $m\mid n$, in some order. In particular, if $n=35$, then $$\{d_1, \dots, d_r\}=\{1,4,6,24\},$$ which is not totally ordered by divisibility.

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    $\begingroup$ The smallest degree for which the second question admits a negative answer is $21$. An example is $X^{21}−1$ over the quadratic subfield of the cyclotomic field $\mathbb Q(\zeta_7)$. The degrees will be $1,2 , 3, 3, 6, 6$. $\endgroup$ Commented Dec 2 at 15:02
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    $\begingroup$ Thanks to you both! This is helpful. $\endgroup$ Commented Dec 2 at 19:18
  • $\begingroup$ @BenWilliams Thank you. Regarding the first question, my guess is that the answer is "yes" for $K=\mathbb{Q}$, but "no" in general. $\endgroup$ Commented Dec 2 at 19:21

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