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I asked this question in a similar form on math.se here, where it has been unanswered for a little over a week some work on it can be found there. The motivation for this question was another question asked on math.se.

Fix a prime $p$ and $L$ a number field. Let $f(t) \in L[t]$ be irreducible such that $\deg f(t)=p^{j}$ and $L \subset K$ a field extension of degree $p^k$. What can we say about how $f(t)$ factors over $K[t]$? In particular is it true that $f(t)$ takes a root or has an irreducible factor of order $p^n$ for some $n$?

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    $\begingroup$ @Michael, I'm not quite sure what you're getting at $[K(c):K]$ is not a prime power for ever root. Consider $L=\mathbb Q$, $K=\mathbb Q(\sqrt[9]{2})$ and $x^{27}-2=(x^3 - \sqrt[9]{2})(x^6 + \sqrt[9]{2}x^3 + \sqrt[9]{2}^2)(x^{18} + (\sqrt[9]{2})^3x^9 + (\sqrt[9]{2})^6)$ over $K[x]$. $\endgroup$
    – Jacob Schlather
    Commented Feb 5, 2013 at 19:57
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    $\begingroup$ The answer to your question is no. Let $f(t)\in L[t]$ be any degree-$7$ polynomial whose Galois group $G$ is the simple group of order $168$. Then $G$ contains two conjugacy classes of index-$7$ subgroups. All groups in one of these conjugacy classes have a fixed point, but the groups in the other class have orbits of sizes 3 and 4. So if $K$ is the subfield of the splitting field of $f$ fixed by one of these index-$7$ subgroups of $G$ from the second class, then $f(t)$ factors in $K[t]$ into irreducibles of degrees $3$ and $4$. $\endgroup$
    – Michael Zieve
    Commented Feb 5, 2013 at 21:22
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    $\begingroup$ To see $3+4$ easily, note that $G \cong GL_3(\mathbb{F}_2)$. One index $7$ subgroup is the stabilizer of a nonzero vector in $\mathbb{F}_2^3$, the other is the stabilizer of a hyperplane. If we use the first group to define the map $G \to S_7$, and take orbits for the other, then we get two orbits: the $3$ nonzero vectors in the hyperplane and the $4$ nonzero vectors out of it. $\endgroup$
    – David E Speyer
    Commented Feb 5, 2013 at 22:05
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    $\begingroup$ @Mike: There are infinitely many primes $p$ for which there is no counterexample of degree $p$: By Burnside a transitive group $G$ of degree $p$ is as subgroup of $AGL(1,p)$, or it is doubly transitive. In the first case, any subgroup of index $p$ fixes a point. As to the second case, there are infinitely many $p$ such that $G=A_p$ or $S_p$. (Follows from the classification of the $2$-transitive groups.) However, in $A_p$ or $S_p$, a subgroup of index $p$ fixes a point. Actually, one can show that all counterexamples of prime degree come from $PSL(n,q)\le G\le P\Gamma L(n,q)$ for $n\ge3$ ... $\endgroup$
    – Peter Mueller
    Commented Feb 5, 2013 at 22:39
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    $\begingroup$ @Mike: No, whenever $p=(q^n-1)/(q-1)$ is a prime for $n\ge3$ (as it happens for Mersenne primes), you have an example like your example (for the same reason as in David's comment). As any finite group is a Galois group over some number field, there isn't a field theoretic obstruction. $\endgroup$
    – Peter Mueller
    Commented Feb 5, 2013 at 23:10

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Just for concreteness, here's an explicit counterexample along the lines of my comment. Let $L=\mathbb{Q}$ and let $f(t)=t^7 - 7t+3$ and $g(t)=t^7-14t^4+42t^2-21t-9$. Then $f(t)$ and $g(t)$ are irreducible in $L[t]$. Now let $K=\mathbb{Q}(c)$ where $g(c)=0$, so $[K:L]=7$. Then $f(t)$ factors in $K[t]$ into irreducibles of degrees $3$ and $4$. For instance, this can be verified using the following Magma code, which can be used on the free Magma online calculator:

_<t>:=PolynomialRing(Rationals());
f:=t^7-7*t+3;
g:=t^7-14*t^4+42*t^2-21*t-9;
IsIrreducible(f);
K<c>:=NumberField(g);
_<x>:=PolynomialRing(K);
Factorization(Evaluate(f,x));
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