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Let $p$ and $q$ be two distinct primes. For a field $F$, assume that $\deg(\alpha, F)=p$. Is it necessarily true that $\deg(\alpha^q, F)=p$? Is there any counterexample?

It is not an exercise problem although it looks very simple. Does anyone know the answer?

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    $\begingroup$ What makes you say it's not an exercise problem (or do you mean that you didn't happen to find it in a textbook)? You have $F\subseteq F(\alpha^q)\subseteq F(\alpha)$ with the degree of $F(\alpha)/F$ being the prime $p$. So what you're asking is whether it is possible for $\alpha^q$ to be in $F$. Suppose it is. Let $f(X)\in F[X]$ be the irreducible poly of $\alpha$ over $F$. Note that $g(X)=X^q-\alpha^q\in F[X]$ has $\alpha$ as a root, so $f(X)\mid g(X)$. Hence $f(X)$ is a product of terms of the form $X-\zeta\alpha$ with the $\zeta$ being primitive $q$'th roots of unity. Does that help? $\endgroup$
    – Joe Silverman
    Commented Oct 12, 2015 at 11:20
  • $\begingroup$ Well, I simply thought that It is not an easy problem. I guess that you are giving a counterexample. As you mentioned, $ f(X)$ is a product of $p$ terms of the form $X-xi \alpha$. Hence one needs to show that some product of $p$ terms of the form $X-xi \alpha$ is irreducible over $F$. How can you show it? $\endgroup$
    – user81389
    Commented Oct 12, 2015 at 11:30
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    $\begingroup$ Please do not post to Mathematics and this site at the same time, especially not without mentioning it. math.stackexchange.com/questions/1476194/… $\endgroup$
    – user9072
    Commented Oct 12, 2015 at 12:43
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    $\begingroup$ @Joe Silverman: the problem is that the primitive qth roots of unity are not in the ground field. This question is more delicate than it seems and I don't know why it's on hold. A collegue of mine asked a question closely related to this in a Galois Theory final exam and several students handed in solutions like yours ;-) but in fact $\alpha^q$ can be in $F$! (as I found out when I was marking the question...). Just let $\alpha$ be a primitive $q$th root of unity and let $F$ be an appropriate field with $p$ a divisor of $q-1$. $\endgroup$
    – eric
    Commented Oct 12, 2015 at 19:33
  • $\begingroup$ @eric, I wonder whether the only counterexamples are those coming from $q$th roots of unity. $\endgroup$
    – Gerry Myerson
    Commented Oct 12, 2015 at 22:26

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