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suppose there is a polynomial $f$ in $\zeta_p$ (root of unity) with coefficients in $K$, $\text{char}(K)>0$. Are there conditions when $f(\zeta_p) \in K$?

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Yes, lots. For example: $f \in K$. I don't really understand the question. What are you actually interested in? –  Qiaochu Yuan Aug 6 '12 at 17:13
    
I am refering to the following statement in one other post: "Inside the algebraic closure of Z/qZ there is a ζp, a primitive pth root of unity (assuming p≠q). Any polynomial in ζp is also in the algebraic closure." When is one of those polynomials in Z/qZ itself? Exactly when raising it to the qth power does not change it. I would like to have a general condition for this, not only when $K=\mathbb{Z} / q \mathbb{Z}$ –  Matt Aug 6 '12 at 17:35
    
Could you link us to the statement you're referring to? It doesn't make sense to say a polynomial is in a field. –  JSchlather Aug 6 '12 at 17:40
    
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The question is for conditions on the coefficients of $f\in K[x]$ such that $f(\zeta_p)\in K$. –  anon Aug 6 '12 at 17:50
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1 Answer

up vote 2 down vote accepted

Suppose that $f(\zeta_p) = k \in K$. Then $f(\zeta_p) - k = 0,$ and so $f(X) -k $ is a polynomial which has $\zeta_p$ as a zero, i.e. is divisible by the minimal polynomial of $\zeta_p$ over $K$.

Thus $f(\zeta_p) \in K$ if and only if $f(X) \equiv \text{ a constant } \bmod$ the minimal poly. of $\zeta_p$ over $K$.


This is a fairly tautological answer, but I'm not sure what else one should say, since the minimal polynomial is a factor of $(X^p -1)/(X-1)$ which depends very much on the particular choice of the field $K$.

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Well, the fact that a minimal polynomial exists is not a tautology: you use the fact that $K[X]$ has a division algorithm. –  Qiaochu Yuan Aug 6 '12 at 18:24
    
Ok, thank you, I just hoped, there would be some conditions, which do not depend on $K$. –  Matt Aug 6 '12 at 18:36
    
@Matt: Dear Matt, Your hope wasn't really reasonable, in that the answer to your question is "every $f$" in and only if $\zeta_p \in K$, which is something that clearly depends on $K$. Regards, –  Matt E Aug 6 '12 at 18:39
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