suppose there is a polynomial $f$ in $\zeta_p$ (root of unity) with coefficients in $K$, $\text{char}(K)>0$. Are there conditions when $f(\zeta_p) \in K$?
General conditions for $f \in K[\zeta_p] \Rightarrow f(\zeta_p) \in K$
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abstract-algebra
polynomials
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0Yes, lots. For example: $f \in K$. I don't really understand the question. What are you actually interested in? – 2012-08-06
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0I 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 $K=\mathbb{Z} / q \mathbb{Z}$ – 2012-08-06
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0Could you link us to the statement you're referring to? It doesn't make sense to say a polynomial is in a field. – 2012-08-06
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0http://math.stackexchange.com/questions/175117/definition-of-k-conjugacy-classes?answertab=votes#tab-top – 2012-08-06
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1The question is for conditions on the coefficients of $f\in K[x]$ such that $f(\zeta_p)\in K$. – 2012-08-06
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0ok, sorry, I changed it – 2012-08-06
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0The corresponding statement for general fields is “exactly when no element of $\operatorname{Gal}(K[\zeta_p]/K)$ changes it.” This is just a simple result in Galois theory: the base field is the fixed field of the Galois group. – 2012-08-06
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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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0Well, the fact that a minimal polynomial exists is not a tautology: you use the fact that $K[X]$ has a division algorithm. – 2012-08-06
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0Ok, thank you, I just hoped, there would be some conditions, which do not depend on $K$. – 2012-08-06
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0@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, – 2012-08-06