If $\zeta= \zeta_{n}$, how does one count the homomorphisms $f:\mathbf{Q}(\zeta)\rightarrow \mathbf{Q}(\zeta)$?
Counting endomorphisms of $\mathbf Q(\zeta _{n})$
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abstract-algebra
ring-theory
galois-theory
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2I don't think it would hurt to say what $\zeta_n$ is and what your thoughts are so far. This is more or less the same as [this question](http://math.stackexchange.com/questions/24056/splitting-field-of-xn-1-over-mathbbq) modulo some field theory, which I'll try to explain. – 2011-12-22
1 Answers
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Let $k = \mathbf Q(\zeta)$. Since it is the splitting field of $X^n - 1$, $k$ is Galois over $\mathbf Q$. Homomorphisms of rings should respect $1$, which implies that $f$ is injective and restricts to the identity on $\mathbf Q$. By looking at dimensions over $\mathbf Q$, it is also surjective. So finding the number of $f$ is the same as determining \[ \#\operatorname{Gal}(k/\mathbf Q) = [k : \mathbf Q] = \text{degree of the minimal polynomial of $\zeta$}. \] There's a sketch of and reference to a proof that this number is $\varphi(n)$ in this answer.