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Splitting field of $x^{n}-1$ over $\mathbb{Q}$

Let $A$ denote the splitting field of $x^n-1$ over $\mathbb{Q}$, so what is the dimension of $A$ as a linear space over $\mathbb{Q}$? (in this question, $n$ is a positive integer)

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marked as duplicate by Chris Eagle, Asaf Karagila, Zev Chonoles Dec 13 '11 at 3:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

It is $\Phi(n)$ where $\Phi$ is the Euler $\Phi$-function. – Matt Sep 13 '11 at 16:06
See this previous question: [Splitting field of $x^n-1$ over $\mathbb{Q}$][1] [1]:… – Ben Blum-Smith Sep 13 '11 at 16:07

The degree is $\varphi(n)$, where $\varphi$ is the Euler phi function. Note that if $\zeta$ is a primitive $n$-th root of unity, then we have a map $$ \operatorname{Gal}(\mathbf Q(\zeta)/\mathbf Q) \to (\mathbf Z/n\mathbf Z)^* $$ sending $\sigma$ to the class of an integer $i$ for which $\sigma(\zeta) = \zeta^i$. You can check that this is well-defined and injective — the hard part is showing surjectivity. The usual way of doing this is to show that the $n$-th cyclotomic polynomial is irreducible by reducing it mod $p$ and exploiting the Frobenius automorphism. You can see a proof in, for example, Keith Conrad's notes on cyclotomic fields.

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