Let $n$ and $d$ be positive integers, with $d\ge 2$ square-free. It is well known that $\Phi_n=\Phi_n(x)$, the $n$-th cyclotomic polynomial, is irreducible over $\mathbb{Q}$. However, as the simple example $$ \Phi_8(x) = x^4 + 1 = (x^2 + x\sqrt{2} + 1)(x^2 - x\sqrt{2} +1) $$ shows, $\Phi_n$ may not be irreducible over a real quadratic number field $\mathbb{Q}(\sqrt{d})$. My question: Is is possible to characterize those $(n,d)$ for which $\Phi_n$ is irreducible over $\mathbb{Q}(\sqrt{d})$? Answers and/or pertinent references will be much appreciated.
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6$\begingroup$ If ${\mathbf Q}(\sqrt{d}) \subset {\mathbf Q}(\zeta_n)$ then $\Phi_n(x)$ is reducible over $\mathbf Q(\sqrt{d})$, so to be irreducible we need ${\mathbf Q}(\zeta_n) \cap {\mathbf Q}(\sqrt{d}) = \mathbf Q$. If $E/F$ and $L/F$ are finite Galois extensions s.t. $E \cap L = F$ then $[EL:L] = [E:F]$. Thus $[{\mathbf Q}(\sqrt{d})(\zeta_n):{\mathbf Q}(\sqrt{d})] = \varphi(n)$ if ${\mathbf Q}(\sqrt{d}) \not\subset {\mathbf Q}(\zeta_n)$, so in that case $\Phi_n(x)$ is irreducible over $\mathbf Q(\sqrt{d})$. Do you know how to decide when a specific quadratic field is in specific a cyclotomic field? $\endgroup$– KConradCommented Feb 20, 2015 at 0:33
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Given prime number $p$, whether the $p$ the cyclotomic field contains Q$[\surd p]$ or Q[$i\surd p$] depends on whether $p\pmod 4$ is 1 or 3 respectively. This follows from the calculation of the discriminant of the cyclotomic field which is easy as there is an obvious integral basis. For a general cyclotomic field generated by $n$ th root of unity, $n$ not a prime, look at each prime divisor of $p$ and apply the above.
