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I'm reviewing my notes, and I'm not fully understanding an argument that the 8th cyclotomic polynomial is irreducible in $\mathbb{Q}[x]$. Here's the online of the argument

$\Phi_8(x)=\dfrac{x^8-1}{\Phi_1(x)\Phi_2(x)\Phi_4(x)}=\dfrac{x^8-1}{x^4-1}=x^4+1$.

This can be factored into: $(x-\zeta_8)(x-\zeta_8^3)(x-\zeta_8^5)(x-\zeta_8^7)$ where $\zeta$ is a primitive root of unity. The next part of the argument is what I'm not seeing.

$(x-\zeta_8)(x-\zeta_8^3)(x-\zeta_8^5)(x-\zeta_8^7)=(x^2-(\zeta_8 +\overline{\zeta_8})x+1)(x^2-(\zeta_8^3+\overline{\zeta_8^3})x+1)=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$

Since $\sqrt{2} \notin \mathbb{Q}$, the polynomial isn't reducible in $\mathbb{Q}$.

Any clarification is greatly appreciated. Thanks

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Let $F$ be the real subfield of $\mathbb{Q}(\zeta_8)$. By grouping the roots into conjugate pairs, it's clear that the stated factorization is the prime factorization of $\Phi_8(x)$ over $F$.

The prime factorization of $\Phi_8(x)$ over $F$ is a refinement of its prime factorization over $\mathbb{Q}$ — i.e. the prime factorization over $F$ come from taking the prime factorization over $\mathbb{Q}$ and then factoring each factor over $F$.

Thus, the only possibilities for the prime factorization over $\mathbb{Q}$ are either the displayed factorization or that it is irreducible... and the displayed factorization is clearly not defined over $\mathbb{Q}$.

Actually, $F$ is itself a Galois extension of $\mathbb{Q}$, so the prime factorization over $\mathbb{Q}$ can be obtained by grouping the conjugate factors and multiplying them together, once again showing irreducibility.

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If the polynomial is reducible, then either it can be factored as a product of degree $1$ and degree $3$ polynomials or it can be factored as a product of two degree $2$ polynomials.

The former case cannot hold because there is no rational root for this polynomial. For the latter you already have the factorization, but the factors do not belong to $\mathbb{Q}[x]$. Hence even this factorization doesn't work. Thus irreducible.

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Another argument is to apply Eisenstein's criterion with $f(x)=x^4+1$ for $$f(x+1)=(x+1)^4+1=x^4+4x^3+6x^2+4x^2+2.$$ Hence $\Phi_8(x)=x^4+1$ is irreducible over $\mathbb{Q}$. The other factorisation you gave is not over $\mathbb{Q}$; this would rather work with $$ x^4+4=(x^2-2x+2)(x^2+2x+2). $$

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