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I want to find degree of the splitting field of $x^4 -1$ over $\mathbb Q$.

$\mathbb Q[\omega]$ would be a splitting field as the 4th roots of unity will be $\omega^i , i= 0,1,2,3$.

Now $x^4 -1$ can be written as $(x^2 +1)(x-1)(x+1)$

So what will be $[\mathbb Q(\omega) : \mathbb Q]$ now that the polynomial is reducible over $\mathbb Q[\omega]$ ?

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The degree is $2$ since it is generated by $i$ such that $i^2+1=0$.

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  • $\begingroup$ That's it? It doesn't matter that the polynomial is reducible over it? $\endgroup$ – The Doctor Sep 13 '17 at 14:28
  • $\begingroup$ It is the factor of $(x+1)(x-1)$ which have roots in $\mathbb{Q}$ and $x^2+1$ which is irreducible. $\endgroup$ – Tsemo Aristide Sep 13 '17 at 14:29
  • $\begingroup$ Correct me if I am wrong, the splitting field we want to find should contain solution of $(x^2+1),(x+1)$ and $(x-1)$ in some form. Since $(x^2+1)$ is irreducible and other two have their solutions in the splitting field so we are done? $\endgroup$ – The Doctor Sep 13 '17 at 14:34
  • $\begingroup$ yes, it is something like that $\endgroup$ – Tsemo Aristide Sep 13 '17 at 14:38

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