# Degree of splitting field over $\mathbb Q$

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 $\mathbb Q[\omega]$ ?

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

• That's it? It doesn't matter that the polynomial is reducible over it? – The Doctor Sep 13 '17 at 14:28
• It is the factor of $(x+1)(x-1)$ which have roots in $\mathbb{Q}$ and $x^2+1$ which is irreducible. – Tsemo Aristide Sep 13 '17 at 14:29
• 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? – The Doctor Sep 13 '17 at 14:34
• yes, it is something like that – Tsemo Aristide Sep 13 '17 at 14:38