# Fields4

Show that the splitting field of $f(x)=x^{4}+1\in {\mathbb {Q}}[x]$ is a simple extension of ${\mathbb {Q}}$.
Let $\zeta _{8}={\frac {{\sqrt {2}}}{2}}+{\frac {{\sqrt {2}}}{2}}i$ be a primitive 8th root of unity and note that $\zeta _{8}\,$ is a root of $f(x)=x^{4}+1\,$. (Make sure you understand why this is true.) Further, as $\zeta _{8}^{8}=1$ and $\zeta _{8}^{4}=-1$, note that $\zeta _{8}^{3},\zeta _{8}^{5},\zeta _{8}^{7}$ are all roots of $f(x)\,$ as well. Thus, we have found the four roots of this quartic polynomial, so the splitting field of $f\,$ is $E={\mathbb {Q}}(\zeta _{8},\zeta _{8}^{3},\zeta _{8}^{5},\zeta _{8}^{7})$. As $E\,$ is a field, all the integral powers of $\zeta _{8}\,$ must be elements of it, so we may write $E={\mathbb {Q}}(\zeta _{8})$, showing that $E\,$ is a simple extension of ${\mathbb {Q}}$.