Fields4

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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}}.

Abstract Algebra