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What is a splitting field $E$ for $f(x)=x^n+a$ over the field $K$ of characteristic zero?

If I put $g(x)=x^{2n}-a^2=(x^n-a)(x^n+a)$. The splitting field $F$ of $g(x)$ is $K(\sqrt[n]{a},\alpha )$ where $\alpha$ is a $2n$th root of unity. The roots of $(x+a^n)$ are such that $\sqrt[n]{a}e^{\pi ik}=-a$ which happens for $k=1,3,5,...,2n-1$. Hence we have the roots $S=\{\sqrt[n]{a} \alpha ^k\}$, $k=1,3,5,...,2n-1$. I would say that $E=F$ since $E\subseteq F$ and since $\alpha \in E$ we must have that all powers of $\alpha$ lies in $E$?

This is not the case however because my teacher said that the splitting field $E$ of $x^4+2$ is $K(\sqrt[4]{2},i)$. He argued that $\{\alpha ^k\}$, $k=1,3,5,7$ lies in $K(\sqrt[4]{2},i)$. Since $\alpha =\frac{1+i}{\sqrt{2}}$ and $(\sqrt[4]{2})^2=\sqrt{2}\in E$ and hence $\alpha \in E$. By this constructions, doesn't all eight roots of unity lie in $E=K(\sqrt[4]{2},i)$? Making $E$ a splitting field $F$ of $x^8-4$? It doesn't make sense to me how $K(\sqrt[4]{2},i)\neq K(\sqrt[4]{2},\alpha)$. I know it can not be the case however that the two fields are the same, because they have minimal polynomials of different degree over $K(\sqrt[4]{2})$ that is, $x^2+1$ and $x^4+1$ so they can't be equal.

I am very confused by this so please clarify!!!

harajm
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1 Answers1

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The splitting field of $f(x) = x^n-a\in K[x]$ is $K(\zeta_n, a^{1/n})$, where $a^{1/n}$ is any root of $f(x)$ and $\zeta_n$ is an $n$th root of unity in an algebraic closure of $K$. Clearly the splitting field must contain $a^{1/n}$ and you can check that $\zeta_n^k a^{1/n}$ are also roots of $f(x)$ directly. On the other hand, if $\alpha$ and $\beta$ were both roots of $f(x)$, then $\alpha^n/\beta^n = a/a = 1$, hence their ratio is an $n$th root of unity.

To answer your question specifically, for $f(x) = x^n + a$, we have $K(\zeta_n, (-a)^{1/n})$.

Hope that helps.

tghyde
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