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Let $K = \mathbb{Q}(\sqrt[11]{7},i)$. Is there a polynomial over $\mathbb{Q}$ whose splitting field is $K$?

I am not too sure of my answer. It is as follows:

It suffices to show that $K/\mathbb{Q}$ is not normal. Hence, we can conclude that $K$ is not the splitting field of any polynomial in $\mathbb{Q}[t]$. Consider the polynomial $f(t) = t^{11} - 7$ in $\mathbb{Q}[t]$. $f(t)$ is irreducible over $\mathbb{Q}$, but it has a root $\sqrt[11]{7}$ in $K$. However, $f(t)$ does not split completely in $K$. We take a root of $f(t)$ over $\mathbb{C}$ where $f(t)$ splits, such as $\sqrt[11]{7}e^{\frac{2\pi{}i}{11}}$. However, $\sqrt[11]{7}e^{\frac{2\pi{}i}{11}} \notin K$. Hence, $K/\mathbb{Q}$ is not normal and $K$ is not the splitting field of any polynomial in $\mathbb{Q}[t]$.

Any advice will really help. Thank you!

Noob4398
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  • Yes, $f(t)$ is the minimal polynomial for $\sqrt[11]{7}$ over $\mathbb{Q}$, so the crux is showing it does not split in $K$. You might want to search "roots of unity" for some perspective. – hardmath Sep 24 '16 at 16:22
  • The answer is not because in order to have $K$ splitting field it must contains all the conjugates of $\sqrt[11]{7}$ which is not the case. – Piquito Sep 24 '16 at 18:52

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