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Prove that $\mathbb{Q}(\sqrt[4]{2}, i)$ is a splitting field over $\mathbb{Q}$.

I know basic Group, Ring, and Field theory, and I've read the definition of a splitting field, yet I still have no idea where to start on this one.

Is it asking me to show that any polynomial in $\mathbb{Q}$ will split over $\mathbb{Q}(\sqrt[4]{2}, i)$?

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    No, the splitting field for a polynomial $f(x)$ is the smallest field in which $f(x)$ splits. A splitting field $E$ is a field extension of a field $F$ such that there is some polynomial $f(x)$ for which $E$ is the splitting of $f$. – 2017-01-05
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    So the question is asking you to show that there is a polynomial $f(x)\in\mathbb Q[x]$ for which $\mathbb Q(\sqrt[4]2,i)$ is the smallest field in which $f$ splits. – 2017-01-05

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It is the splitting field of $X^4-2$. Let $F$ be the splitting field of $X^4-2$. It contains $\sqrt[4]2$ and $i\sqrt[4]2$ which are roots of $X^4-2$. this implies that it contains $(\sqrt[4]2)^3$ and $i\sqrt[4]2(\sqrt[4]2)^3=2i$, so it contains $Q(\sqrt[4]2,i)$. Since $Q(\sqrt[4]2,i)$ contains all the roots of $X^4-2$, it is its splitting field.

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    Thanks! So in general, to solve a problem like this, it looks like you would need to know what the polynomial is. How did you know it was $x^4 - 2$? – 2017-01-05
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    experience,.... – 2017-01-05
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    @setholopolus If an extension $ L/K $ of fields is a normal extension generated as $ L = K(\alpha_1, \alpha_2, \ldots, \alpha_k) $, then $ L/K $ is the splitting field of the product $ f_1 f_2 \ldots f_k $, where each $ f_i $ is the minimal polynomial of $ \alpha_i $ over $ K $. In this case, we don't need the $ X^2 + 1 $ factor coming from $ i $. – 2017-01-05