What is the splitting field of $t^4+2$?

Question

I currently beginning the study if Galois theory but my understanding of the construction of splitting fields could be better. So I must ask if I could see the steps in constructing the splitting field of $t^4+2$.

Answer 1

A splitting field of a polynomial must be computed over a field, and the choice of that field affects the resulting splitting field, since a splitting field is a field extension.

For instance, the splitting field of $t^4 + 2$ over $\mathbb R$ is $\mathbb C = \mathbb R(i)$. The splitting field of that same polynomial over $\mathbb Q$ is something else (hinted by an above comment). And the splitting field of that same polynomial over $\mathbb F_7$ is again something else.

Answer 2

First You must determine that the polynomial is considered to be in $\mathbb{Q}$ or $\mathbb{R}$ or any other field.

I suppose that we're going to find the splitting field of $t^4+2\in\mathbb{Q}[t]$.

Now you know $\mathbb{C}$ is an extension of $\mathbb{Q}$. This is a good start that every polynomial splits in $\mathbb{C}$.

Then let's try to split it! $$t^4+2=(t-\sqrt[4]{2}\omega)(t-\sqrt[4]{2}\omega^3)(t-\sqrt[4]{2}\omega^5)(t-\sqrt[4]{2}\omega^7);$$ where $\omega=e^{\frac{\pi i}{4}}$.

Now it's sufficient to form the field $E=\mathbb{Q}(\sqrt[4]{2}\omega,\sqrt[4]{2}\omega^3,\sqrt[4]{2}\omega^5,\sqrt[4]{2}\omega^7)=\mathbb{Q}(\sqrt[4]{2}\omega,\omega^2)=\mathbb{Q}(\sqrt[4]{2}\omega,i)$

The last line is true as @Mariano Suárez-Alvarez♦ has mentioned in the comments.

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#EDITED

It's definitely wrong that $\mathbb{Q}(\sqrt[4]{2}\omega,\sqrt[4]{2}\omega^3,\sqrt[4]{2}\omega^5,\sqrt[4]{2}\omega^7)=\mathbb{Q}(\sqrt[4]{2}\omega)$, as @Jyrki Lahtonen♦ has stated in the comment below.