NBHM 2020 PART A Question 12 Solution

Let $k$ be the field obtained by adjoining to the field $\Bbb Q$ of rational numbers the roots of the polynomial $x^4−2$. Let $k^{'}$ be the field obtained by adjoining to $k$ the roots of the polynomial $x^4 + 2$. What is the degree of $k^{'}$ over $k$?
Solution:
The roots of the polynomial $x^4 - 2 = (x^2-\sqrt 2)(x^2+\sqrt 2)$ are $\pm \sqrt[4]2$ and $\pm \sqrt[4]2\,i$. Similarly the roots of the polynomial $x^4 + 2 = (x^2-\sqrt 2\,i)(x^2+\sqrt 2\,i)$ are $\pm \sqrt[4]2 \sqrt i$ and $\pm \sqrt[4]2\,i\,\sqrt i (= \pm \sqrt[4]2\,i^{\frac{3}{2}})$.

We claim that $[k^{'}:k] = 1$. Equivalently, we will prove that $$k = k^{'}.$$ 
To prove this, we will prove that all the roots of $x^4 + 2$ are in $k$. To prove this, from the above-given list of roots of $x^4 + 2$, it is enough to prove that $\sqrt i \in k$. [Since we already have $i, \sqrt[4]2 \in k$.]
Now, one can check that $\sqrt i = \frac{1}{\sqrt 2} + i \frac{1}{\sqrt 2}$ which is a primitive 8th root of unity. Since $\sqrt[4] 2 \in k$, its square $\sqrt 2 \in k$ and the reciprocal $\frac{1}{\sqrt 2} \in k$. Since, we already have $i \in k$, this shows that $\sqrt i = \frac{1}{\sqrt 2} + i \frac{1}{\sqrt 2} \in k$  and the proof is done. 
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