NBHM 2020 PART A Question 5 Solution

Let $p(x)$ be the minimal polynomial of $\sqrt{2}+\sqrt{-2}$ over the field $\Bbb Q$ of rational numbers. Evaluate $p(\sqrt 2)$.
Solution:
We shall find the minimal polynomial of $\sqrt{2}+i\sqrt{2}$ explicitly.
Let $x = \sqrt{2}+i\sqrt{2}$ then $x-\sqrt 2= i\sqrt 2$  and squaring both sides we get $x^2  + 4 = -2 \sqrt 2 x$. This not a polynomial over $\Bbb Q$, so again we square both sides we get $$p(x) = x^4 + 16 = 0.$$ This is a monic polynomial of degree 4 over $\Bbb Q$ satisfied by $\sqrt{2}+i\sqrt{2}$ and hence it has to be the minimal polynomial of $\sqrt{2}+i\sqrt{2}$ . Now substituting $x = \sqrt 2$ in the above polynomial we get $p(\sqrt 2) = 20$.

Why the degree of the minimal polynomial has to be $4$?
We have $\sqrt{2}+\sqrt{-2} = \sqrt{2}+i\sqrt{2}$ and this element belong to $\Bbb{Q}(\sqrt 2,i)$ which is a degree 4 extension. Because $$[\Bbb{Q}(\sqrt 2,i): \Bbb Q] = [\Bbb{Q}(\sqrt 2,i): \Bbb Q(\sqrt 2)] \cdot [\Bbb{Q}(\sqrt 2): \Bbb Q] = 2 \times 2 = 4.$$
The proper subfields of $\Bbb{Q}(\sqrt 2,i)$ are $\Bbb{Q}(\sqrt 2)$, $\Bbb{Q}(i)$, $\Bbb{Q}(\sqrt 2i)$ and $\Bbb{Q}$. So $\Bbb{Q}(\sqrt 2,i)$ is the smallest field containing $\sqrt{2}+i\sqrt{2}$ and hence $$\Bbb{Q}(\sqrt 2,i) = \Bbb{Q}(\sqrt{2}+i\sqrt{2})$$ and the degree of minimal polynomial of $\sqrt{2}+i\sqrt{2}$ is equal to $4$.

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