Compute the conjugates of an element in a given field extension

We saw in this previous exercise that i \in \mathbb{Q}(i + \sqrt{2}). Express i as a polynomial in \theta = i + \sqrt{2} and find the conjugates of i over \mathbb{Q}(i+\sqrt{2}).


We showed here that i + \sqrt{2} is a root of q(x) = x^4 - 2x^2 + 9, and we showed here that q(x) is irreducible. Evidently, q(x) = (x - i - \sqrt{2})(x - i + \sqrt{2})(x + i - \sqrt{2})(x + i + \sqrt{2}), giving the conjugates of i + \sqrt{2}.

Using some linear algebra, we see that i = \frac{1}{6} \theta + \frac{1}{6} \theta^3. (WolframAlpha agrees.) Let p(x) = x/6 + x^3/6. Evidently, p(\theta) = p(i - \sqrt{2}) = i and p(-i + \sqrt{2}) = p(-i-\sqrt{2}) = -i. Hence the conjugates of i over \mathbb{Q}(i+\sqrt{2}) are \pm i.

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