## 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 $\pm i$.