## Show that a given extension of QQ contains i

Show that $i \in \mathbb{Q}(\sqrt{2} + i)$.

Recall that in our proof that $F(\alpha,\beta) = F(\theta)$ for some $\theta$, we found that $\theta$ may be taken to be $\alpha + c\beta$ so long as $\alpha + c\beta \neq \alpha^\prime + c\beta^\prime$ for any conjugates $\alpha^\prime$ of $\alpha$ and $\beta^\prime$ of $\beta$. The minimal polynomials of $\sqrt{2}$ and $i$ over $\mathbb{Q}$ are $x^2 - 2$ and $x^2+1$, and their conjugates are $\pm\sqrt{2}$ and $\pm i$, respectively. Evidently, $\sqrt{2} - i$, $-\sqrt{2}+i$, and $-\sqrt{2}-i$ are all distinct from $\sqrt{2}+i$. Thus $\mathbb{Q}(\sqrt{2},i) = \mathbb{Q}(\sqrt{2}+i)$. In particular, $i \in \mathbb{Q}(\sqrt{2} + i)$.