In one exercise I am supposed to find the conjugate of $\sqrt{2}+i$ over $\mathbb{Q}$.
I found the answer by finding irr$( \sqrt{2}+i,\mathbb{Q})$, and then solving the polynomial finding all the roots(this is the definition of conjugates), this gave the correct answer, but in the solutions manual, they use a trick I do not understand.
They say that:
The conjugate are $\sqrt{2}+i,\sqrt{2}-i,-\sqrt{2}+i,-\sqrt{2}-i$. This is clear beacuse $\mathbb{Q}(\sqrt{2}+i)=(\mathbb{Q}(\sqrt{2}))(i)$.
Can someone say how this is clear and why it follows from the fact that: $\mathbb{Q}(\sqrt{2}+i)=(\mathbb{Q}(\sqrt{2}))(i)$?
By what you mention in your comments, the conjugates of $(\sqrt{2}+i)$ can be computed if we knew the image of $(\sqrt{2}+i)$ under every homomorphism $$ \varphi : \mathbb{Q}(\sqrt{2}+i) \to \overline{\mathbb{Q}} $$ Since $$ \mathbb{Q}(\sqrt{2}+i) = \mathbb{Q}(\sqrt{2})(i) $$ we need to determine $\varphi(i)$ and $\varphi(\sqrt{2})$. However, since $\varphi$ is a field homomorphism $$ \varphi(\sqrt{2})^2 = \varphi(2) = 2 $$ since any field homomorphism must fix $\mathbb{Q}$ (this takes a short proof). However, $2$ has exactly 2 square roots in $\overline{\mathbb{Q}} \subset \mathbb{C}$, so $$ \varphi(\sqrt{2}) = \pm \sqrt{2} $$ Similarly, $$ \varphi(i) = \pm i $$ and so the list of possible conjugates is $$ \{\pm\sqrt{2}\pm i\} $$ To prove that each of these is, in fact, a conjugate of $(\sqrt{2}+i)$, you need the fact that the number of such homomorphisms is precisely equal to the degree $$ [\mathbb{Q}(\sqrt{2}+i):\mathbb{Q}] $$ My answer is long enough already, so let me know if this helps and if you have any questions.