Give all extensions of the mapping to an isomorphic mapping

Question

Let $ E = \mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}). $ It can be shown that $ [E : \mathbb{Q}] = 8. $ For each given isomorphic mapping of a subfield of $ E, $ give all extensions of the mapping to an isomorphic mapping of $ E $ onto a subfield of $ \overline{\mathbb{Q}}. $ Describe the extensions by giving values on the generating set $ \{\sqrt{2}, \sqrt{3}, \sqrt{5} \} $ for $ E $ over $ \mathbb{Q}. $

I am having trouble with this exercise since my answers are different from what my teacher gave, so could someone point out where I got wrong and suggest a way to do them?

$ 1) \; \iota: \mathbb{Q}(\sqrt{2}, \sqrt{15}) \to \mathbb{Q}(\sqrt{2}, \sqrt{15}), $ where $ \iota $ is the identity map. I know that the extension $ \tau $ must follow the rules $ \tau(a) = \iota(a), $ so I must have $ \tau(\sqrt{2}) = \iota(\sqrt{2}) = \sqrt{2}, \tau(\sqrt{3}) = \iota(\sqrt{3}) = \sqrt{3}, $ and $ \tau(\sqrt{5}) = \iota(\sqrt{5}) = \sqrt{5}, $ but the answer they have is $ \tau(\sqrt{2}) = \sqrt{2}, \tau(\sqrt{3}) = -\sqrt{3}, $ and $ \tau(\sqrt{5}) = -\sqrt{5}. $

$ 2) \; \sigma: \mathbb{Q}(\sqrt{2}, \sqrt{15}) \to \mathbb{Q}(\sqrt{2}, \sqrt{15}), $ where $ \sigma(\sqrt{2}) = \sqrt{2} $ and $ \sigma(\sqrt{15}) = -\sqrt{15}. $ I know that the extension $ \tau $ must follow the rules $ \tau(a) = \sigma(a), $ so I must have $ \tau(\sqrt{2}) = \sigma(\sqrt{2}) = \sqrt{2}, \tau(\sqrt{3}) = \sigma(\sqrt{3}) = -\sqrt{3}, $ and $ \tau(\sqrt{5}) = \sigma(\sqrt{5}) = \sqrt{5} $ or $ \tau(\sqrt{2}) = \sigma(\sqrt{2}) = \sqrt{2}, \tau(\sqrt{3}) = \sigma(\sqrt{3}) = \sqrt{3}, $ and $ \tau(\sqrt{5}) = \sigma(\sqrt{5}) = -\sqrt{5}, $ which match with the answer key.

$ 3) \; \psi_{\sqrt{30}, \; -\sqrt{30}}: \mathbb{Q}(\sqrt{30}) \to \mathbb{Q}(\sqrt{30}). $ I don't know how to proceed for this question.

Answer 1

In $(1)$, there you extend $\iota$ to either the identity on $E$, or to the map $\tau$ with $\tau(\sqrt3)=-\sqrt3$ and $\tau(\sqrt5)=-\sqrt5$. One simple reason for this is that then $$\sqrt{15}=(-\sqrt3)( -\sqrt5)=\tau(\sqrt3)\tau(\sqrt5)=\tau(\sqrt3\sqrt5)=\tau(\sqrt{15})$$ and this is clearly an automorphism of $E$.

$(3)$ Assuming that $\psi$ is the map that sends $\sqrt{30}$ to $-\sqrt{30}$ and $\sqrt2$, $\sqrt5$ and $\sqrt3$ must all be mapped either identically onto itself or its negative, we obtain $$-\sqrt{30}=\tau(\sqrt{30})= \tau(\sqrt2)\tau(\sqrt3)\tau(\sqrt5)$$ Therefore, each extension $\tau$ of $\psi$ must either map all three of $\sqrt2$, $\sqrt5$ and $\sqrt3$ to its negative, or exactly one. There are thus exactly four extensions which will be $\psi_{\sqrt\alpha,-\sqrt\alpha}:E\rightarrow E$ where $\alpha\in \{2,3,5,30\}$.

You might wonder if these are all the members, since for example $\sqrt6$ must also be mapped identically or to its negative and since $$-\sqrt{30}=\tau(\sqrt{30})= \tau(\sqrt6)\tau(\sqrt5)$$ the extension fixing $\sqrt5$ must either be $\psi_{\sqrt2,-\sqrt2}$ or $\psi_{\sqrt3,-\sqrt3}$. Similarly for the other cases.