In my notes there is the following example:
$$\mathbb{Q}(\sqrt{2}) \overset{\widetilde{\sigma}}{\longrightarrow}\mathbb{R}\\ | \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \mathbb{Q} \overset{\sigma=id : q \mapsto q}{\rightarrow}\mathbb{R}$$
$p(x)=Irr(\sqrt{2}, \mathbb{Q})=x^2-2 \in \mathbb{Q}[x]$
$p^{\sigma}=x^2-2 \in \mathbb{R}[x]$ has two different roots in $\mathbb{R}$ : $ \pm \sqrt{2}$
So there are two embeddings $\widetilde{\sigma} : \mathbb{Q} (\sqrt{2}) \rightarrow \mathbb{R}$ :
- $\widetilde{\sigma}(\sqrt{2})=\sqrt{2}$ so $\widetilde{\sigma} ( \xi)=\xi, \forall \xi \in \mathbb{Q}(\sqrt{2})$
- $\widetilde{\sigma}(\sqrt{2})=-\sqrt{2}$ so $\widetilde{\sigma}(q_o+q_1 \sqrt{2})=q_0-q_1\sqrt{2}$
Could you explain me how we found these two embeddings??
Do we take $\widetilde{\sigma}(\sqrt{2})$ because it is an embedding from $\mathbb{Q}(\sqrt{2})$?? And does it stand that $\widetilde{\sigma}(\sqrt{2})=\pm \sqrt{2}$ because these at the solutions of $Irr(a, \mathbb{Q})$ in $\mathbb{R}$??
Also I haven't understood at the following:
$$\widetilde{\sigma}(\sqrt{2})=\sqrt{2} \text{ so } \widetilde{\sigma} ( \xi)=\xi, \forall \xi \in \mathbb{Q}(\sqrt{2})$$ $$\widetilde{\sigma}(\sqrt{2})=-\sqrt{2}\text{ so } \widetilde{\sigma}(q_o+q_1 \sqrt{2})=q_0-q_1\sqrt{2}$$
how we conclude that $\widetilde{\sigma} ( \xi)=\xi, \forall \xi \in \mathbb{Q}(\sqrt{2})$ and $\widetilde{\sigma}(q_o+q_1 \sqrt{2})=q_0-q_1\sqrt{2}$ ??
We are looking for a field embedding $\tilde{\sigma}:\mathbb{Q}(\sqrt{2})\hookrightarrow\mathbb{R}$, where $$\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}|a,b\in\mathbb{Q}\}.$$Note that $\sqrt{2}$ here should be thought of as an abstract element which satisfies $\sqrt{2}^2=2$, rather than the real positive number to which we relate as $\sqrt{2}$.
We know that the restriction $\tilde{\sigma}|_\mathbb{Q}$ has to be the given embedding $\sigma$, and since a field homomorphism preserves addition and multiplication, $\tilde{\sigma}$ is uniquely determined once we fix $\tilde{\sigma}(\sqrt{2})$. Since there are exactly two elements $\alpha,\beta\in\mathbb{R}$ which satisfy $\alpha^2=\beta^2=2$, our two choices are $\tilde{\sigma}(\sqrt{2})=\alpha$ and $\tilde{\sigma}(\sqrt{2})=\beta$.