MHB How Do We Find the Embeddings of $\mathbb{Q}(\sqrt{2})$ in $\mathbb{R}$?

[mathmari]

Hey!

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 howwe found these two embeddings?? (Wondering)

 

[Fallen Angel]

Hi,

Those morphisms sends roots of the polynomial of the primitive element into roots of the "image of the polynomial" (the polynomial with coefficients the image of the initial coefficients).

And the morphisms are defined by the image of the elements of a base (In this case, $\{1, \sqrt{2}\}$), as you know, these morphism needs to be the identity when restricted over the base field, so $\tilde{\sigma}(1)=1$ and you got two options to choose the image of $\sqrt{2}$ that are the two roots of $x^{2}-2$, i.e. $\tilde{\sigma}(\sqrt{2})=\pm \sqrt{2}$