I am trying to understand a problem from my book regarding field extensions and fields generated from sets.
I have shown the set $B_0 = \lbrace (0,0),(1,0) \rbrace$ to generate a field $Q$ which consists of all elements of the form $(a,0)$, where $a \in \mathbb{Q}$. This field is then isomorphic to $\mathbb{Q}$ with the map
$\sigma : Q \to \mathbb{Q} \quad $ , $\quad \sigma [(a,0)] = a.$
Suppose now we add an element $(1,\sqrt{3}) $ to $B_0$ such that we get $B_1 = \lbrace (0,0),(1,0),(1,\sqrt{3}) \rbrace $. What field would $B_1$ generate? My book claims that this field is $\mathbb{Q}(\sqrt{3})$, but i have trouble seeing this. In similar way to how we found $\sigma$ for $B_0$, what would an isomorphism look like for the field generated by $B_1$?
Thank you very much!
If we call the field generated by $B_1$ something, say $F$, then the map \begin{align*}F &\to \mathbb{Q}(\sqrt{3})\\(x,y\sqrt{3})&\mapsto x+y\sqrt{3} \end{align*} is an isomorphism.
To prove that:
Firstly, it is well-defined, as $x$ and $y$ are rational for all elements of $F$, so $x+y\sqrt{3}$ is, indeed, in $\mathbb{Q}(\sqrt{3})$.
Secondly, it's clearly a field homomorphism (check everything if you're so inclined).
Thirdly, it's injective, since if $x + y \sqrt{3} = 0$ with $x$ and $y$ rational, then $x = y = 0$.
Finally, it's surjective, since we can get all $x + y\sqrt{3}$ in this way, as the image of $(x,y\sqrt{3}) = y(1,\sqrt{3})+(x-y)(1,0)$.