Quadratic extensions are obtained by adjoining a square root

Question

A few questions about this proof (found in Artin's "Algebra", 2nd Ed.):

• Is it true that we can't say that $F(\alpha)=K$ (and hence need to consider the discriminant) because $\alpha^2$ doesn't have to lie in $F$?

• The part with choosing the value of a square root in an arbitrary field looks a little murky to me. Is the story here that in any field $K$, $\sqrt{D}$ is defined to be an element $r\in K$ such that $r^2=D$ in $K$? And in any field, there are two choices because if $r$ is one such element, than $-r$ is another?

Answer 1

1. Yes, its an arbitrarily chosen $\alpha$, e.g. $1+\sqrt{2}\in \mathbb{Q}(\sqrt{2})$.
2. Yes as well because we are working with a field, hence it must have additive inverses and is chosen such that $r^2=D$.