If $E/F$ is a field extension of degree $2$ and $\text{char}(F)\ne 2$ then $E=F(\sqrt a)$ for some $a\in E$

Question

Let $E/F$ be a field extension of degree $2$ and $\text{char}(F)\ne 2$.

1. Show that $E=F(\sqrt a)$ for some $a\in F$.
2. Given some $0\ne a,b\in F$, under what conditions $F(\sqrt a)=F(\sqrt b)$?

My attempt:

We know that if $E/F$ is a field extension of a prime degree then $E/F$ is simple. Thus we have $E=F(a)$ for some $a\in E$. Let $f=\text{irr}(a,F)$.

Answer 1

For the first part; let $x\in E$ with $x\notin F$. Then $1$, $x$ and $x^2$ are linearly dependent over $F$ because $1,x,x^2\in E$ and $\dim_FE=2$. This means there exist $a,b,c\in F$ such that $$ax^2+bx+c=0.$$ Of course $a\neq0$ because $x\notin F$ and so we get $$x^2+\tfrac bax+\tfrac ca=0.$$ Because $\operatorname{char}F\neq2$ we also have $y:=x+\tfrac{b}{2a}\in E$ and this element satisfies $$y^2+\tfrac{c}{a}-\tfrac{b^2}{4a^2}=0,$$ meaning that $y=\sqrt{\tfrac{b^2-4ac}{4a^2}}$, and so we have $E=F(\sqrt{\alpha})$ for $\alpha:=\tfrac{b^2-4ac}{4a^2}$.

For the second part; note that $F(\sqrt{\alpha})=F(\sqrt{\beta})$ for nonzero $\alpha,\beta\in F$ if and only if there exist $a,b\in F$ such that $\sqrt{\beta}=a+b\sqrt{\alpha}$. In this case $$\beta=(a+b\sqrt{\alpha})^2=a^2+\alpha b^2+2ab\sqrt{\alpha},$$ where $\beta\in F$, and so $ab=0$ because $\operatorname{char}F\neq2$. If $b=0$ then $\sqrt{\beta}=a\in F$ and so $$F(\sqrt{\alpha})=F(\sqrt{\beta})=F(a)=F,$$ meaning that $\alpha$ and $\beta$ are both squares in $F$. If $a=0$ then $\sqrt{\beta}=b\sqrt{\alpha}$ and hence $\beta=b^2\alpha$. In both cases we see that $\tfrac{\alpha}{\beta}\in F$ is a square.

Conversely, if $\tfrac{\alpha}{\beta}\in F$ is a square, say $\tfrac{\alpha}{\beta}=c^2$, then $\sqrt{\alpha}=c\sqrt{\beta}$ and so $$F(\sqrt{\alpha})=F(c\sqrt{\beta})=F(\sqrt{\beta}),$$ which shows that $F(\sqrt{\alpha})=F(\sqrt{\beta})$ if and only if $\tfrac{\alpha}{\beta}$ is a square in $F$.