Prove that there exists an element $a$ of $F$ such that $K\cong F(\sqrt a).$

Question

How to solve this problem?

Let $F$ be a field of odd characteristic and $K$ is a field extension over $F$ of degree $2$. Prove that there exists an element $a$ of $F$ such that $K\cong F(\sqrt a).$

Attempt:

Here $[K:F]=2$. Assume that there exists no $a\in F$ such that $K\cong F(\sqrt a).$ Then consider the tower $[K:F]=[K:F(\sqrt a)][F(\sqrt a):F]$.

Since $\sqrt a$ is a root of the polynomial $x^2-a$ over $ F$. Hence $[F(\sqrt a):F]=2\implies [K:F(\sqrt a)]=1\implies K=F(\sqrt a)$

Answer 1

The problem with your attempt is that you are assuming that $K$ contains a square root of $a$ for some $a\in F$ which is not a square in $F$. This assumption is essentially what you're trying to prove.

Instead, if $[K:F]=2$ then choose some $\alpha\in K$ such that $\alpha\not\in F$. Then the set $\{1,\alpha,\alpha^2\}$ is linearly dependent over $F$, and since $\alpha\not\in F$ (so $\alpha$ and $1$ are linearly independent over $F$) it follows that there is a quadratic polynomial $f(x)=x^2+bx+c\in F[x]$ such that $f(\alpha)=0$.

Finally, since $F$ doesn't have characteristic $2$, you can complete the square to find an element $a\in F$ such that $K=F(\sqrt{a})$.