Consider the polynomial $f(X)=X^{4}+aX^{2}+b \in F[X]$ such that $f$ is irreducible over $F$.
Suppose $\alpha$ is a root of $f(X)$ and let $K=F(\alpha)$. Prove that there exists a subfield $L$ of $K$ such that $[L:F]=2$.
How do I prove this claim? Any help will be appreciated. Thanks.
Set $\beta = \alpha^2,$ and notice that $\beta$ solves $X^2 + aX + b.$ So $L = F(\beta)$ has degree either 1 or 2.
But if $L$ had degree 1, then $\beta \in F$. We can use this to factor $X^4 + aX^2 + b$ as follows, contradicting its irreducibility. Let $\beta' = -a - \beta.$
Since $\beta^2 + a\beta + b = 0,$ plugging in $a = -\beta'-\beta$ we get $\beta^2 -\beta\beta' -\beta^2 + b = 0,$ so that $\beta\beta' = b.$
So now factor your polynomial as
$$(X^2 - \beta)(X^2 - \beta') = X^4 + aX^2 + \beta\beta' = X^4 + aX^2 + b,$$ contradicting its irreducibility. Therefore $\beta\not\in F,$ and so $[L : F] = 2.$