Characteristic infinite field

Question

Do you know how to solve these questions?

$K$ is a field of characteristic $0$.

Let $L$ be an extension of $K$ with $[L : K] = 2$. Prove that there exists $u \in L$ such that $L = K(u)$ and $u^2 \in K$.

Let $M$ be an extension of $\mathbb{Q}$ with $[M : \mathbb{Q}] = 4$. Prove that there exists a field $L$ such that $\mathbb{Q} ⊂ L ⊂ M$ and $[L : \mathbb{Q}] = 2$ if and only if there exists $u \in M$ such that $M = \mathbb{Q}(u)$ and the minimal polynomial of $u$ is of the form $X^4 + aX^2 + b$ with $a, b \in \mathbb{Q}$.

Answer 1

Hint:

$[L:K]=2$ means that any element in $L$ can be written as $af+bg$, where $a,b\in K$ and $f,g\in L$, that is, $$L=\left<f,g\right>_K$$ what can you say about these generators?

Answer 2

Take $v\in L$, $v\notin K$. Then $K(v)=L$, because $[L:K(v)][K(v):K]=[L:K]=2$ and $[K(v):K]>1$.

Let $x^2+px+q$ be the minimal polynomial of $v$; then, completing the square, $$ \left(v+\frac{p}{2}\right)^2=\frac{p^2}{4}-q\in K $$ Can you tell what's a good $u$?

^{Note: you just need the characteristic to be $\neq 2$.}

For the second part, suppose there exists $L$ with $\mathbb{Q}\subset L\subset M$ and $[L:\mathbb{Q}]=2$. By the previous part, there is $v\in L$ with $L=\mathbb{Q}(v)$ and $v^2\in\mathbb{Q}$. But now $$ [M:L]=\frac{[M:\mathbb{Q}]}{[L:\mathbb{Q}]}=\frac{4}{2}=2 $$ Apply again the first part.

The converse follows from considering that the minimal polynomial is biquadratic.