I was wondering if somebody could help me out with solving the following:
Let $f(x)$ be an irreducible polynomial of degree $4$ with rational coefficients. Let $α$ be a root of $f$ , say in $\Bbb C$, and let $L = \Bbb Q(α)$.Let $K$ be the splitting field of $f(x)$ over $L$ (hence over $\Bbb Q$). Assume also that $L$ contains a quadratic subfield $L ⊃ M ⊃ Q$, where $M/Q$ is a degree $2$ extension. Assume further that $K$ properly contains $L$. Prove that in this case $K$ has degree $8$ over $Q$; determine the Galois group. (Hint: show that α solves a quadratic polynomial $g(x) ∈ M[x]$. How are $g$ and $f$ related?)
I have been able to show that $L$ over $M$ has degree $2$ as $L$ over $\Bbb Q$ has degree $4$, and $M$ over $Q$ has degree $2$. What is missing is to show $K$ over $L$ also has degree $2$. I know it must have degree $\geq 2$ as $L$ is a proper subfield of $K$. I also know why there is a polynomial $g$ of degree $2$ in $M[x]$ with $\alpha$ as a root (since degree of $L$ over $M$ is $2$, and $\alpha$ is in L hence $1, \alpha, \alpha^2$ forms a linearly dependent set hence solves a quadratic. Also I know that $g$ must be irreducible over $\Bbb Q$, as else $\alpha$ would be in $M$, contracting the degree of $L$ over $M$. So then I know $g$ divides $f$ in $M[x]$.
Could anyone help me finish this problem and also confirm my arguments? Thanks in advance
If you know that $g$ divides $f$, we must have $f=gh$ say. $h$ cannot split over $L$, since $L$ is not the splitting field of $f$. Since $h$ is quadratic it splits over an extension of $L$ of degree 2, hence $[K:L]=8$ as you've got all 4 roots of $f$.