Let $f(x)=x^4+ax^2+b \in K[x]$ (With char $K\neq 2$) be irreducible with Galois group $G.$
If $b$ is not a square in $K$ and $b(a^2-4b)$ is a square in $K$, then $G \cong \mathbb{Z}_4$.
My attempt:
I know if $f$ (irreducible) separable quartic polynomial with Galois group $G$. Let $\alpha, \beta, \gamma$ be the roots of the resolvent cubic of $f$ and let $m=[K(\alpha, \beta, \gamma):K]$ then
$m=2\Leftrightarrow G\cong D_4, \text{or } G\cong \mathbb{Z}_4$; in this case $G\cong D_4$ if $f$ is irreducible over $K(\alpha, \beta, \gamma)$ and $G\cong \mathbb{Z}_4$, otherwise.
So the resolvent cubic is
$$h(x)=x^3-2ax^2+(a^2-4b)x=x(x^2-2ax+(a^2-4b))$$
then $\alpha=0$, $\beta=a+\sqrt b$, $\gamma=a-\sqrt b$ since $b$ is n ot a square in $K$ we can conclude $K(\alpha, \beta, \gamma)=K(\sqrt b)$ then $m=2$ but I have some troubles to proof $f$ is reducible over $K(\sqrt b)$
Futhermore if neither $b$ nor $b(a^2-b)$ is a square in $K$ then $G\cong D_4$... in this case $f$ need to be irreducible over $K(\sqrt b)$ but I have troubles here too :(
My main question is how can I prove $f$ is reducible/irreducible over $K(\sqrt b)$ thanks :).
EDIT: To decide if $f$ is irreducible over $K(\sqrt b)$ we set up a hypothetical factorization $$f(x)=(x^2+px+q)(x^2+rx+s)$$ and read off the algebraic conditions imposed on the coefficients: $$r+p=0, \ ps+qr=0,\ qs=b , \ s+q+pr=a$$ Therefore $r=-p$, $s=q=\pm\sqrt b$, $p=\sqrt{\pm2\sqrt b -a}$ and $r=-\sqrt{\pm2\sqrt b -a}.$ Then $$f(x)=(x^2+(\sqrt{\pm2\sqrt b -a})x+\sqrt b)(x^2-(\sqrt{\pm2\sqrt b -a})x+\sqrt b)$$ Any suggestions to continue demonstrating the irreducibility/reducibility of $f$?
