subfield of quartic extension in characteristic $2$ Let $\mathbb{F}_2(t)$ be the function field over $\mathbb{F}_2$.
Let $\mathbb{F}_2(t)(\alpha)$ be an extension of degree $4$ of $\mathbb{F}_2(t)$, where the minimal polynomial of $\alpha$ over $\mathbb{F}_2(t)$ is of the form
$$
\alpha^4+a\alpha^2 + b=0 \,\,\,\, (a,b\in\mathbb{F}_2(t)).
$$
Simple Observation:

*

*$\mathbb{F}_2(t)(\alpha)$ is inseparable extension of $\mathbb{F}_2(t)$;


*$\mathbb{F}_2(t)(\alpha^2)$ is separable closure of $\mathbb{F}_2(t)$  (degree $2$) inside $\mathbb{F}_2(t)(\alpha)$.
Question: Does $\mathbb{F}_2(t)(\alpha)$ always contain an intermediate field $L$, such that $[L:\mathbb{F}_2(t)]=2$ and $L$ is inseparable over $\mathbb{F}_2(t)$?
 A: Note that if $a=0$, your conclusions are not correct, so I will assume $a\neq 0$.
An inseparable element in $L$ has degree $2$, so its minimal polynomial has the form $X^2-d, d\in F^\times$ non square (the coefficient is $0$, otherwise we would get a separable polynomial).
So, we have to look at equations of the form $(u+v\alpha+w\alpha^2+s\alpha^3)^2=d$. Few computations show that we get $(a^2
s^2+aw^2+bs^2+v^2)\alpha^2+abs^2+bw^2+u^2=d$.
Hence $a^2s^2+aw^2+bs^2+v^2=0 \ \ $ $(1)$ and $abs^2+bw^2+u^2=d \ \ $ $(2)$.
Note that every element of $\mathbb{F}_2(t)$ may be written in a unique way as $x_0^2+tx_1^2$.
Write $a=a_0^2+ta_1^2,b=b_0^2+tb_1^2$.
Then $(as+v+a_0w+b_0s)^2+t(a_1w+b_1s)^2=0$.
You then get $a_1w=b_1s$, $v=a_0w+as+b_0s$.

*

*If $a_1=0$, that is $a$ is a square, then $b_1\neq 0$ (otherwise $X^4+aX^2+b$) would be reducible), so $s=0$ and $w$ is arbitrary, and $d=bw^2+u^2$. You can then choose $w=1$ and $u=0$, and $d=b$ since $b$ is not a square.


*If $a_1\neq 0$, then $w=b_1s/a_1$.
Plugging in $(2)$, we get $(ab+b_1/a_1)s^2+u^2=d$.
If $ab+b_1/a_1$ is a square, $d$ is always a square and $\beta$ does not exists. If $ab+b_1/a_1$ is not a square, you can take $s=1$ and $u=0$.
So your $L$ exists if and only if $a,b$ are choosen such that $X^4+aX^2+b$ is irreducible and either $a$ is a square, or $a$ is not a square and  $ab+b_1/a_1$ is not square.
An infinite family of counterexamples can be constructed as follow: take $a=b=\pi$ an irreductible polynomial.
Then $X^4+aX^2+b$ is $\pi$-Eisenstein, $a$ is not a square but $ab+b_1/a_1=\pi^2+1=(1+\pi)^2$ is a square.
A: Sure, $\alpha^2+a^{1/2}\alpha+b^{1/2}=0$ gives that $[\Bbb{F}_2(t^{1/2},\alpha):\Bbb{F}_2(t^{1/2})]=2$ which implies that $\Bbb{F}_2(t,\alpha)=\Bbb{F}_2(t^{1/2},\alpha)$
