Let $\alpha$ a root of $X^3+X^2+1 \in \mathbb{F}_2[X]$ and $K=\mathbb{F}_2(\alpha)$. Suppose $f$ is an irreducible polynomial in $K[X]$ of degree 4. Let $\beta$ be a root of $f$, and $L$ a splitting field of $f$ over $K$. What is the number of elements of $L$?
I would say, $f$ is irreducible, so $deg(\beta) = 4$. Is this claim correct?
If it is, $[L:K] = deg(\beta) = 4$, so $L$ has $8^4 = 4096$ elements? (since $K$ has 8 elements). Is this correct reasoning?
Furthermore, how many intermediate fields does the extension $L/K$ have?
I would say 2, a field with $8^2= 64$ elements and a field with $8^3 = 512$ elements. But this is intuitively and I don't know how to prove this?
$$\text{HINT}$$ Because of $g(x)=x^3+x^2+1\ne0$ for the only two elements of $\mathbb F_2$, the cubic $g$ is irreducible over $\mathbb F_2$ and, since $\alpha^3=\alpha^2+1$ (obviously$1=-1$ ), each element of $\mathbb F_2(\alpha)$ has the form $$x=a_0+a_1\alpha+a_2\alpha^2$$ It follows that $\mathbb F_2(\alpha)$ has $2\cdot2\cdot2=8$ elements (in general all finite extension of degree $n$ de $\mathbb F_p$ has $p^n$ elements).
Now you have $$f(x)=a_0+a_1x+a_2x^2+a_3x^4\space\text {with } f(\beta)\ne0$$ where each coefficient $a_i$ can take eight distinct values. Therefore the extension $K(\beta)$, like above, has $8\cdot8\cdot8\cdot8=2^{12}$ elements. This is not the end because $K(\beta)$ is not the required splitting field. Can you finish the answer guided by lines above?