Basis of finite field as combination of bases of intermediate fields Let $F=\mathbb{F}_q$ be a finite field with field extensions $K=\mathbb{F}_{q^m}$ and $L=\mathbb{F}_{q^n}$ with $dm=n$ and let $\{a_1,\dots,a_n\}$ be a basis of $K$ over $F$ and $\{b_1,\dots,b_d\}$ be a basis of $L$ over $K$. Then (and this is standard proof for the degree multiplicativity formula) the set $$\{a_ib_j \mid i=1,\dots,n;j=1,\dots,d\}$$ is a basis of $L$ over $F$. Now my question is: Does the converse hold, i.e. is every basis of $L$ over $F$ of this form? Now I think this does not hold by a simple counting argument of the bases but I seem to be too dense to find it.
 A: consider $F=\Bbb{F}_2$, $L = \Bbb{F}_{16}\cong\Bbb{F}_2[x]/(x^4+x+1)$, and $K=L^{<(\cdot)^4>}=\{\overline{a+b(x+x^2)}|a,b\in \Bbb{F}_2\}\cong \Bbb{F}_4$ (i.e. K is the subfield of $\Bbb{F}_{16}$ isomorphic to $\Bbb{F}_4$ found by using Galois theory. Here, it is the subfield fixed by the square of the Frobenius automorphism). (Also, note: We can think of $F$ as being a subfield of the other fields here even though it is not literally a subfield. If you like we can work with $F = L^{<(\cdot)^2>})=\{\overline{0},\overline{1}\}$) Then $\overline{1},\overline{x},\overline{x^2},\overline{x^3}$ is a basis for $K/F$. Suppose that $a_1,a_2$ is a basis for $K/F$ and $b_1,b_2$ is a basis for $L/K$ that satisfy that $\{\overline{1},\overline{x},\overline{x^2},\overline{x^3}\}=\{a_ib_j|i=1,2;j=1,2\}$. Without loss of generalization, $a_1b_1=\overline{1}$. Then, since $a_1\in K$, $b_1\in K$. Then $a_2b_1\in K$. But the only basis element in $\{\overline{1},\overline{x},\overline{x^2},\overline{x^3}\}$ that is in $K$ is $\overline{1}$ which is a contradiction.
