Showing that a set is a basis of a field as a vector space over a subset of that field Let $K \subseteq L \subseteq F$ be fields and assume that $\{\alpha_1,\ldots,\alpha_m\}$ is a basis of $F$ as a vector space over $L$ and $\{\beta_1,\ldots,\beta_n\}$ is a basis of $L$ as a vector space over $K$.
Prove that $\{\alpha_i\beta_j : 1\leq i \leq m; 1 \leq j \leq n\}$ is a basis of $F$ as a vector space over $K$.
Would someone please explain to me what exactly it is that I need to show? My lecturer hasn't given us any resources on how to do this and I am quite lost as on how to start this proof. I will be grateful for any tips and hints. Just a brief guideline on how to start this proof would be great. Thanks in advance.
 A: You need to show the following two things.
a. Linear combinations of the elements of $\{a_ib_j\}_{i,j}$ span $F$.
b. $\{a_ib_j\}_{i,j}$ are linearly independent.
For a., Let $f\in F$. Then as $\{b_1,\ldots,b_n\}$ is a basis of $F_L$, then 
$$
f=\ell_1 b_1+\cdots+\ell_n b_n, \quad \ell_i\in L.
$$
As $\{a_1,\ldots,a_m\}$ is a basis of $L_K$, then 
$$
\ell_i=k_{1i}a_1+\cdots+k_{mi}a_m,\quad k_{ji}\in K
$$
and hence
$$
f=\sum_{i,j}k_{ij}a_ib_j.
$$
For b., assume that
$$
\sum_{i,j}c_{ij}a_ib_j=0.
$$
Then
$$
\sum_{j}b_j\left(\sum_{i}c_{ij}a_i\right)=0,
$$
and as the $b_j$'s are linearly independent over $L$, then
$$
\sum_{i}c_{ij}a_i=0, \quad j=1,\ldots,n,
$$
and as the $a_i$'s are linearly independent over $K$, then 
$$
c_{ij}=0,
$$
for all $i,j$.
A: 
"$\{\alpha_1,\ldots,\alpha_m\}$ is a basis of $F$ as a vector space over $L$."

This means every member of $F$ can be written in exactly one way as a linear combination of members of $\{\alpha_1,\ldots,\alpha_m\}$, the coefficients in the linear combination being members of $L$.

"$\{\beta_1,\ldots,\beta_n\}$ is a basis of $L$ as a vector space over $K$."

This means every member of $L$ can be written in exactly one way as a linear combination of members of $\{\beta_1,\ldots,\beta_m\}$, the coefficients in the linear combination being members of $K$.

"$\{\alpha_i\beta_j : 1\leq i \leq m; 1 \leq j \leq n\}$ is a basis of $F$ as a vector space over $K$."

This means every member of $F$ can be written in exactly one way as a linear combination of members of $\{\alpha_i\beta_j : 1\leq i \leq m; 1 \leq j \leq n\}$, the coefficients in the linear combination being members of $K$.
A: Let $K = \mathbb{Q}$, $L$ the extension $\mathbb{Q}(\sqrt{2})$, and $F=\mathbb{Q}(\sqrt{2},\sqrt{3})$. Then the set $\{1,\sqrt{3}\}$ forms a basis over $L$, meaning that the coefficients used are taken from $L$, but $L$ is also a vector space over $\mathbb{Q}$ with basis $\{1,\sqrt{2}\}$. Your statement now says that $F$ has basis $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ using coefficients in $\mathbb{Q}$.
