Proof. Let $\alpha _1, \ldots , \alpha _ n \in F$ be an $E$-basis for $F$. Let $\beta _1, \ldots , \beta _ m \in E$ be a $k$-basis for $E$. Then the claim is that the set of products $\{ \alpha _ i \beta _ j, 1 \leq i \leq n, 1 \leq j \leq m\}$ is a $k$-basis for $F$. Indeed, let us check first that they span $F$ over $k$.

By assumption, the $\{ \alpha _ i\}$ span $F$ over $E$. So if $f \in F$, there are $a_ i \in E$ with

$f = \sum \nolimits _ i a_ i \alpha _ i,$

and, for each $i$, we can write $a_ i = \sum b_{ij} \beta _ j$ for some $b_{ij} \in k$. Putting these together, we find

$f = \sum \nolimits _{i,j} b_{ij} \alpha _ i \beta _ j,$

proving that the $\{ \alpha _ i \beta _ j\}$ span $F$ over $k$.

Suppose now that there existed a nontrivial relation

$\sum \nolimits _{i,j} c_{ij} \alpha _ i \beta _ j = 0$

for the $c_{ij} \in k$. In that case, we would have

$\sum \nolimits _ i \alpha _ i \left( \sum \nolimits _ j c_{ij} \beta _ j \right) = 0,$

and the inner terms lie in $E$ as the $\beta _ j$ do. Now $E$-linear independence of the $\{ \alpha _ i\}$ shows that the inner sums are all zero. Then $k$-linear independence of the $\{ \beta _ j\}$ shows that the $c_{ij}$ all vanish. $\square$

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