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The Stacks project

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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