The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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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