The Stacks project

Proof. Choose a basis $\alpha _1, \ldots , \alpha _ n$ of $K$ as a vector space over $F$. By Lemma 9.12.11 (and a tiny omitted argument) the set $\mathop{\mathrm{Hom}}\nolimits _ F(K, L)$ has $n$ elements, say $\sigma _1, \ldots , \sigma _ n$. In particular, the two sides have the same dimension $n$ as vector spaces over $L$. Thus if the map is not an isomorphism, then it has a kernel. In other words, there would exist $\mu _ j \in L$, $j = 1, \ldots , n$ not all zero, with $\sum \alpha _ j \otimes \mu _ j$ in the kernel. In other words, $\sum \sigma _ i(\alpha _ j)\mu _ j = 0$ for all $i$. This would mean the $n \times n$ matrix with entries $\sigma _ i(\alpha _ j)$ is not invertible. Thus we can find $\lambda _1, \ldots , \lambda _ n \in L$ not all zero, such that $\sum \lambda _ i\sigma _ i(\alpha _ j) = 0$ for all $j$. Now any element $\alpha \in K$ can be written as $\alpha = \sum \beta _ j \alpha _ j$ with $\beta _ j \in F$ and we would get

which contradicts Lemma 9.13.3. $\square$