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The Stacks project

Lemma 9.13.4. Let K/F and L/F be field extensions with K/F finite separable and L algebraically closed. Then the map

K \otimes _ F L \longrightarrow \prod \nolimits _{\sigma \in \mathop{\mathrm{Hom}}\nolimits _ F(K, L)} L,\quad \alpha \otimes \beta \mapsto (\sigma (\alpha )\beta )_\sigma

is an isomorphism of L-algebras.

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

\sum \lambda _ i\sigma _ i(\alpha ) = \sum \lambda _ i\sigma _ i(\sum \beta _ j \alpha _ j) = \sum \beta _ j \sum \lambda _ i\sigma _ i(\alpha _ j) = 0

which contradicts Lemma 9.13.3. \square


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