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

is an isomorphism of $L$-algebras.

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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