Lemma 9.16.6. Let $L/K$ be a finite extension. Let $M/L$ be the normal closure of $L$ over $K$. Then there is a surjective map

of $K$-algebras where the number of tensors can be taken $[L : K]_ s \leq [L : K]$.

Lemma 9.16.6. Let $L/K$ be a finite extension. Let $M/L$ be the normal closure of $L$ over $K$. Then there is a surjective map

\[ L \otimes _ K L \otimes _ K \ldots \otimes _ K L \longrightarrow M \]

of $K$-algebras where the number of tensors can be taken $[L : K]_ s \leq [L : K]$.

**Proof.**
Choose an algebraic closure $\overline{K}$ of $K$. Set $n = [L : K]_ s = |\mathop{\mathrm{Mor}}\nolimits _ K(L, \overline{K})|$ with equality by Lemma 9.14.8. Say $\mathop{\mathrm{Mor}}\nolimits _ K(L, \overline{K}) = \{ \sigma _1, \ldots , \sigma _ n\} $. Let $M' \subset \overline{K}$ be the $K$-subalgebra generated by $\sigma _ i(L)$, $i = 1, \ldots , n$. Then $M'$ is a field since any $K$-subalgebra of $\overline{K}$ is a field. Any $K$-algebra map from $M'$ to $\overline{K}$ permutes the $\sigma _ i$ so sends $M'$ into and onto $M'$. By construction the field $M'$ is generated by conjugates of elements of $\sigma _1(L)$. Having said this it follows from Lemma 9.15.5 that $M'$ is normal over $K$ and that it is the smallest normal subextension of $\overline{K}$ containing $\sigma _1(L)$. By uniqueness of normal closure we have $M \cong M'$. Finally, there is a surjective map

\[ L \otimes _ K L \otimes _ K \ldots \otimes _ K L \longrightarrow M', \quad \lambda _1 \otimes \ldots \otimes \lambda _ n \longmapsto \sigma _1(\lambda _1) \ldots \sigma _ n(\lambda _ n) \]

and note that $n \leq [L : K]$ by definition. $\square$

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