Lemma 10.45.3. Let $K/k$ be a finitely generated field extension. There exists a diagram

$\xymatrix{ K \ar[r] & K' \\ k \ar[u] \ar[r] & k' \ar[u] }$

where $k'/k$, $K'/K$ are finite purely inseparable field extensions such that $K'/k'$ is a separable field extension. In this situation we can assume that $K' = k'K$ is the compositum, and also that $K' = (k' \otimes _ k K)_{red}$.

Proof. By Lemma 10.42.4 we can find such a diagram with $K'/k'$ separably generated. By Lemma 10.44.2 this implies that $K'$ is separable over $k'$. The compositum $k'K$ is a subextension of $K'/k'$ and hence $k' \subset k'K$ is separable by Lemma 10.42.2. The ring $(k' \otimes _ k K)_{red}$ is a domain as for some $n \gg 0$ the map $x \mapsto x^{p^ n}$ maps it into $K$. Hence it is a field by Lemma 10.36.19. Thus $(k' \otimes _ k K)_{red} \to K'$ maps it isomorphically onto $k'K$. $\square$

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