Lemma 15.92.2. Let $K$ be a field. Let $A$ be a $K$-algebra. There exists a maximal weakly étale $K$-subalgebra $B_{max} \subset A$.

**Proof.**
Let $B_1, B_2 \subset A$ be weakly étale $K$-subalgebras. Then $B_1 \otimes _ K B_2$ is weakly étale over $K$ and so is the image of $B_1 \otimes _ K B_2 \to A$ (Lemma 15.92.1). Thus the collection $\mathcal{B}$ of weakly étale $K$-subalgebras $B \subset A$ is directed and the colimit $B_{max} = \mathop{\mathrm{colim}}\nolimits _{B \in \mathcal{B}} B$ is a weakly étale $K$-algebra by Lemma 15.91.14. Hence the image of $B_{max} \to A$ is weakly étale over $K$ (previous lemma cited). It follows that this image is in $\mathcal{B}$ and hence $\mathcal{B}$ has a maximal element (and the image is the same as $B_{max}$).
$\square$

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