Lemma 15.105.2. 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 15.105.1). 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 15.104.14. 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 $B_{max}$). $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)