For an algebra $A$ over $K$ we write $B_{max}(A/K)$ for the maximal weakly \'etale $K$-subalgebra of $A$. Similarly we write $B_{max}(A'/L)$ for the maximal weakly \'etale $L$-subalgebra of $A'$ if $A'$ is an $L$-algebra. Since $B_{max}(A/K) \otimes_K L$ is weakly \'etale over $L$ (Lemma \ref{lemma-base-change-weakly-etale}) and since $B_{max}(A/K) \otimes_K L \subset A \otimes_K L$ we obtain a canonical injective map $$ B_{max}(A/K) \otimes_K L \to B_{max}((A \otimes_K L)/L) $$ The lemma states that this map is an isomorphism. \medskip\noindent To prove the lemma for $L$ and our $K$-algebra $A$, it suffices to prove the lemma for any field extension $L'$ of $L$. Namely, we have the factorization $$ B_{max}(A/K) \otimes_K L' \to B_{max}((A \otimes_K L)/L) \otimes_L L' \to B_{max}((A \otimes_K L')/L') $$ hence the composition cannot be surjective without $B_{max}(A/K) \otimes_K L \to B_{max}((A \otimes_K L)/L)$ being surjective. Thus we may assume $L$ is algebraically closed. \medskip\noindent Reduction to finite type $K$-algebra. We may write $A$ is the filtered colimit of its finite type $K$-subalgebras. Using Lemma \ref{lemma-properties-of-max-weakly-etale-subalgebra} we see that it suffices to prove the lemma for finite type $K$-algebras. \medskip\noindent Assume $A$ is a finite type $K$-algebra. Since the kernel of $A \to A_{red}$ is nilpotent, the same is true for $A \otimes_K L \to A_{red} \otimes_K L$. Then $$ B_{max}((A \otimes_K L)/L) \to B_{max}((A_{red} \otimes_K L)/L) $$ is injective because the kernel is nilpotent and the weakly \'etale $L$-algebra $B_{max}((A \otimes_K L)/L)$ is reduced (Lemma \ref{lemma-class-weakly-etale-over-field}). Since $B_{max}(A/K) = B_{max}(A_{red}/K)$ by Lemma \ref{lemma-properties-of-max-weakly-etale-subalgebra} we conclude that it suffices to prove the lemma for $A_{red}$. \medskip\noindent Assume $A$ is a reduced finite type $K$-algebra. Let $Q = Q(A)$ be the total quotient ring of $A$. Then $A \subset Q$ and $A \otimes_K L \subset Q \otimes_A L$ and hence $$ B_{max}(A/K) = A \cap B_{max}(Q/K) $$ and $$ B_{max}((A \otimes_K L)/L) = (A \otimes_K L) \cap B_{max}((Q \otimes_K L)/L) $$ by Lemma \ref{lemma-properties-of-max-weakly-etale-subalgebra}. Since $-\otimes_K L$ is an exact functor, it follows that if we prove the result for $Q$, then the result follows for $A$. Since $Q$ is a finite product of fields (Algebra, Lemmas \ref{algebra-lemma-total-ring-fractions-no-embedded-points}, \ref{algebra-lemma-minimal-prime-reduced-ring}, \ref{algebra-lemma-Noetherian-irreducible-components}, and \ref{algebra-lemma-Noetherian-permanence}) and since $B_{max}$ commutes with products (Lemma \ref{lemma-properties-of-max-weakly-etale-subalgebra}) it suffices to prove the lemma when $A$ is a field. \medskip\noindent Assume $A$ is a field. We reduce to $A$ being finitely generated over $K$ by the argument in the third paragraph of the proof. (In fact the way we reduced to the case of a field produces a finitely generated field extension of $K$.) \medskip\noindent Assume $A$ is a finitely generated field extension of $K$. Then $K' = B_{max}(A/K)$ is a field separable algebraic over $K$ by Lemma \ref{lemma-properties-of-max-weakly-etale-subalgebra} part (6). Hence $K'$ is a finite separable field extension of $K$ and $A$ is geometrically irreducible over $K'$ by Algebra, Lemma \ref{algebra-lemma-make-geometrically-irreducible}. Since $L$ is algebraically closed and $K'/K$ finite separable we see that $$ K' \otimes_K L \to \prod\nolimits_{\sigma \in \Hom_K(K', L)} L,\quad \alpha \otimes \beta \mapsto (\sigma(\alpha)\beta)_\sigma $$ is an isomorphism (Fields, Lemma \ref{fields-lemma-finite-separable-tensor-alg-closed}). We conclude $$ A \otimes_K L = A \otimes_{K'} (K' \otimes_K L) = \prod\nolimits_{\sigma \in \Hom_K(K', L)} A \otimes_{K', \sigma} L $$ Since $A$ is geometrically irreducible over $K'$ we see that $A \otimes_{K', \sigma} L$ has a unique minimal prime. Since $L$ is algebraically closed it follows that $B_{max}((A \otimes_{K', \sigma} L)/L) = L$ because this $L$-algebra is a field algebraic over $L$ by Lemma \ref{lemma-properties-of-max-weakly-etale-subalgebra} part (6). It follows that the maximal weakly \'etale $K' \otimes_K L$-subalgebra of $A \otimes_K L$ is $K' \otimes_K L$ because we can decompose these subalgebras into products as above. Hence the inclusion $K' \otimes_K L \subset B_{max}((A \otimes_K L)/L)$ is an equality: the ring map $K' \otimes_K L \to B_{max}((A \otimes_K L)/L)$ is weakly \'etale by Lemma \ref{lemma-weakly-etale-permanence}.