Fix typo

Thanks to Andrea Panontin
https://stacks.math.columbia.edu/tag/0A02#comment-8656

more-algebra.tex

@@ -3055,7 +3055,7 @@ \section{Henselization of pairs}
 and $A^h$-algebra map $\sigma : A' \to A^h/I^h$. Then there exists a
 $B \in \mathcal{C}$ and an \'etale ring map $B \to B'$ such that
 $A' = B' \otimes_B A^h$. See Algebra, Lemma \ref{algebra-lemma-etale}.
-Since $A^h/I^h = A/IB$, the map $\sigma$ induces an $A$-algebra
+Since $A^h/I^h = A/I$, the map $\sigma$ induces an $A$-algebra
 map $s : B' \to A/I$. Then $B'/IB' = A/I \times C$ as $A/I$-algebra,
 where $C$ is the kernel of the map $B'/IB' \to A/I$ induced by $s$.
 Let $g \in B'$ map to $(1, 0) \in A/I \times C$. Then $B \to B'_g$