Three v's should be w's

Thanks to A
https://stacks.math.columbia.edu/tag/0748#comment-3184

brauer.tex

@@ -228,11 +228,11 @@ \section{Lemmas on algebras}
 If $n = 1$, then we win because $v_1 \otimes 1 \in W$.
 If $n > 1$, then we see that for any $c \in K$
 $$
-c v - v c = \sum\nolimits_{i = 2, \ldots, n} v_i \otimes (c k_i - k_i c) \in W
+c w - w c = \sum\nolimits_{i = 2, \ldots, n} v_i \otimes (c k_i - k_i c) \in W
 $$
 and hence $c k_i - k_i c = 0$ by minimality of $n$.
 This implies that $k_i$ is in the center of $K$ which is $k$ by
-assumption. Hence $v = (v_1 + \sum k_i v_i) \otimes 1$ contradicting
+assumption. Hence $w = (v_1 + \sum k_i v_i) \otimes 1$ contradicting
 the minimality of $n$.
 \end{proof}