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Three v's should be w's

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aisejohan committed May 16, 2018
1 parent baf2ff9 commit 81b665b813b9e1bfe7fefd102f9c77e63713eae5
Showing with 2 additions and 2 deletions.
  1. +2 −2 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}

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