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Fix mistake in dualizing

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aisejohan committed Nov 13, 2017
1 parent 42e2c6f commit 69b31626c35672540bc04331a31c712381c06797
Showing with 11 additions and 4 deletions.
  1. +11 −4 dualizing.tex
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@@ -4415,10 +4415,17 @@ \section{Upper shriek algebraically}
be the $R$-linear map which picks off the coefficient
of $x^i$ with respect to the given basis. Then
$\delta_0, \ldots, \delta_{d - 1}$ is a basis for $\Hom_R(B, R)$.
Finally, $x^i \delta_{d - 1} = \delta_{d - 1 - i}$ for $i \leq d - 1$.
Hence $\Hom_R(B, R)$ is a principal $B$-module, and by looking
at ranks we conclude that it is a free $B$-module of rank $1$
(with basis element $\delta_{d - 1}$).
Finally, for $0 \leq i \leq d - 1$ a computation shows that
$$
x^i \delta_{d - 1} =
\delta_{d - 1 - i} + b_1 \delta_{d - i} + \ldots + b_i \delta_{d - 1}
$$
for some $c_1, \ldots, c_d \in R$\footnote{If
$f = x^d + a_1 x^{d - 1} + \ldots + a_d$, then
$c_1 = -a_1$, $c_2 = a_1^2 - a_2$, $c_3 = -a_1^3 + 2a_1a_2 -a_3$, etc.}.
Hence $\Hom_R(B, R)$ is a principal $B$-module with generator
$\delta_{d - 1}$. By looking
at ranks we conclude that it is a rank $1$ free $B$-module.
\end{proof}
\begin{lemma}

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