Lemma 49.2.2. The module (49.2.0.1) is well defined, i.e., independent of the choice of the factorization.

Proof. Let $B', B'', B'''$ be as in Lemma 49.2.1. We obtain a canonical map

$\omega ''' = \mathop{\mathrm{Hom}}\nolimits _ A(B''', A) \otimes _{B'''} B \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ A(B', A) \otimes _{B'} B = \omega '$

and a similar one involving $B''$. If we show these maps are isomorphisms then the lemma is proved. Let $g \in B'$ be an element such that $B'_ g \to B_ g$ is an isomorphism and hence $B'_ g \to (B''')_ g \to B_ g$ are isomorphisms. It suffices to show that $(\omega ''')_ g \to \omega '_ g$ is an isomorphism. The kernel and cokernel of the ring map $B' \to B'''$ are finite $A$-modules and $g$-power torsion. Hence they are annihilated by a power of $g$. This easily implies the result. $\square$

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