The Stacks project

Lemma 47.16.3. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring. Let $F$ be an $A$-linear self-equivalence of the category of finite length $A$-modules. Then $F$ is isomorphic to the identity functor.

Proof. Since $\kappa $ is the unique simple object of the category we have $F(\kappa ) \cong \kappa $. Since our category is abelian, we find that $F$ is exact. Hence $F(E)$ has the same length as $E$ for all finite length modules $E$. Since $\mathop{\mathrm{Hom}}\nolimits (E, \kappa ) = \mathop{\mathrm{Hom}}\nolimits (F(E), F(\kappa )) \cong \mathop{\mathrm{Hom}}\nolimits (F(E), \kappa )$ we conclude from Nakayama's lemma that $E$ and $F(E)$ have the same number of generators. Hence $F(A/\mathfrak m^ n)$ is a cyclic $A$-module. Pick a generator $e \in F(A/\mathfrak m^ n)$. Since $F$ is $A$-linear we conclude that $\mathfrak m^ n e = 0$. The map $A/\mathfrak m^ n \to F(A/\mathfrak m^ n)$ has to be an isomorphism as the lengths are equal. Pick an element

\[ e \in \mathop{\mathrm{lim}}\nolimits F(A/\mathfrak m^ n) \]

which maps to a generator for all $n$ (small argument omitted). Then we obtain a system of isomorphisms $A/\mathfrak m^ n \to F(A/\mathfrak m^ n)$ compatible with all $A$-module maps $A/\mathfrak m^ n \to A/\mathfrak m^{n'}$ (by $A$-linearity of $F$ again). Since any finite length module is a cokernel of a map between direct sums of cyclic modules, we obtain the isomorphism of the lemma. $\square$

Comments (2)

Comment #1692 by Sándor Kovács on

Typo: Near the end of the proof "lenghth" should be "length".

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