The Stacks project

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

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. $\square$