The following claim is from a paper [On the moduli spaces of bundles on K3 surfaces, I, p. 358] of Mukai. Consider an artinian module $\mathrm{M}$ over a local ring, and let $\mathrm{M}_0$ be the submodule of all $x\in\mathrm{M}$ annhilated by the maximal ideal of the local ring. Then every endomorphism of $\mathrm{M}$ preserves $\mathrm{M}_0$, and the natural map $$\mathrm{Hom}(\mathrm{M},\mathrm{M})\rightarrow\mathrm{Hom}(\mathrm{M}/\mathrm{M}_0,\mathrm{M}/\mathrm{M}_0)$$ is surjective. I can't quite see why this map is surjective - this is not justified in the paper, so I may be missing something obvious here (a comment may be enough).
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$\begingroup$ "preserves $M_0$" means "maps $M_0$ into $M_0$"? $\endgroup$– YCorCommented May 19, 2021 at 22:56
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1$\begingroup$ You might provide the precise reference and page. $\endgroup$– YCorCommented May 19, 2021 at 22:57
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$\begingroup$ @YCor exactly. I have included the reference. $\endgroup$– ssxCommented May 20, 2021 at 16:11
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This is false. Take $A=k[x,y]/(x^2,y^2,xy)$ and $M=\omega_A$. Then $\operatorname{Hom}(\omega,\omega)$ has length 3 and $\operatorname{Hom}(M/M_0,M/M_0)$ has length 4, so the natural map cannot be surjective.
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2$\begingroup$ @SWS Isn't Mohan's example also a counterexample for the "if and only if" part of that? $\endgroup$ Commented May 20, 2021 at 17:11