# When The FP-injective cover of module is surjective

Definition of cover of module is: Let $$R$$ be a ring and let $$\mathcal{S}$$ be any class of $$R$$-modules. Then for any $$R$$-module $$M$$, the homomorphism $$\beta:S\longrightarrow M$$ is called an $$\mathcal{S}$$-precover of $$M$$ if $$S\in \mathcal{S}$$ and $$\beta^{*}=Hom(F',\beta):Hom(S',S)\longrightarrow Hom(S',M)$$ is surjective for every $$S'\in \mathcal{S}$$. An $$\mathcal{S}$$-precover $$\beta:S\longrightarrow M$$ is called an $$\mathcal{S}$$-cover, if for every homomorphism $$f:S\longrightarrow S$$ such that $$\beta\circ f=\beta$$, $$f$$ is an automorphism.

and the defintion of FP-injective module is : A module $$H$$ is said to be FP-injective if $$Ext^{1}_{R}(P,H)=0$$ for each finitely presented module $$P$$.

The Question is let $$M\longrightarrow E(M)$$ be the injective envelope of $$M.$$ Consider the exact sequence $$0\longrightarrow M\longrightarrow E(M)\longrightarrow E(M)/M\longrightarrow 0,$$ we have $$E(M)\longrightarrow E(M)/M$$ is an FP-injective precover of $$E(M)/M$$. Also, since $$R$$ is coherent, $$E(M)/M$$ has an FP-injective cover $$L\longrightarrow E(M)/M$$. Then we obtain the following commutative diagram such that all rows are exact:

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I did not understand how deduced this commutative diagram ? and what FP-injective cover is surjective ?

## 1 Answer

it happens iff R Is FC ring by an accepted paper , "coherent ring and absolutely pure preprovers " https://www.researchgate.net/publication/319304138_Coherent_rings_and_absolutely_pure_covers

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