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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:

enter image description here

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

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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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    $\begingroup$ Welcome to Math.SE and thank you for your answer. Please add some key information on your post (instead of a link) to make it searchable for other users around here. $\endgroup$ – Ertxiem - reinstate Monica Apr 17 '19 at 3:01

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