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Let M$M$ be an R$R$-module,where R$R$ is a hereditary  (or cohomological dimension less or equal to 1).Take E(R)$E(R)$ to be injective hull of R$R$, then we have the essential extension i:R^I--->E(R)^I$i:R^I\rightarrowtail E(R)^I$ (product I$I$ times)

and and we also have p:R^I--->>M$p:R^I\twoheadrightarrow M$ is epimorpshim. Then I take the push forward of these two morphisms i$i$ and p$p$, denote the push out by (N,f,g)$(N,f,g)$,where f:M---->N where $f:M\to N$ and g:E(R)^I--->>N)$g:E(R)^I\twoheadrightarrow N$.

itIt is clear that N$N$ is an injective module  (because it is image of E(R)^I$E(R)^I$ and R$R$ is hereditary) and M--->N and $f:M\to N$ is injective morphism. However, N$N$ is not necessarily injective hull of M$M$ because in general, essential extension does not commutes with colimit.

My question is: can we give some conditions to R$R$ or other extra conditions to make N is injective Hullhull of M.$M$?

In general, I know it is not true, but it seems that it gives a approximation of "functor":M--->E(M) $M \to E(M)$

Let M be an R-module,where R is a hereditary(or cohomological dimension less or equal to 1).Take E(R) to be injective hull of R, then we have the essential extension i:R^I--->E(R)^I (product I times)

and we also have p:R^I--->>M is epimorpshim. Then I take the push forward of these two morphisms i and p, denote the push out by (N,f,g),where f:M---->N and g:E(R)^I--->>N)

it is clear that N is an injective module(because it is image of E(R)^I and R is hereditary) and M--->N is injective morphism. However, N is not necessarily injective hull of M because in general, essential extension does not commutes with colimit.

My question is can we give some conditions to R or other extra conditions to make N is injective Hull of M.

In general, I know it is not true, but it seems that it gives a approximation of "functor":M--->E(M)

Let $M$ be an $R$-module,where $R$ is a hereditary  (or cohomological dimension less or equal to 1).Take $E(R)$ to be injective hull of $R$, then we have the essential extension $i:R^I\rightarrowtail E(R)^I$ (product $I$ times) and we also have $p:R^I\twoheadrightarrow M$ is epimorpshim. Then I take the push forward of these two morphisms $i$ and $p$, denote the push out by $(N,f,g)$, where $f:M\to N$ and $g:E(R)^I\twoheadrightarrow N$.

It is clear that $N$ is an injective module  (because it is image of $E(R)^I$ and $R$ is hereditary) and $f:M\to N$ is injective morphism. However, $N$ is not necessarily injective hull of $M$ because in general, essential extension does not commutes with colimit.

My question is: can we give some conditions to $R$ or other extra conditions to make N is injective hull of $M$?

In general, I know it is not true, but it seems that it gives a approximation of "functor" $M \to E(M)$

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Shizhuo Zhang
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When is the essential extension commutes with colimits(or push forward)

Let M be an R-module,where R is a hereditary(or cohomological dimension less or equal to 1).Take E(R) to be injective hull of R, then we have the essential extension i:R^I--->E(R)^I (product I times)

and we also have p:R^I--->>M is epimorpshim. Then I take the push forward of these two morphisms i and p, denote the push out by (N,f,g),where f:M---->N and g:E(R)^I--->>N)

it is clear that N is an injective module(because it is image of E(R)^I and R is hereditary) and M--->N is injective morphism. However, N is not necessarily injective hull of M because in general, essential extension does not commutes with colimit.

My question is can we give some conditions to R or other extra conditions to make N is injective Hull of M.

In general, I know it is not true, but it seems that it gives a approximation of "functor":M--->E(M)