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Let $R$ be a commutative ring, $M$ an $R$-module and $r\in R$. If $f:M\to M$ defined by $f(m)=mr$ is an injective $R$-module endomorphism, then show that the mapping $\text{Hom}_R(M,E)\stackrel{r}{\longrightarrow}\text{Hom}_R(M,E)$ is surjective, where $E$ is the injective cogenerator of $R$.

I have tried to search for injective generators and what I know is that

(1) an injective $R$-module $E$ is called an {\it injective cogenerator} of $R$ if, for every $R$-module $M$ and for every non-zero $m\in M$, there is a homomorphism $\phi:M\to E$ such that $\phi(m)\neq0$.

(2) since $R$ is commutative, $\text{Hom}_R(M,E)$ is also an $R$-module whose elements are maps $\phi\in \text{End}_R(M)$.

However, I cannot figure out the surjectivity of that multiplication map in $\text{Hom}_R(M,E)\stackrel{r}{\longrightarrow}\text{Hom}_R(M,E)$ comes about.

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    $\begingroup$ Surjectivity holds simply because $E$ is an injective $R$-module. $\endgroup$ Aug 1, 2020 at 14:08
  • $\begingroup$ How is the map $\text{Hom}_R(M,E)\stackrel{r}{\longrightarrow}\text{Hom}_R(M,E)$ defined? Is it like $f\mapsto r\cdot f, f\in \text{Hom}_R(M,E)$? $\endgroup$
    – mariam
    Aug 1, 2020 at 16:28

1 Answer 1

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Exercise 2.19 (Rotman): the functors $\operatorname{Hom}_R(-, -)$ preserve multiplication. Explicitly, if $\mu_r : M \to M$ is the multiplication map $\mu_r(m) = r \cdot m$ from a module $M$ over a commutative ring $R$ to itself, then for any $R$-module $N,$ the induced map $\operatorname{Hom}(\mu_r) : \operatorname{Hom}_R(M, N) \to \operatorname{Hom}_R(M, N)$ is the multiplication map $\varphi \mapsto r \cdot \varphi.$

Considering that $E$ is an injective cogenerator of $R,$ it is by definition an injective $R$-module, and this implies that $\operatorname{Hom}_R(-, E)$ is an exact functor, i.e., $\operatorname{Hom}_R(-, E)$ is right exact.

Consequently, the exact sequence $$0 \to M \xrightarrow{r \cdot} M \to \frac{M}{rM} \to 0$$ gives rise to the exact sequence $$0 \to \operatorname{Hom}_R \biggl(\frac{M}{rM}, E \biggr) \to \operatorname{Hom}_R(M, E) \xrightarrow{r \cdot} \operatorname{Hom}_R(M, E) \to 0,$$ as $\operatorname{Hom}_R(-, E)$ is contravariant. But this says that $\operatorname{Hom}_R(M, E) \xrightarrow{r \cdot} \operatorname{Hom}_R(M, E)$ is surjective.

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