Let $R$ be a ring. An injective left $R$-module $E$ is called an injective cogenerator if $Hom_R(M,E) \not =0$ for $0 \not =M \in R-Mod$.

I have seen that $\mathbb{Q}/ \mathbb{Z}$ is an injective cogenerator of $\mathbb{Z}-Mod$(here $\mathbb{Q}$ denotes the ring of rational numbers and $\mathbb{Z}$ denotes the ring of integers): Since $\mathbb{Q}/ \mathbb{Z}$ is divisible as $\mathbb{Z}$-module and all divisible $\mathbb{Z}$-modules are injective, we have $\mathbb{Q}/ \mathbb{Z}$ is injective. Then how to get that $\mathbb{Q}/ \mathbb{Z}$ is a cogenerator?

  • $\begingroup$ Can you show that $\hom_{\mathbb Z}(A,\mathbb Q/\mathbb Z)$ is nonzero when $A$ is finitely generated? $\endgroup$
    – Aaron
    Jun 26, 2017 at 10:51
  • $\begingroup$ Not quite. Since $\mathbb Z$ is a PID, every ideal of $\mathbb Z$ is of the form $a\mathbb Z$, and the map sending $1$ to $a$ is an isomorphism between $\mathbb Z$ and $a\mathbb Z$. Alternatively, every subgroup of a free group is free. In either case, $\ker f$ is free, so $\hom(\ker f,\mathbb Q/\mathbb Z)$ will be a direct product (not necessarily sum if no longer finitely generated) of copies of $\mathbb Q/\mathbb Z$. However, we aren't dealing with vector spaces, and so even if the "rank" were the same, injective and surjective maps wouldn't necessarily be the same thing. $\endgroup$
    – Aaron
    Jun 26, 2017 at 19:46
  • $\begingroup$ For example, the map $\mathbb Q/\mathbb Z \to \mathbb Q/\mathbb Z $ defined by $x\mapsto 2x$ is surjective (because $\mathbb Q/\mathbb Z $ is divisible) but not injective (because $1/2$ maps to $0$). $\endgroup$
    – Aaron
    Jun 26, 2017 at 20:24
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Aaron
    Jun 27, 2017 at 1:40
  • $\begingroup$ @Aaron Actually, we don't need to show that $Hom(A,Q/Z) \not =0$ for all finitely generated modules $A$. We just only need to show that $Hom(S, Q/Z) \not = 0$ for all simple Z-modules $S$. We can get this by the way we use above: Since $S$ is simple ,there is an epimorphism $f: Z \rightarrow S$. Then since $kerf$ is a left ideal of $Z$, it it of the form $kerf =bZ$ for some $b \in Z$. Now the natural embeding $i : bZ \rightarrow Z$ can't induce an isomorphism $Hom(i,Q/Z): Hom(Z, Q/Z) \rightarrow Hom(bZ,Q/Z)$. For example, take $f:Z \rightarrow Q/Z$ as f(x)=x/b, then the composition $f i =0$ $\endgroup$ Jun 27, 2017 at 6:43


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.