4
$\begingroup$

We know that $\mathbb{Q}/\mathbb{Z} $ is injective. Not only that $0 \to A \to B \to C \to 0$ is exact iff $0 \to Hom(C,\mathbb{Q}/\mathbb{Z}) \to Hom(B,\mathbb{Q}/\mathbb{Z})\to Hom(A,\mathbb{Q}/\mathbb{Z}) \to 0$ is exact.

My question is :

Do these statements hold good if $\mathbb{Z}$ is replaced by a general domain $R$ and $\mathbb{Q}$ is replaced by the quotient field $Q$ of $R$ ?

It is easy to see that $Q/Z$ is injective if $R$ is a PID and hence the exactness of the first sequence implies the exactness of the second, but the reverse implication is not clear.

$\endgroup$
  • $\begingroup$ It also holds for Dedekind domains. I suspect amongst commutative noetherian domains, it is only the Dedekind domains. $\endgroup$ – Jack Schmidt May 6 '14 at 19:09
9
$\begingroup$

There is a notion of injective cogenerator of the category $\mathrm{Mod}$-$R$: An injective cogenerator is an injective $R$-module $C$ such that for every nonzero $R$-module $M$, there is a nonzero homomorphism $M\rightarrow C$, i.e. $\mathrm{Hom}_R(M,C)\neq 0$.

As it turns out, injective cogenerators are precisely the modules satisfying the property you are interested in.

See the answer to this question for the fact that injective cogenerators have the property you are interested in, i.e. the functor $\mathrm{Hom}_R(-,C)$ preserves and reflects exact sequences.

Conversely, assume that $C$ is an $R$-module such that $\mathrm{Hom}_R(-,C)$ preserves and reflects exact sequences. Then $C$ is injective. It remains to show that $C$ is a cogenerator.

Let $M$ be an $R$-module. Consider a sequence $$0\rightarrow 0 \rightarrow 0 \rightarrow M \rightarrow 0.$$

Applying $\mathrm{Hom}_R(-,C)$, we obtain a sequence $$0\rightarrow\mathrm{Hom}_R(M,C) \rightarrow \mathrm{Hom}_R(0,C) \rightarrow \mathrm{Hom}_R(0,C) \rightarrow 0 \;\;. $$

If $\mathrm{Hom}_R(M,C)=0$, the latter sequence is exact, so the former sequence is exact. Thus, $M=0$.

Now, in the commutative Noetherian ring case, there is a structure theorem saying that every injective module is a direct sum of indecomposable injective modules, which are of the form $E(R/P)$, where $P$ is a prime ideal of $R$ and $E(-)$ denotes taking the injective hull. From this it follows that the module

$$\bigoplus_{P \in m\mathrm{Spec}(R)}E(R/P)$$

is an injective cogenerator.

In the PID case, it can be verified that for $p \in R$ a prime, the injective hull $E(R/pR)$ is, up to isomorphism, the submodule $R[(1/p)]/R$ of $Q/R$ (with the module $R/pR$ embedded via $r+pR \mapsto \frac{r}{p}+R$; also note that if $R=\mathbb{Z}$ and $p$ is a prime, this gives you an isomorphic copy of the Prüfer group $\mathbb{Z}_{p^{\infty}}$). Thus, we have

$$Q/R=\left(\sum_{pR \in m\mathrm{Spec}(R)}R[(1/p)]\right)/R=\sum_{pR \in m\mathrm{Spec}(R)}\left(R[(1/p)]/R\right)$$

and it is easily seen (from $R$ being UFD) that the sum is direct. So the module $Q/R$ is indeed an injective cogenerator in this case.

$\endgroup$
  • $\begingroup$ Very nice answer! $\endgroup$ – Hanno Dec 5 '14 at 9:20

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.