# Is a finitely generated torsion module over PID an injective module?

Since I often draw wrong conclusions, I need a confirmation. My argument is that for nonzero prime $\mathfrak{p}\subset A$, $A/\mathfrak{p}$ is injective being a field (hence divisible), and $A/p^{k}$ is its quotient hence again is injective, and the f.g. torsion module is a finite direct sum of such quotients hence is injective. I couldn't find any book that mentions this fact and I'm wondering I might have made a mistake somewhere. Thanks!

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$A/\mathfrak{p}^k$ is not a quotient of $A/\mathfrak{p}$, it's the other way around. Since $\mathfrak{p}^k\subseteq \mathfrak{p}$, then $(A/\mathfrak{p}^k)/(\mathfrak{p}/\mathfrak{p}^k)\cong A/\mathfrak{p}$. Also, fields are not necessarily divisible: fields of finite characteristic are not divisible modules. No finite field is a divisible $\mathbb{Z}$-module (and for that matter, neither is $\mathbb{Z}/n\mathbb{Z}$ for any $n\gt 1$). – Arturo Magidin Jan 29 '11 at 23:43
Z/2Z is a divisible Z/2Z module, after all it is a field and you can divide by any nonzero field element. But "nonzero" for the field and "nonzero" for the ring might differ. Z/2Z is not a divisible Z-module, because it is not divisible by 2! – Jack Schmidt Jan 30 '11 at 0:49
You have an argument... But have you considered any example at all? – Mariano Suárez-Alvarez Jan 30 '11 at 1:17
Every torsion module over a 1-dimensional domain (like a PID) is a direct sum of its p-torsion parts. Be careful that injective hulls can decompose quite a lot, even if the original module is indecomposable. However, in the case of torsion modules over a PID, the decomposition is very similar, so your intuition is probably fine there. – Jack Schmidt Jan 30 '11 at 3:26
The Z-module Z/2Z is not divisible by 2, but if you tried to divide it by 2, you'd get Z/4Z. 2Z/4Z ≅ Z/2Z, so that's good. However, Z/4Z is still not divisible by 2. If you tried to divide it by 2, you'd get Z/8Z. 2Z/8Z ≅ Z/4Z. However, ... If you keep this up you finally get to the module Z[1/2]/Z, the Prüfer quasi-cyclic 2-group. It is an injective module, and is the injective hull of Z/2Z. – Jack Schmidt Jan 30 '11 at 3:29

One way to check statements about injective modules is to see what's happening for $\mathbb{Z}$-modules -- i.e., abelian groups. (In general, it's a good idea to test out statements of the form "does this property of modules/rings/spaces/... imply this other property?" by considering relatively simple, well-understood examples. But I don't want to preach too much...)

For $R = \mathbb{Z}$, there is the following nice criterion (due, I believe, to R. Baer): a $\mathbb{Z}$-module $M$ is injective iff it is a divisible group: i.e., for all $n \in \mathbb{Z}^+$, the map $x \mapsto nx$ is surjective. (Here I am writing the group law on $M$ additively, as is traditional.)

From this it is easy to see that any nonzero injective $\mathbb{Z}$-module must be infinite. Indeed, if $M$ is finite, then multiplication by $n = \# M$ on $M$ is identically zero.

For that matter, no (nonzero) finitely generated $\mathbb{Z}$-module is injective. The simplest example of an injective $\mathbb{Z}$-module is probably $(\mathbb{Q},+)$ (i.e., the additive group of the rational field). Since quotients of divisible groups are divisible, also $\mathbb{Q}/\mathbb{Z}$ is divisible. The first example is torsion-free, and the second example is torsion. These examples give the flavor of the general case.

Added: Actually Baer's Criterion extends to all modules over PIDs as follows: for a ring $R$, an $R$-module $M$ is divisible if for all $r \in R \setminus \{0\}$, the endomorphism $r \cdot$ of $M$ is surjective. (This is a direct generalization of the notion of divisible $\mathbb{Z}$-module.) Then it follows from another basic result of Baer that a module over any PID is injective iff it is divisible. I learned this on wikipedia (which is getting better and better at math, I must say): see here.

Note that applying this criterion of Baer immediately gives the following fact: if $R$ is a PID of characteristic $0$ (equivalently, the natural map $\mathbb{Z} \rightarrow R$ is an injection), then an injective $R$-module must also be an injective $\mathbb{Z}$-module. In particular no finite $R$-module is injective.

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You can have an exact sequence $0 \rightarrow Z/2Z \rightarrow Z/4Z \rightarrow Z/2Z \rightarrow 0$, where the maps are inclusion and projection. Then this sequence does not split, since $Z/4Z$ is not isomorphic to $Z/2Z \oplus Z/2Z$. Therefore $Z/2Z$ is not injective.

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