Forgive me for this naive question.

We consider the following lemma and its proof in Lang's algebra, Third Ed., published 1999, Chap. 20, section 4, page 784.

Every module is a submodule of an injective module.

For a module $M$, Lang defines its dual to be $M$^ = $Hom(M, \mathbb{Q}/\mathbb{Z})$.

This "dual", if anything, is an algebraic version of the Pontrjagin dual of a topological group $G$, which is $Hom(G, \bf{S}^1)$, where we consider the continuous homomorphisms. $\mathbb{Q}/\mathbb{Z}$ is the torsion part of $\bf{S}^1$, and it seems this restriction is done for preventing too big a dual.

The proof goes on and shows that any module is a submodule of an injective module.

However, is there any conceptual explanation of why this works, and why this mysterious appearance of a dual which looks so much like the Pontrjagin dual?

  • $\begingroup$ Not much of an answer to your question, but FWIW: on p. 40 of my commutative algebra notes math.uga.edu/~pete/integral.pdf I define $M^{\vee}$ and call it the Pontrjagin dual (yes, even the spelling is the same!). This is to distinguish it from the "R-dual" $M^* = \operatorname{Hom}_R(M,R)$ which comes up e.g. in the study of projective modules. $\endgroup$ – Pete L. Clark Feb 23 '11 at 14:13
  • $\begingroup$ Thank you for a question that had been bothering me for a while. Rotman (An introduction to homological algebra) does this too, and $\frac{Q}{Z}$ also appears magically. $\endgroup$ – Bruno Stonek Feb 23 '11 at 14:34
  • $\begingroup$ A small point: the $\mathbb Q/\mathbb Z$ is a convenient name for a perhaps-more-meaningful categorical characterization, as (ascending union) colim$_N {1\over N}\mathbb Z/\mathbb Z$, or, equivalently as a colim of $\mathbb Z/N\mathbb Z$. This may help, slightly, explain the special-ness. $\endgroup$ – paul garrett Jul 18 '11 at 23:29

Why this works? Because $\mathbb{Q}/\mathbb{Z}$ is an injective cogenerator in the category of abelian groups.

The mysterious appearance can be explained if you look at it from a more general perspective. This would be Morita duality, which was motivated by classical Pontryagin duality (locally compact Hausdorff topological groups), and it replaces the role of the circle group with an injective cogenerator.

In the category Set of sets and mappings, for example, the two point set $\{0,1\}$ is an injective cogenerator. The category Gr of groups and group homomorphisms on the other hand doesn't have any cogenerator.


Let me sketch a proof of the result (which I would guess is the one Lang gives) with some motivation that I hope resonates with your question.

There is a general adjunction, if $A \to B$ is a ring homomorphism and $M$ and $N$ are $A$- and $B$-modules respectively, of the form $Hom_A(N,M) = Hom_B(N,Hom_A(B,M)).$ This shows that if $M$ is injective over $A$, then $Hom_A(B,M)$ is injective as a $B$-module, and also that if $N \hookrightarrow M$ is an embedding of $A$-modules, then the induced map $N \to Hom_A(B,M)$ is also an embedding.

Thus it suffices to construct enough injectives in the category of $\mathbb Z$-modules; the above considerations then produce enough injectives for modules over any ring.

Now $I$ is injective if and only if $Hom(\text{--},I)$ is exact, which is already a kind of Pontrjagin duality property: it gives an exact functor reversing arrows. Furthermore, we always have a double duality map $M \to Hom(Hom(M,I))$, and another feature of Pontrjagin duality is that this map should be an isomorphism. Let's here consider the weaker property that it is injective (so we at least have a non-degenarate duality, although not necessarily perfect).

Suppose then that $I$ is some injective module over $\mathbb Z$, with the property that the double duality map $M \to Hom(Hom(M,I),I)$ is injective for all $M$ (i.e. with the property that it gives rise to some kind of Pontrjagin duality). Then given this $I$, if we take a free module surjecting onto $Hom(M,I),$ and then apply $Hom(\text{--},I)$ and compose with the double duality map, we get an embedding from $M$ into a product of copies of $I$, and hence an embedding of $M$ into an injective.

Thus constructing enough injectives is reduced to finding such $I$, and here we are guided by the idea that we are looking for an object that establishes some kind of (at least weak form of) Pontrjagin duality.

Now in order for the double duality to be injective, one needs that for any $m \in M$, there is a map from $\langle m \rangle$ to $I$ that is non-zero on $m$ (since this will extend to all of $M$ by injectivity of $I$). So we need $I$ to be injective, i.e. divisible, and to contain torsion elements of arbitrary order. Thus we could take $I = {\mathbb Q}/{\mathbb Z}$, as in Lang, or also the circle group $S^1$. (But the latter has infinite order elements which are not necessary, so ${\mathbb Q}/{\mathbb Z}$ is more economical. Since it also has a more algebraic/less analytic feel than the circle, this is probably why it is more commonly used in such arguments.)

  • $\begingroup$ You mean we are guided by some idea that we are looking for a dual like the Pontrjagin dual, though it may not be fully possible to recover the original from the dual? Can you please elaborate a bit there? It was what had me thinking and prompted this question. $\endgroup$ – Anweshi Feb 5 '10 at 15:11

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