Let $M$ and $N$ be $R$-modules for some ring $R$. There is a standard result involving the computation of $\text{Ext}^n(M,N)$, using projective resolutions, which says that you can always choose a projective resolution such that the maps you get from the projective resolution to the extension are always surjective (this is useful when deriving equivalences between extensions which define the same element of $\text{Ext}^n(M,N)$). The result is not difficult to prove; it basically amounts to tacking on a large enough projective module (via direct sum) to certain of the projective objects in the resolution so that the maps are always onto.

I would instead (for reasons I won't go into) prefer to compute $\text{Ext}^n(M,N)$ using injective resolutions. My question: does an analogous result (exchanging "projective object" for "injective object", and "surjective map" for "injective map") for the above hold? That is, is there some way that I can always choose an injective resolution so that the maps I get from a given extension to the constituent injective objects are always one-to-one?

Since the aforementioned result involves tacking on certain direct sums to a projective resolution, if an analogy exists, I imagine it would take the form of taking certain sub-objects of an existing injective resolution, or possibly reduction to an injective envelope (of something). But either way, I can't see it.

If it makes any difference, the modules I am interested in are finite dimensional representations for an algebraic group over a field (of arbitrary characteristic).

Thanks in advance for any help.


1 Answer 1


I'm afraid the answer is NO! At least in the full generality in which the question is stated, there is pretty easy to disprove it.

Let consider the standard injective resolution of the abelian group $\mathbf Z$: $$0\to{\mathbf Z}\to{\mathbf Q}\to\mathbf{Q}/{\mathbf Z}\to 0$$. Now if $$0\to{\mathbf Z}\to E^0\to E^1\to\cdots$$ is another injective resolution, we construct in the usual manner the group homomorphisms ${\mathbf Q}\to E^0$ and ${\mathbf Q}/{\mathbf Z}\to E^1$, making commutative the squares which occur. It is clear that the map ${\mathbf Q}\to E^0$ is injective (${\mathbf Q}$ being the injective envelope of ${\mathbf Z}$) and the image of the map ${\mathbf Q}/{\mathbf Z}\to E^1$ is torsion, so the map $E^0\to E^1$ can't be injective.

  • 2
    $\begingroup$ George, the question wasn't if there is an injective resolution with injective differentials, but if there is an injective resolution $I$ of a bounded complex $X$ (the extension in Mike's terminology) such that the maps $X_i \to I_i$ are injective. $\endgroup$
    – tj_
    Nov 7, 2011 at 21:53
  • $\begingroup$ I'm afraid I stated the problem in more general terms than I intended to, but your reply will I believe nonetheless point me towards the answer I am looking for. Much appreciated. $\endgroup$ Nov 10, 2011 at 6:04

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