Injective modules are of course just projective modules in the opposite category, so it seems to me that the question really is "why is the opposite of a module category more complicated than a module category?" Probably this is because the opposite of a module category is almost never itself a module category (see this MO question). It embeds into a module category by Freyd-Mitchell, but this is quite noncanonical.

For the sake of concreteness, by Pontrjagin duality $\text{Ab}^{op}$ itself is equivalent to the category of compact (Hausdorff) abelian groups (which embeds into $\text{Ab}$ itself but this is not too useful of an embedding for our purposes). An injective abelian group dualizes to a projective object in $\text{Ab}^{op}$, and it is not so straightforward as in a module category to find a projective object here. The simplest nontrivial thing that deserves to be called a free object is the Bohr compactification of $\mathbb{Z}$ (which dualizes to $S^1$ with the discrete topology). The injective abelian group $\mathbb{Q}/\mathbb{Z}$, as a filtered colimit of the groups $\mu_n$, dualizes to a cofiltered limit giving the profinite integers $\hat{\mathbb{Z}}$.

This gives one way to find an injective resolution of an abelian group by following a recipe exactly analogous to the free module recipe, but in $\text{Ab}^{op}$: find a projective resolution of its Pontrjagin dual by products of copies of the Bohr compactification of $\mathbb{Z}$, then dualize it!

**Edit:** Steven Landsburg makes the following comment below:

But I thought that another part of your answer was that *in the categories we often choose to look at* (module categories) projectives might look simpler, though in the opposite categories it's the injectives that look simpler. That leaves the question of why it's the projectives that are simpler in the categories we're naturally led to look at.

My revised revised answer is that free objects are projective, and free objects are simpler in the categories we're naturally led to look at.

Let $C$ be a category and let $U : C \to \text{Set}$ be a faithful functor. If $U$ has a left adjoint $F$, then $U$ preserves monomorphisms, so the monomorphisms in $C$ are precisely the maps which are injective on underlying sets. Thus to find projective objects it suffices to find objects with the lifting property. Now, for a set $S$, $F(S)$ clearly has the lifting property (hence is projective) because $S$ has the lifting property in $\text{Set}$ (in other words, is projective).

The categories we're naturally led to look at, such as module categories, are usually concretely defined so are equipped with a faithful functor to $\text{Set}$, and the corresponding objects $F(S)$ usually exist and provide a plentiful supply of projectives in $C$. (Without the functors $U$ and $F$ it is not clear why $C$ should have any projective objects whatsoever, and it might not: for an abelian category example, take $C = \text{FinAb}$.)

On the other hand, the opposite of a category, even one equipped with a faithful functor to $\text{Set}$, doesn't itself come equipped with a faithful functor to $\text{Set}$ (instead it comes equipped with a faithful functor to $\text{Set}^{op}$, which is harder to understand) and finding one with a left adjoint (as in the Pontrjagin duality example above) might be difficult.

nota functor, whereas the "free abelian group on the elements of $M$" $F(M)$isfunctorial. So perhaps this qualifies as a "deep" reason: if $M$ is an $R$-module, the "natural way to produce a projective that surjects onto $M$" is $F(M) \to M$, which is functorial. The "natural way to produce an injective module into which $M$ injects" is $M \hookrightarrow I(M)$, which isnotfunctorial (and hence, not really "natural" at all, in the technical sense). However, I don't know about the existence of a "better" natural injective. $\endgroup$ – Charles Staats Jul 27 '12 at 16:03