The notion of uniform or Goldie dimension is something I’ve only seen discussed for categories of modules, but I believe the theory works the same way in any Grothendieck category $\mathcal C$. Recall that the uniform dimension of $c \in C$ is the supremum of the $n$ such that there exists a monomorphism $c_1 \oplus \cdots \oplus c_n \to c$ with all $c_i$ nonzero. Equivalently it’s the number of factors in a finest direct sum decomposition of the injective hull of $c$.

The uniform dimension is subadditive in short exact sequences and monotonic in subobjects but not necessarily in quotients. That is, if $0 \to d \to c \to c/d \to 0$ is a short exact sequence, then $dim(d) \leq dim(c) \leq dim(d)+ dim(c/d)$ but in general we do not have $dim(c/d) \leq dim(c)$. For example when $\mathcal C = Mod(\mathbb Z)$ is the category of abelian groups, we have $dim(\mathbb Z)=1$ but $dim(\mathbb Z/6)=2$.

However, this pathological example seems attributable to the fact that $Mod(\mathbb Z)$ has more than one indecomposable injective. In the category $\mathbb Z_{(p)}$ of $p$-local abelian groups for a prime $p$ it seems that maybe we do have that the dimension of $c$ is at least the dimension of any quotient? More generally, I wonder:

Question: Let $\mathcal C$ be a Grothendieck category with a unique indecomposable injective. Then for $c \in \mathcal C$ and $c/d$ a quotient, is $dim(c) \geq dim(c/d)$?

I’m happy to understand special cases like assuming $\mathcal C$ is a category of modules, or alternatively assuming $\mathcal C$ is locally coherent or something.