Motivation for spectrum of an Abelian category

Question

In his book Noncommutative Algebraic Geometry and Representations of Quantized Algebras, Rosenberg defines (III.1.2 on page 111) the spectrum of an Abelian category $\mathbf{A}$ in the following way. He first defines a preorder $\succ$ on $\operatorname{Ob}(\mathbf{A})$ by declaring that

$V\succ W$ iff $W$ is a subquotient of a coproduct of finitely many copies of $V$.

Let us write $V\sim W$ iff $V\succ W$ and $W\prec V$ (my notation, not his---he just says "equivalent with respect to $\succ$"). He then defines

$$\operatorname{Spec}(\mathbf{A}):=\left\{ V\in \operatorname{Obj}(\mathbf{A}):V\neq 0\text{ and }V\sim W\text{ for all nonzero subobjects }W\text{ of }V\text{.}\right\}$$

My question is simply "Why?": of all the possible definitions one might write down, why this one?

It is not at all clear to me what the intuition for this definition is supposed to be. What does this actually 'mean'?

Answer 1

The basic motivation for this definition is that if $\mathbf{A}$ is the category of modules over a commutative ring $R$, then the $\sim$-equivalence classes of $\operatorname{Spec}(\mathbf{A})$ are naturally in bijection with the spectrum of the ring $R$. So this is a generalization of the spectrum of a commutative ring that can be defined for any abelian category.

Indeed, suppose $R$ is a commutative ring and $\mathbf{A}$ is the category of $R$-modules. Then for any $V\in\operatorname{Spec}(\mathbf{A})$, $V\sim W$ for every nonzero cyclic submodule $W$ of $V$ and $W\in\operatorname{Spec}(\mathbf{A})$ as well, so it suffices to consider cyclic modules. Note that for cyclic modules $R/I$ and $R/J$, we have $R/I\prec R/J$ iff $J\subseteq I$. So if $R/I$ is in $\operatorname{Spec}(\mathbf{A})$, then no nonzero element of $R/I$ can have an annihilator which is larger than $I$ (since then it would generate a cyclic submodule $R/J$ such that $R/J\not\sim R/I$). This is equivalent to $I$ being a prime ideal. So if $R/I$ is in $\operatorname{Spec}(\mathbf{A})$, then $I$ is a prime ideal. Conversely, if $I$ is prime, then every nonzero cyclic submodule of $R/I$ is isomorphic to $R/I$ and it follows that every nonzero submodule of $R/I$ has a submodule isomorphic to $R/I$, and so $R/I\in \operatorname{Spec}(\mathbf{A})$.

To sum up, then, the $\sim$-equivalence classes of $\operatorname{Spec}(\mathbf{A})$ are in bijection with the prime ideals of $R$, with each prime ideal $I$ corresponding to the module $R/I$ which is an element of $\operatorname{Spec}(\mathbf{A})$.