If $text{Mod}_{R}$ and $text{Mod}_{S}$ are equivalent, then $R$ and $S$ have the same simple modules.

Question

Let $R$ and $S$ be (unital, associative) rings. If $text{Mod}_{R}$ and $text{Mod}_{S}$ are equivalent categories, then is it true that $R$ and $S$ have the same amount of isomorphism classes of simple modules?
Since an equivalence of categories preserves 'all categorical properties': an equivalent question would be? Is being a simple module a categorical property? Intuitively, I would guess so because Schur's lemma allows us to say something about morphisms from and to a simple module but I'm struggling with making everything precise.

Eric Wofsey · Accepted Answer

Yes. A simple module is an object $A$ of $mathrm{Mod}_R$ which is not a zero object such that every monic morphism $Bto A$ from any other object $B$ either factors through a zero object or is an isomorphism. (In concrete terms, this says any injective homomorphism to $A$ is either $0$ or surjective, so every submodule of $A$ is either trivial or all of $A$.) Each part of this definition is preserved by an equivalence of categories, so an equivalence of categories $mathrm{Mod}_Rto mathrm{Mod}_S$ sends simple modules to simple modules.

Correct answer by Eric Wofsey on January 14, 2021

If $text{Mod}_{R}$ and $text{Mod}_{S}$ are equivalent, then $R$ and $S$ have the same simple modules.

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