A ring (say unital for simplicity) is semiprimitive (or Jacobson semisimple) if its Jacobson radical is trivial, or equivalently it has faithful semisimple module. Semiprimitivity is a Morita invariant, meaning that if $R$ and $S$ are unital rings and the categories of (left) $R$-modules and $S$-modules are equivalent, then $R$ is semiprimitive iff $S$ is semiprimitive. In some sense, this should say that semiprimitivity is a categorical notion, i.e., there should be some way to define semiprimitivity for abelian categories (maybe with some extra structure like enough projectives or something) so that $R$ is semiprimitive iff the category of $R$-modules is semiprimitive.

The proof of Morita invariance doesn't go through categorical notions. One first has that if $R$ is Morita equivalent to $S$, then $S$ must be isomorphic to the endomorphism ring of a finitely generated projective $R$-module. Therefore, $S$ looks like $eM_n(R)e$ for some idempotent $e$. If $M$ is a faithful semisimple $R$-module, then one easily check $eM^n$ with the obvious module structure is a faithful semisimple $eM_n(R)e$-module.

The reliance on $M_n(R)$ in this proof (and hence on free $R$-modules) bothers me particularly because free $R$-modules cannot be defined categorically without specifying an underlying set functor.

My question is whether there is some purely categorical definition of semiprimitivity that would make sense for abelian categories (perhaps with some extra properties like enough projectives) and that doesn't required using the ring theoretic definition. In other words, I don't want to say that an abelian category is semiprimitive iff the endomorphism ring of every projective object is semiprimitve in the usual sense.

My motivation is that I have a particular family of rings for which I am trying to decide semiprimitivity. I have shown that the module categories for these rings are equivalent to more geometrically defined abelian categories and would like to use the latter, which I understand better, to check semiprimitivity,


2 Answers 2


I believe this answers your question, although it may not be that enlightening. Basically, you can rephrase the existence of a faithful semisimple $R$-module in category-theoretic language as follows. First, an $R$-module $M$ is faithful iff $M$ is a cogenerator in $\mbox{Mod-}R$ for the subcategory of (finitely-generated) projective modules; i.e., for every nonzero map $f : P \rightarrow P'$ between projectives $P$ and $P'$ there exists a map $g : P' \rightarrow M$ such that $gf \neq 0$. Thus the ring $R$ is semiprimitive iff $\mbox{Mod-}R$ contains a semisimple object $M$ that is a cogenerator for the subcategory of projective objects.

  • $\begingroup$ Thanks. This may help. Do I need this for all projective modules or can I restrict to a set of projective modules which form a set of generators? $\endgroup$ Commented May 27, 2014 at 19:37
  • $\begingroup$ here you are using that projectiles generate right? $\endgroup$ Commented May 28, 2014 at 1:21
  • $\begingroup$ It is enough to know that M cogenerates the full subcategory {R}, which then implies that it also cogenerates Proj-R. Likewise, it would suffice if M cogenerates a set of projective modules that generate Mod-R. $\endgroup$
    – Alex Dugas
    Commented May 28, 2014 at 22:47

The answer is yes, and this works in the full generality of ringoids aka additive categories.

In this setting module categories are just the enriched free cocompletion, the regular representation is the Yoneda embedding, Morita equivalence is the same as their completions under absolute colimits being equivalent, etc etc. All this agrees with the standard notions in the one-object case of course, and surely you know all this, but I just wanted to set the context.

Then, you can define (two-sided) ideals as essentially classes of morphisms closed under local sum and pre- and post-composition (they also have a nice categorical interpretation); quotient under an ideal works exactly the same here (and it is probably some sort of 2-colimit). Now, you can mimic the definition of the Jacobson radical in this setting and everything works; a nicer, more categorical characterization/definition is

$rad(A)$ is the largest ideal such that the quotient map reflects isomorphisms

which is very much in the spirit of what the Jacobson radical means from a ring-theoretic point of view, and of course coincides with it in the one object case. It could probably help in your case; it was first defined like this by Kelly.

Being primitive (having a simple faithful module) can be defined exactly the same once one realizes that being faithful as a module is exactly the same as being so as a functor $M \colon R \to \mathbf{Ab}$. So defining $R$ an $\mathbf{Ab}$-enriched category to be semiprimitive when it has a faithful semisimple module, we have that this is equivalent to any of

  • $rad(R) = 0$
  • for each arrow $r$ in $R$ there is a simple module with $Mr \neq 0$

Even more, the Jacobson density theorem generalizes as easily as all this to

$R$ semiprimitive iff there exists a division ringoid $D$ and a faithful, noether-full functor $R \to \mathbf{Mod}_D$

where Noether-full means essentially full on Noetherian subobjects.

All this and more is in this greatly under-appreciated paper of Ross Street:

  • Ideals, radicals, and the structure of additive categories - Ross Street :: link

If one day I'd need to teach ring theory to undergraduates, I'll do it based on it :)

  • $\begingroup$ Thanks for this. I did know of ringoids and radicals of categories but didn't think of it in this context. Is it clear R is semiprimitive iff R-mod is semiprimitive though? This is necessary for this to really answer my question. $\endgroup$ Commented May 27, 2014 at 12:20
  • $\begingroup$ mmm I think it should be true. One implication (and the one you're interested in, I think) is trivial though: if R-Mod is semiprimitive compose the faithful Noether-full functor with Yoneda to get one for R. In the other direction, extension along Yoneda of R --> Mod_D looks like faithful Noether-full but is not immediate (to me at least) $\endgroup$ Commented May 28, 2014 at 0:30
  • $\begingroup$ After thinking a bit about it, in the case of a ring, the radical at the regular rep being 0 should imply the ring being semiprimitive $\endgroup$ Commented May 28, 2014 at 1:14

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