Given an algebraically closed field $k$ and a finitely generated commutative $k$-algebra $A$, all simple modules over $A$ are 1-dimensional

  • What is the analogous statement for symmetric monoidal $k$-linear categories?
  • What is the analogous statement for braided monoidal $k$-linear categories?

We can assume the category is Abelian and the product functor is right exact in both variables

I expect something of the sort "any simple (in some sense) module category is equivalent to $Vect$", although I have no idea how braided and symmetric are different in this repsect


2 Answers 2


Consider C = Rep(G) for a finite group G. This is a symmetric tensor category. Any subgroup $H \subset G$ yields a module category Rep(H) with the action given by restricting and then tensoring. This module category is simple in any appropriate sense, and in no interesting sense is it "1-dimensional."

  • $\begingroup$ How does it look like in the 4-category of braided categories, monoidal bimodule categories, bimodule categories, functors and natural transformations? Is the 1-morphism $\text{Rep}H$ Morita equivalent to $\text{Vect}$? $\endgroup$ Oct 21, 2015 at 8:08
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    $\begingroup$ @Turion: No, global dimension is higher Morita invariant, so Rep(H) and Vect are not higher Morita equivalent as tensor categories (let alone in a way compatible with the action). $\endgroup$ Oct 21, 2015 at 17:47

If A is a commutative Noetherian A algebra in Vec, then A-mod is not typically equivalent to Vec. First, if it's not a semisimple algebra then there are interesting non-simple modules, and even in the semisimple case you don't just have one 1-dimensional A-module. So I don't think your question makes sense as currently written.

At any rate, you can have a commutative semisimple finite dimensional algebra A in a braided tensor category with simple A-modules that aren't one-dimensional. For example, your tensor category is U_q(sl_2) at an appropriate root of unity, your commutative algebra is the sum of the two 1-dimensional objects out at the ends, and the category of modules is quite interesting and goes by the name D_2n (the easiest place for you to see more about this is probably http://arxiv.org/abs/math/0101219). I'm not totally sure about the symmetric case.

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    $\begingroup$ Noah you completely misunderstood my intention. Probably I need to reformulate. I am not asking about algebra objects in symmetric monoidal categories. I'm trying to categorify the statement i.e. consider module categories over symmetric monoidal categories. A module category $\mathcal{M}$ over a symmetric monoidal category $\matcal{A}$ is a category equipped with a linear (or even right exact) bifunctor $\otimes: \mathcal{A} \times \mathcal{M} \rightarrow \mathcal{M}$ satisfying the appropriate conditions $\endgroup$
    – Vanessa
    Feb 18, 2012 at 17:08

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