Question

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

Answer 1

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."

Answer 2

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.