Module categories over symmetric/braided monoidal categories Given an algebraically closed field $k$ and a finitely generated commutative $k$-algebra $A$, all simple modules over $A$ are 1-dimensional


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*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
 A: 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."
A: 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.
