Example of tensor category with non-simple unit $J\to \mathbb{1} \to Q$ and suitably extension $Q\to M\to J$ Edit: Thanx very much to Neil Strickland for quickly explaining to us that the following cannot be realized over finite commutative $\mathbb{C}$-algebras, as I had originally asked.
I know that there is a finite tensor category (from minimal models) with the following relations that seem rather strange to me. In particular it seems now one cannot realize it as modules over a ring. 
Does anyone know an algebraic situation, where something similar occurs? (maybe in a derived category?). 
A non-simple unit object $\mathbb{1}$ 
$$0\to J\to \mathbb{1} \to Q \to 0$$
$$J\otimes Q=\{0\}$$
$$Q\otimes Q=Q$$
(for modules over a ring $R,\otimes_R$, this means $J$ is an ideal with $J^2=J$, thus $R$ often splits, see below). 
such that $J\otimes J$ is an extension the-other-way-around $$M:=J\otimes J$$
$$0\to Q\to M \to J \to 0$$
(for modules over a ring $R,\otimes_R$ the product $J\otimes J$ cannot be larger then $J$) 
and which acts somewhat like a second identity
$$M\otimes M=M $$
$$M\otimes Q=\{0\}$$ 
$$M\otimes J= M$$
Any hints what some of these situations are called in literature are also very welcome. 
Thanx very much for your help in advance! 
 A: In general, if $N$ is a finitely generated $S$-module, and $J$ is an ideal with $JN=N$, it is a standard fact that there exists $u\in J$ with $(1-u)N=0$.  (This is one incarnation of Nakayama's Lemma.)  In your context everything is finite-dimensional and therefore finitely generated.  The condition $R/I\otimes_RI=0$ is equivalent to $I=I^2$, so there exists $e\in I$ with $(1-e)I=0$.  In particular $(1-e)e=0$ so $e$ is idempotent.  This means that there is a splitting $R=R_0\times R_1$ with $e=(0,1)$ and $I=0\times R_1$ so $R/I=R_0$.  Now for any $R$-module $M$ the conditions $M\otimes_R(R/I)=0$ and $M\otimes_RI=M$ are equivalent and just mean that $M=0\times M_1$ for some $R_1$-module $M_1$.  Your final condition $M\otimes_RM=M$ is then equivalent to $M_1\otimes_{R_1}M_1$.  You can make this true by taking $M_1=R_1$, and I think that that is the only possibility.  But anyway, as $R/I=R_0\times 0$ and $M=0\times M_1$ we have $\text{Hom}_R(R/I,M)=0$, so there can be no short exact sequence $R/I\to M\to I$.
