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 $J^2=J$).

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 $J$) 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!

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!

I am looking for examples of finite-dimensional commutative $\mathbb{C}$-algebra $R$ with an ideal $I$ such that $$R/I\otimes_R I=\{0\}$$ and an $R$-module extension $M$ "the-other-way-around": $$0\to R/I\to M\to I \to 0 $$ which has the exact opposite vanishing tensor products and acts as "second identity" as follows: $$M\otimes_R M=M $$ $$M\otimes_R R/I=\{0\}$$ $$M\otimes_R I= M$$ $$I\otimes_R I=M$$ The question might become trivial from certain perspectives, but I would be very happy if you can provide such an example or indicate how to construct a family. Any hints how these conditions appear in literature are also very welcome. Or I am also happy if you can explain to me why this is impossible (then my next question would be using bimodules over non-commutative algebras...)

Thanx very much for your help in advance!

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 $J^2=J$).

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 $J$) 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!

I am looking for examples of finite-dimensional commutative $\mathbb{C}$-algebra $R$ with an ideal $I$ such that $$R/I\otimes_R I=\{0\}$$ and an $R$-module extension $M$ "the-other-way-around": $$0\to R/I\to M\to I \to 0 $$ which has the exact opposite vanishing tensor products and acts as "second identity" as follows: $$M\otimes_R M=M $$ $$M\otimes_R R/I=\{0\}$$ $$M\otimes_R I= M$$ The question might become trivial from certain perspectives, but I would be very happy if you can provide such an example or indicate how to construct a family. Any hints how these conditions appear in literature are also very welcome. Or I am also happy if you can explain to me why this is impossible (then my next question would be using bimodules over non-commutative algebras...)

Thanx very much for your help in advance!

I am looking for examples of finite-dimensional commutative $\mathbb{C}$-algebra $R$ with an ideal $I$ such that $$R/I\otimes_R I=\{0\}$$ and an $R$-module extension $M$ "the-other-way-around": $$0\to R/I\to M\to I \to 0 $$ which has the exact opposite vanishing tensor products and acts as "second identity" as follows: $$M\otimes_R M=M $$ $$M\otimes_R R/I=\{0\}$$ $$M\otimes_R I= M$$ $$I\otimes_R I=M$$ The question might become trivial from certain perspectives, but I would be very happy if you can provide such an example or indicate how to construct a family. Any hints how these conditions appear in literature are also very welcome. Or I am also happy if you can explain to me why this is impossible (then my next question would be using bimodules over non-commutative algebras...)

Thanx very much for your help in advance!