# 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!

• My guess is that there is a mistake in the thought process that led you to this precise list of conditions. Perhaps you could outline your argument? – Neil Strickland Jul 6 '18 at 10:37
• Obviousely, thank you alot for the help: I am interested in the category with conjectured (!) tensor product in arXiv:0905.0916. On p.17-19 they describe the effect I asked about for the Virasoro algebra, p.38 they describe the same for the so-called logarithmic models, both by explicit lists of tensor products. There, the unit object is called V, the subobject J is called V(2), the quotient Q is called V(0), the extension the-other-ways-around is called V*. Respecitvely W,W(0),W(2),W*. – Simon Lentner Jul 7 '18 at 7:47
• I myself think I understand quite well also in more general settings the quotient W(0)\mapsto 0, which is described as quantum group representations. But I do not understand the effect by which the W(0) enters, and this is what I ask about. – Simon Lentner Jul 7 '18 at 7:50

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 $R/I\to M\to I$.

• As you say, $M=R/I\oplus I$. But, then $M\otimes R/I=R/I\otimes R/I\neq 0$. I am a bit confused. – Mohan Jul 5 '18 at 21:52
• Sorry, I got the last bit backwards. I have corrected it. – Neil Strickland Jul 5 '18 at 22:00
• Thank you very much! The Nakayama-Argument with the splitting idempotent was unknown to me. I understand now, why this is not possible...But this leaves me puzzled: I know this category exists, yet it appearently cannot be realized as modules over a ring....? (I'll edit accordingly) – Simon Lentner Jul 6 '18 at 7:21