# Modules satisfying $M\oplus M \cong M \otimes M$

The queston was inspiried by the somewhat silly notion of whether one can "solve" equations of modules, with the sum of two $R$-modules being $\oplus$ and multiplication being $\otimes$.

When I asked our lecturer he mentioned that the right setting for the question might be algebraic K-theory, where $K_0(R)$ is something like the free module of isomorphism classes of finitely generated $R$ modules with direct sum serving as the group operation. Using tensor products of modules as the multiplication, this gives us a ring (This is what I recall, but I know nothing about algebraic K-theory besides what I just wrote, so it might be horribly wrong).

One can see that the zero module satisfies the condition and vector spaces of dimension $2$ and of infinite dimension over $k$ also satisfies the condtion. The question thus is not very specific: A "complete" classification seems very diffcult, but maybe other classes of modules satisfy the aforementoned condition.

Lastly, if anyone could shed some more light on the algebraic K-theory perspective that would certainly be much appreciated (if there is any light to be shed in this direction)!

• For example, when $k$ is a field, $K_0(k) \cong \mathbf{Z}$ with the isomorphism given by $[k] \leftrightarrow 1$. For finitely generated modules over $k$ it is also clear that $M \oplus M \cong M \otimes M$ iff $M \cong 0$ or $k^2$. For any PID, the structure theorem gives an isomorphism $K_0(R) \cong \mathbf Z$ where all modules of the same rank are equivalent in $K_0$. This shows that $K_0$ has more relations just than $[A] = [B]$ if $A \cong B$. Mar 23, 2018 at 17:26

So the idea of looking at this through algebraic $K$-theory is a good one.

Let us start with the basic setup (or rather, the setup I decided to go with for this answer): Let $\mathcal{C}$ be a tensor category (i.e. an additive category with a tensor product, and things satisfy all sorts of nice compatibility conditions that I won't bother to write up here). Let us further assume that $\mathcal{C}$ has finitely many indecomposable objects.

The Grothendieck ring of $\mathcal{C}$ is then the free abelian group with basis given by the indecomposable elements and a multiplication given by the tensor product.

The condition that $M\otimes M \cong M\oplus M$ then translates to $M^2 = 2M$ in the Grothendieck ring.

So far so good. But are there any interesting examples where this happens (outside the trivial examples mentioned)?
Yes, yes there is. In fact, an example is given by an extremely important category, namely the category of Soergel bimodules for a Coxeter group. I am not going to go into detail with how these are defined, but I will say a little about what this looks like on the Grothendieck group, which happens to be a very familiar thing: It is the group algebra of the Coxeter group. To make this more concrete, let us from now on just consider the group $S_3 = \langle s,t\mid s^2 = t^2 = (st)^3 = 1\rangle$.

The indecomposable Soergel bimodules provides us with a basis for $\mathbb{Z}[S_3]$ consisting of the elements $1, 1 + s, 1 + t, 1 + s + t + st, 1 + s + t + ts, 1 + s + t + st + ts + sts$, and in this basis, we precisely have that the elements $C_s = 1 + s$ and $C_t = 1 + t$ satisfy $C_s^2 = 2C_s$ (and the same for $t$). I invite the reader to write up what happens when you square the other elements.

(This basis also happens to have another name: It is the so-called Kazhdan-Lusztig basis, and it plays a fundamental role in the representation theory of Lie algebras).

If $R$ is a commutative Noetherian ring and $M$ a finitely generated module, this question has a nice answer. Notice that you may replace $R$ by $R/\mathrm{Ann} M$, since nothing changes and thus we may assume that $M$ is faithful. Then I claim that $M$ is a rank two projective module over $R$, if not zero.

If $M\neq 0$, we may localize and assume that $M\neq 0$ and $R$ is local with maximal ideal $\mathfrak{m}$. Let $r$ be the minimal number of generators of $M$. Then we have a minimal presentation $R^m\to R^r\to M\to 0$ and tensoring with $M$, we get $M\otimes R^m\to M\otimes R^r\to M\otimes M\to 0$. Minimality of this sequence implies that the minimal number of generators of $M\otimes M$ is $r^2$ which must be $2r$ and thus $r=2$, since $r\neq 0$.

Now let $0\to K\to R^2\to M\to 0$ be a presentation. We will show that $K=0$ and then we are done. Again, tensor with $M$ to get, $K\otimes M\to M\oplus M\to M\otimes M\to 0$. This says the image of $K\otimes M$ in $M\oplus M$ must be zero. Let $(a,b)\in R^2$ be in $K$. Then the image of this after tensoring is the collection of elements of the form $(am,bm)\in M\oplus M$ and since these must be zero, we see that $am=bm=0$ for all $m$ and thus $a=b=0$, since annihilator of $M$ was assumed to be zero.