Representation dimension of a special algebra

Question

Hi,

I'm reading the following paper: http://fma2.math.uni-magdeburg.de/~holm/ARTIKEL/holm-hu-23-05.pdf

I've come across a piece of information, which I don't understand, and wanted to ask, if I overlook something.

Let $A_0:=k[x,y]/\langle x^2,xy,y^2\rangle$.

Let $A_0^0=\langle \tilde{1} \rangle$ be the regular $A_0$ left module, that means diagrammatically, that $A_0^0$ consists of 3 points (the basis vectors), which we will call $\tilde{1}$, $\tilde{x}$ and $\tilde{y}$, and 3 edges ($A_0$ operates via left musltiplication).

Let $DA_0^0$ be the module consisting of the vertices $\tilde{x},\tilde{y}$ and $\tilde{xy}$ and again 3 edges, which correspond to the action of $A_0$.

Let $U_0$ be a module, which is isomorphic to $k$.

Let $U_1$ be the module consisting of 2 basis vectors and 1 edge, which symbolizes the multiplication by $y$.

Let $X$ be the module consisting of 2 basis vectors and 1 edge, which symbolizes the multiplication by $x$.

Now, define the Auslander generator $M_0:=A_0^0\oplus DA_0^0\oplus U_0\oplus U_1\oplus X$.

The aim is to show that rep.dim $(k[x,y]/\langle x^2,y^{n+2}\rangle)$ is $\leq$ 3 by showing that rep.dim $(k[x,y]/\langle x^2,xy^{n+1},y^{n+2}\rangle)$ is $\leq$ 3.

For every indecomposable direct summand $T$ of $M$ a suitable exact sequence $0\rightarrow K\rightarrow N_1 \rightarrow T$ with $N_1\in$ add($M$), which has the following property, is constructed:

(*) Every homomorphism from an indecomposable summand of $M_0$ to $T$ factors through $N_1$.

Applying the functor Hom$(M_0,$_$)$, we get another short exact sequence:

$0\rightarrow (M_0,K)\rightarrow (M_0,N_1) \rightarrow (M_0,T) $.

If the property (*) is fullfilled, then the cokernel of $(M_0,N_1)\rightarrow (M_0,T)$ is one-dimensional, and therefore simple. Thus, we get the initial part of a projective resolution of $E_T$, whereupon $E_T$ is the simple module corresponding to the projective module $(M_0,T)$.

Now, the (*) property says, that every homomorphism from $A_0^0$ to $A_0^0$, except for the multiples of the identity on $A_0^0$ has to factor through $N_1$, which is the radical here. We have rad$(A_0^0)=k\oplus k=U_0\oplus U_0$.

If we consider the homomorphism from $A_0^0$ to $A_0^0$, which sends $\tilde{1}$ to $\tilde{1}+\tilde{x}$, and sends $\tilde{x}$ to $\tilde{x}$ and $\tilde{y}$ to $\tilde{y}$, then we have found an isomorphism, which doesn't factor through the radical, but is no multiple of the identity.

There seem to be analogue counter-examples, if we consider other isomorphisms (other indecomposable summands of the Auslander Generator M_0).

Questions: Did I understand the (*)-property right? Do I overlook something?

Thanks for the help.

Answer 1

I don't think (*) is correctly copied from the paper. The corresponding claim in the paper is that every morphism from an indecomposable summand of $M$ except for the identity morphism from $T$ to $T$ factors through $N_1$.

I think a precisely correct statement is that, for any direct summand $T$ of $M$, Hom($T$,$T$) can be split as a vector space into the multiples of the identity map and a complementary subspace consisting of morphisms which factor through $N_1$. This seems to be sufficient for the argument; I expect it is what the authors meant by what they wrote.