# Representation dimension of a special algebra

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.

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Maybe it's a matter of my cultural background, nothing absolute, but I think it is not very polite to post this question here. I think you should contact first the authors and see what they have to say. If they don't answer you, or their answer doesn't satisfy you, then you may come and bring the question. The question can also be phrased without using the word error, which is very rude and may not be the case. Keep the word for when the question is settled, if settled in the negative. – Fernando Muro Dec 15 '12 at 15:21
Sorry. You're right. I was unpolite. I edited my question. – Bernhard Boehmler Mar 2 '13 at 22:32

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.