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).



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> > Questions: Did I understand the (\*)-property right? Do I overlook something?
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Thanks for the help.



Representation dimension of a special algebra
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. 


