# How do I show this isomorphism between an opposite endomorphism ring and a module over path algebras?

So I'm working with quivers with relations.

Let $\Gamma$ be a quiver with vertices $\Gamma_0 = \{1,2,...,n\}$ and let $\rho$ be a set of relations in $\Gamma$ with $J^t \subset \langle\rho\rangle \subset J^2$. Let $\Lambda = k\Gamma /\langle p\rangle$. The ideal $J$ in $\Lambda$ is genrated by the arrows in the quiver.

I want to show that $\text{End}_\Lambda(\Lambda e_i)^{\text{op}}$ is isomorphic to $e_i \Lambda e_i$, where $e_i$ is the trivial path in the quiver going from vertex $i$ to $i$. That is, $e_i : i \rightarrow i$ for $i \in \Gamma_0$.

I want to define a homomorphism of rings $\phi : \text{End}_\Lambda (\Lambda e_i)^{\text{op}} \rightarrow e_i \Lambda e_i$ which I could then show is an isomorphism, but I'm having troubles even setting up said homomorphism in a smart way.

I tried defining $\phi(f) = f(e_i)$, but I got stuck down the line. Then I tried $\phi(f) = f(m)$ for some fixed $m \in \Lambda e_i$, but this didn't get me very far either. I'm looking for any hints on how to define this in a good way, or even just a thought process for handling this type of problem.

If $A$ is an arbitrary unital ring, and $e\in A$ an idempotent, you can define a map $$\Phi\colon\operatorname{End}_A(Ae)\to eAe:f\mapsto ef(e)e.$$

This is clearly an additive map. Note $f(e)\in Ae$, $f(e)=f(e)e$, since $e$ is a right identity in $Ae$. So, $$\Phi(gf)=egf(e)e=eg(f(e))e=eg(f(e)e)e=ef(e)g(e)e=ef(e)e^2g(e)e=\Phi(f)\Phi(g).$$

So $\Phi$ is a ring anti-homomorphism. Replacing the domain with the opposite ring gives you a usual ring homomorphism.

Note $ef(e)e=ef(e)=f(e^2)=f(e)$, so $\Phi$ is injective, since an $A$-map on $Ae$ is uniquely determined by its image on $e$. For surjectivity, let $a\in eAe$. Define an $A$-map $\theta\colon Ae\to Ae$ by $\theta(be)=ba$. It is well-defined, since if $be=ce$, then since $a=ae=ea$, $$\theta(be)=ba=bea=cea=ca=\theta(ce).$$ Then $\Phi(\theta)=e\theta(e)e=\theta(e)=a$, so $\Phi$ is surjective, hence invertible.

This is true for any idempotent $e$ in any ring $A$, the basic ingredients being the following: (1) For any $A$-module $M$ you have $M=eM\oplus (1-e)M$, so the projection defines a canonical isomorphism $eM\cong M/(1-e)M$. (2) $eM = \text{Ann}_M(1-e)$. Applying this and the universal property of the quotient module, you get $$\text{Hom}_A(Ae,M)\cong\text{Hom}_A(A/(1-e)A,M)\cong\text{Ann}_M(1-e)=eM$$ for any $A$-module $M$. Also, making all steps explicit shows that the map is given by $\varphi\leftrightarrow\varphi(e)$.

Putting $M=A$ in the above yields $\text{End}_A(Ae)\cong eAe$ as abelian groups via $\varphi\mapsto\varphi(e)$. Can you figure out the part about the ring structure yourself, using $\varphi(e)=\varphi(e)e$ since $\varphi(e)\in Ae$? Let me know if you have trouble.

• @Auclair: Do you have any questions? – Hanno Jun 3 '17 at 9:20