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I am having trouble verifying the following (this is self-study):

There is an isomorphism between the center of a ring $A$ and the ring of endomorphisms of the identity functor of the category of (right) $A$-modules.

The map $\Psi\ \colon Z(A) \to \mathrm{End}(1_{Mod\ A})$ sends $a \in Z(A)$ to multiplication by $a$ (let's write $\Psi(a) = \theta^{a}$). This poses no trouble.

It seems reasonable to define the inverse map $\Xi\ \colon \mathrm{End}(1_{Mod\ A}) \to Z(A)$ as sending $ \eta \in \mathrm{End}(1_{Mod\ A})$ to $\eta_A(1)$, since from this we readily get $\Xi \circ \Psi (a) = \Xi(\theta^a)=\theta^{a}_{A}(1)=1a=a$.

However I can't see why $\Psi \circ \Xi = 1_{\mathrm{End}(1_{Mod\ A})}$. Unfolding, we have to prove that $\Psi \circ \Xi (\eta) = \theta^{\eta_A(1)} = \eta$, that is, for any right $A$-module $L$, we need to have $\theta^{\eta_A(1)}_L = \eta_L$. I'm at a loss on this. I tried to find the right components on which to use naturality (that's how I succeeded to prove that $\eta_A(1)$ does lie in $Z(A)$, which wasn't obvious to me from the outset), but to no avail.

Some googling led me to this blog post by Q. Yuan (see the "sub-example") where the direction of the map $\Xi$ is deduced from the fact that $A$ is a generator of $\mathrm{Mod}\ A$. However I can't quite see explicitly why (in the perspective adopted there, I believe this is just rephrasing the issue I'm having) the lifting of a central element to an endomorphism of $1_{\mathrm{Mod}\ A}$ is inverse to the composition of the two isomorphisms with the natural injection $\mathrm{End}(1_{\mathrm{Mod}\ A}) \to Z(\mathrm{End}_A(A))$.

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I think you're supposed to use the Yoneda lemma and the fact that the forgetful functor $U : \textbf{Mod}_A \to \textbf{Set}$ is represented by the free (right) $A$-module $A$. –  Zhen Lin Jul 13 '12 at 10:46
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1 Answer 1

up vote 3 down vote accepted

As suggested in the comments, break it into pieces.

Step 1. The identity functor is the same as the functor $\text{Hom}_A(A, \text{_})$ so you just need to know the endomorphisms of this latter functor. By Yoneda, the endomorphisms of this functor are the same as the endomorphisms of $A$; this is cheating a little since Yoneda is about the functor to sets and not the functor to $A$-modules, so you have to chase through the proof of Yoneda to see that $A$-linear endomorphisms of this functor correspond to $A$-linear endomorphisms of $A$.

Step 2. The $A$-linear endomorphisms of $A$ are $Z(A)$. This step is easy; clearly $Z(A)$ gives rise to $A$-linear endomorphisms of $A$ and any endomorphism is determined by $f(1)$, which must be central.

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Thank you countinghaus and Zhen Lin. Most helpful :-) –  befunctored Jul 14 '12 at 6:09
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