Let $R$ be a ring (with unity) and let $E = \text{End}_{\text{Ab}}(R)$ be the ring of endomorphisms of $R$'s underlying abelian group. There is an injective ring homomorphism $\lambda: R \to E$ given by allowing $R$ to act on itself by left multiplication.
An exercise in Algebra: Chapter 0 by Paolo Aluffi asks the student to show that this map induces an isomorphism of the centers $Z(R)$ and $Z(E)$. I can show that an element of $Z(E)$ must be a left multiplication by an element of $Z(R)$, but somehow I can't see why an element of $Z(R)$ is necessarily mapped to an element of $Z(E)$. Clearly the image $\lambda_r$ of some $r \in Z(R)$ must commute with all the other left multiplication endomorphisms, but I can't convince myself that there won't be some other endomorphism with which it won't commute.
Why, then, is it the case that $\lambda(Z(R)) \subset Z(E)$?
[Some notes:
- The hint in the textbook makes it clear that this is the "easy" direction, so I am sure the answer is not difficult, BUT...
- The textbook has not yet introduced modules OR matrices, which makes me believe that this proposition can be verified by hand (without new machinery).]
UPDATE: It appears that this claim is false. Is it almost true? Anyone know what the author might have intended here?
UPDATE 2: I checked with the author, who thanked us for bringing the error and counterexample to his attention. An erratum is being added to his website.