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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:

  1. The hint in the textbook makes it clear that this is the "easy" direction, so I am sure the answer is not difficult, BUT...
  2. 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.

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    About Update 2: Outstanding attitude of Jon, Jack, and the author! Good example of the usefulness of MSE!2011-09-29

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Let R be the finite field with 4 elements. Then E is the 2×2 matrix ring over the field with 2 elements. Z(E) is the set of scalar matrices, isomorphic to the field with 2 elements. Z(R) = R is of course of order 4.

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    @Jon: The exercise looks pretty wrong to me. If$R$is a field not equal to its prime field K, then$E$is is the ring of K-endomorphisms of R, with center isomorphic to K. Another reason it must be wrong is that Z(E) only depends on the additive group of R, but there are lots of ring structures on the same additive group, some commutative, and some not. For instance if R1 is 2×2 matrices over K and R2 is K×K×K×K, then E is 4×4 matrices over K, and Z(R1) = Z(E) = K, but Z(R2) = R2 has dimension 4 over K. It basically claims the center of a ring is independent of its multiplication.2011-09-27