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.

  • $\begingroup$ Dear Jon: It seems to me you didn't mention the author's name. If such is the case, I suggest that you edit your question. If you do, you could change "Alegbra" to "Algebra". $\endgroup$ – Pierre-Yves Gaillard Sep 27 '11 at 4:19
  • $\begingroup$ @Pierre-Yves Gaillard: Thanks for the suggestion. I have added the author's name. The book is part of the Graduate Studies in Mathematics series. $\endgroup$ – Jon Sep 27 '11 at 12:46
  • $\begingroup$ Dear Jon: Thanks! By the way, I voted for your question (and for Jack's answer). It shows that you read books seriously. I'm also realizing that my suggestion to name the author could be misinterpreted since a mistake has been pointed out. But on the other hand, everybody makes mistakes, and the best thing that can happen to a mistake is to be corrected. $\endgroup$ – Pierre-Yves Gaillard Sep 27 '11 at 17:02
  • 1
    $\begingroup$ I have not had time to see what might be true; nor even to verify the other direction that Jon already proved, which seemed interesting to me. I think a "true" version would need to pay attention to S-module structure for S some large subring of R (not just S=Z for abelian group). The "double centralizer" theorem might be something along these lines. $\endgroup$ – Jack Schmidt Sep 27 '11 at 17:06
  • 1
    $\begingroup$ About Update 2: Outstanding attitude of Jon, Jack, and the author! Good example of the usefulness of MSE! $\endgroup$ – Pierre-Yves Gaillard Sep 29 '11 at 5:09

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.

  • 2
    $\begingroup$ Well that's strange. I can't see any error in what you wrote; have I misunderstood the exercise? Here is the text exactly as it appears: "2.17. Let $R$ be a ring, and let $E = \text{End}_{\text{Ab}}(R)$ be the ring of endomorphisms of the underlying abelian group $(R, +)$. Prove that the center of $E$ is isomorphic to the center of $R$." $\endgroup$ – Jon Sep 27 '11 at 2:12
  • 1
    $\begingroup$ @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. $\endgroup$ – Jack Schmidt Sep 27 '11 at 14:23
  • $\begingroup$ +1 Yesterday I was wrestling with the exercize and had serious doubts. Today I find your counterexample and am relieved. $\endgroup$ – drhab Jun 11 at 8:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.