(This is exercise 10 of chapter 2 of Atiyah and Macdonald.)

The exercise starts by asking me to prove that if $A^n\cong A^m$ then $n=m$ for any nonzero ring A. I managed to do that (by tensoring with the residue field of a maximal ideal to get some nice vector spaces) - however, I can't seem to reconcile why the following is not a counterexample.

Let $A$ denote $\mathbb Z^{\aleph_0}$, the ring of sequences of integers. Then the map $\phi:A^2\to A$ given by interleaving sequences is an isomorphism. (If two sequences interleaved are zero, then both sequences were zero. Hence, $\ker\phi=0$ Any sequence can be decomposed into its even and odd subsequences, in which case applying $\phi$ will reconstruct the original. Therefore $\mathrm{im}\ \phi=A$.) Nonetheless, 2>1. What am I missing?

  • 9
    $\begingroup$ $\phi$ is an isomorphism of $\mathbb{Z}$-modules, but it isn't an isomorphism of $A$-modules. If you run through the proof of your exercise in this case you'll notice that e.g. the first factor of $A$ acts differently in the two modules. $\endgroup$ Sep 5, 2012 at 3:47
  • $\begingroup$ Presumably "Noetherian" is needed here to ensure the existence of a maximal ideal? Or is there a different, non-Noetherian proof? $\endgroup$
    – Will Sawin
    Sep 5, 2012 at 5:30
  • 6
    $\begingroup$ Will, maximal ideals are always there! :-) $\endgroup$ Sep 5, 2012 at 5:37
  • $\begingroup$ Ah! So $\phi$ doesn't respect our scalar action of $A$. Thank you very much! $\endgroup$ Sep 5, 2012 at 15:27
  • 1
    $\begingroup$ Xander, your proof of the exercise is absolutely correct, and Qiaochu's comment explains why the map $\varphi$ is not an $A$-module isomorphism. However, I wanted to point out that for non-commutative rings this result fails miserably. For example, if $A$ is the endomorphism ring of an infinite dimensional vector space then $A^2\cong A$. The idea in proving this isomorphism is very similar to the map you took above! $\endgroup$ Sep 5, 2012 at 15:28


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