# Maps of free modules over a ring [closed]

(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?

-

## closed as off topic by Qiaochu Yuan, Chris Gerig, Fernando Muro, Andreas Blass, Steven LandsburgSep 5 '12 at 16:02

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here. If this question can be reworded to fit the rules in the help center, please edit the question.

$\phi$ 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. –  Qiaochu Yuan Sep 5 '12 at 3:47
Presumably "Noetherian" is needed here to ensure the existence of a maximal ideal? Or is there a different, non-Noetherian proof? –  Will Sawin Sep 5 '12 at 5:30
Will, maximal ideals are always there! :-) –  Mariano Suárez-Alvarez Sep 5 '12 at 5:37
Ah! So $\phi$ doesn't respect our scalar action of $A$. Thank you very much! –  Xander Flood Sep 5 '12 at 15:27
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! –  Pace Nielsen Sep 5 '12 at 15:28