# 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?

• $\phi$ 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-Álvarez 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