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Let $A$ be a commutative unital ring and $M$ an $A$-module. Suppose that $M\oplus A \cong A\oplus A$. Then is $M\cong A$?

We have that both $M\oplus A$ and $A\oplus A$ are biproduct for $(A, A)$ and $(M, A)$, so actually short exact sequences. Further $A\oplus A$ is the free $A$-module of rank two. However I can't conclude. Maybe I have to use some extension stuff from homological theory or there are some apparent counter-examples that I'm not able to figure out.

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What is your background in multilinear algebra? – KCd Jan 16 '13 at 18:11
I have some familiarity with tensor algebras and their famous quotients, but only in a really theoretical way (as rappresentable functors). The one in the OP is an exercise from a introductive course in homological algebra. – Andrea Gagna Jan 16 '13 at 20:24
up vote 5 down vote accepted

Since you wrote in a comment that you know about tensor products and their quotients, here is a hint. Oh, I should say first that I think this is a great problem (I used it myself when teaching a graduate algebra course) because it is a striking application of multilinear algebra when it's hard to see how to even get started using more direct tools, as far as I know.

Hint: Using exterior powers, there is a Kunneth formula $$ \Lambda^2(M_1 \oplus M_2) \cong \Lambda^2(M_1) \oplus (M_1 \otimes_A M_2) \oplus \Lambda^2(M_2) $$ for any two $A$-modules $M_1$ and $M_2$. (There is a corresponding formula for higher exterior powers, but it's not needed here.) Take the second exterior power of both sides of the isomorphism $M \oplus A \cong A \oplus A$ and use the Kunneth formula. Look closely at what you have. Then take second exterior powers again and look closely at what you have.

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This can (as I’m sure you know) be unwound into a very elementary direct proof, with neat formulas that seem to be pulled out of a hat. Write the map $M \oplus A \to A \oplus A$ as a $2 \times 2$ matrix $\Phi$, whose left column entries are from $M^*$ (and right col from $A$). Similarly, write the inverse map as a matrix $\Psi$, with top row entries from $M$. Now $\det \Phi$ (the usual formula) is an element of $M^*$, so a map $M \to A$; and $\det \Psi$ is an element of $M$, so (by multiplication) a map $A \to M$. These are mutually inverse just by the usual multiplicativity of determinant. – Peter LeFanu Lumsdaine Jan 16 '13 at 22:59
Incidentally, from this direct approach using determinants of matrices, it’s easy to see that the fact generalises to work from the hypothesis that $M \oplus A^n \cong A^{n+1}$ for any $n \geq 0$. This can also of course be seen using Kunneth formulas and exterior powers — but the calculations are much less familiar. – Peter LeFanu Lumsdaine Jan 16 '13 at 23:40
I actually never thought about a more elementary way to solve this problem, since the exterior power method is what first came to mind and it worked so nicely. I agree with the "pulled out of a hat" aspect, but it's nice to see that it can expressed using less technology. I wonder whether the first person who ever solved this problem did it by matrices or by exterior powers! – KCd Jan 16 '13 at 23:59
Great solution! @PeterLeFanuLumsdaine: that is the approach Weibel presents at the very begenning of his K-Book. – Andrea Gagna Jan 17 '13 at 18:14

The phrase you're looking for is that of a stably free module. Your question is answered in the affirmative in Keith Conrad's (as always) very helpful notes on stably free modules here. In particular, note that it's possible to have $A\oplus M\cong A\oplus A\oplus A$ without having $M\cong A\oplus A$, but this is a minimal counterexample: If $A\oplus M\cong A\oplus A$, then $A\cong M$.

Below here is an answer to a different question than the one being asked, which I'm leaving in only because of a comment that it was helpful.

Nope! Take $A=\mathbb{Z}$ and $M$ the direct sum of coutably many copies of $\mathbb{Z}$. Then $M\oplus A\cong M\cong M\oplus M$, but $M\not\cong A$.

On the other hand, for some classes of rings (e.g., $A=\mathbb{Z}$, and I think Dedekind domains) it's true that you can "cancel" finitely-generated modules in the way that you want (that is, if one of the factors is the ring itself).

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Thanks for using this example. Maybe if I see it often enough I will actually remember it for future reference. – Chris Leary Jan 16 '13 at 17:50
Yes, it was a typo. Thank you! I know this conter-example, but pay attention to the fact that I actually request $A\oplus M$ to be isomorphic to the direct (bi)product of the ring, i.e., $A\oplus A$. – Andrea Gagna Jan 16 '13 at 17:52
@user58504: Actually, as it stands. I've answered the wrong question. Did you mean $M\oplus A\cong A\oplus A$ or $M\oplus A\cong M\oplus M$? – Cam McLeman Jan 16 '13 at 17:53
Very helpful! Thank you very much, Cam. – Andrea Gagna Jan 16 '13 at 20:06
Another beautiful reference I've just found is the first chapter of the book(-in-progress) "Algebraic K-Theory" of Weibel: – Andrea Gagna Jan 16 '13 at 21:01

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