# Thread: M+A = A+A implies M = A?

1. ## M+A = A+A implies M = A?

Let $\displaystyle A$ be a commutative unital ring and $\displaystyle M$ an $\displaystyle A$-module. Suppose that $\displaystyle M\oplus A \cong A\oplus A$. Is it true that necessary $\displaystyle M\cong A$?

2. ## Re: M+A = A+A implies M = A?

The answer is affirmative and contained in these beautiful notes (click) by Brian Conrad. A proof an be find in Weibel's K-Book (click).

3. ## Re: M+A = A+A implies M = A?

It doesn't really require a very complicated proof. In any ring, not just "commutative" or "unary" (in fact, since there is no mention of multiplication, you don't even need a ring), we have additive inverses. Adding the additive inverse of A to both sides immediately results in M= A.

4. ## Re: M+A = A+A implies M = A?

Originally Posted by HallsofIvy
It doesn't really require a very complicated proof. In any ring, not just "commutative" or "unary" (in fact, since there is no mention of multiplication, you don't even need a ring), we have additive inverses. Adding the additive inverse of A to both sides immediately results in M= A.
I really think you misunderstood the question in the OP. Maybe I had to specify that with $\displaystyle -\oplus -$ I mean the (bi)product (in the abelian category) of $\displaystyle A$-module, where $\displaystyle A$ is supposed to be commutative and unital in order to easily satisfy the IBP.