M+A = A+A implies M = A?

**Curu** · January 16th 2013, 07:51 AM

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

**Curu** · January 16th 2013, 11:19 AM

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).

**HallsofIvy** · January 16th 2013, 01:08 PM

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.

**Curu** · January 16th 2013, 01:17 PM

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 $-\oplus -$ I mean the (bi)product (in the abelian category) of $A$ -module, where $A$ is supposed to be commutative and unital in order to easily satisfy the IBP.

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