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

• Jan 16th 2013, 07:51 AM
Curu
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\$?
• Jan 16th 2013, 11:19 AM
Curu
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).
• Jan 16th 2013, 01:08 PM
HallsofIvy
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.
• Jan 16th 2013, 01:17 PM
Curu
Re: M+A = A+A implies M = A?
Quote:

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.