Understanding Algebras: The Relationship Between Rings and Modules

[Artusartos]

"Let R be a commutative ring. We say that M is an algebra over R, or that M is an R-algebra if M is an R-module that is also a ring (not necessarily commutative), and the ring and module operations are compatible, i.e., r(xy) = (rx)y = x(ry) for all x, y \in M and r \in R."

I'm not really sure why the second equality is true, because it implies commutativity and the definition tells us that an R-module is not necessarily commutative, right?

Thank you in advance

 

[HallsofIvy]

No, it does NOT imply commutativity because x and y are not commuted. What is says is that it doesn't matter if you multiply the real number "r" by x or by y before you multiply the two module members.

 

[Artusartos]

No, it does NOT imply commutativity because x and y are not commuted. What is says is that it doesn't matter if you multiply the real number "r" by x or by y before you multiply the two module members.

Thanks, but I cannot see any ring/module operation that would permit that. We know that associativity and distributivity hold (and of course others too, but they aren't related to this case). But I can't see how these basic operations would imply what you said.

 

[micromass]

The definition does that $rx = xr$, but that's for $r\in R$ and $x\in M$. The definition does not state that $xy = yx$ for $x,y\in M$. So $M$ is not commutative in that sense.

 

[Artusartos]

The definition does that $rx = xr$, but that's for $r\in R$ and $x\in M$. The definition does not state that $xy = yx$ for $x,y\in M$. So $M$ is not commutative in that sense.

Thanks a lot.