# The definition of an algebra

1. Jun 12, 2013

### 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?

2. Jun 12, 2013

### HallsofIvy

Staff Emeritus
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.

3. Jun 12, 2013

### Artusartos

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.

4. Jun 12, 2013

### micromass

Staff Emeritus
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.

5. Jun 13, 2013

### Artusartos

Thanks a lot.