Definition of an algebra over a noncommutative ring - MathOverflow

Definition of an algebra over a noncommutative ring

2010-04-19T21:54:26Z

I've tried in vain to find a definition of an algebra over a noncommutative ring. Does this algebraic structure not exist? In particular, does the following definition from http://en.wikipedia.org/wiki/Algebra_(ring_theory) make sense for noncommutative $R$?

Let $R$ be a commutative ring. An algebra is an $R$-module $A$ together with a binary operation $$ [\cdot,\cdot]: A\times A\to A $$ called $A$-multiplication, which satisfies the following axiom: $$ [a x + b y, z] = a [x, z] + b [y, z], \quad [z, a x + b y] = a[z, x] + b [z, y] $$ for all scalars $a$, $b$ in $R$ and all elements $x$, $y$, $z$ in $A$.

So, is there a common notion of an algebra over a noncommutative ring?

Answer by captain obvious for Definition of an algebra over a noncommutative ring

2010-04-19T22:05:09Z

Why not just: $A$ is a ring together with a ring homomorphism $R\to A$?

Answer by mathphysicist for Definition of an algebra over a noncommutative ring

2010-04-19T22:50:23Z

Unfortunately that's not exactly what you want but in this paper the authors define Lie algebras over noncommutative rings.

Answer by Manny Reyes for Definition of an algebra over a noncommutative ring

2010-04-19T23:23:18Z

[Edit: Thanks to Harry Gindi for pointing out that I'm only extending the notion of an associative algebra.]

Here is a somewhat "brute force" approach to define an associative algebra over a general ring. Let's say that a ring $A$ with a fixed ring homomorphism $f\colon R\to A$ is centrally generated over $R$ (with respect to $f$) if $A$ is generated as a ring by the image of $R$ and a subset $X\subseteq A$ such that every element of the image of $R$ commutes with every element of $X$.

Then it's clear that whenever $R$ is commutative, a ring homomorphism $f\colon R\to A$ makes $A$ into an $R$-algebra if and only if $A$ is centrally generated over $R$ with respect to $f$.

Answer by Ben Webster for Definition of an algebra over a noncommutative ring

2010-04-20T00:26:06Z

I would argue that this notion doesn't have one natural generalization. One obvious one is what I would call an $R$-bimodule algebra, that is, an algebra which is an R-bimodule in such a way that left multiplication of $A$ commutes with right multiplication of $R$ and vice versa. If you only have one of these actions, you would have an R-left or right module algebra.

On some level, you can't really expect there to be one correct generalization; which one is right depends on the context. If you have an example, pick the definition that fits your example, and if you don't have an example, why worry?

Answer by Zoran Škoda for Definition of an algebra over a noncommutative ring

2010-04-20T03:51:08Z

The commutative notion of an (associative or not) algebra $A$ over a commutative ring $R$ has two natural generalization to the noncommutative setup, but the one you list with defined left $R$-linearity in both arguments is neither of them; in particular your multiplication does not necessarily induce a map from the tensor product, unless the image of $R$ is in the center. Most useful is the notion of an $R$-ring $A$ (or a ring $A$ over $R$), which is just a monoid in the monoidal category of $R$-bimodules: in other words the multiplication is a map $A\otimes A\to A$ which is left linear in first and right linear in the second factor. If we drop the associativity for the multiplication all works the same way, but I do not know if there is a common name (maybe descriptive like magma internal to the monoidal category of $R$-bimodules; or one may try a rare term nonassociative $R$-ring).

In the commutative case, the mutliplication is both left and right linear in each factor, what is here possible only if $R$ maps into the center of $A$. (Edit: I erased here one additional nonsense sentence clearly written when tired ;) ). Thus the two useful concepts in the noncommutative case are $R$-rings (possibly nonassociative!) and, well, the subclass with that property: $R$ maps into $Z(A)$, deserving the full name of "algebra". There is also a notion of $R$-coring, which is a comonoid in the monoidal category of $R$-bimodules, generalizing the notion of an $R$-coalgebra to a noncommutative ground ring.

Edit: I suggest also this link.