# Definition of an algebra over a noncommutative ring

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?

-
As usual, "correct" depends on what you want to do with the concept... There is no absolute notion of correctness of definitions. – Mariano Suárez-Alvarez Apr 19 '10 at 21:58
You're right. I'll fix that. – Rasmus Bentmann Apr 19 '10 at 21:59
Of course the definition does make sense. WIth that definition, though, R is not an algebra over itself in the obvious way in general... you may or may not want it to be, though. – Mariano Suárez-Alvarez Apr 19 '10 at 22:02
As I said, it depends on what you want to do: there are sensible options... You can decide, for example, that an $R$-algebra is just a morphism $R\to A$ , or sort of more generally that $A$ is to be an $R$-bimodule such that multiplication decends to a morphism of $R$-bimodules $A\otimes_RA\to A$. I'm sure people can come up with other variants! – Mariano Suárez-Alvarez Apr 19 '10 at 22:10
To paraphrase my remote ancestor (Humpty Dumpty), words mean whatever you define them to mean. But where does a definition lead? – Jim Humphreys Apr 19 '10 at 22:14

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.

-
This is a great answer. Thank you! – Rasmus Bentmann Apr 20 '10 at 6:32
But, "the image of $R$" makes sense only if $A$ is unital, right? – Rasmus Bentmann Apr 20 '10 at 6:35
I think that Chris' comment to Ben's answer below also applies here. – Manny Reyes Apr 20 '10 at 22:04
Apart from wrong phrasing about the center, I was anticipating Chris's comment precisely with the discussion of left vs light linearity in each factor. Namely, if the scalars are in the center of $A$, then left and right linearity is interchangeable, what automatically takes care of that kind of comment. I added a John Baez's discussion in the link in the corrected edit above. – Zoran Skoda Apr 21 '10 at 18:02

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?

-
The troublesome thing with the R-bimodule definition is that when you restrict to R a commutative algebra, it doesn't seem (at least to me) to reproduce the usual notion of R-algebra. Specifically if R is a commutative algebra, then left modules are the same as right modules, but these are not the same as bimodules. There is no reason for the left and right module actions to be "the same". – Chris Schommer-Pries Apr 20 '10 at 13:01
Well, right, there's no way to make the left and right actions "the same" when the ring is not commutative. You could require the ring to have an anti-involution and then try to do so, but that sounds pretty horrible. – Ben Webster Apr 20 '10 at 14:47

[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 $f$.

-
There's no reason why an algebra should be a ring. Most aren't, in fact! – Harry Gindi Apr 19 '10 at 23:28
Oops, thanks for the pointer, Harry. I usually live in "associative-land," so I immediately (and erroneously) assumed the question also lived there. – Manny Reyes Apr 19 '10 at 23:40
It should also be clear that I'm living in "unital-land." – Manny Reyes Apr 19 '10 at 23:41

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

-
The problem with that definition is that it is not a generalization of an algebra over a commutative ring because here R is mapped to Z(A). – user717 Apr 19 '10 at 22:07
Well, there are very interesting examples of things that people do want to be $R$-algebras in which $R$ non-central... – Mariano Suárez-Alvarez Apr 19 '10 at 22:18
But then I would call $A$ an "$R$-Ring" (I'm thinking of the comma category Rng/R) rather than an "$R$-algebra". But, again, I have no idea... – user717 Apr 19 '10 at 22:22
@Arminius: "... here $R$ is mapped to $Z(A)$." Why? Who decided? That does not feel natural to me at all. – captain obvious Apr 20 '10 at 0:10
@captain obvious: Well, the one who defined what an algebra over a commutative ring is decided this! Take a look at your wikipedia link: ...is a morphism $f:R \rightarrow A$ such that the image of $f$ is contained in the centre of $A$. But I think this makes sense. – user717 Apr 20 '10 at 7:03

This is a follow-up to the answer by Zoran Skoda, applicable to the case where one requires "algebras" to be associative and have a unit.

I'd like to point out that a monoid $(A,\mu,\eta)$ internal to the monoidal category of $R$-bimodules is exactly the same thing as a ring $A$ along with a ring homomorphism $\eta:R\rightarrow A$. And a morphism $(A,\mu,\eta)\rightarrow (A',\mu',\eta')$ is exactly the same thing as a ring morphism $A\rightarrow A'$ that makes the triangle with $R$ commute.

In other words, this viewpoint can be greatly simplified if we just view an "algebra over $R$" as an object $\eta:R\rightarrow A$ in the category of rings under $R$. One can require that $\eta:R\rightarrow A$ have central image, or not, depending on one's needs.

-
How does this apply to a standard ring homomorphism $\mathbb{C} \to \mathbb{H}$ to the quaternions? That is to say: how are you regarding $\mathbb{H}$ as a monoid in $\mathbb{C}$-bimodules? – Todd Trimble Oct 15 '15 at 4:35
What goes wrong if you do it in the obvious way? $\mathbb{H}$ is a $\mathbb{C}$-$\mathbb{C}$-bimodule via the standard ring homomorphism $\mathbb{C}\rightarrow\mathbb{H}$. The multiplication map on it is left-$\mathbb{C}$-linear and right-$\mathbb{C}$-linear and middle-$\mathbb{C}$-linear. Perhaps I'm missing something; I'll give this another look tomorrow. – ebrahim Oct 15 '15 at 7:13
It still looks fine to me. It seems perhaps the worry is that $hch'\neq chh'$ for certain complex numbers $c$ and quaternions $h,h'$, so that the multiplication fails to be $\mathbb{C}$-bilinear on the left. But that doesn't seem to me to be a requirement for a multiplication $\mathbb{H}\otimes_\mathbb{C}\mathbb{H}\rightarrow \mathbb{H}$. – ebrahim Oct 15 '15 at 20:09
My sincere apologies; it was my confusion. – Todd Trimble Oct 15 '15 at 20:50

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

-