[JadeNB]

> More generally an algebra A, over a ring R, an R-algebra, is a ring A equipped with a map Hom(A,Z(R)).

I don't think that's the usual definition of an algebra. For example, it would mean that there is no difference between an algebra over a non-commutative ring and over its centre, which seems weird; and it clashes with the usual habit to regard every non-0 commutative ring as a non-trivial ℤ-module, whereas, for example, the only homomorphism ℤ/2ℤ → ℤ is the trivial one.

I would expect rather the datum of an R-algebra structure on a ring A to be a ring homomorphism R → End_{gp}(A). EDIT: Now that I think of it, maybe got your A and R mixed up and meant the more restrictive definition, whereby the ring homomorphism I mention is supposed to factor through R → Z(A) → End_{gp}(A)? I'd call this more restricted notion, at least over a unital ring R, a unital algebra A (but often people want implicitly to assume unital-ness).

[joppy]

I think that usually when people say “algebra over a ring” they assume that ring to be commutative, so that the word “bilinear” in “bilinear multiplication” is useful. It’s possible to define an algebra over a non-commutative ring as a bimodule (rather than left module or right module) equipped with a bilinear multiplication, but I have rarely seen this used.

The definition the parent poster used (or intended to use, but wrote the wrong way around, I believe) was that an algebra over a non-commutative ring is just an algebra over its commutative centre. (In which case, we’re still really just talking about algebras over commutative rings).

[JadeNB]

But the definition doesn't work even for commutative rings; as I mention, it says that the only ℤ-module structure on ℤ/2ℤ is the trivial one, which is not the usual understanding of the term. I agree that, if you switch A and R in Hom(A, Z(R)), then an element of the Hom space Hom_{ring}(R, Z(A)) makes A into an R-algebra, but I would argue it's not the only way; there's a map Hom_{ring}(R, Z(A)) -> Hom_{ring}(R, End_{gp}(A)), but it need not be surjective if the rings aren't assumed unital. Consider, for example, a polynomial ring R = k[t] and its ideal A = tR, which has a natural structure of an R-algebra.

[hypersoar]

At long last, my disastrous and scarring grad-school experience in representation theory can save the day!

First, regarding the OP: Having spent many years studying algebra, I don't find the hierarchy of axioms to be very useful in thinking about these things. Sure, you can think of a field as a "commutative ring with inverses", but rings and fields present themselves so differently that this connection doesn't end up being all that useful. Fields are not rich enough on their own to support much interest. You'll find them mostly as building blocks rather than powerful tools in and of themselves. Ditto for modules and vector spaces. Sure, a module is "like a vector space but over a ring", but vector spaces are so boring by themselves that they show up mostly as scaffolding. The study of modules, on the other hand, is its own branch of mathematics. It's much more useful to think of them in terms of what you actually do with them.

Now, on to definitions. The following few paragraphs are all very small-minded and look far more complicated than they actually are. It all encodes pretty much what you'd expect.

If you want to define algebras over commutative rings, you need to start with left- and right-algebras. A left-algebra is an abelian group A equipped with a map \phi: R -> End(A). The abelian group structure defines the addition in the algebra, and the map defines the left-multiplication: if r \in R, and a \in A, then you define a times r as \phi(r)(a), where \phi(r) is an endomorphism on A.

A right-algebra is the same, only the map is from R to the opposite ring of End(A), where the opposite ring is the one you get by just reversing the multiplication. You need to do this because associativity demands that you compute ((a)r)s, where a \in A, r,s \in R, by first acting on a with r, then by s. But with the usual conventions of composition of functions, \phi(r) \circ \phi(s) means you first "do" s, then r. So you need to flip it. Working with left- and right-algebras is a pain in the butt because you have to carry around a ton of left-right nonsense.

A bialgebra (in the literature I read) is a an abelian group that is both a left- and right-algebra. A central bialgebra is one where the left and right multiplication are the same, which is not a given. Noncentral bialgebras are especially annoying, mostly because you have to figure out how to do pre-subscripts in LaTeX so you can write nonsense like "_R M_S".

Obviously, all of these things collapse if R is commutative. Noncommutative ring theory requires a special kind of patience. And don't even get me started on noncommutative geometry.

[JadeNB]

You are speaking with a representation theorist, too. It seems to me that we agree on the definition. (I agree that I was writing as if `R` were automatically commutative, after having made a big fuss about the possibility that it wasn't. Incidentally, if you feel insufficiently scarred, you might like to expand your stable of algebras: there is the notion of a coalgebra, which is dual to that of an algebra; and I believe that the usual notion of bialgebra is of a ring equipped with the structure both of an algebra and a coalgebra: https://en.wikipedia.org/wiki/Bialgebra ; but maybe it's different in the world of algebras over non-commutative rings, which is not my speciality. Then among the bialgebras are the Hopf algebras, etc.)