# What is an $R$-algebra?

In the following, assume that rings are rings with unity.

Here is the definition of $$R$$-algebra from Wikipedia:

Let $$R$$ be a commutative ring and $$(M,+,\cdot)$$ an $$R$$-module.

Assume $$\ast$$ is a binary operation on $$M$$, such that:

1. $$x\ast (y+z)= x\ast y + x\ast z$$
2. $$\forall x,y,z \in M.(x+y)\ast z = x\ast z + y\ast z$$
3. $$\forall r,s \in R,x,y \in M.(rx)\ast (sy)=(rs)(x\ast y) (r,s\in R, x,y\in M)$$

then $$(M,+,\cdot,\ast)$$ is called an $$R$$-algebra.

This definition has the same form as my definition for "algebra over a field". Note that this definition does not require $$\ast$$ to be associative.

Here is an equivalent definition given in Dummit&Foote (original form is given using a ring homomorphism):

Let $$(M,+,\cdot)$$ be a ring and $$R$$ a commutive ring.

If $$M$$ is an $$R$$-module and the multiplication on $$M$$ is bilinear, then $$M$$ is called "an $$R$$-algebra".

Note that this definition requires $$\ast$$ to be associative.

What is the usual definition for $$R$$-algebra?

References

Abstract Algebra, Dummit and Foote, 3rd edition, page 355.

• I think this depends on context, really. In some places, an $R$-algebra is usually associative. To avoid confusion, you can always talk about "associative $R$-algebra", which actually happens quite often.
– HSN
Oct 12 '14 at 16:28

It only gets more ambiguous as you go along. Somewhere down the rabbit hole you learn that "algebras" might also mean general systems like monoid, loops, and quasi-groups. The point is that calling something "an algebra" is a context dependent definition. Authors should be responsible to explain what assumption they intend for their meaning of "an algebra".

Even so there are some top-level popular choices.

• When in doubt, default to associative. This is going to upset those who think of Lie algebras the most, but it is clear that to get to Lie theory you first would have been around associative algebras. So on that measure I think the correct default is to say products are associative and if they are not then you should, as an author, point that out very early on. For example, the AMS subject classifications list "nonassociative algebra" to emphasize that the algebras are possibly not associative. (Not non-associative means "not required to be associative" so it is larger class, not a complementary class of algebra).
• When dealing with modules. If you have module, especially if everything is a vector space, you have some ring $K$ of scalars for you modules. So often we want everything in sight to remain linear, or bilinear, or multilinear, with respect to $K$. In the context it is most common to encounter $K$-algebra as a $K$-module $A$ equipped with a distributive product $*:A\times A\to A$ that is $K$-bilinear. That way the natural things you do, say left multiply $L_a(x)=ax$, right multiply $R_a(x)=xa$, remain $K$-linear. Now as for the associative assumption that depends on the modules you are studying. Example, if you modules are associative modules then sure, assume associative $*$. But if you have Lie modules then you would not want $*$ to be associative.

• When dealing with finite or combinatorial forms of algebra. Here algebra probably means something more like generic names for products of any sort, say monoids, loops, etc. The name comes from some correspondences that arrose early on which showed combinatorial methods explained the working of the algebra (e.g. Magnus' treatment of free groups he called "Combinatorial Group Theory")

Some observations to make about the $K$ in a $K$-algebra (both associative and non-associative).

• Every distributive product is $\mathbb{Z}$-bilinear and so every ring is a $\mathbb{Z}$-algebra.

• Every bilinear product has a unique maximal ring $K$ of scalars over which it is bilinear (unique upto natural isomorphism). This is known as the centroid and can be computed by solving a system of linear equations. So long as your product has no degeneracy ($A*A=A$ and $a*A=0$ implies $a=0$, e.g. if $A$ has a $1$ or is simple) then the centroid is commutative. So it is common to assume $K$ is commutative to start with.

With those two points it means that insisting that something is a $K$-algebra instead of ring is mostly for convenience. You simply cannot make a distributive product that is not an algebra. You are just deciding to make that point evidently clear.

The most general definition of an algebra is as follows. Fixing a commutative ring $$(A, +, \cdot)$$ then by $$A$$-algebra one means an ordered system $$(E, +, \cdot, \cdot)$$ where:

1. The subjacent structure $$(E, +, \cdot)$$ given by only the first three components of the above system is an $$A$$-module, in other words $$+$$ is a binary operation on $$E$$ rendering it into an abelian group and $$\cdot: A \times E \to E$$ is an external operation of $$A$$ on $$E$$ by endomorphisms of $$(E, +)$$, obeying the axioms defining the algebraic structure known as module.
2. The final rightmost component $$\cdot: E \times E \to E$$ is an internal binary operation required to be bilinear with respect to $$A$$. This explicitly means that: $$(\forall x, y, z)(x, y, z \in E \Rightarrow x(y+z)=xy+xz \wedge (x+y)z=xz+yz)$$

in other words distributivity of the internal multiplication with respect to addition and also that: $$(\forall \lambda, x, y)(\lambda \in A \wedge x, y \in E \Rightarrow (\lambda x)y=x(\lambda y)=\lambda(xy))$$

which is the condition of homogeneity of the external multiplication with respect to the internal one.

Note that in general, apart from satisfying the axioms stated above, the internal multiplication on $$E$$ can be as pathological as they could get, not being necessarily associative or not necessarily possessing an identity element.

It is true though that such a structure can be rather arid and in order to allow for further theoretical developments one will need to impose further conditions on the algebras being studied (as Algeboy points out). However, depending on the context of study, structures which are not necessarily associative/commutative/unital etc can very naturally arise, so it does make sense to have a very broad acceptation as a starting point.