TITLE: What is an $R$-algebra?
QUESTION [12 upvotes]: 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:

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

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.

REPLY [2 votes]: 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:

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.
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.