I have seen two different definitions from several sources of an algebra over a ring. From Wikipedia:

  • Let R be a commutative ring. An R-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:

Bilinearity:

[$\alpha x + \beta y, z] = \alpha [x, z] + \beta [y, z]$, $\quad [z, \alpha x + \beta y] = \alpha [z, x] + \beta [z, y]$

for all scalars $\alpha, \beta$ in R and all elements x, y, z in A *

I have also frequently seen (this one's from Virginia):

Let R be a commutative ring. An R-algebra is a ring A which is also an R-module such that the multiplication map A $\times$ A $\to$ A is R-bilinear, that is, r $\ast$ (ab) = (r $\ast$ a) b = a $\cdot$ (r $\ast$ b) for any $a, b \in A$ $r \in R$ where $\ast$ denotes the R-action on A.


I'm trying to prove they are equivalent. I am fine with everything apart from proving that, if definition 1 is satisfied, then multiplication in A is associative. Unless this property is what determines whether the algebra is associative or not?

  • Associativity is an extra condition that is sometimes omitted from the definition of algebra. – Zhen Lin Oct 24 '14 at 11:02
  • 1
    @MattBurrows I don't see how Definition(2) implies Definition(1). From $r*(ab)=(r*a)b=a(r*b)$ how do you prove$\quad [ \alpha x + \beta y, z] = \alpha [x, z] + \beta [y, z]$ and $\quad [z, \alpha x + \beta y] = \alpha [z, x] + \beta [z, y]$? – Babai Apr 24 '16 at 22:02
  • 1
    @Babai, I think the op assumes in definition 2 that multiplication distributes over addition. – R_D May 2 '16 at 16:46
  • Another definition is a follows: a ring $A$ is an $R$-algebra, if there exists a homomorphism from $R$ to $A$ which sends the multiplicative identity of $R$ to that of $A$. – User0112358 Jul 26 '16 at 9:01

Associativity may or may not be an axiom, depending on your context. Requiring associativity results in a strictly smaller class of objects.

While associative algebras are common, the study of nonassociative algebras is also full of important topics (Lie theory is a good example.)

A basic example to keep in mind is $\Bbb R^3$ with the cross product. That makes an algebra that isn't associative.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.