2
$\begingroup$

I am struggling to find a definition of an associative algebra.

Wikipedia (http://en.wikipedia.org/wiki/Associative_algebra) says

Let $R$ be a fixed commutative ring. An associative ''R''-algebra is an additive abelian group $A$ which has the structure of both a ring and an $R$-module in such a way that ring multiplication is $R$-bilinear:

$r\cdot(xy) = (r\cdot x)y = x(r\cdot y)$

for all $r ∈ R$ and $x, y ∈ A$. We say $A$ is unital if it contains an element 1 such that

$1 x = x = x 1$

for all $x ∈ A$. Note that such an element 1 must be unique if it exists at all.

If $A$ itself is commutative (as a ring) then it is called a commutative $R$-algebra.

I have also tried to find something in books as Serge Lang's Algebra and Undergraduate Algebra, but I could not find anything with respect to associative algebras.

Is multiplication of an associative algebra also commutative (because Wikipedia says "Let $R$ be a fixed commutative ring.")?

Thank you very much for your help!

$\endgroup$
  • 1
    $\begingroup$ No, and in general it is not. It is best to think of such a structure as a vecotorspace (and if your more inclined, a module) that just happens to have a notion of multiplication. Then the first axiom just expresses that the actions respect each other. An example you could look at are of-course function algebras, fields over their characteristic, or one of the classics, the quaternions (which are not commutative!). Algebras are actually very nice, for example, Commutative Rings are just $Z$-algebras (you can define algebras independent of rings, so this is actually meaning-ful). Hope this help $\endgroup$ – Pax Kivimae Jul 5 '14 at 13:29
  • $\begingroup$ Oh, also, lay off lang, get Artin, it's far better. $\endgroup$ – Pax Kivimae Jul 5 '14 at 13:30
4
$\begingroup$

We can reexpress the axioms you gave this way:

  1. A is an associative ring
  2. A is an R module for a commutative ring R
  3. The ring product is compatible with the module product

These three things describe an R-algebra A.

R being commutative does not imply A is also commutative. A good first example is a matrix ring over a field F. For n by n matrices, n>2, the matrix ring is noncommutative even though it is an F algebra.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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