# Thread: Algebra over a field

1. ## Algebra over a field

Hi all, I'm trying to understand the structure of an "algebra over a field".

The definition I get from wolfram is

"Formally, an algebra is a vector space V over a field F with a multiplication."

So if I was to break down this down....

A vector space has two operations
+:V x V --> V (addition)
. :F x V --> V

and so if A was an algebra over F, would it mean it has three operations?
+:A x A --> A (addition)
. :A x V --> A
* : A x A --> A (multiplication)

2. Originally Posted by superbaros
Hi all, I'm trying to understand the structure of an "algebra over a field".

The definition I get from wolfram is

"Formally, an algebra is a vector space V over a field F with a multiplication."

So if I was to break down this down....

A vector space has two operations
+:V x V --> V (addition)
. :F x V --> V

and so if A was an algebra over F, would it mean it has three operations?
+:A x A --> A (addition)
. :A x V --> A
* : A x A --> A (multiplication)
Yes, exactly. The name "Algebra" for this structure, by the way, comes from the fact that the set of functions, in x, say, over a field, form a vector space with ordinary addition and "scalar multiplication" but also have a natural product: f*g(x)= f(x)*g(x). (Here, f(x) and g(x) are members of the field so the multiplication on the right is the field multiplication.)

Originally Posted by HallsofIvy
The name "Algebra" for this structure, by the way, comes from the fact that the set of functions, in x, say, over a field, form a vector space with ordinary addition and "scalar multiplication" but also have a natural product: f*g(x)= f(x)*g(x). (Here, f(x) and g(x) are members of the field so the multiplication on the right is the field multiplication.)
Sorry.... I'm not quite following this...

Further, I'm a bit confused about the second part of the definition that states,

"The multiplication must be distributive and, for every and must satisfy
"

The distributive part I understand, but is the following bit a 'general rule' and not for example, an associative or commutative law?

4. Originally Posted by superbaros

Sorry.... I'm not quite following this...

Further, I'm a bit confused about the second part of the definition that states,

"The multiplication must be distributive and, for every and must satisfy
"

The distributive part I understand, but is the following bit a 'general rule' and not for example, an associative or commutative law?
I'm puzzled by that myself! For example, the set of all polynomials form an algebra over the real numbers, but if $\displaystyle f(x)= x^2$, it does NOT follow that f(xy)= $\displaystyle x^2y^2$= f(x)y= x^2y[/tex] or $\displaystyle xf(y)= xy^2$. Also. you say "$\displaystyle x,y \in V$. What is V? x and y should be in a field, not a vector space. Please check to see if you copied that correctly.

5. The original definition from

Algebra -- from Wolfram MathWorld

gives

Formally, an algebra is a vector space over a field with a multiplication. The multiplication must be distributive and, for every and must satisfy

I'm still thinking about the first bit you just said.

6. Originally Posted by superbaros

Further, I'm a bit confused about the second part of the definition that states,

"The multiplication must be distributive and, for every and must satisfy
"

The distributive part I understand, but is the following bit a 'general rule' and not for example, an associative or commutative law?
this is just saying that two multiplications that we have in V are compatible. a better way to explain this is that we want the multiplication map $\displaystyle V \times V \mapsto V$ to be $\displaystyle F$ bilinear, i.e.

$\displaystyle 1) \ (au+v)w=a(uw) + vw,$

$\displaystyle 2) \ u(av + w)=a(uv) + uw,$

for all $\displaystyle a \in F, \ u,v,w \in V.$ now if you in 1) and 2) put $\displaystyle a=1_F,$ you'll get the distributivity law and if you, in 1), put $\displaystyle v=0$ you'll get $\displaystyle (au)w=a(uw).$ also if you, in 2), put $\displaystyle w=0,$ you'll get

$\displaystyle u(av)=a(uv).$ so the "bilinearlity" condition gives us your condition, i.e. $\displaystyle a(uv)=u(av)=(au)v,$ for all $\displaystyle a \in F$ and $\displaystyle u,v \in V.$ conversely, if we have distributivity and your condition, then

we'll have bilinearity.

but al these don't help you that much to understand what an algebra over a field is. there's an equivalent definition of an algebra over a field, which is easier and gives a better view:

an algebra over a field $\displaystyle F$ is a ring $\displaystyle A$ (with unity) which contains a copy of $\displaystyle F$ in its center. this is very simple, isn't it? so, containing a copy of $\displaystyle F$ in the center means there exists an injective

ring homomorphism from $\displaystyle F$ to $\displaystyle Z(A),$ the center of $\displaystyle A.$ now we actually can identify $\displaystyle F$ with its copy and so we may assume that $\displaystyle F \subseteq Z(A).$ then when multiplying elements of $\displaystyle A$ by an element

$\displaystyle a \in F,$ we can move $\displaystyle a$ anywhere we want because $\displaystyle F$ is in the center of $\displaystyle A.$ this also explains $\displaystyle a(uv)=u(av)=(au)v.$