There are two different definition of algebra:
The first one is given as follows :
Let $R$ be a commutative ring,$R-$algebra is a ring $A$ with ring homomorphism $f:R\to A$ (that mapping identity to identity), such that $f(R)$ is contained in the center of $A$.
Then we can make the $R$-algebra into $R$-module by defining scalar product $r\cdot a = f(r)a$.
If $R$ is a field then the above definition makes it a vector space.
There is another definition for $K$-algebra ($K$ is a field):
Let $K$ be a field, and let $A$ be a vector space over $K$ equipped with an additional binary operation from $A \times A$ to $A,$ denoted here by $\cdot$ (i.e. if $\mathbf{x}$ and $\mathbf{y}$ are any two elements of $A, \mathbf{x} \cdot \mathbf{y}$ is the product of $\mathbf{x}$ and $\mathbf{y})$. Then $A$ is an algebra over $K$ if the following identities hold for all elements $\mathbf{x}, \mathbf{y}, \mathbf{z} \in A,$ and all elements (often called scalars) $a$ and $b$ of $K$ :
- Right distributivity: $(\mathbf{x}+\mathbf{y}) \cdot \mathbf{z}=\mathbf{x} \cdot \mathbf{z}+\mathbf{y} \cdot \mathbf{z}$
- Left distributivity: $\mathbf{z} \cdot(\mathbf{x}+\mathbf{y})=\mathbf{z} \cdot \mathbf{x}+\mathbf{z} \cdot \mathbf{y}$
- Compatibility with scalars: $(a \mathbf{x}) \cdot(b \mathbf{y})=(a b)(\mathbf{x} \cdot \mathbf{y})$.
which is given in wiki here
Are these two definition conside?
I can show the first definition satisfy the second one,the bilinear map defined in the second one with three axiom ,such that first two is given by Ring structure $A$,and the third one is given by $K=R$ lies in the center of $A$.
But it seems the second definition even need not to make $A$ ring ,due to associative and identity is not given
Best Answer
The second one is more general, as it allows one to go on to define Lie algebras, Jordan algebras, Poisson algebras and other non-associative algebras in addition to associative algebras.
The first one determines only associative algebras. It defines a subclass of the class defined by the second definition.
(Overlooked the field/ring difference too. You can extend the second to be over commutative rings as well, which is what I was thinking of while writing above. A great deal of work on algebras focuses on the scalar ring being a field, but using more general rings has also been extremely useful.)