Definition of R-Algebra

Question

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

I seems the second definition is more general than the first one ? — yi li, Jan 03 '21 at 03:03
The first comes from Dummit & Foote which is the best modern reference for this level of algebra. With associativity and unity (multiplicative identity) then the two would coincide. — Oliver Kayende, Jan 03 '21 at 03:06
Thanks, I also think the first one is much easier to understand — yi li, Jan 03 '21 at 03:08
Someone asked almost the same question 6years ago right down to the references. But they’re asking “what’s the usual definition” and not “are these the same thing” so I guess not a duplicate. — rschwieb, Jan 03 '21 at 03:26
I mean they definitely don't coincide as the second definition requires the base ring to be a field. But if $R$ is a field, they coincide. — pancini, Jan 03 '21 at 03:27
As a fun aside, using the first definition, a $\Bbb Z$-algebra is just the usual definition of a ring. — pancini, Jan 03 '21 at 03:28
@ElliotG the first definition uses the term ring in it, so it would be a circular definition. But I wholeheartedly agree that all rings with identity can be called $\mathbb Z$ algebras, which was the spirit of your comment. — rschwieb, Jan 03 '21 at 03:33

rschwieb · Accepted Answer · 2021-01-03T03:30:25.687

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.)

Definition of R-Algebra

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