What is $R$-algebra and do I need to understand $R$-modules for it? I was given the following definition of $R$-algebra:

Let $R$ be a commutative ring. An $R$-algebra is a ring $A$ (with $1$) together with a ring homomorphism $f : R \to A$ such that

*

*$f(1_R) = 1_A$;

*$f(R) \subset  Z(A)$, where $Z(A)$ is the center of $A$.

The pair $(A, f)$ will also be called an $R$-algebra.

As far as I can see from googling, it's not the usual way to define it. Almost everyone does it via $R$-module.
I'm fairly new to algebra, in fact, I'm taking a second linear algebra course and the professor digresses quite a bit. So if possible, I'd like to understand what $R$-algebra is without getting too much into other topics.
Is it possible to develop a feeling for $R$-algebra without first developing it for $R$-module? Or would it be rather unnatural? E.g. one can define a vector space by listing tons of axioms but if we define groups first, the definition of a vector space becomes more compact, elegant, and structured, yet one can comfortably work with vector spaces (by keeping $\mathbb R^n$ in mind) without explicitly understanding groups.
So if it's not too harmful to stick to the above definition:

*

*What are your favorite examples of $R$-algebras?

*What role does the ring homomorphism play?

*Why do we need $f(R)$ to be in the centralizer of $A$?

 A: Here are some statements you might find helpful.
If you consider only commutative rings with 1, as is the case for anything related to algebraic geometry, you can see that any ring homomorphism $R\rightarrow S$ makes $S$ into an $R$-algebra. I don't deal with, nor do I know anyone who deals with noncommutative rings, so I won't address the condition with the centralizer, which of course is trivial for commutative rings.
$R$-algebras are a natural definition, and it's good to have a name for the idea. Polynomial rings $A = R[x_1,\ldots,x_n]$ are all $R$-algebras. In this case, the ring homomorphism is just the injection $R\hookrightarrow R[x_1,\ldots,x_n]$. Sometimes you want to make a distinction between the elements of $A$ which are "scalars" from the rest, so it might be useful to think about $A$ as an $R$-algebra, as opposed to just a ring (here, you should think of $R$ as being the scalars). For example, you might only want to consider $R$-algebra homomorphisms between $R$-algebras, as opposed to just ring homomorphisms. ($R$-algebra homomorphisms must be $R$-linear. In the case with polynomial rings over $R$, they must be the identity on $R$).
In the case of polynomial rings over a field $k$, it's sort of "obvious" that you should think of the elements of $k$ inside $k[x_1,\ldots,x_n]$ as the scalars, but sometimes, when looking at say $k[x,y]$, you might want to only consider the "$y$-part", in which case you might want to view it as a $k[x]$-algebra, via the injective homomorphism $k[x]\hookrightarrow k[x][y] = k[x,y]$.
Basically, the ring homomorphism determines how the "ring of scalars" (ie, $R$), acts on the $R$-algebra. Of course it doesn't have to be injective, but in many natural situations it is.
Any finitely presented $R$-algebra has the form $R[x_1,\ldots,x_n]/\langle f_1,\ldots,f_m\rangle$, where the $f_i$'s are in $R[x_1,\ldots,x_n]$.
This probably won't make much sense to you, but to me, $R$-algebras are locally what every morphism looks like in algebraic geometry.
