I' m studying over Brauer group. But I have a problem in starting point. What is the central algebra?
Defination of center of $\mathbb{k}$-algebra S as following ( Noncommutative Algebra, Farb & Dennis, p:86,1991)
$ Z(S) = \{x\in S : x.s=s.x \, \forall s\in S\}$ that is, Z(S) is just the center of S considered as a ring. Note that for an algebra S over $\mathbb{k}$, it is always true that $\mathbb{k} \subset Z(S)$.
But why? By the defination Z(S) must be a subset of S. How we say $\mathbb{k}\subset S$? For this I think defined new map $\alpha:\mathbb{k} \rightarrow S$, $k\rightarrow k.1_S$ (with module operation) If we think this way, we say $\alpha(\mathbb{k})\subset Z(S)$. Is this what definition want to say?
If explained to what is $\mathbb{k} \subset Z(S)$ , it is easy to reach the definition of central algebra.
Yes, there is an abuse of notation going on here and your suggested interpretation is correct.
There is a natural embedding of $\mathbb{k}$ into $S$ whenever $S$ is a $\mathbb{k}$-algebra - namely, via the map $$x\mapsto x 1_S,$$ where $1_S$ is the multiplicative identity in $S$. Generally we conflate $\mathbb{k}$ with its image under this embedding.
So more precisely, a $\mathbb{k}$-algebra $S$ is central iff $$Z(S)=\{x 1_S: x\in\mathbb{k}\}.$$ For example, $\mathbb{C}$ as an $\mathbb{R}$-algebra is not central since it is fully commutative but the relevant inclusion $\mathbb{R}\hookrightarrow\mathbb{C}$ is not surjective. Meanwhile, the quaternions $\mathbb{H}$ do form a central $\mathbb{R}$-algebra since the center is the set of quaternions of the form $a+0i+0j+0k$, which is exactly the image of the canonical embedding of $\mathbb{R}$ into $\mathbb{H}$.