I'm following notes which state that the data of an $A$-Algebra $B$ is the same as a ring homomorphism $A\to B$. I'm wondering if this is the case, since what were missing is how the $A \times B\to B$ relation works, which after given the ring homomorphism $A\to B$ can be interpreted as:
for $a\in A$ and $b\in B$, $ab$ is given $F(a)b$ where $F$ is the above ring homomorphism?
If you take all the algebra axioms into account you see that $F$ has to map elements of $A$ into the center of $B$. The action $A\times B \to B$ of $A$ on $B$ can then be obtained from the multilpication $\mu$ on $B$ via $$ A \times B \overset{F \times id_B}\longrightarrow B \times B \overset{\mu} \longrightarrow B.$$ Or in formulas: $a \cdot b = F(a)b$, where $\cdot$ denotes the action of $A$ on $B$ and concatenation denotes the product of $B$.