Commutative algebra over a ring

Remarks

Example: for $R = \Z$ and $A = \Z/n\Z$, the reduction map $\Z \to \Z/n\Z$ is a ring homomorphism, so $\Z/n\Z$ is a $\Z$-algebra. Example: the polynomial ring $R[x_1, \ldots, x_n]$ with the inclusion of constants is an $R$-algebra. Non-example: the ring $M_2(\R)$ of $2 \times 2$ real matrices receives the ring homomorphism $c \mapsto c I$ from $\R$. But $M_2(\R)$ is not commutative, so it is an associative $\R$-algebra and not a commutative one. Equivalent form: a commutative $R$-algebra is a commutative ring $A$ with an $R$-module structure such that $r(ab) = (ra)b$ for all $r \in R$ and $a, b \in A$; the structure map is then $r \mapsto r \cdot 1$. The notion of a finite type algebra and the tensor product of algebras are built on this definition.