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An algebra over a commutative ring $A$ is finitely generated if it is a quotient ring of a polynomial ring $A[t_1,\dots ,t_n]$ in finitely many variables. Equivalently, finitely many elements generate it under addition, multiplication, and the action of $A$. The induced map of affine schemes $\mathrm{Spec}B\to \mathrm{Spec}A$ is a morphism of finite type exactly when this holds. Finite generation as an algebra does not require finite generation as an $A$-module.

The localization of a ring $R_r$ is generated over $R$ by the single element $1/r$, since every fraction can be written asThe presentation of a localization at one element isTo verify the presentation even when $R$ has zero divisors, observe that $r$ is invertible in $R[t]/(rt-1)$ with inverse $t$. Hence the map from $R$ to this quotient ring extends uniquely to $R_r$. This extension and the displayed map are inverse ring homomorphisms: they agree on $R$ and send the designated inverse of $r$ to the designated inverse of $r$.

Thus $R_r$ is a finitely generated algebra over a ring, and the criterion for a morphism of finite type between affine schemes applies to $u:\mathrm{Spec}{R}_r\to \mathrm{Spec}R$. There is just one affine scheme in the required cover of the source, with one algebra generator. Therefore $u$ is of finite type. If $r$ is nilpotent, the same presentation is the zero ring, and the source is the empty scheme; the conclusion still holds.

The localization of a ring $A$ at the powers of $f$ is generated over $A$ by $1/f$. The presentation $A_f\cong A[t]/(ft-1)$ follows from the universal property of localization: both sides universally receive a ring homomorphism from $A$ making $f$ invertible. This proof works with zero divisors and includes the zero ring when $f$ is nilpotent. In particular, a principal open subscheme of an affine scheme has an inclusion which is a morphism of finite type.