= Solution
{c}
The <localization of a ring> $R_r$ is generated over $R$ by the single element $1/r$, since every fraction can be written as
$$
\frac{a}{r^n}=\phi(a)(1/r)^n.
$$
The <presentation of a localization at one element> is
$$
\boxed{R_r\cong R[t]/(rt-1),\qquad t\longmapsto1/r.}
$$
To 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 $u:\operatorname{Spec}R_r\to\operatorname{Spec}R$. There is just one <affine scheme> in the required cover of the source, with one algebra generator. Therefore \b[$u$ 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.
Back to article page
