Schemes Subschemes and Morphisms Locally of Finite Type — Examples

The ideals $(t)$ and $(t^2)$ in $k[t]$ have the same radical $(t)$, so their quotient-spectrum closed immersions have the same underlying closed subset $V(t)$.

The ring $k[t]\mathrm{/}(t)$ is reduced, whereas the nonzero class of $t$ in $k[t]\mathrm{/}(t^2)$ is nilpotent. Their quotient rings are therefore not isomorphic, so [F1] gives distinct closed subscheme structures with the same support.

Let $X={\coprod }_{m\ge 1}{\mathbf{A}}_k^1$ and map it to $\mathrm{Spec}k$. Each component is an affine finite-type chart, so the map is locally of finite type.

The inverse image of the one-point base is covered by the open components, and no finite subfamily covers $X$. Thus the map is not quasi-compact; by [F1] it is not finite type.

The quotient map $A\to A\mathrm{/}I$ is generated as an $A$-algebra by the empty set, so it is of finite type and its affine scheme morphism is finite type.

The source is the single point corresponding to $I$. Every affine target neighbourhood contains some $D(f)$ with $f\mathrm{\notin }I$. Modulo $I$, such an $f$ is a nonzero scalar. The vector space $I\mathrm{/}{I}^2$ has basis given by the classes of the $x_i$, and localization at $f$ leaves it infinite-dimensional because $f$ acts on it by that nonzero scalar. Hence $I_f\mathrm{/}{I}_f^2\cong (I\mathrm{/}{I}^2{)}_f$ is not finitely generated over $A_f\mathrm{/}{I}_f\cong k$, so $I_f$ itself is not finitely generated. Thus $A_f\to A_f\mathrm{/}{I}_f$ is not finitely presented on any target neighbourhood of the source point. By [F2], the affine morphism is not locally of finite presentation and therefore not of finite presentation.

Let $Z\hookrightarrow X$ be a closed subscheme through which $j$ factors, and let $\mathcal{I}$ be its ideal sheaf. On a nonempty affine open $V=\mathrm{Spec}A$, every $a\in \mathcal{I}(V)$ restricts to zero on the dense open $U\cap V$.

Choose a nonempty principal open $D(f)\subseteq U\cap V$. By [F1], $A$ is a domain and $f\mathrm{\ne }0$; the equality $a\mathrm{/}1=0$ in $A_f$ gives $f^{n}a=0$ for some $n$, hence $a=0$. Thus $\mathcal{I}{\mathrm{\mid }}_V=0$.

The same holds on every nonempty affine open, while the empty opens carry only the zero ideal. Hence $\mathcal{I}=0$, so every closed factorization of $j$ contains $X$ itself. By [F2], the smallest such factorization is $X$.