Past exam of the introduction to schemes course of the University of Oxford / 2017 / 2

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.

Put $C=\overline{\{z\}}$, where the closure is taken in the underlying topological space of $X$. It is nonempty because it contains $z$.

Suppose $C=A\cup B$ for two closed subsets of $C$ in the subspace topology. Since $C$ is a closed subset of $X$, both $A$ and $B$ are also closed subsets of $X$. The point $z$ lies in one of them, say $A$. By the defining minimality of the closure, every closed subset of $X$ containing $z$ contains $C$. Hence $C\subseteq A$, so $A=C$.

Thus $C$ cannot be expressed as the union of two proper closed subsets, proving that $C$ is an irreducible closed subset. This proves closure of a point is irreducible in every topological space; the scheme assumption is unnecessary. Equivalently, $z$ is a generic point of its own closure.

Yes: every scheme is a Kolmogorov space. Choose an affine open subscheme $V=\mathrm{Spec}A$ containing $x$. If $y\notin V$, take $U=V$.

Otherwise, the distinct points $x,y\in V$ correspond to distinct prime ideals ${\mathfrak{p}}_x,{\mathfrak{p}}_y\subseteq A$. There is an element in one of these prime ideals and not the other. After interchanging the names of the two points if necessary, chooseThe principal open subschemecontains $y$ and does not contain $x$. It is an open subset of $V$, hence an open subset of $X$ because $V$ is open in $X$. This establishes $(\ast )$ in every case, proving schemes are Kolmogorov spaces.