Affine Schemes and the Structure Sheaf — Examples

The only ideal of the zero ring contains $1=0$, hence is not proper.

Thus $\mathrm{Spec}(0)$ has no points and is empty.

By [F1], its global sections are $0$.

A nonzero prime ideal has a least positive member and division shows it is $(p)$ for a rational prime $p$; conversely [F1] and [F2] make every $(p)$ prime, while $(0)$ is prime because $\mathbb{Z}$ is a domain.

The ideals $(p)$ are maximal and therefore are closed points.

The closure of $(0)$ is $V((0))=\mathrm{Spec}\mathbb{Z}$, so it is generic.

These are exactly the claimed generic and closed points.

Since $(1,0)(0,1)=0$, a prime is uniquely either $\mathfrak{p}\times B$ or $A\times \mathfrak{q}$ for a prime of one factor.

The two families are disjoint open-and-closed sets, and projections give homeomorphisms with the two factor spectra by [F1].

Their localizations are $A_{\mathfrak{p}}$ and $B_{\mathfrak{q}}$, so these homeomorphisms identify the structure sheaves.

Hence the spectrum is the claimed disjoint union of locally ringed spaces.

Localizing $k[t]$ at $t$ adjoins $t^{-1}$, giving $k[t{]}_t\cong k[t,t^{-1}]$.

By [F1], $D(t)\cong \mathrm{Spec}(k[t{]}_t)$.

Combining the two identifications proves the example.

Every element of $D$ has a unique form $a+bϵ$ with $a,b\in k$ and ${ϵ}^2=0$. Such an element is nilpotent exactly when $a=0$, since its image under $D\to D\mathrm{/}(ϵ)\cong k$ must then be nilpotent in a field, and every $bϵ$ is square-zero. Thus $\mathrm{Nil}(D)=(ϵ)$ and $D\mathrm{/}\mathrm{Nil}(D)\cong k$. By [F2], $\mathrm{Spec}D$ is homeomorphic to $\mathrm{Spec}k$, which has one point by [F1].

The first scheme is reduced because a field has no nonzero nilpotents. The second is not reduced: $ϵ$ is nonzero in $D$ because the polynomial $ϵ$ does not belong to $({ϵ}^2)$, but its square is zero.

Scheme isomorphisms preserve affine coordinate rings up to isomorphism and hence reducedness, so these schemes are not isomorphic.

Send a $k$-algebra map $\alpha :k[t]\to R$ to $\alpha (t)\in R$.

Given $r\in R$, [F1] supplies the unique map with $t\mapsto r$.

The assignments are inverse and commute with postcomposition, so the bijection is natural.

If $f,g\in k[t]$ are nonzero, the leading coefficient of $fg$ is the nonzero product of their leading coefficients; hence $k[t]$ is a domain and $(0)$ is a point of ${\mathbb{A}}_k^1$. By [F1], its closure is all of $\mathrm{Spec}k[t]$.

Evaluation at $0$ identifies $k[t]\mathrm{/}(t)$ with the domain $k$, so $(t)$ is prime and strictly contains $(0)$. Thus the closure in step 1.1 is not a singleton, and $(0)$ is not closed. quotient-domain criterion

A relative $k$-valued coordinate representing a point $\mathfrak{p}$ is a [F1, step 2.1, algebra] $k$-algebra map $k[t]\to k$ whose kernel is $\mathfrak{p}$. Such a map fixes $k$, hence is surjective and has maximal kernel. By [F1] its closure is the set of primes containing it, which is the singleton consisting of that maximal ideal. Its kernel is therefore closed, whereas $(0)$ is not, so the generic point $(0)$ has no relative $k$-valued coordinate.