Zariski Topology on Prime Spectra — Examples

The displayed identity is a unit expression of the form required by [L1], with $f_1=2$ and $f_2=3$.

Applying [L1] gives $\mathrm{Spec}(\mathbb{Z})=D(2)\cup D(3)$. Concretely, every prime ideal of $\mathbb{Z}$ omits at least one of $2$ and $3$, because no prime ideal contains their linear combination $1$.

Thus one explicit unit-ideal expression produces a finite distinguished-open cover.

By [A1], the points of $\mathrm{Spec}(\mathbb{Z})$ are $(0)$ and the prime ideals $(p)$. Since $(0)\subseteq (p)$ for every prime number $p$, fact [L1] shows that every $(p)$ is a specialization of $(0)$. Thus $(0)$ is the unique generic point.

Fact [A1] says that the maximal ideals of $\mathbb{Z}$ are exactly the ideals $(p)$, so [L2] shows that the closed points of the spectrum are exactly the points $(p)$.

For a nonzero integer $n$, fact [L3] says that $D(n)$ consists of the prime ideals that do not contain $n$. The point $(0)$ never contains a nonzero integer, and $(p)$ contains $n$ exactly when $p$ divides $n$. Hence $D(n)=\{(0)\}\cup \{(p):p\nmid n\}$.

Therefore $\mathrm{Spec}(\mathbb{Z})$ has one generic point, closed points $(p)$, and the distinguished opens described above.

By [L1], $(2)$ is a specialization of $(0)$. Equivalently, [L2] records the containment $(0)\subseteq (2)$.

By [A1], every open neighborhood of $(2)$ contains $(0)$. Hence no open neighborhood of $(0)$ can be disjoint from an open neighborhood of $(2)$. So the two distinct points cannot be separated by disjoint open sets.

Therefore $\mathrm{Spec}(\mathbb{Z})$ is not Hausdorff.

Fact [A1] gives the full point set $\mathrm{Spec}(R)=\{(0),(x)\}$. Since $(0)\subseteq (x)$, fact [L1] shows that $(x)$ is a specialization of $(0)$, so $(0)$ is the generic point.

By [A1], the ideal $(x)$ is maximal. Therefore [L2] makes $(x)$ the unique closed point of the spectrum.

Thus the specialization poset of $\mathrm{Spec}(k[x{]}_{(x)})$ is the two-point chain $(0)⇝(x)$.

The element $e=(1,0)$ satisfies $e^2=e$, so [L1] gives a clopen partition $\mathrm{Spec}(R)=D(e)\bigsqcup D(1-e)$.

By [A1], the only prime ideals are $(0)\times k$ and $k\times (0)$. The ideal $k\times (0)$ contains $e$, while $(0)\times k$ does not; hence $D(e)=\{(0)\times k\}$. Similarly $D(1-e)=\{k\times (0)\}$.

Therefore the coordinate idempotents split the product spectrum into two clopen singleton pieces.

By [L2] and [L3], ${\mathrm{Supp}}_{\mathbb{Z}}(M)={\bigcup }_{p\text{ prime}}{\mathrm{Supp}}_{\mathbb{Z}}(\mathbb{Z}\mathrm{/}p\mathbb{Z})={\bigcup }_{p\text{ prime}}\{(p)\}=\{(p):p\text{ prime}\}$.

This support is closed under specialisation by [L1]: each point $(p)$ is already closed, so the only specialisation of $(p)$ is itself.

The set from step 1.1 is not Zariski-closed. Indeed, if it were closed, its complement would be an open neighborhood of the missing point $(0)$. By [L5], that neighbourhood contains some distinguished open $D(n)$ with $(0)\in D(n)$. Since $(0)\in D(n)$, the integer $n$ is nonzero by [L4]. Now [L4] says that $D(n)$ also contains every closed point $(q)$ with $q\nmid n$. Choosing such a prime $q$, we get $(q)\in D(n)$ and $(q)$ lies in the support from step 1.1, contradicting disjointness.

Therefore $M$ has specialization-closed support that is not Zariski-closed.