Describe \(\Spec \ZZ\), and show that it is a final object for the category of schemes: each scheme \(X\) admits a unique morphism to \(\Spec \ZZ\).
Solution
1. The prime ideals of \(\mathbb Z\) are\[ (0) \qquad\text{and}\qquad (p) \]for prime integers \(p\).
The ring \(\mathbb Z\) is a principal ideal domain. A nonzero prime ideal is therefore generated by an irreducible element, and the irreducible elements of \(\mathbb Z\) are precisely the primes up to sign. Thus every nonzero prime ideal is \((p)\) for a prime \(p\). Since \(\mathbb Z\) is an integral domain, \((0)\) is prime as well.
2. Hence, as a set,\[ \boxed{ \operatorname{Spec}\mathbb Z =\{(0)\}\cup\{(p):p\text{ prime}\}. } \]The point \((0)\) is the generic point, and every \((p)\) is closed.
The closure of a point \(\mathfrak p\) in an affine spectrum is \[ V(\mathfrak p) =\{\mathfrak q:\mathfrak q\supseteq\mathfrak p\}. \] Thus \[ \overline{\{(0)\}} =V(0) =\operatorname{Spec}\mathbb Z, \] so \((0)\) is generic.
For a prime \(p\), \[ V(p)=\{(p)\}, \] because the only prime ideal containing the maximal ideal \((p)\) is itself. Hence \((p)\) is closed.
3. The proper closed subsets of \(\operatorname{Spec}\mathbb Z\) are finite sets of closed points.More precisely, for \(0\ne n\in\mathbb Z\),\[ \boxed{ V(n)=\{(p):p\mid n\}. } \]
A prime ideal \(\mathfrak p\) belongs to \(V(n)\) exactly when \[ n\in\mathfrak p. \] The generic prime \((0)\) contains \(n\) only when \(n=0\). For a closed point \((p)\), \[ n\in(p) \iff p\mid n. \] Only finitely many primes divide a nonzero integer. Conversely, every closed subset is \(V(I)\) for an ideal \(I=(n)\), since every ideal of \(\mathbb Z\) is principal.
4. The local rings are\[ \boxed{ \mathcal O_{\operatorname{Spec}\mathbb Z,(0)}=\mathbb Q, \qquad \mathcal O_{\operatorname{Spec}\mathbb Z,(p)}=\mathbb Z_{(p)}. } \]
For every affine scheme, \[ \mathcal O_{\operatorname{Spec}A,\mathfrak p}=A_{\mathfrak p}. \] At \((0)\), localizing \(\mathbb Z\) at all nonzero integers gives its fraction field \(\mathbb Q\). At \((p)\) one obtains the localization \[ \mathbb Z_{(p)} =\left\{\frac ab:p\nmid b\right\}. \]
5. For every scheme \(X\), morphisms\[ X\longrightarrow\operatorname{Spec}\mathbb Z \]are naturally in bijection with ring homomorphisms\[ \mathbb Z\longrightarrow\Gamma(X,\mathcal O_X). \]
This is Hartshorne II.2.4 applied to the ring \(A=\mathbb Z\): \[ \operatorname{Hom}_{\mathrm{Sch}}(X,\operatorname{Spec}\mathbb Z) \cong \operatorname{Hom}_{\mathrm{Ring}} (\mathbb Z,\Gamma(X,\mathcal O_X)). \]
6. There is exactly one unital ring homomorphism\[ \mathbb Z\longrightarrow R \]for every ring \(R\).
A unital ring homomorphism must send \[ 1\longmapsto1_R. \] Then additivity forces \[ n\longmapsto n\cdot1_R \] for every positive integer \(n\), and similarly for negative integers. Thus the homomorphism is uniquely determined and always exists.
7. Consequently, every scheme has a unique morphism to \(\operatorname{Spec}\mathbb Z\):\[ \boxed{ \operatorname{Hom}_{\mathrm{Sch}}(X,\operatorname{Spec}\mathbb Z) \text{ is a singleton}. } \]Hence \(\operatorname{Spec}\mathbb Z\) is the final object in the category of schemes.
Combine 5 and 6.
8. Q.E.D.
Steps <1>1–<1>4 describe the spectrum, and <1>5–<1>7 prove its terminal property.