Example 26.15.2. Consider the rule which associates to every scheme $T$ the set $F(T) = \Gamma (T, \mathcal{O}_ T)$. We can turn this into a contravariant functor by using for a morphism $f : T' \to T$ the pullback map $f^\sharp : \Gamma (T, \mathcal{O}_ T) \to \Gamma (T', \mathcal{O}_{T'})$. Given a ring $R$ and an element $t \in R$ there exists a unique ring homomorphism $\mathbf{Z}[x] \to R$ which maps $x$ to $t$. Thus, using Lemma 26.6.4, we see that

$\mathop{\mathrm{Mor}}\nolimits (T, \mathop{\mathrm{Spec}}(\mathbf{Z}[x])) = \mathop{\mathrm{Hom}}\nolimits (\mathbf{Z}[x], \Gamma (T, \mathcal{O}_ T)) = \Gamma (T, \mathcal{O}_ T).$

This does indeed give an isomorphism $h_{\mathop{\mathrm{Spec}}(\mathbf{Z}[x])} \to F$. What is the “universal family” $\xi$? To get it we have to apply the identifications above to $\text{id}_{\mathop{\mathrm{Spec}}(\mathbf{Z}[x])}$. Clearly under the identifications above this gives that $\xi = x \in \Gamma (\mathop{\mathrm{Spec}}(\mathbf{Z}[x]), \mathcal{O}_{\mathop{\mathrm{Spec}}(\mathbf{Z}[x])}) = \mathbf{Z}[x]$ as expected.

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