This is unpacking Definition 2.9

By Proposition 2.5 and Proposition 2.8, open immersions are locally of finite type and composition of morphisms locally of finite type is again locally of finite type, so

is a morphism locally of finite type as well.

Consider the only open affine cover \(\{ \operatorname {Spec}\mathbb {C}\} \) of \(\operatorname {Spec}\mathbb {C}\), its pullback cover is \(\{ X_{\mathbb {C}} := X \times _{\operatorname {Spec}\mathbb {C}} \operatorname {Spec}\mathbb {C}\} \) where \(X_{\mathbb {C}}\) can be covered by \(U_{i}\times _{\operatorname {Spec}\mathbb {C}}\operatorname {Spec}\mathbb {C}\) where \(U_{i}\) runs over the collection of all affine open sets. Then the conclusion follows from Proposition 2.7.

Since \(U\) is open, its complement \(U^{\complement }\) is of the form \(V(I)\) for some ideal \(I\). Since \(A\) is a finite \(\mathbb {C}\)-algebra, \(I\) is the span of \(\{ f_{1},\dots , f_{n}\} \) for some \(f_{i} \in A\). Thus \(V(I)\) is \(\bigcap _{i}V(f_{i}\cdot R)\) hence \(U\) is \(\bigcup _{i}D(f_{i})\).

Since \(\left(X, \mathcal{O}_{X}\right)\) is of finite type, it has a   finite affine covering \(\mathcal{U}_{i} = \{ U_{1},\dots , U_{n}\} \). It is sufficient to show that every open cover of \(U_{i} \cap V\) has a finite subcover ¹ . In another word, we only need to show if \(\left(X, \mathcal{O}_{X}\right) \cong \left(\operatorname *{Spec}{A}, \widetilde{A}\right)\) is affine and \(V\) is an open subset of \(X\), then \(V\) is quasicompact. Since \(V\) is a finite  union of \(D(f_{1}),\dots , D(f_{n})\) for some \(f_{i}\)’s in \(A\), we only need to show that \(D(f)\) is quasicompact. Since \(D(f)\) is affine, it is quasicompact.

We have a ring homomorphism \(\phi \), just need to check it commutes with algebra map.

\(\operatorname {Spec}\phi \) is injective since \(\phi \) is surjective. We need to check that every open set in \(\operatorname {Spec}S\) is an inverse image of some open set in \(\operatorname {Spec}R\). Since \(\operatorname {Spec}{S}\) has a basis of basic open sets \(D(f)\)’s where \(f \in S\), we only need to check \(D(f)\) is the inverse image of some open set in \(\operatorname {Spec}R\). Since \(\theta \) is surjective, we know that \(\theta (r) = f\) for some \(r \in R\). Then \(D(f) \subseteq \operatorname {Spec}S\) is the inverse image of \(D(r) \subseteq \operatorname {Spec}R\).

From corollary 1.3, we know that for each closed point \(\mathfrak m\), i.e. a maximal ideal, there is a unique \(\phi _{\mathfrak m} : A \to \mathbb {C}\) whose kernel is \(\mathfrak m\). Conversely, for any \(\phi : A \to \mathbb {C}\), \(\ker \phi \) is certainly a prime ideal ² . Since \(\phi \) is surjective ³ , its kernel is maximal.

It remains to show that \(\mathfrak m \mapsto \phi _{\mathfrak {m}}\) and \(\phi \mapsto \ker \phi \) are inverse to each other. But this follows from the uniqueness from corollary 1.3: Let \(\mathfrak {m}\) be a maximal ideal, then the \(\ker \phi _{\mathfrak {m}}\) is exactly \(\mathfrak m\) by definition of \(\phi _{\mathfrak m}\); on the other hand, if \(\phi \) is an algebra homomorphism then \(\phi \) and \(\phi _{\ker \phi }\) are both algebra homomorphism that has kernel \(\ker \phi \), hence must be equal.

Let \(x \in X\) be a closed point and an affine open neighbourhood of \(x \in U \cong \left(\operatorname *{Spec}{A}, \widetilde{A}\right)\) where \(A\) is a finite \(\mathbb {C}\)-algebra. Thus the \(x\) corresponds to a morphism \(\Phi _{A}\) between \(\left(\operatorname *{Spec}{\mathbb {C}}, \widetilde{\mathbb {C}}\right)\) and \(\left(\operatorname *{Spec}{A}, \widetilde{A}\right)\) by corollary 2.22; we define \(\Psi _{x}\) to be the composition of

Moreover, \(\Psi _{x}\) does not dependent on the choice of affine neighbourhood \(\operatorname {Spec}A\): suppose \(x \in \operatorname {Spec}A \cap \operatorname {Spec}B\), then \(\operatorname {Spec}A \cap \operatorname {Spec}B\) admits an open covering of spectra of finitely generated \(\mathbb {C}\)-algebras by Proposition 2.13. Thus we can find a finitely generated \(\mathbb {C}\)-algebra \(C\) such that \(\operatorname {Spec}C \subseteq \operatorname {Spec}A \cap \operatorname {Spec}B\).

where \( (\_ : \operatorname {Spec}C \hookrightarrow \operatorname {Spec}A) \circ \Phi _{C}\) is exactly \(\Phi _{A}\) and \((\_ {: \operatorname {Spec}C \hookrightarrow \operatorname {Spec}B}) \circ \Phi _{C}\) is exactly \(\Phi _{B}\) by Corollary 2.22; thus both composition in the commutative square above is \(\Psi _{x}\), in another word, \(\Psi _{x}\) is independent from the choice of affine neighbourhood.

On the otherhand, if we are given a morphism \(\Psi : \left(\operatorname *{Spec}{\mathbb {C}}, \widetilde{\mathbb {C}}\right)\to \left(\operatorname *{Spec}{A}, \widetilde{A}\right)\), let us denote \(x\) to be the image of the unique point in \(\operatorname {Spec}\mathbb {C}\) under \(\Psi \); we want to show that \(x\) is a closed point. Since affine open set forms a basis, we only need to check that, for any affine open \(\left(\operatorname *{Spec}{A}, \widetilde{A}\right) \hookrightarrow \left(X, \mathcal{O}_{X}\right)\), \(x\) is closed in \(\operatorname {Spec}A\). We consider the factorisation of \(\Psi \):

where \(\psi \) is a \(\mathbb {C}\)-algebra homomorphism \(A \to \mathbb {C}\) such that \(\operatorname {Spec}{\psi } = \Psi |_{\operatorname {Spec}A}\), hence by Corollary 2.22, we have \(x\) is closed in \(\left(\operatorname *{Spec}{A}, \widetilde{A}\right)\). The two construction above is bijection is verified as the following:

1. Let \(x\) be a closed point, then it corresponds to \(\Psi _{x}\), but the image of the unique point in \(\operatorname {Spec}\mathbb {C}\) under \(\Psi _{x}\) is \(x\);

2. if \(\Phi \) is a morphism \(\left(\operatorname *{Spec}{\mathbb {C}}, \widetilde{\mathbb {C}}\right)\to X\) and denote the unique image as \(x\), \(\Phi \) factors through affine open neighbourhood of \(x\) hence it is \(\Psi _{x}\) because \(\Psi _{x}\) does not dependent on the choice of affine neightbourhood.

Let \(x\) be a closed point in \(X\), then \(x\) corresponds to a unique \(\Psi _{x} = (\psi _{x}, \psi _{x}^{*}) : \left(\operatorname *{Spec}{\mathbb {C}}, \widetilde{\mathbb {C}}\right) \to X\) such that \(\psi _{x}(\star ) = x\) where \(\star \) is the unique point of \(\operatorname {Spec}\mathbb {C}\). The composite \(\Phi \circ \Psi _{x}\) is a morphism \(\left(\operatorname *{Spec}{\mathbb {C}}, \widetilde{\mathbb {C}}\right)\to Y\) thus \(\Phi \circ \Psi _{x}(\star )\) is closed in \(Y\), since \(\Phi \circ \Psi _{x}(\star ) = \Phi (\Psi _{x}(\star )) = \Phi (x)\), we conclude that \(\Phi (x)\) is a closed point in \(Y\).

Write \(\Phi = (\phi , \phi ^{*})\), note that \(\phi \) is necessarily an embedding \(X \hookrightarrow Y\) thus \(\operatorname {Max}\left\{ \phi \right\} \) being the restriction of \(\phi \) must be embedding as well:

where the vertical arrows and \(\phi \) are embeddings so \(\operatorname {Max}\left\{ \phi \right\} \) is embedding as well.

Let \(\mathfrak {m} \subseteq R\) be a maximal ideal inside the image of \(\operatorname {MaxSpec}\theta \), i.e. there exists a maximal ideal \(\mathfrak {p} \subseteq S\) such that \(\theta ^{-1}\mathfrak {p}=\mathfrak {m}\). \(\mathfrak {m}\) corresponds to the unique algebra homomorphism \(\phi _{\mathfrak {m}} : R \to \mathbb {C}\) whose kernel is \(\mathfrak {m}\) and \(\mathfrak {p}\) corresponds to the unique algebra homorphism \(\psi _{\mathfrak {p}} : S \to \mathbb {C}\) whose kernel is \(\mathfrak {p}\). Thus \(\theta ^{-1}\mathfrak {p} = \mathfrak {m}\) precesily when \(\psi _{\mathfrak {p}} \circ \theta = \phi _{\mathfrak {m}}\); and this happens precisely when \(\psi _{\mathfrak {p}}\) annaliates the kernel of \(\theta \).