Let A be a ring and S be a multiplicative subset of A

1. Localization and Spectrum All rings are assumed to be commutative with identity.

Proposition 1.1. Let A be a ring and S be a multiplicative subset of A. The map A → S⁻¹A induces a homeomorphism

Spec(S⁻¹R) → {p ∈ Spec A : S ∩ p = ∅}

where topology of the right hand side is the subspace topology of Spec A. The inverse map is given by p → S⁻¹p.

Proof. Let ϕ : A → S⁻¹A be the map sending a to a/1. Then we have a continuous map Spec ϕ : Spec(S⁻¹A) → Spec A. For simplicity, denote Spec ϕ by h. Let p⁰ be a prime ideal in S⁻¹A. Then ϕ⁻¹p⁰ is a prime ideal in A so that ϕ⁻¹(p⁰) ∩ S = ∅. If not, there exists f ∈ ϕ⁻¹(p⁰) ∩ S. Then f ∈ S and f /1 ∈ p⁰. Since f ∈ S, 1/f ∈ S⁻¹A. This implies that 1/1 ∈ p⁰, i.e. S⁻¹A = p⁰ which is absurd because p⁰ is a prime ideal. Hence Im h ⊂ {p ∈ Spec A : S ∩ p = ∅}. Conversely if p ∈ {p ∈ Spec A : S ∩ p = ∅}, then ϕ(p) = S⁻¹p is a prime ideal in S⁻¹A. This is because the localization of an integral domain is an integral domain and hence S⁻¹A/S⁻¹p ∼= S⁻¹(A/p) is an integral domain.

Moreover, p = ϕ⁻¹(S⁻¹p). Therefore p ∈ Im h. We find Im h = {p ∈ Spec A : S ∩ p = ∅}.

Let h⁰ : Im h → Spec(S⁻¹R) by p → S⁻¹p. For p ∈ Im h, h ◦ h⁰(p) = h(S⁻¹p) = ϕ⁻¹(S⁻¹p) = p and for any p⁰, h⁰ ◦ h(p⁰) = h⁰(ϕ⁻¹p⁰) = S⁻¹(ϕ⁻¹p⁰) = p⁰ by definition.

Hence h⁰ is the inverse of h. Now, we only need to show that h is an open mapping.

Let D(t/s) be a standard open subset in Spec(S⁻¹A). Let us show that h(D(t/s)) = D(t) ∩ Im h. Suppose p ∈ D(t) ∩ Im h. Then p ∩ S = ∅ and t 6∈ p. Then t/s 6∈ p⁰ = ϕ(p).¹ This shows that p⁰ ∈ D(t/s). In other words, p = h(p⁰) ⊂ h(D(t/s))) Therefore D(t)∩Im h ⊂ h(D(t/s)). Suppose that p ∈ h(D(t/s)). Then p ∈ Im h and there is p⁰ ∈ D(t/s) so that p= ϕ⁻¹(p⁰). Since p ∈ Im h, p ∩ S = ∅. Since p⁰ ∈ D(t/s), t/s 6∈ p⁰. Now, we want to show p∈ D(t). Suppose not. t ∈ p. Then t/s ∈ p⁰ which leads to the contradiction that t/s 6∈ p⁰. Therefore t 6∈ p and hence p ∈ D(t). We conclude that

h(D(t/s)) = D(t) ∩ Im h.

This shows that h is an open mapping. 

Corollary 1.1. Let A be a ring and f ∈ A. Then we obtain a homeomorphism Spec A_f → D(f ).

Proof. Let ϕ : A → A_f be the localization and h : Spec A_f → Spec A be its induced map. Then Im h = {p ∈ Spec A : p ∩ Sf = ∅}, where Sf = {fⁿ : n ≥ 0}. By definition, Im h = D(f ). Using Proposition 1.1, Spec A_f → D(f ) is a homeomorphism.

 Proposition 1.2. Let A be a ring and I be an ideal. Then the quotient map A → A/I induces a homeomorphism

Spec(A/I) → V (I) ⊂ Spec A.

Proof. The bijection

{prime ideals of A/I} ←→ {prime ideals of A containing I.}

1If t/s ∈ p⁰, then t/s = t⁰/s⁰for t⁰/s⁰∈ S⁻¹p. Hence there is s⁰⁰∈ S so that s⁰⁰(ts⁰− st⁰) = 0. Since t⁰∈ p, s⁰⁰st⁰∈ p and s⁰⁰s⁰t is thus in p. Since S ∩ p = ∅, s⁰s⁰⁰6∈ S. Since p is a prime, we obtain t ∈ p which leads to a contradiction that t 6∈ p.

1

2

implies that the continuous map h : Spec(A/I) → V (I) is a bijection. Here h is the induced map of the quotient map.

Let us prove that that map is an open mapping. Claim h(D(s + I)) = D(s) ∩ V (I),

where s is any representative of s + I. Suppose p ∈ D(s) ∩ V (I). Then s 6∈ p and p contains I. Then s + I 6= I. Because s 6∈ p, s + I 6∈ p/I. Hence p/I ∈ D(s + I), we see that p ∈ h(D(s + I)). We find D(s) ∩ V (I) ⊂ h(D(s)). Conversely, if p ∈ h(D(s + I)), then s + I 6∈ p/I. This implies that s 6∈ p and hence p ∈ D(s). We see that p ∈ D(s) ∩ V (I). We obtain h(D(s + I)) ⊂ D(s) ∩ V (I).