By Papaioannou A.

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Finally, given any element a/s ∈ (S −1 A)G , we define a map φ : (S −1 A)G −→ (S G )−1 AG by letting a/s → Σ(a)/Σ(s), where Σ = σ1 ◦ σ2 ◦ · · · ◦ σn (we assume G = {σi }1≤i≤n ). We see that this map is well defined (Σ(aΣ(s)) being stable under G), and it’s furthermore surjective (any element a/s gets hit my its image) and injective (because of our definition of Σ, we will have Σ(a/s) = Σ(a)/Σ(s), whence injectivity). This completes the proof that φ is an isomorphism. 13 Let p1 , p2 ∈ P and let x ∈ p1 .

The converse follows in the same fashion, since V (p) ⊇ V (q) ⇔ p ⊆ q for any two prime ideals p and q of A. 50 CHAPTER 6. 1 We note that Σ has maximal ideals and this follows by a typical Zorn’s Lemma argument. Given such a maximal ideal a, assume that x, y ∈ / a, but xy ∈ a. Then, b = a + (x) strictly contains a, therefore it must be finitely generated; say b = a0 + (x), where a0 is also finitely generated. Note that a + (x) = a0 + (x) implies a0 ⊆ a. We claim it also implies a = a0 + x(a : x).

If Spec(B) is a Noetherian space, then the converse is also true. Indeed, let V (p) ⊆ Spec(B); then, by the equivalent condition (c) of chapter 5, exercise 10, we have that V (q) ⊆ f ∗ (V (p)), where q is merely the restriction of p in A. Then, we would like to show that the map is also injective, so that the closed set V (p) is mapped to a closed set. If the inclusion was strict, then we would have the infinite strictly descending chain (the pi arise from the going-up property): f ∗−1 (V (q)) ⊇ V (p) ⊇ V (p1 ) ⊇ V (p2 ) ⊇ .