By Hermann Stahl

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Pn }. By our inductive hypothesis, there is an ideal b and a minimal decomposition b = q1 ∩ q2 ∩ · · · ∩ qn−1 , such that each qi is pi -primary. If b ⊆ S)pn (0) and p were any minimal ideal contained in pn , then Spn ⊆ Sp (0), which implies b ⊆ Sp (0). Taking the radicals of both sides yields q1 ∩ q2 ∩ · · · ∩ qn−1 ⊆ p, hence pi ⊆ p, for some 1 ≤ i ≤ n by chapter 1, hence pi = p by the minimality of p. But this is a contradiction, since p may be minimal, but no pi is supposed to be. This implies that b Spn (0) and therefore there is a pn -primary ideal qn such that b qn .

44 CHAPTER 5. INTEGRAL DEPENDENCE AND VALUATIONS (ii) ⇒ (iii) This direction is obvious, since the single point-set must meet the set of all maximal ideals, which are of course the closed points in Spec(A). (iii) ⇒ (i) Let p be any prime ideal which is not maximal; this will correspond to a point in Spec(A), which, as a set, is not closed. The intersection of all maximal ideals of A that strictly contain p also contains p. If the converse were not true, then there would be a non-empty set F ⊆ A such that F is contained in any maximal ideal m of A which contains p, but F is not contained in p.

If (a1 , a2 , . . , an ) ∈ Vm , then the previous relation implies that m ⊆ t1 − a1 , t2 − a2 , . . , tn − an ), which is absurd, lest m = (t1 − a1 , t2 − a2 , . . , tn − an ). This completes the proof. 18 Let x1 , x2 , . . , xn generate B as a k-algebra. We will proceed by induction on n. If n = 1, then the result is obvious since B is a field. Therefore assume n > 1 and that the result holds for n − 1 generators. Let A = k[x1 ] and let K = k(x1 ) be the field of fractions of A. Then, by the inductive hypothesis, B is a finite algebraic extension of K, therefore, the generators x − 2, x3 , .

### Abriss einer Theorie der algebraischen Funktionen einer Veraenderlichen by Hermann Stahl

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