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Sample text

Assuming that JA = 0 and given any v ∈ B, we wish to construct a maximal ideal of B that doesn’t contain v. By applying exercise 21 to the ring Bv and its subring A (we use the usual embedding A −→ Bv and this is injective, because A is an integral domain) we obtain an s ∈ A − {0} such that, if Ω is an algebraically closed field and f : A −→ Ω doesn’t vanish at s, then f can be extended to a homomorphism B −→ Ω. Let m be a maximal ideal of A such that s ∈ / A, and let k = A/m be the residue field.

By renumbering the xi ’s, if necessary, we may assume that x1 , x2 , . . , xr are algebraically independent over k and each of the xr+1 , . . , xn are algebraic over k[x1 , x2 , . . , xr ]. Now we apply induction to the difference n − r; if n = r, then there is nothing to be shown, so assume that the proposition holds for n−1 generators and n > r. In this case, the generator xn is algebraic over k[x1 , . . , xn−1 ], hence there exists a polynomial f with coefficients in k such that f (x1 , x2 , .

C’) ⇒ (b’) This is the dual of the ’(c) ⇒ (a)’ statement above; we can thus prove it in exactly the same fashion, reversing the arrows. 11 Since f : A −→ B is flat, the induced map f ∗ : Spec(Bq ) −→ Spec(Aq ) is surjective (where q is a prime ideal of B and p is an ideal of A that lies over it in A), by chapter 3, exercise 18. Therefore, by exercise 10, f has the going-down property. 40 CHAPTER 5. 12 It’s obvious that A is integral over AG ; for, given a ∈ A, a is a root of the monic polynomial P (x) = G σ∈G (x − σ(a)).