6 1. Aﬃne and Projective Varieties monic with respect to Xn and its leading term is Xc n with c = j1dn−1 + j2dn−2 + · · · + jn−1d + jn. The monic polynomial g(X) := h(x1 − xdn−1,x n 2 − xdn−2,...,x n n−1 − xd,X) n then satisfies g(xn) = f(x1,...,xn) = 0, which gives that the ring C[x1,x2,...,xn] = C[x1 − xdn−1,x n 2 − xdn−2,...,x n n−1 − xd,xn] n is integral over the ring R1 := C[x1 − xn dn−1 , x2 − xn dn−2 , . . . , xn−1 − xn].d If the ring R1 is isomorphic to the ring of polynomials C[X1,...,Xn−1], then the proof is finished. Otherwise, by repeating the process we obtain a ring R2 generated over C by n − 2 elements over which R1 is integral and, by transitivity of integral dependence, R as well . Since R is finitely generated over C, in a finite number of steps we obtain the result. The case in which r = 0, i.e. R is integral over C, is not excluded. Remark 1.1.9. Let us consider the ring R in Proposition 1.1.8 and let us assume that it is an integral domain. If we denote by K the fraction field of R, we have that the elements y1,...,yr form a transcendence basis of K over C hence the transcendence degree of K over C is equal to r. Corollary 1.1.10. Let S ⊂ R be two finitely generated domains over C. Then there exist elements f ∈ S and x1,...,xr ∈ R, algebraically indepen- dent over Sf such that Rf is integral over Sf [x1, . . . , xr]. Proof. Let K ⊂ L be the fields of fractions of S and R. Denote by R the localization of R with respect to the multiplicative system S∗ of the nonzero elements of S. We apply Proposition 1.1.8 to the finitely generated K- algebra R . There exist elements x1,...,xr in R algebraically independent over K such that R is integral over K[x1,...,xr]. It is clear that the elements xi can be chosen in R as the denominators appearing are invertible in K. Now R is finitely generated over S and each generator satisfies an integral dependence relation over K[x1,...,xr]. By choosing a common denominator f for the coeﬃcients of all these relations, we obtain that Rf is integral over Sf[x1,...,xr]. For the proof of the Hilbert’s Nullstellensatz, we use the following propo- sition, sometimes called “weak Nullstellensatz”.

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