Noether Normalisation
نویسنده
چکیده
Proof. Let A = k[s1, . . . , sn] where s1, . . . , sn ∈ A. We assume that A is not integral over k, in which case at least one of the si is not algebraic over k. If the set {s1, . . . , sn} is algebraically independent, then we are done. Otherwise assume that sn is algebraic over k[s1, . . . , sn−1] (by relabeling if necessary). Let f(x1, . . . , xn) be a nonzero polynomial with f(s1, . . . , sn) = 0 and let F be the homogenous part of f of highest degree. If the degree of F is e, then for any constants λ1, . . . , λn−1 ∈ k the polynomial f(x1 + λ1xn, . . . , xn−1 + λn−1xn, xn) expands to
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