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2), sn , hence s, satisfies an equation of integral dependence with coefficients in I. 2 Lemma Let R be an integral domain with fraction field K, and assume that R is integrally closed. Let f and g be monic polynomials in K[x]. If f g ∈ R[x], then both f and g are in R[x]. Proof. In a splitting field containing K, we have f (x) = i (x−ai ) and g(x) = j (x−bj ). Since the ai and bj are roots of the monic polynomial f g ∈ R[x], they are integral over R. The coefficients of f and g are in K and are symmetric polynomials in the roots, hence are integral over R as well.

8). (6) ⇒ (3): By hypothesis, M = M2 , so we may choose t ∈ M \ M2 . But then t + M2 is a generator of the vector space M/M2 over the field R/M. Thus R(t+M2 )/M2 = M/M2 . By the correspondence theorem, t + M2 = M. Now M(M/(t)) = (M2 + (t))/(t) = M/(t), so by NAK, M/(t) = 0, that is, M = (t). ♣. 1)] excludes the trivial valuation v(a) = 0 for every a = 0. 12 Corollary The ring R is a discrete valuation ring if and only if R is a local PID that is not a field. In particular, since R is a PID, it is Noetherian.

Then there is an equation of the form αn + cn−1 αn−1 + · · · + c1 α + c0 = 0 with the ci in V . We must show that α ∈ V . If not, then α−1 ∈ V , and if we multiply the above equation of integral dependence by α−(n−1) , we get α = −cn−1 − cn−2 α−1 − · · · − c1 α−(n−2) − c0 α−(n−1) ∈ V. 5. If I and J are ideals of V , then either I ⊆ J or J ⊆ I. Thus the ideals of V are totally ordered by inclusion. Suppose that I is not contained in J, and pick a ∈ I \ J (hence a = 0). If b ∈ J, we must show that b ∈ I.

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