Discrete Valuation Rings

Characterizing Discrete Valuation Rings
Matematics 601
If F is a field and v is a discrete valuation on F, then Ov = {a ∈ F ∗ | v(a) ≥ 0} ∪ {0} is a ring with quotient field F . This ring is called a discrete valuation ring (DVR). In other words, a ring R is a DVR if there is a discrete valuation v on the quotient field F of R, such that R = {a ∈ F ∗ | v(a) ≥ 0} ∪ {0}. The following theorem givesthe main ring theoretic information about DVR’s. Theorem 1. Let R be a local Noetherian integral domain with dim(R) = 1. If m is the unique maximal ideal of R and k = R/m, then the following conditions are equivalent. 1. R is a DVR. 2. R is integrally closed 3. The maximal ideal m of R is principal. 4. dimk (m/m2 ) = 1. 5. Every proper nonzero ideal of R is a power of m. Proof. Before we get intothe proof of the equivalences of these conditions, we point out two things. First, mn+1 = mn for each n. For, if mn+1 = mn , then the finitely generated R-module M = mn would satisfy mM = M, so M = 0 by Nakayama. This is false, so √ √ mn+1 = mn . Second, if I is any nonzero ideal of R, then I = m, since I is the intersection of all prime ideals containing I, and since R is local and 1-dimensional,the only nonzero prime ideal of R is m. From this we can show that mn ⊆ I for some n. To see why, suppose √ m = (x1 , . . . , xr ). Since I = m, there are integers ni with xni ∈ I. If n = ni , then mn is i s1 sr si = n. For each such generator, there is some generated by the elements x1 · · · xr with si i with si ≥ ni , so xi ∈ I, hence the product xs1 · · · xsr ∈ I. This means mn ⊆ I. 1 r We nowbegin the series of implications to prove this theorem. 1 → 2 : Let F be the quotient field of R. Suppose α ∈ F is integral over R. Then αn + a1 αn−1 + · · · + a0 = 0 for some ai ∈ R. If R = Ov for some valuation v, then v(ai ) ≥ 0 / for each i. Suppose α ∈ R. Then v(α) < 0. From the polynomial equation above we get 1

2 → 3 : Let a ∈ m with a = 0. By the second comment above, we have mn ⊆ (a)for some n. Choose n minimal, so that mn−1 (a). Take b ∈ mn−1 − (a), and let x = a/b. / / Then x−1 ∈ R since b ∈ (a), so x−1 is not integral over R. The conditions mn ⊆ (a) and −1 n−1 b∈m imply that x m ⊆ R. If x−1 m = R then m = (x) is principal, and we would then be done. The only other possibility is for x−1 m ⊆ m since x−1 m is an ideal of R. Suppose this happens. Let u1 , . . . , ur begenerators for m (recall R is Noetherian, so m is finitely generated). Then x−1 ui = rij uj for some rij ∈ R. This means (rij − δ ij x−1 )ui = 0 for each i, hence det (rij − δ ij x−1 ) annihilates each uj . Since we are in an integral domain, this forces det (rij − δ ij x−1 ) = 0. Expanding this out we see that x−1 is integral over R, a contradiction. Thus x−1 m ⊆ m is false, hence m = (x) from theprevious case. 3 → 4 : Suppose m = (x). Then as an R-module, m is generated by x, hence m/m2 is generated by x + m2 . But then, m/m2 is generated as an R/m-module by x + m2 , so dimk (m/m2 ) ≤ 1. But, m/m2 = 0 by the first comment above, so dimk (m/m2 ) = 1.

α = −(an−1 + an−2 α−1 + · · · + a0 α−n+1 ). The right hand side is in R since v(α−1 ) > 0, so α−1 ∈ R. This shows α ∈ R, so R is integrallyclosed.

4 → 5 : Let x + m2 be a generator for the R/m-vector space m/m2 . Then m = Rx + m2 = Rx + m · m. Therefore, by Nakayama’s lemma (with M = m), we see m = (x). Let I be any proper nonzero ideal of R. Then I ⊆ m since m is the unique maximal ideal of R. Let r be maximal with I ⊆ mr , so I mr+1 . There is such an r, since if I ⊆ mr for every r, then by the second comment above, mn ⊆ I for somen, which forces mn = mn+s for each s ≥ 0. This is false, by the first comment above. Take y ∈ I with y ∈ mr+1 . Then y = axr for some / r+1 / a ∈ R, since m = (x). Moreover, since y ∈ m , the element a ∈ m, hence a is a unit in R. / r r r Thus, (y) = (x ) = m , and since (y) ⊆ I ⊆ m , this forces I = mr to be a power of m. 5 → 1 : Let F be the quotient field of R, and suppose every ideal of R is a...