6. Valuation rings, Prüfer rings, Dedekind domains.

6.1 Definition. A commutative ring R is called a valuation ring if for every two elements a, b ∈ R either a | b or b | a holds.

The definition of valuation ring is equivalent to saying that the principal ideals of R are totally ordered with respect to inclusion. In fact, more is true.

6.2 Lemma. Let R be a valuation ring. Then the ideals of R are totally ordered with respect to inclusion. In particular, R is a local ring.

Proof. Let I, J be ideals of R and suppose I ⊈ J. Let i ∈ I \ J. Then for all j ∈ J, j |/i, and therefore i | j. We have shown J ⊆ iR ⊆ I. □

Although the above definition of valuation ring is phrased for commutative rings in general, we will only be interested in valuation rings that are integral domains.

6.3 Remark. Let V be an integral domain contained in a field K. It is easy to see that V is a valuation ring with quotient field K if and only if for every u ∈ K either u ∈ V or u^{−1} ∈ V .

6.4 Remark. If V is a valuation ring with quotient field K and R a ring with V ⊆ R ⊆ K then R is also a valuation ring. This is an easy consequence of the preceeding remark.

The unique maximal ideal of a valuation domain V then consists of all elements u in the quotient field K with u^{−1} ∉ V . Moreover, if M is this unique maximal ideal of V then for every u ∈ K exactly one of the alternatives u ∈ M or u^{−1} ∈ V holds.

∗ ∗ ∗ Valuations ∗ ∗ ∗

6.5 Definition. Let (G, +) be an Abelian group and ≤ an order relation compatible with the group operation (i.e., g ≤ g′ and h ≤ h′ implies g + h ≤ g′ + h′). Then (G, +, ≤) is called an ordered group. If, moreover, ≤ is a total ordering of G (i.e., for all g, h ∈ G, either g ≤ h or h ≤ g holds) then (G, +, ≤) is called a totally ordered group.

Notation: we will use g < h for “g ≤ h and g ≠ h” and g ≥ h for h ≤ g.

6.6 Definition. Let K be a field and (G, +) a totally ordered group. v: K^{∗} → G (where K^{∗} = K \ {0}) is called a valuation on K if

(1) v(ab) = v(a) + v(b) and

(2) v(a + b) ≥ min(v(a), v(b)). The subgroup Γ_{v} = Im(v) of G is called the valuation group of v. It is

customary to extend the definition of v to all of K by adding an infinity element to G and setting v(0) = ∞. With the conventions ∞ + ∞ = ∞, g + ∞ = ∞ and g ≤ ∞ for all g ∈ G, (1) and (2) still hold and

(3) v(a) = ∞ ⟺ a = 0

Some easy consequences of the definition of a valuation v:

(i) v(1) = 0 (ii) v(a^{−1}) = −v(a) (iii) v(−a) = v(a) (iv) v(a − b) ≥ min(v(a), v(b)) (iv) v(a) ≠ v(b) ⟹ v(a + b) = min(v(a), v(b)).

If v is a valuation on K then R_{v} = {k ∈ K | v(k) ≥ 0} is a local ring with maximal ideal M_{v} = {k ∈ K | v(k) > 0}. (Clearly, the units of R_{v} are exactly those field elements with both k and k^{−1} in R_{v}, which are the elements of R_{v} with v(k) = 0.) R_{v} is a valuation ring, since for all a, b ∈ R_{v} either v(a) ≤ v(b) or v(b) ≤ v(a) and v(a) ≤ v(b) means b/a ∈ R_{v} and therefore a | b in R_{v}. We will show that every valuation domain arises from a valuation in this way.

6.7 Lemma. Let R be a valuation domain with quotient field K then the R-submodules of K are totally ordered by inclusion

Proof. Let I, J ⊆ K be R-modules and I ⊈ J. Let i ∈ I \ J. Then for all j ≠ 0 in J either i/j ∈ R or j/i ∈ R, but i/j ∈ R is impossible, since it would imply i = j(i/j) ∈ J. Therefore for all j ≠ 0 in J we have j/i ∈ R, which implies j = i(j/i) ∈ iR ⊆ I. Thus I ⊈ J implies J ⊆ iR ⊆ I. □

6.8 Proposition. Let R be a valuation domain with quotient field K. Then G = {zR | z ∈ K, z ≠ 0} is totally ordered by

yR ≤ zR ⟺ yR ⊇ zR

and

yR ⋆ zR := yzR

defines an Abelian group structure on G compatible with the ordering. Moreover,

v: K^{∗} → G v(z) = zR

is a valuation on K with R = {k ∈ K | v(k) ≥ 0}.

Proof. Easy. □

∗ ∗ ∗ Valuation rings and integral closure ∗ ∗ ∗

6.9 Definition. A ring is called Bézout ring if every finitely generated ideal is principal.

6.10 Remark. It is clear that every valuation ring is a Bézout ring. In particular, every valuation ring is a GCD-ring. Therefore every valuation domain is integrally closed. (This is of course easy to show directly.)

We have seen that valuation domains are integrally closed. It follows that arbitrary intersections of valuation domains contained in a field are integrally closed. We shall see that conversely, every integral closure of a domain is an intersection of valuation domains.

Before showing this, however, we note in passing that valuation rings are characterized among local rings by the Bézout ring property.

6.11 Lemma. A local ring is a valuation ring if and only if it is a Bézout ring.

Proof. Every valuation ring is a Bézout ring. Conversely, let R be a local Bézout ring and a, b ∈ R. We must show a | b or b | a. By dividing a and b by their greatest common divisor, we may assume gcd(a, b) = 1 and therefore Ra+Rb = R. This shows that a and b are not both in the unique maximal ideal of R, hence one of them is a unit and therefore divides the other. □

6.12 Lemma. Let R ⊆ T be commutative rings, I an ideal of R and u ∈ T a unit. If IR ≠ R then IA ≠ A for either A = R[u] or for A = R[u^{−1}].

Proof. Suppose otherwise. Then there exist a_{i} and b_{i} ∈ I such that

(∗)  1 = a_{0} + a_{1}u + … + a_{n}u^{n}
(∗∗)  1 = b_{0} + b_{1}u^{−1} + … + b_{m}u^{−m}

where n, m ∈ ℕ are minimal and w.l.o.g. n ≥ m.

Multiplication of (∗∗) with u^{n} gives

(1 − b_{0})u^{n} = −b_{1}u^{n−1} − … − b_{m}u^{n−m}.

By multiplying (∗) with (1 − b_{0}) and substituting for (1 − b_{0})u^{n} from the above formula we can express 1 as an I-linear combination of powers u^{k} with 0≤k≤n−1, contradicting the minimality of n. □

6.13 Lemma. Let R ⊆ K be an integral domain contained in a field and I an ideal of R. If I ≠ R then there exists a valuation domain V with quotient field K such that R ⊆ V and IV ≠ V .

Proof. Let S be the set of pairs (R_{α}, I_{α}) where R_{α} is a ring with ideal I_{α} ≠ R_{α} and R ⊆ R_{α} as well as I ⊆ I_{α}. We order S by (R_{α}, I_{α}) ≤ (R_{β}, I_{β}) if and only if both R_{α} ⊆ R_{β} and I_{α} ⊆ I_{β}. By Zorn’s Lemma, S has a maximal element (V, J). We know IV ≠ V , since I ⊆ J. Now let u be an element of K, we must show u ∈ V or u^{−1} ∈ V . Suppose not. Since by a previous lemma the extension of J is a proper ideal of either V [u] or V [u^{−1}], either (V [u], JV [u]) or (V [u^{−1}], JV [u^{−1}]) is an element of S strictly greater than (V, J), a contradiction. □

6.14 Theorem. Let R be an integral domain with quotient field K and R′ the integral closure of R. Then R′ = ⋂_{R⊆V ⊆K} V , where V ranges over all valuation domains between R and K.

Proof. ⋂_{R⊆V ⊆K} V contains R, is integrally closed (as an intersection of integrally closed domains) and therefore contains R′. Conversely, let z ∈ ⋂_{R⊆V ⊆K} V ; we must show z is integral over R. Suppose not. Let z = u^{−1}. Then u^{−1} ∉ R′ implies u^{−1} ∉ R[u]. Since u is not a unit in R[u], uR[u] ≠ R[u] and there exists a valuation domain V containing R[u] with uV ≠ V . This contradicts z = u^{−1} ∈ V . □

∗ ∗ ∗ Invertible ideals and Prüfer rings ∗ ∗ ∗

6.15 Definition. Let R be an integral domain with quotient field K and I an R-submodule of K.

• I is called a fractional ideal of R if there exists a non-zero r ∈ R such that rI ⊆ R.

• The inverse of I is defined as I^{−1} = {z ∈ K | zI ⊆ R}.

• I is called invertible if I^{−1}I = R. (Note that I^{−1}I ⊆ R is automatic.)

In the context of invertible ideals, fractional ideals are usually just called ideals, “fractional” being understood. No confusion should result.

6.16 Remark.

1) Every finitely generated R-submodule of K is a fractional ideal. (We can multiply I by a common denominator of the generators.)

2) Every non-zero principal fractional ideal is invertible. (The inverse of Ra is Ra^{−1} and RaRa^{−1} = R.)

6.17 Proposition. Let I be an R-submodule of K. If I is invertible then I is finitely generated.

Proof. I^{−1}I = R, so there exist a_{i} ∈ I^{−1} and b_{i} ∈ I with ∑_{i−1}^{n} a_{i}b_{i} = 1. We claim the b_{i} generate I. Let c ∈ I. c = 1c = ∑_{i−1}^{n} a_{i}cb_{i} =. Now a_{i} ∈ I^{−1} implies a_{i}c ∈ R and we have c = ∑_{i−1}^{n} r_{i}b_{i} with r_{i} ∈ R. □

6.18 Definition. A commutative ring R with only finitely many maximal ideals is called semi-local.

6.19 Proposition. Let R be a semi-local domain. Then every invertible ideal is principal.

Proof. Let M_{1}, …, M_{n} be the maximal ideals of R. By II^{−1} = R, there exist a_{i} ∈ I and b_{i} ∈ I^{−1} such that a_{i}b_{i} ∉ M_{i}. By the Chinese Remainder Theorem, there exists u_{i} ∈ R with u_{i} ∉ M_{i} and u_{i} ∈ ⋂_{j≠i} M_{j}. Let v = ∑_{i=1}^{n} u_{i}b_{i}. Then vI ⊆ R is an ideal of R and not contained in any M_{i}, since va_{i} ∉ M_{i}. Therefore vI = R. This shows I = v^{−1}R. I is principal. □

To be able to consider localizations of fractional ideals we need to extend the construction of rings of fractions to modules:

6.20 Definition. Let R be a commutative ring, S ⊆ R a multiplicative set and M an R-module. The S^{−1}R-module S^{−1}M is defined as follows: Its elements are the equivalence classes on S × M w.r.t. the equivalence relation

(s, m) ∼ (s′, m′) :⟺ ∃t ∈ S t(sm′ − s′m) = 0.

The equivalence class of (s, m) is denoted by \frac{m}{s}. Addition and scalar multiplication are defined by:

\frac{m}{s} + \frac{m′}{s′} = \frac{s′m + sm′}{ss′}  and  \frac{r}{s} · \frac{m}{s′} = \frac{rm}{ss′}

If S = R \ P for a prime ideal P of R then S^{−1}M is denoted M_{P} and called the localization of M at P.

6.21 Remark.

(i) The zero-element of S^{−1}M is \frac{0}{s} for any s ∈ S, and \frac{m}{s} = 0 iff there exists a t ∈ S with tm = 0.

(ii) The map f: M → S^{−1}M given by f(m) = \frac{sm}{s} (or f(m) = \frac{m}{1} , if 1 ∈ S) is an R-module homomorphism with Kerf = {m ∈ M | ∃s ∈ S sm = 0}.

6.22 Lemma. If f: M → N is an R-module homomorphism, then

S^{−1}f: S^{−1}M → S^{−1}N S^{−1}f(\frac{m}{s}) := \frac{f(m)}{s}

is an S^{−1}R-module homomorphism. Also,

(i) S^{−1}id_{M} = id_{S^{−1}M} (ii) S^{−1}f ∘ S^{−1}g = S^{−1}(f ∘ g) (iii) Ker S^{−1}f = S^{−1}Kerf; Im S^{−1}f = S^{−1}Imf

(Furthermore, S^{−1} is an exact functor, i.e. it maps exact sequences of Rmodules to exact sequences of S^{−1}R-modules.)

6.23 Proposition. Let I be an invertible ideal of a domain R, S ⊆ R a multiplicative subset of R. Then S^{−1}I is an invertible ideal of S^{−1}R.

Proof. Easy. □

6.24 Theorem. Let I be a finitely generated ideal of an integral domain R. Then I is invertible if and only if I_{M} is principal for every maximal ideal M of R.

Proof. If I is invertible, we have seen that I_{M} is invertible and hence principal for every maximal ideal M of R. Conversely, assume every I_{M} is principal. Suppose I is not invertible then II^{−1} ⊆ M for some maximal ideal M of R. I_{M} is principal, and we may choose a generator in I: I_{M} = iR_{M}, with i ∈ I. Let I = Ri_{1}+…+Ri_{n}. Then there exist r_{k} ∈ R and s_{k} ∈ R \ M with i_{k} = \frac{r_{k}}{s_{k}} i. Set s = s_{1} … s_{n}. Then si^{−1}i_{k} ∈ R for every generator i_{k} of I, and therefore si^{−1} ∈ I^{−1}. This implies s = si^{−1}i ∈ I^{−1}I ⊆ M, a contradiction to s ∈ R \ M. □

6.25 Theorem. Let R be an integral domain. Then the following are equivalent.

(1) Every finitely generated non-zero ideal of R is invertible.

(2) R_{P} is a valuation domain for every prime ideal P of R.

(3) R_{M} is a valuation domain for every maximal ideal M of R.

6.26 Definition. A domain satisfying one (and hence all) of the equivalent conditions in the preceding theorem is called Prüfer domain.

Proof. (1 ⇒ 2) We show that R_{P} is Bézout. Let \frac{r_{1}}{s_{1}} , …, \frac{r_{n}}{s_{n}} ∈ R_{P}, r_{i} ∈ R, s_{i} ∈ R \ P. Since R is Prüfer, I = r_{1}R + … r_{n}R is invertible, and therefore I_{P} = \frac{r_{1}}{s_{1}} R_{P} +…+ \frac{r_{n}}{s_{n}} R_{P} is invertible. Since R_{P} is local, invertible implies principal. (2 ⇒ 3) Obvious. (3 ⇒ 1) If I is a finitely generated ideal of R then I_{M} is principal (as a finitely generated ideal in the Bézout domain R_{M}) for all maximal ideals M of R. Therefore I is invertible. □

6.27 Theorem. Let R be a Prüfer domain with quotient field K. If V is a valuation ring with R ⊆ V ⊆ K then V = R_{P} for some prime ideal P of R.

Proof. Let M be the maximal ideal of V and P = M ∩ R. For any s ∈ R \ P, s ∉ M implies s^{−1} ∈ V . Therfore R_{P} ⊆ V .

Suppose V ⊈ R_{P}. Let v ∈ V \ R_{P}. Since R_{P} is a valuation domain, v^{−1} ∈ R_{P}. This means v = \frac{s}{r} with r ∈ R, s ∈ R \ P. We must have r ∈ P, since otherwise v ∈ R_{P}. Now s = rv ∈ P V ⊆ M, a contradiction. □

∗ ∗ ∗ Discrete valuation rings and Dedekind domains ∗ ∗ ∗

6.28 Theorem. Let R be a local integral domain with maximal ideal M, and not a field. The following are equivalent:

(1) R is Noetherian and integrally closed and dim R = 1.

(2) R is Noetherian and M is principal.

(3) R is a unique factorization domain with just one (up to units) irreducible element.

(4) R is a principal ideal domain.

Proof. (1 ⇒ 2) Let m ∈ M with m ≠ 0 and mR ≠ M and consider the ring T = R/mR. mR not being prime, T is not a domain. The set N ≠ {0} of zero-divisors of T is a union of prime ideals (as the complement of the saturated multiplicative set of non-zero-divisors), therefore N equals the unique non-zero prime ideal of T, N = M/mR. By the universal zero-divisor lemma, there exists a ∈ R \ mR such that aM ⊆ mR. Now \frac{a}{m} ∈ M^{−1} \ R and \frac{a}{m}M is an ideal of R. We will show \frac{a}{m}M = R (which implies that M = \frac{m}{a}R is a principal ideal). Suppose otherwise; then \frac{a}{m}M ⊆ M and therfore R[\frac{a}{m}]M ⊆ M. Now M is a faithful R[\frac{a}{m}]-module and finitely generated as an R-module, which means \frac{a}{m} is integral over R and therefore in R; a contradiction. (2 ⇒ 3) The acsending chain condition implies that every non-zero element is a product of irreducible elements (the ascending chain condition for principal ideals suffices for that). Also, every irreducible element is prime, and there is (up to multiplication by units) only one irreducible element. (Since the unique maximal ideal M of R is principal, it is the only principal ideal maximal among proper principal ideals, and therefore the only ideal generated by an irreducible ideal. So there is up to multiplication by units only one irreduclble element and it generates M. Now M is also a prime ideal, wherfore the unique (up to units) irreducible element is also prime). Together this implies a unique factorization domain with just one ireducible element.

(3 ⇒ 4) and (4 ⇒ 1) are easy. □

6.29 Definition. A domain satisfying one (and hence all) of the equivalent conditions in the preceeding theorem is called discrete valuation domain (DVR), or, more precisely discrete valuation domain of rank 1.

6.30 Proposition. A domain is a DVR if and only if it is the valuation ring of a valuation with value group isomorphic to (ℤ, +).

Proof. Exercise. □

6.31 Theorem. Let R be an integral domain. Then the following are equivalent (1) Every non-zero ideal of R is invertible.

(2) R is Noetherian and integrally closed and dim R ≤ 1.

(3) R is either a field or a Noetherian domain s.t. for every maximal ideal M of R, R_{M} is a DVR.

6.32 Definition. A domain satisfying one (and hence all) of the equivalent conditions in the preceding theorem is called a Dedekind ring.

Proof. (1 ⇒ 3) Invertible implies finitely generated, so R is Noetherian. Let M be a maximal ideal of R. Every ideal of R_{M} is of the form I_{M} for an invertible ideal I of R and hence is principal. R_{M} is a local principal ideal domain and therefore a DVR. (3 ⇒ 1) For every non-zero ideal I of R, every localization I_{M} at a maximal ideal is principal, since R_{M} is a principal ideal domain. (3 ⇒ 2) R is integrally closed as the intersection of its localizations R_{M}, which are integrally closed as valuations rings. The Krull dimension of R is the supremum of dim R_{M}, where M ranges through the maximal ideals of R and therefore dim R ≤ 1. (2 ⇒ 3) If dim R ≤ 1 and R is not a field, then dim(R) = 1. For every maximal ideal M of R, the localization R_{M} is local, Noetherian, integrally closed and one-dimensional and therefore a DVR by the previous Theorem. □

6.33 Theorem. Let D be a domain. If every non-zero ideal of D is invertible, then every non-zero ideal is expressible as a product of prime ideals.

Proof. Invertible implies finitely generated, so D is Noetherian. Suppose there exist non-zero ideals not expressible as products of prime ideals. The set of such ideals has a maximal element, I. This I is not R (since R is an empty product of prime ideals); nor is I itself prime, so there exists a prime ideal P ⊃ I properly containing I. We have I ⊆ IP^{−1} ⊆ R. If IP^{−1} = I, then (since P is invertible) multiplication by P gives I = IP and by Nakayama’s Lemma, there is an element a ∈ R with aI = 0 and a ≡ 1 mod P, which is impossible in a domain. Therfore IP^{−1} properly contains I and, by maximality of I, is expressible as a product of prime ideals IP^{−1} = P_{1} … P_{m}. Using P P^{−1} = R again, we get I = P_{1} … P_{m}P, a contradiction. □

6.34 Corollary. In a Dedekind ring, every non-zero ideal is uniquely expressible as a product of prime ideals.

Proof. By 6.31, every non-zero ideal in a Dedekind ring is invertible, so 6.33 implies existence of a factorization. If P_{1} … P_{n} = Q_{1} … Q_{m} then P_{1}, being prime, contains one of the Q_{i}, say Q_{1} ⊆ P_{1}. Since D is at most 1-dimensional, it follows that Q_{1} = P_{1} and we can cancel invertible ideals and get P_{2} … P_{n} = Q_{2} … Q_{m}. Inductively we see that the factorizations are the same (up to order). □

6.35 Remark. Prime factorization of ideals characterizes Dedekind rings, even without the assumption of uniqueness: a domain in which every ideal is a product of prime ideals is a Dedekind domain.

∗ ∗ ∗ rank 1 valuation rings ∗ ∗ ∗

6.36 Definition. An ordered group (Γ, +) satisfies the Archimedean axiom if for any a, b ∈ Γ with a > 0 there exists n ∈ ℕ with na ≥ b.

6.37 Proposition. An ordered group satisfies the Archimedean axiom if and only if it is order-isomorphic to a subgroup of (ℝ, +).

Proof. Every subgroup of (ℝ, +) satisfies the Archimedean axiom. Now let (Γ, +) be an ordered group satisfying the Archimedean axiom and fix an element a > 0 of Γ. We will construct an order-preserving injective map f: Γ → ℝ by defining a Dedekind cut for every b ∈ Γ. First consider only elements b > 0 and define

S_{b} = \{\frac{m}{n} ∈ ℚ | m ∈ ℤ, n ∈ ℕ, ma ≤ nb\}.

Note that \frac{m}{n} ∈ S_{b} and \frac{m̃}{ñ} = \frac{m}{n} (with n, ñ > 0) implies that m̃a ≤ ñb. Also, S_{b} ≠ ℚ (choose m < 0) and S_{b} ≠ ∅ (by the Archimedean axiom).

If \frac{m}{n} ∈ S_{b} then every \frac{m′}{n′} ≤ \frac{m}{n} in ℚ is also in S_{b}. (By choosing a common denominator, we may assume n′ = n, which implies m′ ≤ m, and then m′a ≤ ma ≤ nb = n′b.) S_{b} thus defines a unique real number sup S_{b} and we set f(b) = sup S_{b}.

We extend the definition of f to all of Γ by seting f(0) = 0 and f(b) = −f(−b) for b < 0. It is then not hard to check that f is an order-preserving group homomorphism and injective. □

6.38 Proposition. Let v be a valuation on a field K with valuation ring R_{v} and valuation group Γ_{v} ≠ 0. Then Γ_{v} satisfies the Archimedean axiom if and only if dim R_{v} = 1.

Proof. Assume the Archimedean axiom. Γ_{v} ≠ 0 implies that R_{v} is not a field. Suppose dim R > 1. Then there exists a prime ideal P with (0) ⊂ P ⊂ M_{v}. Let a ∈ M_{v} \P and b ∈ P \(0) and set α = v(a), β = v(b). Since α > 0 there exists n ∈ ℕ with nα ≥ β. This means v(a^{n}) ≥ v(b), i.e., \frac{a^{n}}{b} ∈ R_{v} and therfore a^{n} ∈ bR_{v} ⊆ P. We conclude that a ∈ P, a contradiction.

Conversely, assume dim R = 1. R is a one-dimensional local ring, so M_{v} is its only non-zero prime ideal and therefore the radical of every proper non-zero ideal. In particular, for every non-zero a ∈ M_{v} and any b ∈ R_{v} there exists n ∈ ℕ such that a^{n} ∈ bR_{v}, which implies that for every α > 0 in Γ and every β ≥ 0 in Γ there exists n ∈ ℕ with nα = γ + β for some γ ≥ 0, or in other words, nα ≥ β. □

∗ ∗ ∗ Quick and dirty Dedekind ring properties ∗ ∗ ∗

6.39 Lemma. In a Noetherian ring, every non-zero ideal contains a product of non-zero prime ideals.

Proof. Suppose M maximal among the counterexamples. Then M is not prime, so there exist r, s with rs ∈ M but r ∉M and s ∉ M. Since M is striclty contained in M + rR and M + sR, these ideals each contain a product of prime ideals, and so does (M + rR)(M + sR), which is contained in M, a contradiction. □

6.40 Lemma. Let R be a Noetherian ring with dim R = 1 and quotient field K, and (0) ⫋ I ⫋ R a proper ideal of R. Then there exists γ ∈ K \ R with γI ⊆ R.

Proof. Let a ≠ 0 in I then there exist non-zero prime ideals P, P_{1}, …, P_{r} with P ⊇ I ⊇ aR ⊇ P_{1} · . . . · P_{r}, where r is minimal with the property that aR contains a product of r prime ideals. P being prime, it contains one of the P_{i}, say P_{1} ⊆ P and therefore P_{1} = P. So we have

P ⊇ I ⊇ aR ⊇ P · P_{2} · . . . · P_{r}.

Pick b ∈ P_{2} · . . . · P_{r} \ aR (possible by minimality of r) and set γ = b/a. Then γ ∉ R, and for every i ∈ I, iγ = (ib)/a is in R, because ib ∈ P · P_{2} · . . . · P_{r} ⊆ aR. □ 6.41 Theorem. In a Dedekind ring, every non-zero ideal is invertible.

Proof. Let I ≠ (0) and suppose I^{−1}I ⫋ R. Let γ ∈ K \ R with γI^{−1}I ⊆ R, then γI^{−1} ⊆ I^{−1}. This makes I^{−1} a faithful R[γ]-module which is finitely generated as R-module, showing γ ∈ K \ R to be integral over R - a contradiction to R being integrally closed. □