0. MODULES OVER COMMUTATIVE RINGS

E. L. Lady

August 5, 1998

The assumption in this book is that the reader is either a student of abelian group theory who knows the bare minimum of commutative ring theory or a commutative ring theorist whose knowledge of abelian group theory is essentially nil. In this chapter we valiantly attempt to present a quick and dirty summary of the commutative ring theory which will be relevant for what follows and which might not be included in the normal basic graduate algebra course, as well as all the abelian group theory (in disguised form) which is needed to study finite rank torsion free groups and finite rank torsion free modules over dedekind domains. Although most of the results in this chapter are well known, your author has found it hard to track some of them down in standard references, so it seems worthwhile to gather them all together here.

By inference, then, it might seem plausible that this book is accessible to a reader who knows no abelian group theory and almost no commutative ring theory, provided that he knows at least a little non-commutative ring theory, at least enough to be comfortable when encountering the terms “left” and “right.” However it is doubtful that this book would be a good choice of reading matter for such a reader.

proposition 0.1. (1) A vector space over an infinite field cannot be the union of a finite number of proper subspaces.

(2) If p_{1}, …, p_{n} are prime ideals in a commutative ring and 𝔞 is an ideal such that 𝔞 ⊆ ⋃_{1}^{n} p_{i}, then 𝔞 ⊆ p_{i} for some i.

proof: (1) Suppose that V = V_{1} ∪ ⋯ ∪ V_{n} with V_{i} ⊊ V . We may assume wlog that V is not a union of any proper subset of {V_{1}, …, V_{n}}. Thus we may choose v_{1} ∈ V_{1} with v_{1} ∉ V_{2} ∪ ⋯ ∪ V_{n} and v_{2} ∈ V_{2} with v_{2} ∉ V_{1} . Then for any k ≠ 0 ∈ K, kv_{1} + v_{2} ∉ V_{1} ∪ V_{2} so kx + y ∈ V_{i} for i ≥ 3. Since K is infinite and there are only finitely many V_{i}, it follows that there exist distinct k, k′ ∈ K such that kv_{1} + v_{2}, k′v_{1} + v_{2} ∈ V_{i} for some i ≥ 3. But it then follows that v_{1} ∈ V_{i}, a contradiction.

(2) We may assume without loss of generality that 𝔞 is not contained in the union of any proper subset of {p_{1}, …, p_{n}}. Thus there exist a_{i} ∈ 𝔞 with a_{i} ∈ p_{i} and a_{i} ∉ p_{j} for j ≠ i. Then a_{2} ⋯ a_{n} ∈ p_{i} for i > 1 and a_{2} ⋯ a_{n} ∉ p_{1} since p_{1} is prime. If n > 1, let x = a_{1} + a_{2} ⋯ a_{n}. Then x ∈ 𝔞 but x ∉ ⋃_{1}^{n} p_{i}, a contradiction. Thus n = 1 and

𝔞 ⊆ p_{1} . ✓

LOCALIZATION AND GLOBALIZATION. If S is a multiplicative set in a commutative ring R and M is an R-module then S^{−1}M is defined in a well known fashion. We will call S^{−1}M the localization of M with respect to S. If S is the complement of a prime ideal p, we write M_{p} = S^{−1}M. (Some other authors prefer to reserve the term “localization” for the case where S is the complement of a prime ideal.) It is well known that R_{p} is a local ring (or “quasi-local,” some would say) whose unique maximal ideal is pR_{p}. It is also well known that if M is noetherian then S^{−1}M is noetherian. There is a canonical map θ: M → S^{−1}M with θ(m) = m/1 and θ(m) = 0 if and only if sm = 0 for some s ∈ S.

proposition 0.2. For any R-module M and multiplicative set S, S^{−1}M ≈ S^{−1}R ⊗_{R} M. proof: The maps φ: S^{−1}R ⊗_{R} M → S^{−1}M and ψ: S^{−1}M → S^{−1}R ⊗_{R} M given by

φ (\frac{r}{s} ⊗_{R} m) = \frac{rm}{s} and ψ (\frac{m}{s}) = \frac{1}{s} ⊗ m are inverse to each other. ✓

proposition 0.3. If M and N are R-modules and S is a multiplicative set then

(1) Hom_{S^{−1}R}(S^{−1}M, S^{−1}N) = Hom_{R}(S^{−1}M, S^{−1}N) = Hom_{R}(M, S^{−1}N) .

(2) S^{−1}M ⊗_{S^{−1}R} S^{−1}N = S^{−1}M ⊗_{R} S^{−1}N. Furthermore, Hom_{R}(M, S^{−1}N) and S^{−1}M ⊗_{R} S^{−1}N are S^{−1}R-modules. proof: (1) For φ ∈ Hom_{R}(S^{−1}M, S^{−1}N), m ∈ M, r ∈ R, s, s′ ∈ S,

φ (\frac{r}{s} \frac{m}{s′}) = \frac{s}{s} φ (\frac{rm}{ss′}) = \frac{r}{s} φ (\frac{sm}{ss′}) = \frac{r}{s} φ (\frac{m}{s′}) .

Thus every R-linear map from S^{−1}M to S^{−1}N is in fact S^{−1}R-linear. Furthermore, every R-linear map from M to S^{−1}N extends uniquely to a map from S^{−1}M to S^{−1}N.

(2) This follows from the fact that for m ∈ S^{−1}M, n ∈ S^{−1}N, r ∈ R and s ∈ S, the following holds in S^{−1}M ⊗_{R} S^{−1}N:

m ⊗ \frac{rn}{s} = \frac{sm}{s} ⊗ \frac{rn}{s} = \frac{rm}{s} ⊗ \frac{sn}{s} = \frac{rm}{s} ⊗ n. ✓

proposition 0.4. If M is a finitely generated module over a noetherian ring R then for every R-module N and multiplicative set S, the obvious map S^{−1} Hom_{R}(M, N) → Hom_{R}(S^{−1}M, S^{−1}N) is an isomorphism. proof: The “obvious map” is the one that takes φ/s ∈ S^{−1} Hom_{R}(M, N) to the

homomorphism given by \frac{m}{s′} ↦ \frac{φ(m)}{ss′}. This is clearly an isomorphism when M = R. It

then follows that it is an isomorphism when M = R^{t} for t < ∞. Now if M is finitely generated then there is a surjection ε: R^{t} → M for some t < ∞. Since R is noetherian, Ker ε is also finitely generated and we thus get an exact sequence R^{s} → R^{t} → M → 0.

Since localization preserves exactness, applying Hom( , N) and localizing with respect to S then yields a commutative diagram

0 → Hom_{R}(M, N)  ⟶ S^{−1} Hom_{R}(R^{t}, N) ⟶ S^{−1} Hom_{R}(R^{s}, N)
↓  ↓^{≈}  ↓^{≈}
0 → Hom_{R}(S^{−1}M, S^{−1}N) ⟶ Hom_{R}(S^{−1}R^{t}, S^{−1}N) ⟶ Hom_{R}(S^{−1}R^{s}, S^{−1}N)

where the two right-hand vertical maps are isomorphisms. It follows that the left-hand vertical map is also an isomorphism, proving the result. ✓

Note: It is essential here that M be finitely generated and R noetherian (or, more generally, that M be finitely presented). A lot of things in the chapters that follow would become trivialized if Proposition 0.4 were true when M is not finitely generated. (See, in particular, Chapter 9.)

proposition 0.5. Let M, N, P be modules over a commutative ring R.

(1) If m_{1}, m_{2} ∈ M, then m_{1} = m_{2} if and only if m_{1}/1 = m_{2}/1 ∈ M_{𝔪} for all maximal ideals 𝔪.

(2) M = 0 if and only if M_{𝔪} = 0 for all maximal ideals 𝔪.

(3) Suppose that N, P ⊆ M. Then N = P if and only if N_{𝔪} = P_{𝔪} for all maximal ideals 𝔪.

(4) If φ ∈ Hom_{R}(M, N) then φ is monic [epic] if and only if φ_{𝔪}: M_{𝔪} → N_{𝔪} is monic [epic] for all maximal ideals 𝔪.

(5) A sequence M → N → P is exact if and only if the induced sequence M_{𝔪} → N_{𝔪} → P_{𝔪} is exact for all maximal ideals 𝔪.

(6) If M is a submodule of a vector space over the quotient field F of R, then M = ⋂_{p} M_{p}.

warning. M_{𝔪} ≈ N_{𝔪} for all maximal ideals 𝔪 does not, in general, imply that M ≈ N.

corollary 0.6. Chinese Remainder Theorem. Let 𝔞_{1}, …, 𝔞_{n} be ideals in a commutative ring R such that 𝔞_{i} + 𝔞_{j} = R for all i ≠ j. Let M be an R-module. Then

\frac{M}{⋂ 𝔞_{i}M} ≈ ⨁_{1}^{n} \frac{M}{𝔞_{i}M} .

proof: For each k there is an obvious map M/ ⋂ 𝔞_{i}M → M/𝔞_{k}M and therefore there is a map M/ ⋂ 𝔞_{i}M → ⨁ M/𝔞_{i}M. Since 𝔞_{i} +𝔞_{j} = R for k ≠ j, a prime ideal p can contain at most one 𝔞_{i}, and (𝔞_{i}M)_{p} = M_{p} if 𝔞_{i} ⊈ p. Thus if 𝔞_{k} ⊆ p then (⋂_{1}^{n} 𝔞_{i}M)_{p} = 𝔞_{k}M_{p} and M_{p}/𝔞_{i}M_{p} = 0 for i ≠ k. Therefore for all p,

\frac{M}{𝔞_{k}M_{p}} = ( \frac{M}{⋂_{1}^{n} 𝔞_{i}M} )_{p} → (⨁_{1}^{n} \frac{M}{𝔞_{i}M} )_{p} = (\frac{M}{𝔞_{k}M} )_{p}

is an isomorphism, so the result follows from Proposition 0.5. ✓ proposition 0.7. A finitely generated projective module M over a local ring R is free. In fact, if 𝔪 is the maximal ideal in R and m_{1}, …, m_{t} ∈ M are such that the cosets m̄_{1}, …, m̄_{t} are a basis for M/𝔪M as a vector space over R/𝔪, then m_{1}, …, m_{t} are a basis for M. proof: Let 𝔪 be the unique maximal ideal in R. Choose m_{1}, …, m_{t} ∈ M so that the cosets m̄_{1}, …, m̄_{t} are a basis for the vector space M/𝔪M over the field R/𝔪. Let φ: R^{t} → M be defined by φ(r_{1}, …, r_{t}) = ∑ r_{i}m_{i}. It follows easily from Nakayama’s Lemma that φ is surjective. Since M is a projective module, φ splits, so R^{t} = K ⊕ L with K = Ker φ and L ≈ M. Then K is finitely generated. Since φ induces an isomorphism from R^{t}/𝔪R^{t} to M/𝔪M, it follows that K/𝔪K ⊕ L/𝔪L ≈ M/𝔪M. These are finite dimensional vector spaces over the field R/𝔪 and comparing dimensions yields K/𝔪K = 0. Thus K = 0 by Nakayama’s Lemma. Thus φ is monic and hence an isomorphism. ✓

Projective modules over a local ring are free even if not finitely generated, but the proof is much more difficult.

proposition 0.8. A finitely generated module M over a noetherian ring is projective if and only if M_{p} is a free R_{p}-module for all prime ideals p. proof: ( ⇒ ): Using the criterion that projective modules are just the direct summands of free modules, it is easy to see that the localization of a projective R-module at p is a projective module over R_{p}. It then follows from Proposition 0.7 that this localization is a free R_{p}-module.

( ⇐ ): Suppose now that M is finitely generated and for all p, M_{p} is a free R_{p}-module. To show that M is projective one must show that for every surjection φ: X → Y , the induced map φ_{∗}: Hom_{R}(M, X) → Hom_{R}(M, Y) is surjective. By Proposition 0.5, it suffices to prove that for all maximal ideals p, the localized map (Hom_{R}(M, X))_{p} → (Hom_{R}(M, Y))_{p} is surjective. But since M is finitely generated, by Proposition 0.4 and Proposition 0.3 there are natural isomorphisms yielding the following commutative diagram:

(Hom_{R}(M, X))_{p} ⟶ (Hom_{R}(M, Y))_{p}
↓^{≈}  ↓^{≈}
Hom_{R_{p}}(M_{p}, X_{p}) ⟶ Hom_{R_{p}}(M_{p}, Y_{p}) ⟶ 0

where the bottom map is surjective since M_{p} is a projective R_{p}-module. Thus (Hom_{R}(M, X))_{p} ⟶ (Hom_{R}(M, Y))_{p} is a surjection, proving the result. ✓

remark. The hypothesis that M be finitely generated is essential here. In later chapters we will see many examples of non-finitely generated non-projective modules M such that M_{p} is a free R_{p}-module for all prime ideals p. Such modules will be called locally free. (See Chapter 9 in particular.)

COMAXIMAL IDEALS.

definition 0.9. Local Property.

proposition 0.10. Let 𝔞, 𝔟, 𝔠 ⊆ R.

(1) If 𝔞 + 𝔟 = R then 𝔞 ∩ 𝔟 = 𝔞𝔟.

(2) If 𝔞 + 𝔟 = R and 𝔞 + 𝔠 = R, then 𝔞 + (𝔟 ∩ 𝔠) = R.

(3) If 𝔞 + 𝔟 = R then for any n ≥ 1, 𝔞^{n} + 𝔟^{n} = R. proof: In each case the hypothesis and conclusion are “local properties,” i. e. it suffices to prove in (1) that 𝔞_{p} ∩ 𝔟_{p} = 𝔞_{p}𝔟_{p} for all prime ideals p. Thus wlog we may assume that R is a local ring.

(1) & (3) Since 𝔞 + 𝔟 = R, if R is a local ring it follows that 𝔞 = R or 𝔟 = R, If, say, 𝔟 = R then 𝔞 ∩ 𝔟 = 𝔞 ∩ R = 𝔞 = 𝔞R = 𝔞𝔟. Furthermore 𝔞^{n} + 𝔟^{n} = 𝔞^{n} + R = R.

(2) Assuming again that R is a local ring, either 𝔞 = R or 𝔟 = 𝔠 = R. Both cases are

trivial. ✓

proposition 0.11. If 𝔞 + 𝔟 = R then 𝔞 ⊕ 𝔟 ≈ R ⊕ 𝔞𝔟. proof: By Lemma 0.10 the usual short exact sequence

0 → 𝔞 ∩ 𝔟 → 𝔞 ⊕ 𝔟 → 𝔞 + 𝔟 → 0

becomes

0 → 𝔞𝔟 → 𝔞 ⊕ 𝔟 → R → 0,

which splits because R is free. Thus 𝔞 ⊕ 𝔟 ≈ R ⊕ 𝔞𝔟. ✓

ASSOCIATED PRIMES. One of the most basic concepts in abelian group theory is that of the order of an element. And for any abelian group G, one is particularly interested in the set of prime numbers p such that G contains an element of order p. For modules over a commutative ring, the analogous concepts are the annihilator of an element and the set of associated primes of a module.

If M is an R-module and m ∈ M we will write ann M for the annihilator of M and ann m for the annihilator of m. I. e.

ann m = {r ∈ R | rm = 0}
ann M = {r ∈ R | (∀m ∈ M) rm = 0} = ⋂_{m∈M} ann m.

.

lemma 0.12. Let m ∈ M and let p be a prime ideal. Then m/1 ≠ 0 ∈ M_{p} if and only if ann m ⊆ p. proof: In fact, m/1 = 0 if and only if sm = 0 for some s ∉ p, i. e. if and only if

ann m ⊈ p. ✓

definition 0.13. We say that a prime ideal p is an associated prime for M if there exists m ∈ M such that p = ann m.

We write Ass M (sometimes called the assassinator of M) for the set of associated primes for M.

The French call Ass M “l’assassin de M.” A literal translation of this would be “the murderer of M,” or “the killer of M.” Since this sounds much more violent in English than it does in French, English speaking mathematicians have invented the word “assassinator” to translate the French.

lemma 0.14. A prime ideal p is an associated prime for M if and only if M contains a submodule isomorphic to R/p. proof: For m ∈ M, if p = ann(m) then R/p is isomorphic to the cyclic submodule Rm. ✓

proposition 0.15. If p is a prime ideal then Ass R/p = {p}. proof: Clearly p ∈ Ass R/p. If q ∈ Ass R/p then R/p contains a submodule X isomorphic to R/q. But then X is cyclic, i. e. is a principal ideal in the integral domain R/p. Hence X ≈ R/p and q = ann X = ann R/p = p. ✓

For p to be an associated prime of M it is important that we have p = ann m for some m, not merely p ⊆ ann m. It is also important that p be a prime ideal. It is not obvious, however, that there always exist ideals satisfying both of these requirements. In other words, it is not a priori obvious that if M ≠ 0 then Ass M will be non-empty. For noetherian rings, however, this does turn out to be the case. (See Proposition ** below.)

example 0.16. Let R = ℤ.

(1) If p is a non-zero prime number, then (p) ∈ Ass M if and only if M contains an element of order p.

(2) 0 ∈ Ass M if and only if M contains a torsion free element, i. e. an element m with rm ≠ 0 for all r ≠ 0 ∈ ℤ.

We say that a module M over an integral domain R is torsion if for every m ∈ M there exists r ≠ 0 ∈ R such that rm = 0. We say that M is torsion free if no non-trivial submodule of M is torsion, or, equivalently,

(∀m ∈ M) (∀r ∈ R) m ≠ 0 & r ≠ 0 ⇒ rm ≠ 0.

proposition 0.17. Let M be a non-trivial module over a noetherian integral domain R. Then

(1) M is torsion if and only if 0 ∉ Ass M.

(2) M is torsion free if and only if Ass M = {0}. proof: (1) M is torsion if and only if there does not exist an element m ∈ M whose annihilator is the zero ideal.

(2) M is torsion free if and only if the annihilator of every non-trivial element is the zero ideal. ✓

If ann M = 0 then we say that M is faithful. In abelian group theory, a group G which is not faithful as a ℤ-module is called bounded.

Clearly a module which is not faithful is torsion, however the converse is far from true. For instance, with R = ℤ, the module ℚ/ℤ is torsion but also faithful.

A classical theorem in abelian group theory implies that every module over a dedekind domain which is not faithful is a direct sum of cyclic modules. Therefore it is only the faithful torsion modules which have any real interest in abelian group theory.

The following theorem is known as the Krull Intersection Theorem. In abelian group theory terms, it says that a finitely generated torsion free module has no (non-trivial) elements of infinite p-height, for any prime ideal p.

proposition 0.18. If M is a finitely generated torsion free module over a noetherian integral domain and p is any prime ideal then ⋂_{1}^{∞} p^{k}M = 0.

note: Counter-examples abound when M is not finitely generated. For instance, if R = ℤ, M = ℚ, and p is any non-trivial prime ideal then pM = M so ⋂ p^{k}M = M ≠ 0.

lemma 0.19. Suppose that M is a module over a commutative noetherian ring R and let m ≠ 0 ∈ M. Let S be a multiplicative set in R such that S ∩ ann m = ∅. Then there exists p ∈ Ass M such that ann m ⊆ p and S ∩ p = ∅. In fact, if q is any prime ideal such that ann m ⊆ q then there exists p ∈ Ass M with ann m ⊆ p ⊆ q.

proof: Since ann m ≠ R then by Zorn’s Lemma there exist ideals q maximal with respect to the properties ann m ⊆ q and q ∩ S = ∅. We claim that any such ideal must be prime. In fact, if r, s ∉ q then S ∩ (r) + q ≠ ∅ and S ∩ (s) + q ≠ ∅. Thus there exist rx + q_{1}, sy + q_{2} ∈ S, with x, y ∈ R, q_{1}, q_{2} ∈ q. Then rsxy + rxq_{2} + syq_{1} + q_{1}q_{2} ∈ S and since S ∩ q = ∅ it follows that rs ∉ q.

Let q be such a prime ideal. Since R is noetherian, among the ideals p such that p ⊆ q and p = ann rm for r ∈ R and rm ≠ 0, there exist ones maximal with this property. Let p = ann rm be such an ideal. Clearly ann m ⊆ ann rm = p. We claim that p is prime. In fact, if r_{1}r_{2} ∈ p and r_{2} ∉ p then r_{2}rm ≠ 0 and p ⊆ p + (r_{1}) ⊆ ann(r_{2}rm) ⊆ q (why?). By the maximality of p we conclude that p + (r_{1}) = p, i. e. r_{1} ∈ p. This shows that p is prime and since p = ann rm, thus

p ∈ Ass M. ✓

corollary 0.20. If R is noetherian then

Ass M = ∅ ⟺ M = 0.

proof: If M = 0 then clearly Ass M = ∅. Conversely, if M ≠ 0 then there exists m ≠ 0 ∈ M. Thus by Lemma 0.19 there exists p ∈ Ass M. ✓

There are actually occasions when the most economical way of proving that a module M over a noetherian ring is trivial is by proving that Ass M = ∅.

proposition 0.21. If m ∈ M then m = 0 if and only if m/1 = 0 ∈ M_{p} for all p ∈ Ass M. proof: If m ≠ 0 then by Lemma 0.19 there exists p ∈ Ass M with ann m ⊆ p. By

Lemma 0.12, m/1 ≠ 0 ∈ M_{p}. ✓

definition 0.22. The support of an R-module M is defined to be Supp M = {p | M_{p} ≠ 0}.

proposition 0.23. If M is finitely generated then

Supp M = {p | p is prime and p ⊇ ann M}.

proof: If ann M ⊈ p then sM = 0 for some s ∉ p so M_{p} = 0, i. e. p ∉ Supp M.

Now suppose that m_{1}, …, m_{t} is a finite set of generators for M. Let p be a prime ideal with ann M = ⋂_{1}^{t} ann m_{i} ⊆ p. Since p is prime, ann m_{i} ⊆ p for some i. Thus for all s ∉ p, s ∉ ann m_{i} so sm_{i} ≠ 0. Hence m_{i}/1 ≠ 0 ∈ M_{p} so that M_{p} ≠ 0 and

p ∈ Supp M. ✓

example 0.24. Let R = ℤ and M = ℚ/ℤ. M is not a finitely generated module. The ideal 0 is prime and ann M = 0 but 0 ∉ Supp M. In fact, for every m ∈ M there exists s ∉ (0) such that sm = 0. Thus the localization of M at the zero ideal is trivial.

note. By the Localization-Globalization Theorem (Theorem 0.5), M = 0 ⟺ Supp M = ∅.

We have the following variation on Nakayama’s Lemma.

proposition 0.25. If M is a finitely generated R-module and p ∈ Supp M then pM ≠ M. proof: If pM = M then pM_{p} = M_{p}. Considering M as an R_{p}-module (necessarily finitely generated), Nakayama’s Lemma then implies M_{p} = 0, i. e. p ∉ Supp M. ✓ corollary 0.26. Ass M ⊆ Supp M. Conversely, if q ∈ Supp M then there exists p ∈ Ass M with p ⊆ q. proof: If p ∈ Ass M then M contains a submodule M_{1} isomorphic to R/p and so 0 ≠ (M_{1})_{p} ⊆ M_{p}, so p ∈ Supp M.

Now suppose that q ∈ Supp M. Then there exists m ∈ M such that m/1 ≠ 0 ∈ M_{q}. Thus by Lemma 0.12, ann m ⊆ q. Therefore by Lemma 0.19, there exists p ∈ Ass M with

p ⊆ q. ✓

corollary 0.27. If Ass M consists of maximal ideals then Supp M = Ass M.

In particular, if G is a bounded (i. e. not faithful) abelian group, then Corollary 0.27 combined with Proposition FinGenSupp says that Ass G consists of the prime divisors of the smallest positive integer n such that nG = 0. From this we easily derive the fact that if G is finite, then Ass G consists of those primes dividing the order of G. (This is essentially just Cauchy’s Theorem.)

proposition 0.28. (1) Supp_{S^{−1}R} S^{−1}M = { pS^{−1}R | p ∈ Supp M & p ∩ S = ∅ } (2) Ass_{S^{−1}R} S^{−1}M = { pS^{−1}R | p ∈ Ass M & p ∩ S = ∅ }

(3) Ass_{R} S^{−1}M = { p | p ∈ Ass M & p ∩ S = ∅ }.

proposition 0.29. If N ⊆ M then Ass N ⊆ Ass M ⊆ Ass N ∪ Ass M/N. proof: If p ∈ Ass N then p = ann n for n ∈ N and clearly p ∈ Ass M. Now suppose that p ∈ Ass M. Then M contains a submodule M_{1} isomorphic to R/p. If M_{1} ∩ N ≠ 0 then p ∈ Ass M_{1} ∩ N ⊆ Ass N since Ass M_{1} = {p} by Proposition 0.15. On the other hand, if M_{1} ∩ N = 0 then M/N contains the submodule (M_{1} + N)/N ≈ M_{1} ≈ R/p, so

p ∈ Ass M/N. ✓

proposition 0.30. If M is a finitely generated module over a commutative noetherian ring then Ass M is a finite set. (note: In general, Supp M will not be finite.) proof: M is noetherian, thus there exists a submodule N of M maximal with respect to the property that Ass N is finite. Furthermore N ≠ 0 since if p ∈ Ass M then M contains a submodule N′ isomorphic to R/p and by Proposition 0.15, Ass R/p = {p}. If N ⊊ M applying this reasoning to M/N shows that there is a submodule N′′ of M with N ⊊ N′′ such that Ass N′′/N is finite and non-empty. But by Proposition 0.29, Ass N′′ ⊆ Ass N ∪ Ass N′′/N, contradicting the maximality of N. Therefore N = M and Ass M is finite. ✓

A module M is called p-primary if Ass M = {p}. (Somewhat inconsistently, a submodule N of M is sometimes called a p-primary submodule if M/N is p-primary. This is a historical accident due to the fact that the term p-primary comes both from abelian group theory and ideal theory. The second usage will not occur in this book.)

Proposition 0.15 shows that for any prime ideal p, R/p is p-primary. The following proposition shows how to generalize in case p is maximal.

proposition 0.31. If p is a maximal ideal then for all n > 0 the cyclic module R/p^{n} is p-primary.

proof: If p′ is any prime then by Proposition 0.23, p′ ∈ Supp R/p^{n} if and only if p^{n} ⊆ p′ i. e. if and only if p ⊆ p′. Since p is maximal it then follows that Supp R/p^{n} = {p}. Since ∅ ≠ Ass R/p^{n} ⊆ Supp R/p^{n} it follows that Ass R/p^{n} = {p}. ✓

If p is not maximal then R/p^{n} need not be p-primary. For dedekind domains, however, which are the rings of interest in this book, all non-trivial prime ideals are maximal.

proposition 0.32. If p is minimal in Ass M then M_{p} is p-primary.

proof: By Proposition 0.28, Ass M_{p} = {p′ ∈ Ass M | p′ ⊆ p} = {p}. ✓

In abelian group theory, we know that an abelian group G is p-primary for some prime number p if and only if the order of every element of G is a power of p. The analogous characterization for modules over a commutative ring is the following:

proposition 0.33. If R is noetherian and M is an R-module and p is a prime ideal, the following conditions are equivalent:

(1) M is p-primary.

(2) The natural map θ: M → M_{p} is monic and

(∀m ∈ M) (∃k ≥ 1) p^{k}m = 0.

proof: (1) ⇒ (2): Suppose that M is p-primary, i. e. Ass M = {p}. Then by Lemma 0.12, if m ≠ 0 ∈ M, then m/1 ≠ 0 ∈ M_{p}. Therefore θ: M → M_{p} is monic. Now let r ≠ 0 ∈ p and let S = {r^{k} | k ≥ 1}. By Proposition 0.28, p ∉ Ass S^{−1}M ⊆ {p}. Therefore Ass S^{−1}M = ∅ and so S^{−1}M = 0. This means that for each m ∈ M there exists r^{k} ∈ S such that r^{k}m = 0. Since this is true for all r ∈ p and p is finitely generated (because R is noetherian) it then follows easily that p^{k′}m = 0 for some k′.

(2) ⇒ (1): Now suppose the stated conditions hold. Since M → M_{p} is monic, by Proposition 0.29, Ass M ⊆ Ass M_{p}, so it suffices to prove that Ass M_{p} = {p}. Therefore we may assume wlog that M = M_{p}. Now let q ∈ Ass M. Then q ⊆ p and q = ann m for some m. By assumption, for some k, p^{k} ⊆ ann m = q Then p ⊆ q since q is prime. Therefore q = p, showing that M is p-primary. ✓

definition 0.34. We say that r ∈ R is a zero divisor on M if rm = 0 for some m ≠ 0 ∈ M. proposition 0.35. If R is noetherian then ⋃{p | p ∈ Ass M} is the set of elements in R which are zero divisors on M. proof: By definition, r ∈ R is a zero divisor on M if and only if r ∈ ann m for some m ≠ 0 ∈ M. If this is the case then by Lemma 0.19, r ∈ p for some p ∈ Ass M. Conversely, if p ∈ Ass M then pm = 0 for some m ≠ 0 ∈ M, so p consists of zero divisors

on M. ✓

corollary 0.36. If R is noetherian then ⋃{p | p ∈ Ass R} equals the set of zero divisors in R.

MODULES WITH FINITE LENGTH. Henceforth we will assume that the ring R is noetherian.

In abelian group theory, an important role is played by the finite groups. A number of important definitions and theorems in the theory of finite rank torsion free groups have to do with the condition that a certain quotient G/H is finite. (For instance, if G is a finite rank torsion free group and H ⊆ G, then H is quasi-equal to G if and only if G/H is finite.)

For many commutative rings R, there do not exist non-trivial R-modules which are finite as sets. Instead, the appropriate condition is that a module have finite length.

definition 0.37. A module M over a ring R is said to have finite length if and only if it has a composition series

0 = M_{0} ⊊ M_{1} ⊊ … ⊊ M_{ℓ} = M

where each quotient M_{i}/M_{i−1} is a simple module, and in this case we define length M to be the length ℓ of this composition series. The Jordan-Hölder Theorem asserts that length M is independent of the particular composition series.

A module has finite length if and only if it is both noetherian and artinian.

theorem 0.38. If R is a noetherian ring and M is an R-module such that all the associated primes of M are maximal, then Ass M = Supp M and for each p ∈ Ass M, the canonical map M → M_{p} is surjective and M_{p} is isomorphic to the p-primary component of M, i. e. to {m ∈ M | (∃n) p^{n}m = 0}. Furthermore M is the direct sum of its p-primary components. proof: By Theorem 0.5, to show that M → M_{p} is surjective, it suffices to prove that for every maximal ideal 𝔪, the induced map M_{𝔪} → (M_{p})_{𝔪} is surjective. But since p is maximal, if 𝔪 ≠ p then 𝔪 ∉ Ass M_{p} = Supp M_{p} so that (M_{p})_{𝔪} = 0 so M_{𝔪} → (M_{p})_{𝔪} is surely surjective. And if 𝔪 = p then M_{𝔪} → (M_{p})_{𝔪} is essentially the identity map. Likewise this shows that for each maximal ideal 𝔪, the obvious map M_{𝔪} → (⨁_{p∈Ass M} M_{p})_{𝔪} is an isomorphism, showing that M → ⨁_{Ass M} M_{p} is an isomorphism.

Now for given p ∈ Ass M, define a functor F by F(X) = {x ∈ X | (∃n) p^{n}x = 0}. (Later, we will frequently write X[p^{∞}] = F(X).) Since p is maximal, by Proposition 0.28, M_{p} is p-primary, so by Proposition 0.33, F(M_{p}) = M_{p}. On the other hand, for p ≠ q ∈ Ass M then F(M_{q}) = 0 since if x ∈ F(M_{q}) then (∃n) p^{n}x = 0 and by Proposition 0.33, (∃n) q^{n}x = 0. Taking n as the larger of these two, by Proposition 0.10 p^{n} + q^{n} = R and Rx = (p^{n} + q^{n})x = 0 so x = 0. It follows that

F(M) = F (⨁_{Ass M} M_{p}) = M_{p}. ✓

note: When R = ℤ the preceding theorem simply gives us the familiar result that every torsion abelian group (finitely generated or not) is a direct sum of its p-primary components.

Recall that in general, one cannot conclude that two modules M and N are isomorphic from the fact that M_{p} ≈ N_{p} for every prime ideal p. However it follows from Theorem 0.38 that this is the case if Ass M (and consequently Ass N) consists of maximal ideals.

corollary 0.39. If Ass M consists of maximal ideals and M_{p} ≈ N_{p} for every maximal ideal p, then M ≈ N. proof: If M_{p} ≈ N_{p} for every maximal ideal p then this will in fact be true for every prime ideal p. It then follows that Ass N = Ass M. But by Theorem 0.38, M ≈ ⨁_{Ass M} M_{p} and N ≈ ⨁_{Ass N} N_{p}. It follows that M ≈ N. ✓

proposition 0.40. If M is an artinian module then Ass M is finite and consists of maximal ideals. proof: If p ∈ Ass M then M contains a submodule isomorphic to R/p. Thus if M is artinian then R/p is artinian. But an artinian integral domain is a field. Thus p is a maximal ideal.

It now follows from Theorem 0.38 that M ≈ ⨁_{Ass M} M_{p}, and that M_{p} ≠ 0 for all p ∈ Ass M. Since M is artinian, this direct sum can have only finitely many components. Thus Ass M is finite. ✓

theorem 0.41. A module M over a noetherian ring R has finite length if and only if it is finitely generated and Ass M consists of only maximal ideals. proof: ( ⇒ ): If M has finite length then it is certainly noetherian, hence finitely generated. It is also artinian so by Proposition 0.40, all its associated primes are maximal.

( ⇐ ): If M is finitely generated then it is noetherian, so there exists a submodule N of M which is maximal with respect to the property of having finite length. Now if Ass M consists of maximal ideals then by Proposition 0.27, Ass M = Supp M. Since Supp M/N ⊆ Supp M, then Supp M/N consists of maximal ideals and so Supp M/N = Ass M/N. If p ∈ Ass M/N then M/N contains a submodule N_{1}/N isomorphic to R/p (where N ⊆ N_{1} ). Since p is maximal, N_{1}/N is simple and since N has finite length it follows that N_{1} has finite length, contradicting the maximality of N. ThusAss M/N = ∅ so M/N = 0 and M has finite length. ✓

proposition 0.42. Let M have finite length and I = ann M. Then Ass M consists of the prime ideals containing I. Furthermore length M ≥ length R/I. proof: By Corollary 0.27 and Proposition 0.23, Ass M = Supp M and Supp M consists of those prime ideals containing ann R/I = I.

Thus Ass M = Ass R/I and so by Theorem 0.41, R/I has finite length. By Proposition 0.38, M ≈ ⨁ M_{p} and R/I ≈ ⨁(R/I)_{p}. It thus suffices to prove that for each p ∈ Ass M, length M_{p} ≥ length(R/I)_{p}. In other words, we may suppose that R is a local ring and I = p^{k}, where p^{k−1}M ≠ 0. Then for i < k, applying Nakayama’s Lemma to p^{i}M shows that p^{i+1}M ⊊ p^{i}M. Then the series

0 = p^{k}M ⊊ p^{k−1}M ⊊ … ⊊ pM ⊊ M

with length k can be refined to a composition series, so that k = length R/I ≤ length M. ✓

lemma 0.43. Let R be a noetherian local ring with maximal ideal 𝔪. Let E be the injective envelope of R/𝔪. Then the functor Hom_{R}( , E) restricted to the category of finite-length R-modules preserves length. Conversely, if M is an R-module such that Hom_{R}(M, E) has finite length, then M has finite length. proof: Consider first a module S with length 1 (i. e. a simple module). Then S ≈ R/𝔪 (so E is isomorphic to the injective envelope of S). Now if φ ∈ Hom_{R}(S, E), then φ(S) is simple (unless trivial) and φ(S) ∩ (R/𝔪) ≠ 0 since E is an essential extension of R/𝔪. Thus φ(S) = R/𝔪, since both these modules are simple. From this we see that Hom_{R}(S, E) ≈ Hom_{R}(R/𝔪, R/𝔪) ≈ R/𝔪, so that length Hom_{R}(S, E) = 1. Since the functor Hom_{R}( , E) preserves (or, more precisely, reverses) short exact sequences, it now follows by induction that for all R-modules M with finite length, length Hom_{R}(M, E) = length M.

Now suppose that M is an R-module such that Hom_{R}(M, E) has finite length. Consider first the case where M is finitely generated. M contains submodules N such that M/N has finite length (for instance, N = M). By the preceding, for such an N, Hom_{R}(M/N, E) has the same length as M/N. But the surjection M → M/N induces a monomorphism Hom_{R}(M/N, E) → Hom_{R}(M, E), so length M/N ≤ length Hom_{R}(M, E). Thus there is a bound on the length of M/N for those N such that M/N has finite length. But if M/N has finite length and N ≠ 0, then since N is noetherian, it contains a maximal proper submodule N′ and length M/N′ = 1 + length M/N. Therefore eventually we must have N = 0, so that M has finite length.

If M is not finitely generated and Hom_{R}(M, E) has finite length, consider a finitely generated submodule M′ of M. The inclusion map from M′ to M induces a surjection Hom_{R}(M, E) → Hom_{R}(M′, E) (because E is injective). Therefore Hom_{R}(M′, E) has finite length and so M′ has finite length by the preceding paragraph, and in fact the lengths of the finitely generated submodules of M are all bounded by length Hom_{R}(M, E). Furthermore, if M′′ is also finitely generated and M′ ⊊ M′′ then length M′′ > length M′. It follows that there exists a maximally finitely generated submodule of M. But clearly this is only possible if this submodule is M itself. Therefore M is finitely generated and hence, by the preceding paragraph, has finite

length. ✓

theorem 0.44. Let R be a commutative noetherian local ring and let 𝔪 be its unique maximal prime ideal. The following conditions are equivalent:

(1) R is artinian.

(2) R has finite length (as an R-module).

(3) 𝔪 is the only prime ideal in R.

(4) Ass R = {𝔪}.

(5) 𝔪^{k} = 0 for some positive integer k.

(6) The injective envelope E of R/𝔪 is finitely generated.

(7) There exists a finitely generated injective R-module. proof: (2) ⇒ (1): Clear.

(1) ⇒ (4): By Proposition 0.40.

(4) ⇒ (5): If Ass M = {𝔪} then for some k, 𝔪^{k}1 = 0 ∈ R so 𝔪^{k} = 0.

(5) ⇒ (2): For each i from 0 to k, 𝔪^{i}/𝔪^{i+1} is finitely generated and a vector space over the field R/𝔪, hence is a finite-length R-module. Thus by induction, R/𝔪^{i} has finite length. If 𝔪^{k} = 0, this shows that R has finite length.

(5) ⇒ (3): If 𝔪^{k} = 0 and p is a prime ideal, then 𝔪^{k} ⊆ p so 𝔪 ⊆ p. Since 𝔪 is maximal, p = 𝔪.

(3) ⇒ (4): Clear since Ass R ≠ ∅.

(2) ⇔ (6): If E = E(R/𝔪) is finitely generated, then it has finite length, since all its associated primes are maximal. And conversely, if it has finite length then it is certainly finitely generated. But E ≈ Hom_{R}(R, E). Therefore the preceding lemma implies that E has finite length if and only if R has finite length.

(6) ⇒ (7): Clear.

(7) ⇒ (6): Let M be a non-trivial finitely generated injective module and let p ∈ Ass M. Then M contains a copy of R/p. Since R/p is an essential submodule of R_{p}/pR_{p}, M therefore contains a submodule isomorphic to R_{p}/pR_{p}. Therefore R_{p}/pR_{p} is finitely generated. Now let x ∉ p. Then multiplication by x is an isomorphism on R_{p}/pR_{p}. If x ∈ 𝔪, this contradicts Nakayama’s Lemma. Therefore x ∉ 𝔪 and we conclude that p = 𝔪. This shows that 𝔪 ∈ Ass M, and so M contains a submodule isomorphic to R/𝔪. Since M is injective, it thus contains a copy of E = E(R/𝔪). Therefore E is finitely generated.

(6) ⇒ (4) [Alternate proof]: Since Ass E = {𝔪}, it follows that for each e ∈ E, 𝔪^{k}e = 0 for some k. Thus if E is finitely generated, then 𝔪^{k}E = 0 for some k. It now suffices to show that E is faithful in order to conclude that 𝔪^{k} = 0. Suppose in fact, by way of contradiction, that rE = 0 for some r ≠ 0 ∈ R. The principal ideal (r) is isomorphic to R/𝔞, where 𝔞 = ann r. Since 𝔞 ⊆ 𝔪, there is a surjection φ: (r) → R/𝔪 (so φ(r) ≠ 0). Since E = E(R/𝔪), this extends to a map φ: R → E. Then φ(r) = rφ(1) = 0 since rE = 0. This is a contradiction, showing that no such r exists and therefore E is faithful. Since 𝔪^{k}E = 0, it follows that 𝔪^{k} = 0. ✓

corollary 0.45. A commutative noetherian ring is artinian if and only if every prime ideal is maximal (including the zero ideal, if applicable). If this is the case, then R is a finite product of local rings each of which has a unique prime ideal.

RANK-ONE PROJECTIVE MODULES.

proposition 0.46. Let R be an integral domain with quotient field F and let P be as R-submodule of F. Then the following conditions are equivalent:

(1) P is projective.

(2) There exist elements p_{1}, …, p_{n} ∈ P and f_{1}, …, f_{n} ∈ F such that f_{i}P ⊆ R for all i and ∑ f_{i}p_{i} = 1.

(3) There exists a submodule M of F such that MP = R. Furthermore in this case P is generated by p_{1}, …, p_{n}. proof: (1) ⇒ (2): If P is projective then it is a summand of a free module R^{(I)} . Then there exist maps σ: P → R^{(I)} and π: R^{(I)} → P such that πσ = 1_{P} . Localizing at the zero ideal, σ extends to a map σ_{0} : F → F^{(I)} and π to a map π_{0} : F^{(I)} → F. For each i ∈ I let f_{i} be the i^{th} coordinate of σ_{0}(1) and let p_{i} = π_{0}(e_{i}), where e_{i} is the canonical i^{th} basis vector of F^{(I)} . Then the composition of σ_{0} with the projection of F^{(I)} onto the i^{th} coordinate is given by x ↦ f_{i}x. Since this composition maps P into R it follows that f_{i}P ⊆ R. Furthermore since π is given by ∑ y_{i}e_{i} ↦ ∑ y_{i}p_{i}. Then the equation π_{0}σ_{0}(1) = 1 translates to ∑ f_{i}p_{i} = 1. This sum can have only finitely many non-trivial terms, and at this point we can replace I by the finite set of i ∈ I such that f_{i}p_{i} ≠ 0.

(2) ⇒ (3): Given f_{i} and p_{i} as indicated, let M be the submodule of F generated by f_{1}, …, f_{n}. Then clearly MP ⊆ R. But 1 = ∑ f_{i}p_{i} ∈ MP so MP = R. Note also that p_{1}, …, p_{n} generate P since for p ∈ P, we have p = p ∑ f_{i}p_{i} = ∑(f_{i}p)p_{i} and all f_{i}p ∈ R.

(3) ⇒ (2): If MP = R then 1 ∈ MP so there exist f_{i} ∈ M, p_{i} ∈ P with ∑ f_{i}p_{i} = 1. Furthermore for all i, f_{i}P ⊆ MP = R.

(2) ⇒ (1): Given f_{i} ∈ F and p_{i} ∈ P with the indicated property, map P into R^{n} by σ: p ↦ (f_{1}p, …, f_{n}p) and map R^{n} to P by π: (r_{1}, …, r_{n}) ↦ ∑ r_{i}p_{i}. Then πσ(p) = ∑ pf_{i}p_{i} = p1 = p. Thus σ is a split monomorphism and P is a summand of a free module, hence is projective. ✓

Note that the module M above is uniquely determined by P. In fact, if MP = M′P = R then M = MR = MP M′ = RM′ = M′. We write M = P^{−1} and say that the rank-one module P is invertible.

SUBMODULES OF DIRECT SUMS. If W is a dedekind domain (see below) with quotient field Q, then every torsion free W -module with rank at most r is isomorphic to a submodule of Q^{r}. Thus if we had a good theory for submodules of direct sums K_{1} ⊕ ⋯ ⊕ K_{r}, we would completely understand finite torsion torsion free W -modules.

When the direct sum has only two factors, we have the following theorem:

theorem 0.47. Let K and N be R-modules , let κ and ν be the projections of K ⊕ N onto K and N, and let K_{1} ⊆ K_{2} ⊆ K and N_{1} ⊆ N_{2} ⊆ N. Then the submodules M of K ⊕ N such that M ∩ K = K_{1} and M ∩ N = N_{1} and κ(M) = K_{2} and ν(M) = N_{2} are in one-to-one correspondance with the isomorphisms φ from K_{2}/K_{1} to N_{2}/N_{1} . Specifically, such an isomorphism φ corresponds to the inverse image in K ⊕ N of the submodule

{(x, φ(x) | x ∈ K/K_{1}}

of K/K_{1} ⊕ N/N_{1} .

proof: ****

Unfortunately, the value of this theorem is drastically limited because it does not provide us a good way of determining whether or not two submodules of K ⊕ N are isomorphic.

DEDEKIND DOMAINS.

For my own perverse reasons, I will generally use the symbol W rather than R when referring to a dedekind domain. As I now start to develop the concepts needed to define a dedekind domain, I will now start calling my commutative ring W. The quotient field of W will be denoted by Q.

proposition 0.48. Let W be an integral domain with quotient field Q. The following conditions are equivalent:

(1) If q ∈ Q and q satisfies a monic polynomial f(X) ∈ W [X], then q ∈ W.

(2) If W′ is a subring of Q with W ⊆ W′ and W′ is a finitely generated W -module, then W′ = W.

(3) if W′ is a subring of Q with W ⊆ W′ and wW′ ⊆ W for some w ≠ 0 ∈ W, then W′ = W.

proof: **** (This is a very standard result.)

definition 0.49. An integral domain satisfying the conditions of Proposition 0.49 is called integrally closed (in its quotient field).

Using the terminology we will introduce in Chapter 3, condition (3) in Proposition 0.48 says that W is not quasi-equal to any subring of Q properly containing it. lemma 0.50. If W is integrally closed then so is S^{−1}W for every multiplicative set S. proof: Let Q denote the quotient field of W and let q ∈ Q be integral over S^{−1}W and let f ∈ S^{−1}W [X] be a monic polynomial satisfied by q. Let d be the degree of f and let s ∈ S be a common denominator for the coefficients of f. Then s^{d}f(q) = 0 and one sees that sq satisfies a monic polynomial in W [X]. Thus if W is integrally closed, sq ∈ W. It follows that q ∈ S^{−1}W. Thus S^{−1}W is integrally closed. ✓

lemma 0.51. A commutative ring W is integrally closed if and only if W_{p} is integrally closed for all prime ideals p. proof: ( ⇒ ): By Lemma 0.50.

( ⇐ ): Let q ∈ Q be integral over W. Then a fortiori q is integral over each W_{p}. If all W_{p} are integrally closed, then q ∈ ⋂ W_{p} = W (c. f. Proposition 0.*). ✓

proposition 0.52. Unique factorization domains are integrally closed. In particular, principal ideal domains are integrally closed. proof: If W is a UFD with quotient field Q and q ∈ Q and f(q) = 0 where f is monic in W [X], then X − q divides f in Q[X]. By Gauss’s Lemma, X − q ∈ W [X], so

q ∈ W. ✓

definition 0.53. A local principal ideal domain is called a discrete valuation ring.

proposition 0.54. A discrete valuation ring is a maximal proper subring of its quotient field. proof: Let W be a discrete valuation ring and let (p) be the unique maximal ideal in W. By Proposition 0.18, ⋂_{1}^{∞}(p^{k}) = 0 so for each r ≠ 0 ∈ W there exists a largest k with r ∈ (p^{k}). Thus r = up^{k} with k ≥ 0 and u ∉ (p), so u is invertible since (p) is the unique maximal ideal in W. From this we see easily that each non-trivial element of the quotient field Q has the form x = up^{k} where u is invertible in W and k is an integer. If x ∉ W then k < 0.

It now follows that any subring of Q strictly containing W must contain p^{k} for some k < 0, and hence contains p^{nk} for all n ≥ 1. We then see that it contains all elements up^{m} for any integer m and invertible element u of W. In other words, the subring contains all of Q. ✓

proposition 0.55. A local integral domain with only one non-trivial prime ideal is integrally closed if and only if it is a discrete valuation ring. proof: ( ⇐ ): A discrete valuation ring is a principal ideal domain hence a unique factorization domain, hence by Proposition 0.52 is integrally closed.

( ⇒ ): **** ✓

proposition 0.56. An integral domain W is integrally closed if and only if it is the intersection of all the valuation rings on its quotient field which contain it. proof: ( ⇒ ):

( ⇐ ): This follows from a more general result that the intersection of any family of integrally closed subrings of a field Q is integrally closed. In fact, suppose that W = ⋂ W_{i} and each W_{i} is integrally closed. Let q ∈ Q be integral over W. Then it is integral over each W_{i}. Hence q ∈ W_{i} for all i since each W_{i} is integrally closed. Therefore q ∈ W, showing that W is integrally closed. ✓

proposition 0.57. If Q′ is a separable extension of Q and W is integrally closed, then the integral closure W′ of W in Q′ is a finitely generated W -module. proof: ****

proposition 0.58. Let W be an integral domain. The following conditions are equivalent:

(1) W is a principal ideal domain.

(2) All ideals in W are free W -modules.

(3) All finitely generated torsion free W -modules are free.

proof: (3) ⇒ (2): Clear.

(2) ⇒ (1): If r ≠ s ∈ W then r and s are linearly dependent since sr − rs = 0. Thus an ideal in W which is free must have a basis consisting of at most one element, hence be principal.

(1) ⇒ (3): **** ✓

theorem 0.59. Let W be an integral domain with quotient field Q. The following conditions are equivalent:

(1) Every ideal in W is projective.

(2) Every non-trivial ideal of W is invertible.

(3) W is noetherian, integrally closed, and all non-trivial prime ideals are maximal.

(4) W is noetherian and W_{p} is a discrete valuation ring for every non-trivial prime ideal p.

proof: (1) ⇔ (2): By Proposition 0.46 ideals are projective if and only if they are invertible.

Note now that by Proposition 0.46, projective ideals are finitely generated. Hence (1) implies that W is noetherian.

(1) ⇒ (4): The ideals in W_{p} have the form 𝔞_{p}, where 𝔞 is an ideal in W. By hypothesis 𝔞 is projective, so by Proposition 0.8 𝔞_{p} is a free W_{p}-module. Thus all ideals of W_{p} are free, so that W_{p} is a local principal ideal domain, i. e. a discrete valuation ring.

(4) ⇒ (3): To see that W is integrally closed, it suffices by Lemma 0.51 to see that W_{p} is integrally closed for all primes p. But this is true if W_{p} is a discrete valuation ring, since principal ideal domains are integrally closed. Now let p be a prime ideal in W. The prime ideals contained in p correspond to the prime ideals of W_{p}. Since W_{p} is a discrete valuation ring, its only prime ideals are pW_{p} and 0. Thus there are no non-trivial prime ideals strictly contained in p, so p has height one. It follows that all prime ideals of W are maximal.

(3) ⇒ (1): Let 𝔞 be an ideal in W. Since W is noetherian, 𝔞 is finitely generated. Hence by Proposition 0.8, to prove 𝔞 projective it suffices to prove that 𝔞_{p} is a free W_{p}-module for all primes p. But since W_{p} is a principal ideal domain, 𝔞_{p} is in fact

free. ✓

definition 0.60. An integral domain W satisfying the equivalent properties of Theorem 0.59 is called a dedekind domain.

proposition 0.61. If W is a dedekind domain and S a multiplicative set in W, then S^{−1}W is a dedekind domain. proof: ****

proposition 0.62. If 𝔞 is a non-trivial ideal in a dedekind domain W then there are only finitely many prime ideals containing 𝔞 and W/𝔞 has finite length. proof: Since 𝔞 = ann W/𝔞 ≠ 0, 0 ∉ Ass W/𝔞. Thus by Theorem 0.59, all the associated primes of W/𝔞 are maximal. Thus by Proposition 0.23 the prime ideals containing 𝔞 are precisely those in Supp W/𝔞 = Ass W/𝔞. Hence there are only finitely many of them. Since W/𝔞 is finitely generated (in fact cyclic), it thus follows from Theorem 0.41 that W/𝔞 has finite length. ✓

For the remainder of this book, W will denote a dedekind domain and Q will denote its quotient field. Furthermore the term “prime ideal” will be used henceforth to mean a non-trivial prime.

The following proposition is crucial to the study of finite rank torsion free modules over dedekind domains. It is curiously hard to track down in standard references.

proposition 0.63. If W′ is an integral domain and W ⊆ W′ ⊆ Q, then W′ is a dedekind domain and is the localization of W with respect to some multiplicative set S. proof: Let X be the set of prime ideals p of W which survive in W′, i. e. such that pW′ ≠ W′ or, equivalently, W′_{p} ≠ Q. Let S be the set-theoretic complement in W of ⋃_{X} p. Now if s ∈ S and p′ is a prime ideal in W′ then W ∩ p′ ∈ X so s ∉ W ∩ p′ so s ∉ p′ since s ∈ W. Thus s does not belong to any prime ideal of W′ and hence is invertible in W′. Therefore S^{−1}W ⊆ W′.

To prove that S^{−1}W = W′ it suffices to prove that S^{−1}W_{p} = W′_{p} for all prime ideals p of W. If p ∉ X then S^{−1}W_{p} = Q = W′_{p}. But if p ∈ X then W_{p} ⊆ S^{−1}W_{p} ⊆ W′_{p} ⊊ Q. Since W_{p} is a discrete valuation ring, by Proposition 0.54 it is a maximal subring of Q, so S^{−1}W_{p} = W′_{p}. It follows that S^{−1}W = W′. ✓

corollary 0.64. If W′ is a subring of Q containing W and M, N are W′-modules, then Hom_{W′}(M, N) = Hom_{W} (M, N) and M ⊗_{W′} N = M ⊗_{W} N.

proof: By Proposition 0.63, W′ = S^{−1}W for some multiplicative set S. Thus the result reduces to Proposition 0.3. ✓

proposition 0.65. If G is a torsion module over a dedekind domain, then for every prime p the p-primary component of G can be canonically identified with the localization G_{p}. Furthermore G = ⨁ G_{p}.

proof: Immediate from 0.38 since all primes in Ass G are necessarily maximal. ✓

proposition 0.66. If every p-primary component of a finitely generated torsion W -module G is cyclic, then G is cyclic.

proof: By Proposition 0.65, if G is torsion then G ≈ ⨁ G_{p} and if G is finitely generated then there can be only finitely many non-trivial summands in this expression. Now if each G_{p} is cyclic, choose a generator g_{p} for each G_{p}. Then g_{p}/1 = 0 ∈ G_{q} for q ≠ p. It follows that if g = ∑ g_{p}, then the image of g in G_{p} is a generator for all p. It then follows from Proposition 0.5 that g generates G, so G is cyclic. ✓

proposition 0.67. A finitely generated torsion module over a dedekind domain W has finite length and is a finite direct sum of cyclic modules of the form W/p^{k}, for various prime ideals p.

proof: If G is torsion then 0 ∉ Ass G. Since all non-trivial primes in a dedekind domain are maximal, it follows from Theorem 0.41 that G has finite length. Furthermore, since all the associated primes of G are necessarily maximal, by Proposition 0.38, G ≈ ⨁_{Ass G} G_{p}. It thus suffices to prove that each G_{p} is a direct sum of cyclic modules W/p^{k}. But we may think of G_{p} as a W_{p}-module, and by Proposition 0.59 W_{p} is a principal ideal domain, so this is a standard result. ✓

We will prove in Chapter 1 that every finitely generated torsion free module over a dedekind domain is a direct sum of ideals, hence in particular is projective. It then follows immediately that the torsion submodule of any W -module is a direct summand of that module. proposition 0.68. Every non-trivial ideal in a dedekind domain W can be uniquely written as a finite product of prime ideals. proof: Let 𝔞 be a non-trivial ideal in W. Then by Proposition 0.62, W/𝔞 has finite length, so Supp W/𝔞 is finite, which means simply that there are only finite many prime ideals containing 𝔞. Furthermore if p is one of these primes then by Proposition 0.* 𝔞W_{p} = p^{e_{p}}W_{p} for some positive integer e_{p}. If we now write 𝔞′ = p_{1}^{e_{1}} ⋯ p_{n}^{e_{n}}, where p_{1}, …, p_{n} are the primes containing 𝔞 then for every prime p, 𝔞W_{p} = 𝔞′W_{p} since p_{i}W_{p_{j}} = W_{p_{j}} for j ≠ i. Thus by Proposition 0.5, 𝔞 = 𝔞′. ✓

proposition 0.69. If 0 ≠ 𝔟 ⊆ 𝔞 ⊆ W then 𝔞/𝔟 ≈ W/𝔞^{−1}𝔟 ≈ 𝔟^{−1}𝔞/W. In particular, 𝔞/𝔟 is a cyclic W -module. proof: At first sight, this result seems self-evident. For instance, one simply multiplies both the numerator and denominator of 𝔞/𝔟 by 𝔞^{−1} to get W/𝔞^{−1}𝔟. This is not quite valid, however, mostly because 𝔞^{−1} is a set and not an element. This reasoning does work, however, if 𝔞 is a principal ideal.

By Theorem 0.38, 𝔞/𝔟 ≈ ⨁(𝔞/𝔟)_{p}. Thus to see that 𝔞/𝔟 ≈ W/𝔞^{−1}𝔟 it suffices to see that (𝔞/𝔟)_{p} ≈ (W/𝔞^{−1}𝔟)_{p} for all prime ideals p. Therefore we may localize at p. Hence we may suppose that W is a local dedekind domain, and therefore a principal ideal domain. But then the result is easy, as indicated above. ✓

proposition 0.70. A semi-local dedekind domain W is a principal ideal domain. proof: It suffices to see that every prime ideal p is principal. Let p_{1}, …, p_{n} be the finite set of non-trivial prime ideals in W. Since these are maximal ideals, p_{i}W_{p_{j}} = W_{p_{j}} and also p_{i} + p_{j} = W for i ≠ j. It follows that the Jacobson radical of W is J = ⋂ p_{i} = p_{1} ⋯ p_{n} (c. f. Proposition 0.10). Let x ∈ p_{1} with x ∉ p_{1}^{2} . By the Chinese Remainder Theorem, the canonical surjection W ⟶^{≈} ⨁ W/p_{i}^{2} mapping an element w onto the sum of its respective residue classes induces an isomorphism ζ: W/J^{2} ⟶^{≈} ⨁ W/p_{i}^{2} . Thus there exists w ∈ W such that ζ(w) = (x, 1, …, 1). Then wW_{p_{i}} = W_{i} = p_{1}W_{p_{i}} for i ≠ 1 and since W_{p_{1}} is a principal ideal domain one sees that wW_{p_{1}} = p_{1}W_{p_{1}} . Thus by Proposition 0.5, p_{1} = (w). ✓

proposition 0.71. If 𝔞 is a non-trivial ideal in W and w_{1} ≠ 0 ∈ 𝔞 then there exists w_{2} ∈ 𝔞 such that 𝔞 is generated by w_{1} and w_{2} . Furthermore there exists q ∈ Q such that 𝔞^{−1} is generated by 1 and q. proof: Let 𝔟 = (w_{1}) ⊆ 𝔞. Then by Proposition 0.69, 𝔞/𝔟 is cyclic. If w_{1} ∈ 𝔞 is a representative for a generator of 𝔞/𝔟, then 𝔞 is generated by w_{1} and w_{2} .

Likewise, 𝔞^{−1}/W ≈ W/𝔞 is cyclic, and if q ∈ 𝔞^{−1} is a representative for a generator of 𝔞^{−1}/W, then 𝔞^{−1} is generated by 1 and q. ✓ proposition 0.72. If 𝔞 and 𝔟 are non-trivial ideals in a dedekind domain W then there exists an ideal 𝔞′ with 𝔞′ ≈ 𝔞 such that 𝔞′ + 𝔟 = W. proof: Choose a ≠ 0 ∈ 𝔞. Let 𝔠 = a𝔞^{−1} ⊆ W. Now by Proposition 0.71, 𝔠 is generated by 𝔟𝔠 together with an additional element x. Now multiplying the equation

𝔠 = xW + 𝔟𝔠

by 𝔠^{−1} = 𝔞/a yields

W = 𝔠^{−1}𝔠 = x𝔠^{−1} + 𝔟 = \frac{x𝔞}{a} + 𝔟.

But clearly 𝔞x/a ≈ 𝔞. ✓

proposition 0.73. If 𝔞 and 𝔟 are ideals in a dedekind domain W then (1) 𝔞 ⊕ 𝔟 ≈ W ⊕ 𝔞𝔟.

(2) 𝔞 ⊕ 𝔞^{−1} ≈ W ⊕ W. proof: (1) Let 𝔠 = 𝔞 + 𝔟. Then 𝔠^{−1}𝔞, 𝔠^{−1}𝔟 ⊆ 𝔠^{−1}(𝔞 + 𝔟) = W and 𝔠^{−1}𝔞 + 𝔠^{−1}𝔟 = W. Thus by Proposition 0.11, 𝔠^{−1}𝔞 ⊕ 𝔠^{−1}𝔟 ≈ W ⊕ 𝔠^{−2}𝔞𝔟. Thus 𝔞 ⊕ 𝔟 ≈ 𝔠 ⊕ 𝔠^{−1}𝔞𝔟. ****

proposition 0.74. If 𝔞 is an ideal in a dedekind domain W, then there exists an ideal 𝔟 such that 𝔞 ⊕ 𝔟 = W ⊕ W. proof: By Proposition 0.71, 𝔞 is generated by two elements. Hence there exists a surjection W ⊕ W → 𝔞. Since 𝔞 is projective this must split, so that W ⊕ W ≈ 𝔞 ⊕ 𝔟 for some W -module 𝔟, which is necessarily projective. Localizing at the zero ideal yields Q ⊕ Q = 𝔞_{0} ⊕ 𝔟_{0} , so that 𝔟 is isomorphic to a non-trivial projective submodule of Q. Then if d is a common denominator for a set of generators for 𝔟, then 𝔟 ≈ d𝔟 ⊆ R, so 𝔟 is isomorphic to an ideal. ✓

We say that G is divisible if wG = G for every w ≠ 0 ∈ W. (In other words, one can divide an element in G by any non-zero element of W: If g ∈ G and w ≠ 0 ∈ W, then there exists x ∈ G such that wx = g. Note however that x is not necessarily unique unless G is torsion free.)

proposition 0.75. A W -module G is flat if and only if it is torsion free and injective if and only if it is divisible. proof: Over any integral domain, a flat module is torsion free since ****. On the other hand, any module G is the direct union of its finitely generated submodules and if G is torsion free then by Proposition 0.** these finitely generated submodules are projective. Hence a torsion free module G is a direct union of flat submodules, hence is flat.

Now over any integral domain, an injective module is divisible since ****. On the other hand, to say that G is divisible is to say that ****. ✓

If 𝔞 is an ideal of W we will denote the module W/𝔞 by W (𝔞). For any W -module G we will write G[𝔞] to denote {g ∈ G | 𝔞 g = 0}. For a prime ideal p we will write G[p^{∞}] = ⋃_{1}^{∞} G[p^{n}]. This is simply the p-primary component of G as defined in Theorem 0.38. proposition 0.76. If p is a prime ideal, then there is an embedding γ_{k}: W (p^{k}) → W (p^{k+1}) such that the image of γ_{k} is then W (p^{k+1})[p^{k}] = pW (p^{k+1}). proof: By Proposition 0.**, W (p^{k}) ≈ p^{−k}/W and W (p^{l+1}) ≈ p^{−k−1}/W. Now p^{−k} ⊆ p^{−k−1} , and this inclusion then induces an embedding p^{−k}/W → p^{−k−1}/W. The image of this embedding is p(p^{−k−1}/W), and this can also be seen to be

(p^{−k−1}/W)[p^{k}]. ✓

For us, the significance of Proposition 0.76 is that it provides a chain of monomorphisms

W (p) ↣ W (p^{2}) ↣ W (p^{3}) ↣ · · · .

We can then take the direct limit, which we denote by W (p^{∞}):

W (p^{∞}) = lim_{→} W (p^{k}).

proposition 0.77. The following three modules are isomorphic:

(1) W (p^{∞})

(2) The p-primary component of Q/W.

(3) Q/W_{p}.

proof: The p-primary component of Q/W is equal to ⋃_{1}^{∞} p^{−k}/W and by Proposition 0.69, p^{−k}/W ≈ W (p^{k}). Furthermore, the inclusion p^{−k}/W → p^{−k−1}/W corresponds precisely to the embedding W (p^{k}) → W (p^{k+1}) indicated above. Thus the p-primary component of Q/W is isomorphic to W (p^{∞}).

On the other hand, by Proposition 0.**, the p-primary component of Q/W is isomorphic to (Q/W)_{p} = Q_{p}/W_{p} = Q/W_{p}. ✓

proposition 0.78. For each prime p, W (p^{∞}) is isomorphic to the injective envelope of W (p). Every injective W -module is a direct sum of modules isomorphic to W (p^{∞}) (for various p) and to Q. proof: ****

proposition 0.79. If G is a p-primary module, where p is a non-zero prime ideal, and if length G[p] = 1, then either G ≈ W (p^{k}) for some k < ∞ or G ≈ W (p^{∞}). proof: We claim first that G is an essential extension of G[p]. In fact, if N is a non-zero submodule of G then ∅ ≠ Ass N ⊆ Ass G = {p}. Thus there exists n ∈ N with ann n = p. Then 0 ≠ n ∈ N ∩ G[p].

It follows that the injective envelope of G is the same as the injective envelope of G[p]. But since length G[p] = 1, G[p] ≈ W (p). Thus the injective envelope of G is isomorphic to W (p^{∞}).

To finish, it suffices to prove that every non-trivial proper submodule of W (p^{∞}) is isomorphic to W (p^{k}) for some k. In fact, **** ✓ proposition 0.80. If p is a non-trivial prime ideal and G is a reduced p-primary module such that G/pG has finite length, then G has finite length.

proof: ****

THE p-ADIC COMPLETION. One of the most useful tools for understanding the deep structure of finite rank torsion free modules is the concept of completion. If M is a finite rank torsion free module and p is a prime ideal, then we can define the p-adic completion either as the inverse limit of the quotient modules M/p^{n}M or as the set of Cauchy sequences from M with respect to the p-adic topology. Really there is no difference between the two constructions. Since most readers will be familiar with the concept in any case, we give a quick and dirty construction:

For any prime ideal (or even any non-trivial ideal) p of R and any R-module M there exists the p-adic filtration on M: M ⊇ pM ⊇ p^{2}M ⊇ …. By taking this family of submodules as a neighborhood basis at 0, M becomes a topological module. (A neighborhood basis at a point m ∈ M consists of the family of cosets m + p^{n}M, n = 1, 2, ….) This topology is called the p-adic topology on M.

We write p^{∞}M = ⋂_{1}^{∞} p^{k}M and note that p^{∞}M is the closure of 0 in M with respect to the p-adic topology.

For convenience, I will choose definition of the completion which is quick and dirty, although it’s not the most useful way to think about it.

definition 0.81. Let M be a finite rank torsion free W -module. We define the p-adic completion of M to be the submodule M̂ of ∏_{1}^{∞} M/p^{k}M consisting of those sequences (m_{k})_{1}^{∞} ∈ ∏_{1}^{∞} M/p^{k}M such that (∀k) m_{k+1} ≡ m_{k} (mod p^{k}M).

The p-adic completion will be important in this book because of its algebraic properties. The topology will not be of any real relevance. However, for the record, we will give a few topological properties.

If we give each module M/p_{k}M the discrete topology, then ∏_{1}^{∞} M/p^{k}M has an induced product topology. The relative topology that M̂ inherits is called the inverse limit topology. A sequence {x_{t}} in M̂ converges to an element x ∈ M̂ with respect to the inverse limit topology if and only if the sequence converges in each separate coordinate. Since the topology on the modules M/p^{k}M is descrete, this means that if where x_{t} = {m_{tk}}_{1}^{∞} and x = {m_{k}}, then lim_{t→∞} x_{t} = x if and only if for each k, m_{tk} = m_{k} for sufficiently large t.

(This is related to the “finite topology” on End_{R} M as defined by Jacobson if we think of End_{R} M as a submodule of M^{M}.)

proposition 0.82. The inverse limit topology on M̂ is the same as the p-adic topology. M̂ is a complete topological module in this topology. If θ: M → M̂ is the canonical map, then θ(M) is topologically dense in M̂. proof: The neighborhood system at 0 in the inverse limit topology has a basis consisting of those submodules U_{n} consisting of elements whose first n coordinates are zero. Now since the first n coordinates live in M/p^{k}M for k ≥ n, it follows that p^{n}M̂ ⊆ U_{n}. On the other hand, since the sequences in M̂ satisfy the condition m_{n+k} ≡ m_{k} (mod p^{n}), it follows that if m_{r} = 0 for r ≤ n then m_{r} ∈ p^{n}M̂ for all r. Thus U_{n} ⊆ p^{k}M̂, so the inverse limit topology and the p-adic topology are the same.

It seems clear that the p-adic completion is a contrived device to simplify the proofs of certain theorems rather than a concept of intrinsic interest in its own right. There is no natural geometric way of visualizing the p-adic numbers intuitively, and there are no natural real-world applications which can be naturally represented by p-adic numbers.

However in terms of ring theory, there is one way in which the p-adic completion arises in a fairly intrinsic way.

proposition 0.83. If p is a (non-zero) prime in a dedekind domain, then End_{W} W (p^{∞}) ≈ Ŵ_{p}. proof: *****

As an important application of this, we can construct the famous Pontraygin module.

For convenience, suppose that W is a discrete valuation ring and let p be it unique non-trivial prime. We will construct a submodule M of Q ⊕ Q containing W ⊕ W such that M ∩ (Q ⊕ 0) = W ⊕ 0 and M ∩ (0 ⊕ Q) = 0 ⊕ W, and π_{i}(M) = Q. By Theorem 0.47, such a submodule will be given by an isomorphism φ from Q/W to itself. By Proposition 0.77, Q/W ≈ W (p^{∞}), and therefore by Proposition 0.83, φ corresponds to an invertible element α ∈ Ŵ_{p}.

More specifically, M will be the inverse image in Q ⊕ Q of the module

{(x, φ(x)) | x ∈ Q/W } = {(x, αx) | x ∈ Q/W } ⊆ Q/W ⊕ Q/W .

Now x ∈ Q/W can be written as a coset \frac{w}{p^{k}} + W for some w ∈ W and k ≥ 0. Using

this, we will describe M in the following way:

M = \{\frac{(w, αw)}{p^{k}} | w ∈ W, k ≥ 0 \} .

We need to make sense of this, since α ∈ Ŵ_{p} and so αw ∉ W.

For given k, there exists a_{k} ∈ W such that α ≡ a_{k} (mod p^{k}Ŵ_{p}). It follows that \frac{α}{p^{k}}

and \frac{a_{k}}{p^{k}} have the same image in Q/W. Now if φ is the automorphism of Q/W given

by φ(x) = αx, then φ(\frac{w}{p^{k}} + W) = \frac{αw}{p^{k}} + W = \frac{a_{k}w}{p^{k}} + W. Thus we can describe the

Pontryagin module corresponding to α ∈ Ŵ_{p} as

\{\frac{(w, a_{k}w)}{p^{k}} | w ∈ W, k ≥ 0 \} .

proposition 0.84. (1) M̂ is a torsion free R_{p}-module.

(2) M̂/pM̂ ≈ M/pM.

(3) M̂ is reduced but is q -divisible for every prime ideal q ≠ p.

(4) The p-adic topology on M̂ is the same as the topology it inherits as a subspace of ∏_{1}^{∞} M/p^{k}M if the modules M/p^{n}M are considered discrete spaces and ∏_{1}^{∞} M/p^{k}M is given the product topology.

(5) M̂ is a closed subspace of ∏_{1}^{∞} M/p^{k}M.

(6) The diagonal map from M to ∏_{1}^{∞} M/p^{k}M embeds M/p^{∞}M as a pure dense submodule of M̂.

proof: (1) The factors M/p^{k}M are all R_{p}-modules, hence so is ∏_{1}^{∞} M/p^{k}M. And if (m_{k}) ∈ ∏_{1}^{∞} M/p^{k}M satisfies the compatibility conditions m_{n+1} ≡ m_{k} (mod p_{k}M) and r ∈ R_{p}, then (rm_{k}) will also satisfy these conditions. Now suppose (m_{k}) ≠ 0 ∈ M̂. Then for some n, m_{k} ≠ 0 ∈ M/p_{k}M. ****

(2) For k ≥ 1 identify ∏_{k}^{∞} M/p_{k}M with the submodule of ∏_{1}^{∞} M/p^{k}M consisting of sequences (m_{k}) ∈ M̂ such that m_{k} = 0 for n ≤ k. Then the family of subsets ∏_{k}^{∞} M/p_{k}M forms a neighborhood basis at 0 for ∏_{1}^{∞} M/p^{k}M. Thus it will suffice to show that p_{k}M̂ = M̂ ∩ ∏_{k}^{∞} M/p_{k}M. ****

(3) It suffices to see that for any fixed k, the set of (m_{k}) ∈ M̂ satisfying m_{k+1} ≡ m_{k} (mod p^{k}M) is a closed subset of M̂. But this set is the kernel of a continuous map from M̂ into the discrete module M/p^{k}M.

(4) The diagonal map sends m ∈ M to the “constant sequence” (m_{k}) where for all n, m_{k} is the coset m + M/p_{k}M ∈ M/p_{k}M. Clearly this is a homomorphism and we see immediately that its kernel is p^{∞}M. ✓

proposition 0.85. If 0 → M_{1} → M_{2} → M_{3} → 0 is exact then so is 0 → M̂_{1} → M̂_{2} → M̂_{3} → 0.

proof: ****

It follows from Proposition 0.85 that an element of M̂ can be represented as an infinite series ….

In order to say wear and tear on your author’s brain, for the remainder of this section we will restrict attention to modules over a dedekind domain W. As before, Q will denote its quotient field.

proposition 0.86. Ŵ_{p} is a commutative ring and QŴ_{p} is a field.

proposition 0.87. If M is a p-primary W -module, then M has a unique Ŵ_{p}-module structure compatible with its W -module structure. proof: The crucial fact here is that if M is p-primary then for every m ∈ M there exists n (dependent on m, of course) such that p^{n}m = 0. Now consider a sequence r̂ = {r_{k}} ∈ R̂ ⊆ ∏ R/p^{k}R.; We will define r̂m = r_{n}m. This is actually independent of n, provided that p^{n}m = 0 because of the condition that r_{n+k} ≡ r_{n} (mod p^{n}). From this, it is easy to see that all the conditions are satisfied for M to be an R̂-module.

If W is the ring of integers and M has finite rank, then all M/p^{n}M are finite and so M̂ is compact by Tychonof’s Theorem. In general, if p is a maximal ideal we can only say that the M/p^{n}M have finite length. But it still seems intuitively plausible that M̂ would have all the algebraic properties that one would expect of a compact module. In fact, abelian group theorists will recognize that M̂ belongs to the class of “algebraically compact” modules, which are characterized by the following Proposition 0.* below. The proof is a simple consequence of the fact that Ŵ_{p} is the inverse limit of the family of modules W/p_{k}W. What this means is given by the following proposition:

proposition 0.88. Let π_{n}: Ŵ_{p} → W/p_{k}W and π̄_{n}: W/p^{n+1}W → W/p_{k}W be the canonical maps. Let M be a W -module and for each n ≥ 1 let φ_{n}: M → R/p_{k}W. Suppose that for every n, π̄_{n}φ_{n+1} = φ_{n}. Then there exists a unique φ: M → Ŵ_{p} such that φ_{n} = π_{n}φ for all n. proof: ****

corollary 0.89. The map Hom(M, Ŵ_{p}) → Hom(M, W/pW) given by φ ↦ π_{1}φ is surjective. proof: ****

proposition 0.90. If p is a maximal ideal and M̂ is a pure submodule of any (not necessarily torsion free) W -module X, then M̂ is a summand of X. proof: ****

One should not expect that M̂ is a finite rank W -module, except in the trivial case R = R̂. However we have the following:

proposition 0.91. If M is a finite rank torsion free W -module then M̂ is a finite rank torsion free Ŵ_{p}-module. Furthermore, if p is maximal then M̂ is a free Ŵ_{p}-module. proof: ****

ADDITIVE FUNCTORS. It is an absolute prerequisite for much of this book that the reader be comfortable with the ideas of categories and natural transformations. All that your author can do to help with this is to present here a few topics which may seem, from the prospective of a standard graduate algebra course, a bit specialized.

The categories of modules (whether right or left) over a commutative ring, along with almost all of the other categories used in this book, are what are called additive categories. This means that if φ , ψ ∈ Hom(X, Y) then φ + ψ and φ − ψ are defined, and these operators make Hom(X, Y) into an abelian group. Furthermore, it is required that composition of functors be bilinear, i. e. in the same notation if α ∈ Hom(W, X) and β ∈ Hom(Y, Z) then (φ+ψ)α = φα +ψα and β(φ+ψ) = βφ+βψ. Furthermore, any two objects X and Y in these categories have a direct sum X ⊕ Y , which is simultaneously a product and a coproduct. Finally, an additive category always has a zero object 0, analogous to the trivial module in the category of modules over a ring and characterized by the property that Hom(0, X) = Hom(X, 0) = 0 for all objects X.

Essentially, the point of an additive category is that one may use all of the instincts and habits one has learned from working with modules over a ring without getting into trouble. The one exception is that for some of the categories used in this book statements that involve kernels or cokernels, or the word “epimorphism,” may not work exactly the way one thinks they should. This does not cause problems if one simply uses common sense. However for technical reasons it has seemed better to avoid the term “epimorphism” in favor of the safer term “surjection.”

The following examples may be skipped for now by anyone who is confused rather than enlightened by them.

examples 0.92.

(1) For an integral domain R, we can consider the category of torsion free R-modules, i. e. those modules M such that (∀r ∈ R)(∀m ∈ M) r ≠ 0 & m ≠ 0 ⇒ rm ≠ 0. This is an additive category and it causes no special problems unless one has been trained so thoroughly in category theory that one no longer has common sense. For instance if K is the quotient field of R and φ: R → K is the inclusion map, then within the category in question (assuming K ≠ R) it would not make sense to talk about the cokernel of φ, since the cokernel in the usual sense, i. e. K/R, is not torsion free and hence does not exist within the category. Now it turns out that within the category of torsion free modules over R φ does indeed have a cokernel, namely the zero module, and that φ is an epimorphism. However only a bloody-minded fool would think in those terms. Which is why in this book we will avoid the word “epimorphism.”

(2) In Chapter 3 we will introduce a category which is absolutely fundamental to the study of torsion free modules. Namely the category of “torsion free modules under quasi-homomorphisms.” The objects in this category are simply the torsion free modules, but instead of using ordinary homomorphisms we define the morphisms from M to N to be the elements of KHom(M, N), where KHom(M, N) denotes the localization of Hom(M, N) at the zero ideal (or, if one prefers, K ⊗_{R} Hom(M, N)). Thus morphisms in this category are not (usually) homomorphisms from M to N in the customary sense. Nonetheless, the set of morphisms KHom(M, N) is an abelian group, and this is an additive category, with the direct sum of two modules being simply their ordinary direct sum. We will see in Chapter 3 that this category presents no particular problems, and that the way to approach it is simply to forget that there’s anything strange about it and pretend that one is actually in the category from part (1) above.

Unlike “epimorphism,” the word “exact” is too important for us to be able to live without. Fortunately, exactness poses no real conflict here between category theory and common sense. Exactness in the category of torsion free R-modules (Example 0.92 (1)) means exactly what is does for the category of all R-modules. (So that – for what it’s worth – although as noted above the inclusion map R → K is an epimorphism in the category of torsion free modules, the sequence R → K → 0 is not exact in the category.) And we can avoid confusion when working in the category of torsion free modules under quasi-homomorphisms (Example 0.92 (2)) by the simple expedient of using the term “quasi-exact” rather than “exact.”

As Peter Freyd has pointed out, the whole reason one defines categories in the first place is in order to define functors. And the reason that we have made a fuss about the concept of an additive category is not because it itself has any crucial importance in what follows, but because we want to be able to define the concept of an additive functor.

definition 0.93. A functor F between additive categories C and D is called an additive functor if the function it induces from Hom_{C}(C_{1}, C_{2}) to Hom_{D}(F (C_{1}), F (C_{2})) is a homomorphism. In other words F (φ + ψ) = F (φ) + F (ψ).

Additive functors are to additive categories what continuous maps are to topological spaces, or what homomorphisms are to modules. In other words, just about any functor that anyone ever uses between two additive categories and in particular virtually any functor used in this book, is additive. The concept is worth noting, however, because of a couple of useful propositions.

proposition 0.94. An additive functor preserves direct sums. proof: ****

Actually, the converse is also true: A functor between additive categories is additive if and only if it preserves direct sums (see [Freyd], for instance). However this converse is not especially useful, at least not for us.

proposition 0.95. If R is a commutative ring and F is an additive functor from the category of R-modules into the category of abelian groups, then for every R-module M, the abelian group F (M) has a canonical R-module structure, so that from practical purposes one may as well think of F as a functor from the category of R-modules into itself. proof: ****

There is a wonderful bit of black magic which is probably the main reason for discussing all of this at all.

proposition 0.96. If S and T are additive functors from the category of R-modules to some other additive category, and if η: S → T is a natural transformation, and if η_{R}: S(R) → T (R) is an isomorphism, then η_{P} : S(P) → T (P) is an isomorphism for every finitely generated projective module P. Furthermore if S and T preserve arbitrary (i. e. infinite) coproducts, then η_{P} is an isomorphism for all projective modules P. proof: ****

definition 0.97. If C and D are categories and if F: C → D and G: D → C are functors, then we say that F is a left adjoint to G and that G is a right adjoint to F is Hom_{D}(F (C), D) is naturally isomorphic to Hom_{C}(C, G(D)) for pair of objects C ∈ C and D ∈ D. In other words, there exists an isomorphism η: Hom_{D}(F (C), D) ⟶^{≈} Hom_{C}(C, G(D)) such that whenever γ: C → C′ and δ: D → D′ are morphisms in C and D then the following diagrams commute:

Hom_{D}(F (C′), D) ⟶^{F (γ)^{∗}} Hom_{D}(F (C), D)
η↓  ^{η}↓
Hom_{C}(C′, G(D)) ⟶^{γ^{∗}} Hom_{C}(C, G(D))
Hom_{D}(F (C), D) ⟶^{δ_{∗}} Hom_{D}(F (C), D′)
η↓  ^{η}↓
Hom_{C}(C, G(D)) ⟶^{G(δ_{∗})} Hom_{C}(C, G(D′)).

The concept of a pair of adjoint functors seems initially mysterious and technical. However adjoint pairs are actually quite pervasive in module theory. The best way of understanding the concept is by thinking of the pair of functors F = ⊗_{R}M and G = Hom_{R}(M, ) where M is a module over a commutative ring R.

proposition 0.98. Let R be a commutative ring and M an R-module. Let F and G be the functors from the category of R-modules into itself given by F (X) = X ⊗_{R} M and G(Y) = Hom_{R}(M, Y). Then F is a left adjoint to G, i. e. there exist a natural isomorphisms η: Hom_{R}(X ⊗_{R} M, Y) ⟶ Hom_{R}(X, Hom_{R}(M, Y)). proof: For φ ∈ Hom_{R}(X ⊗_{R} M, Y) define η(φ): X → Hom_{R}(M, Y) by η(φ)(x): m ↦ φ(x ⊗ m). It is left to the intrepid reader to check that (1) η(φ) thus defined is an R-linear homomorphism; (2) η itself is an R-linear map, in particular that η(rφ) = rη(φ); (3) that η is an isomorphism, i. e. one-to-one and surjective; and (4) that η is natural in the sense described above. ✓ corollary 0.99. If R′ is an overring of R and P is an R′-module then Hom_{R′}(R′ ⊗_{R} M, P) ≈ Hom_{R}(M, P).

(This is used in Chapter 4.) proof: ****

If F is a left adjoint to G, then in particular for any X in C and Y in D we get isomorphisms η: Hom(F (X), F (X)) ⟶^{≈} Hom(X, GF (X)) and η^{−1} : Hom(G(X), G(X)) ⟶^{≈} Hom(F G(Y), Y). We thus get maps δ = η(1_{F (X)}): X → GF (X) and σ = η^{−1}(1_{G(X)}): F G(Y) → Y . It is easy to check that δ and σ are in fact natural transformations. They are usually not isomorphisms.

Just about the only time we will use δ and σ is in the case of the tensor product and Hom functors. They are “natural” not only in the technical sense, but also in the ordinary usage of the word. The map

δ: X → Hom_{R}(M, M ⊗_{R} X)

is given by δ(x)(m) = m ⊗ x and

σ: M ⊗_{R} Hom_{R}(M, Y) → Y

is given by σ(m ⊗ φ) = φ(m). We sometimes refer to σ here as the “evaluation map.”

These two innocent little natural transformations will play a crucial role in Chapter 9, after making a brief appearance in Chapter 4. One can easily check that the isomorphism η: Hom_{R}(X ⊗_{R} M, Y) → Hom_{R}(X, Hom_{R}(M, Y)) described in the proof of Proposition 0.99 sends φ ∈ Hom_{R}(X ⊗_{R} M, Y) to the map given by the composition

X ⟶^{δ} Hom_{R}(M, M ⊗_{R} X) ⟶^{φ_{∗}} Hom_{R}(M, Y)

and that η^{−1} sends ψ ∈ Hom_{R}(X, Hom_{R}(M, Y)) to the map given by the composition

X ⊗_{R} M ⟶^{ψ⊗_{R}1_{M}} Hom_{R}(M, Y) ⊗_{R} M ⟶^{σ} Y.