Commutative Algebra Seminar Talk

Reeve Garrett January 26, 2015

1 The ideal-adic topology

For the entirety of this talk, let R be a commutative unitary ring, I be an ideal of R, and M be an R-module.

n ∞ Definition 1.1 Consider the family F := {I M}n=0. The topology for which F is a fundamental system of neighborhoods of 0 ∈ M is called the I-adic topology (sometimes called the Krull topology). Recall that a collection V of neighborhoods of a point x is called a fundamental system of neighborhoods of x if for any neighborhood U of x there exists a finite sequence V1,V2, ..., Vn of neighborhoods in V such that V1 ∩ V2 ∩ · · · ∩ Vn ⊆ U. Consequently, all open sets of M are unions of arbitrary numbers of sets of the form b + IsM for b ∈ M and s ∈ N. The I-adic topology makes M a topological R-module, and in the case M = R, R is a topological ring, meaning all the algebraic operations (addition, multiplication, inversion) are continuous - in particular, translation defines a homeomorphism.

Observation 1.2 Since by deﬁnition each InM is open, each coset x + InM (x∈ / InM) is open, so their union, the complement of InM, is open, meaning InM must be closed as well. Thus, if we endow M/InM with the quotient topology, that quotient topology must be the discrete topology.

We now establish some basic facts about this topology.

Lemma 1.3 [5, (16.2), p. 51] A submodule N of the R-module M is an open set in the I-adic topology iﬀ (N : M) := {r ∈ R : rM ⊆ N} contains some power of I. In that case, N is also closed.

Proof. If N is open, then 0 ∈ N implies InM ⊆ N for some n since F forms a neighborhood base of 0. Conversely, if InM ⊆ N, then since N is a submodule, b + InM ⊆ N for any b ∈ N, and N is the union of the open sets b + InM (b ∈ N), proving N is open. Moreover, by the same reasoning as the preceding observation, we see N is closed.

Theorem 1.4 [5, (16.3), p. 51] The closure N of a submodule N of M in M with the I-adic topology is T n T n n≥0(N + I M). In particular, the set N := n≥0 I M, is the closure of {0}. Proof. Each N + InM is an open set and therefore simultaneously closed by the preceding lemma. T n T n Therefore, N ⊆ n≥0(N + I M). Conversely, let x ∈ n≥0(N + I M). Then, x = bn + an with bn ∈ N n n and an ∈ I M for each n ∈ N. Consequently, we see x + I M meets N for any n (with the element x − an). Since the x + InM form a neighborhood base at the point x, we consequently have x ∈ N.

In order for a topological space to be Hausdorﬀ, points must be closed. In fact, we have an equivalence:

Lemma 1.5 [8, p. 252] If the zero of a topological module M is closed set, M is a Hausdorﬀ space.

Proof. Let x 6= y ∈ M, V be a neighborhood of y − x which does not contain 0 ∈ M, and U = x − y + V . Then U is a neighborhood of 0 not containing x − y. Let W be another neighborhood of 0 such that W − W ⊆ U (the existence of such a neighborhood follows from the continuity of (x, y) 7→ x − y). Then x + W and y + W are disjoint neighborhoods of x and y, and thus since x and y were arbitrary, we have M is Hausdorﬀ.

T n Thus, M is Hausdorﬀ in the I-adic topology iﬀ n≥0 I M = 0.

1 Corollary 1.6 [5, (16.4), p. 51] For a submodule N of M, the I-adic topology of M/N is Hausdorﬀ iﬀ N is a closed subset in the I-adic topology of M.

T n Proof. The I-adic topology of M/N is Hausdorﬀ iﬀ n≥0(I M + N)/N) = 0 as just noted, or equiva- T n lently, N = n≥0(N + I M), which is equivalent to N being closed by the theorem.

For N a submodule of M, the I-adic topology of N might not coincide with the subspace topology of N in M. However, when M is a Noetherian module, these topologies do coincide. To prove this, we need an important lemma: the Artin-Rees lemma.

Theorem 1.7 [1, Proposition 4.1.5 and Lemmas 4.1.7 and 4.1.8] If R is a Noetherian ring, M is a ﬁnite R-module, and N is a submodule of M, then there is an integer k depending only on R, M, N, and I such that InM ∩ N = In−k(IkM ∩ N).

Proof. See the cited reference.

Theorem 1.8 [5, (16.5), p. 52] If M is a Noetherian module, then for any submodule N of M, the I-adic topology of N coincides with the topology of N as a subspace of M with the I-adic topology.

Proof. By [5, (3.11), p. 12], for any Noetherian module M over a ring R we have R/(0 : M) is a Noetherian ring and M is a ﬁnite R/(0 : M)-module. Modding out (0 : M) if necessary we may thus assume R is Noetherian. Obviously, InN ⊆ InM ∩ N, so by the Artin-Rees lemma, for some k > 0 and for n > k, InM ∩ N = In−k(IkM ∩ N) ⊆ In−kN, thus proving the assertion.

Remark 1.9 By similar reasoning to the above proof, we see if I and J are two ideals such that for some natural numbers m and n Im ⊆ J and J n ⊆ I, the I-adic topology and J-adic topology coincide. This is a helpful observation that allows us in several cases reduce to the case where I is R’s maximal ideal (if R is local) or Jacobson radical (if R is semi-local).

For the remainder of the section, we will assume M is Hausdorff. In this case, we have a well-defined arithmetic function w : M → N ∪ {∞}, defined by w(x) = sup{k | x ∈ IkM} if x 6= 0 and w(0) = ∞. From this arithmetic function, we get a metric d defined by d(x, y) = e−w(x−y), which induces the I-adic topology. In fact, d is an ultrametric (that is, a metric with a “strong triangle inequality” d(x, z) ≤ sup{d(x, y), d(y, z)}) because w(a − b) ≥ min{w(a), w(b)}. In the case M = R and R is a domain, the quotient field K with the same ultrametric also is a topological ring. Of course, with a metric space, it’s natural to consider Cauchy sequences and their convergence, and thus we arrive at one formulation of the I-adic completion, arguably the most natural formulation.

2 The ideal-adic completion

Given a commutative unitary ring R, ideal I, and R-module M, the I-adic completion of M is often developed from 2 seemingly diﬀerent perspectives. The ﬁrst of these, through Cauchy sequences, is perhaps the most T n elementary and natural. Interestingly, we don’t need the hypothesis that n≥0 I M = 0 for the notion of Cauchy sequence to make sense.

N(n) Deﬁnition 2.1 A Cauchy sequence in M is a sequence {xn} ∈ M N such that xn − xn+i ∈ I M for all i ≥ 0, where N(n) → ∞ as n → ∞.

m Notation 2.2 We write xn → 0 as n → ∞ if for each m ∈ N there exists N ∈ N such that xn ∈ I M for each n ≥ N. In this case, we say {xn} is a null sequence.

Deﬁnition 2.3 A limit of a sequence {xn} is any element y of M such that {xn − y} is a null sequence.

2 0 0 T n Remark 2.4 If {xn} has a limit y, then y is also a limit of the sequence iff y − y ∈ n≥0 I M, and we say y and y0 are equivalent. Definition 2.5 The collection of all equivalence classes of limits of Cauchy sequences of M is called the I-adic completion of M, denoted Mc. Equivalently, if we define two Cauchy sequences {xn} and {yn} to be m equivalent if xn − yn → 0 as n → ∞, i.e. for all m ∈ N there exists N(m) ∈ N such that xn + I M = m yn + I M for n ≥ N(m), we may define Mc as the set of equivalence classes of Cauchy sequences in M according to this equivalence relation. Of course, intuitively, we wish to identify M as a dense subspace of Mc, and since Mc has a natural R-module structure via the latter definition, the easiest way to do so is to assign via a map φ : M → Mc T n assigning x ∈ M to the constant sequence {x}. However, the kernel of this map is n≥0 I M, so we see φ is injective iff M is Hausdorff. In this case, because the ring and module operations of R and M are uniformly continuous, it follows that Rb is a topological ring and Mc is a topological Rb-module and R-module by extending the given operations uniquely (algebraic identities are preserved through passage to limits). This perspective of defining the completion using Cauchy sequences, while intuitive, does not lend itself quite so easily to studying exactness properties of the completion. However, defining the completion using inverse limits does, so we’ll recall the important definitions in the formulation of inverse limits, and we’ll define the completion as an inverse limit and demonstrate that this “new” definition of inverse limit really is the same as the definition just given. Definition 2.6 Let Ω be a directed set, i.e. Ω is endowed with a partial order ≤ which is reflexive and transitive such that any pair of elements has an upper bound. Let {Mi}i∈Ω be a family of modules indexed by the directed set Ω. If for each pair of indices i ≤ j there exists a homomorphism fij : Mj → Mi (fii always being the identity map) that can be composed consistently with other such homomorphisms (i.e. if i ≤ j ≤ k, then fij ◦ fjk = fik), we call the collection of modules and maps an inverse system.

Definition 2.7 Given an inverse system ({Mi}i∈Ω, {fij}i≤j∈Ω), the inverse limit of the inverse system is Q the submodule of i∈Ω Mi given by Y lim M := (a ) ∈ M | a = f (a ) ∀ i ≤ j ∈ Ω ←− i i i∈Ω i i ij j i∈Ω In the case Ω = , we refer to the elements of lim M as coherent sequences. N ←− i The inverse limit comes equipped with projection maps π : lim M → M that satisfy π = f ◦ π j ←− i j i ij j for all i ≤ j. Moreover, these maps give lim M the following universal property: given (Y, (ψ ) ) where ←− i i i∈Ω ψ : Y → M is such that ψ = f ◦ ψ for all i ≤ j, there exists a unique morphism θ : Y → lim M such that i i i ij j ←− i ψi = πi ◦ θ for all i ∈ Ω. θ is defined by θ(y) = (ψi(y))i∈Ω). For more details, including on the construction of inverse limits in other categories, see Section 5.2 of [6]. n We’re only interested in a very special case of this: the case where Ω = N and Mn := M/I M. In this n n−1 case, our “fij” are just compositions of the maps θn : Mn → Mn−1 given by θn(a + I M) = a + I M. Definition 2.8 The I-adic completion of M is (equivalently) M = lim M , c ←− i i where Mi := M/I M. Let’s see why this is this equivalent to our original definition of a completion. We want to identify each equivalence class of Cauchy sequences with a coherent sequence, i.e. a sequence {xn} such that for each n n−1 n−1 we have xn + I M = xn−1 + I M. Recall that a Cauchy sequence in M is a sequence {xn} such that r for every r ∈ N there is a positive integer N(r) such that xn −xm ∈ I M for m, n ≥ N(r). This is equivalent r r to xn + I M = xm + I M for m, n ≥ N(r), so except for its first N(r) terms (which we can replace with whatever we want since we’re only dealing with equivalence classes of Cauchy sequences anyways), we have the first r terms of {xn} satisfy the “coherency relation”, so taking r → ∞, {xn} is essentially (very nearly) a coherent sequence if we consider just its equivalence class. So, coherent sequences and equivalence classes of Cauchy sequences are really the same thing. Now, we demonstrate the utility of defining completions in this manner.

3 Remark 2.9 Of course, when we define a completion, we want the completed space to have a topology such that the original space (that was completed) is dense in the completion, and we want the completion to be complete with respect to the given topology. Sure enough, we have one excellent candidate for a fundamental system of neighborhoods of 0, and we’ll call the topology they define the “inverse limit” topology. Define ∗ ∗ MN (0) := {x = (xn)n∈N ∈ Mc : xn = 0 ∀ n ≤ N} for each N. Then the family {Mn(0)}n∈N forms a fundamental system of neighborhoods of 0 that define the topology of Mc. Naturally, the neighborhood ∗ n ∗ Mn(0) “closes up” I M in Mc too. We also see that Mn(0) is precisely the kernel of the natural (surjective) n Q n ∗ ∼ n projection map πn : Mc → M/I M (recall Mc ⊆ n M/I M). Thus, M/Mc n(0) = M/I M, so taking inverse limits (i.e. completing with respect to this topology), we get Mc is its own completion in this topology.

Of course, a natural question to ask is the following: why would this “inverse limit” topology coincide with the I-adic topology of Mc? It turns out that it doesn’t always coincide, as one might expect, but in some cases (namely the Noetherian ring and ﬁnite module case, as we’ll later see) it does. When R is Noetherian, we have especially nice characterizations of Rb. To obtain these characterizations, ﬁrst, we recall an important result about exact sequences and inverse limits.

0 0 00 00 Theorem 2.10 [1, Theorem 4.2.6] Let {Mn, θn}, {Mn, θn}, and {Mn , θn} be inverse systems of modules and assume the diagram below is commutative with exact rows.

0 fn+1 gn+1 00 0 → Mn+1 → Mn+1 → Mn+1 → 0 0 00 ↓θn+1 ↓θn+1 ↓θn+1 0 fn gn 00 0 → Mn → Mn → Mn → 0 Then, the sequence 0 → lim M 0 → lim M → lim M 00 ←− n ←− n ←− n 0 is exact. If the maps θn are all surjective, then

0 → lim M 0 → lim M → lim M 00 → 0 ←− n ←− n ←− n is exact. Q Proof. Let M = Mn and deﬁne an R-homomorphism dM : M → M by dM ((xn)n∈N) = (xn − 0 00 θ (x )) . The kernel of d is lim M . Similarly, we may deﬁne M , M , d 0 , and d 00 , where the n+1 n n∈N M ←− n M M corresponding inverse limits are their kernels. Moreover, the maps {fn} and {gn} naturally induce maps f : M 0 → M and g : M → M 00. Thus, we obtain the following commutative diagram with exact rows.

f g 0 → M 0 → M → M 00 → 0

↓dM0 ↓dM ↓dM00 f g 0 → M 0 → M → M 0 → 0 Now, applying the Snake lemma, we get an exact sequence

0 → ker dM 0 → ker dM → ker dM 00 → coker dM 0 0 0 and if the θn are all surjective, then dM is surjective, meaning the cokernel of dM 0 is 0. Thus, we have what we want.

In particular, we obtain the following as an easy corollary.

Corollary 2.11 [1, Corollary 4.2.9], [2, Proposition 10.12] Suppose we have an exact sequence of ﬁnitely generated modules over a Noetherian ring R

f g 0 → M 0 → M → M 00 → 0.

4 Then, the sequence of I-adic completions

f g 0 → Mc0 → Mc → Md00 → 0

is exact.

Proof. By the exactness, up to isomorphism we reduce to the case

f g 0 → N → M → M/N → 0.

By the second and third isomorphism theorems, for each n ∈ N,

(M/InM)/(N/(N ∩ InM)) =∼ M/(N + InM) =∼ (M/N)/((N + InM)/N).

Thus, the following sequence is exact for each n ∈ N, 0 → N/(N ∩ InM) → M/InM → (M/N)/((N + InM)/N) → 0.

By Theorem 1.8, the subspace topology of N ⊆ M coincides with the I-adic topology of N, and it’s clear that the (N + InM)/N are the fundamental system of neighborhoods that deﬁne the I-adic topology for M/N, so if we take inverse limits, the previous theorem gives us

0 → Nb → Mc → M/N.\ n n−1 Moreover, the maps θn : M/(N + I M) → M/(N + I M) that deﬁne the inverse system from which M/N\ was obtained are surjective, so we in fact have precisely the short exact sequence we wanted by the previous theorem.

Consequently, we have for N a submodule of a ﬁnite module M over a Noetherian ring R that M/N\ =∼ ˆ ˆ T n M/N. More is true. By the Krull intersection theorem, n≥0 I = 0 so R is Hausdorﬀ in its I-adic topology, and therefore we may consider R as lying inside Rb. So, it’s natural to compare completions with closures in Rb. For this, we have the following.

Theorem 2.12 [4, Theorem 8.11] Let R be a Noetherian ring, I and J be ideals of R, and M be a ﬁnite R-module. Let ˆ denote I-adic completion. Then JMd = JMc, and this is in fact the closure of JM in Mc. Consequently, M/JM\ = M/Jc Mc.

Proof. By the above corollary, JMd is the kernel of Mc → M/JM\ , and this must necessarily be the closure ˆ of JM in M. Indeed, the condition x = (xn)n∈N ∈ Mc belongs to JMd (which can be thought of as meaning each xn can be represented as an element of JM) is equivalent to the condition that x ∈ φ(JM)+ker(πn) for ˆ Q n n each n, where φ : M → M ⊆ n M/I M is the natural map and πn : Mc → M/I M is the natural projection map. But, as we saw before in the remark, the ker πn form the fundamental system of neighborhoods of 0 in Mc which define the linear topology of Mc much like the InM do for M, which by similar reasoning as in T Theorem 1.4 forces the closure of JM in Mc to be n(φ(JM) + ker πn), meaning JMd is indeed the closure of JM (identified with φ(JM)) in Mc. r P It remains to show JMd = JMc. Suppose J = (a1, ..., ar). Define ϕ : M → M by (x1, ..., xr) 7→ aixi. Then, the sequence ϕ µ M r → M → M/JM → 0 where µ is the natural projection map is exact. Taking completions preserves exactness in our context, so we have a new exact sequence

ϕˆ µˆ ([M r) =∼ (Mc)r → Mc → M/JM\ → 0

5 P whereϕ ˆ is given by the same formula (x1, ..., xr) 7→ aixi as ϕ, so since M/JM\ = M/c JMd , we have P JMd = kerµ ˆ = Imϕ ˆ = aiMc = JMc, as desired.

The key to relating the inverse limit topology to the I-adic topology on Mc is the fact that sometimes Mc can be constructed directly from M and Rb as we’re about to show. First, note that it’s easy to consider n n Mc as a Rb module by defining for a = (a1, a2, ...) ∈ Rb (ai ∈ R/I ) and x = (x1, x2, ...) ∈ Mc (xi ∈ M/I M) their product as ax = (a1x1, a2x2, ...) ∈ Mc. Theorem 2.13 [4, Theorem 8.7] Let R be a Noetherian ring, I be an ideal of R, and M be a finite R-module. ∼ Then, M ⊗R Rb = Mc. Hence, if R is I-adically complete, so are all its finite modules. Proof. Over a Noetherian ring, all finite modules are finitely presented (since a finite module over a Noetherian ring is Noetherian, meaning all its submodules are finitely generated), meaning for the finite module M (and any other finite module) we have an exact sequence of the form

0 → Rp → Rq → M → 0

where p and q are positive integers. We just showed completion is an exact functor, and we know that the functor · ⊗R Rb is right exact, so we obtain a commutative diagram

0 → (Rb)p → (Rb)q → Mc → 0 ↑ ↑ ↑ p q R ⊗R Rb → R ⊗R Rb → M ⊗R Rb → 0

where the vertical maps are the natural maps deﬁned the following way: for x ∈ M and a = (an)n∈N ∈ Rb, n assign x ⊗ a to (xan + I M)n∈N, which is in Mc. Since tensor products commute with direct sums and N completions commute with direct sums (intuitively because a given sequence (mk, nk)k∈N ∈ (M ⊕ N) is Cauchy iﬀ (mk) is Cauchy in M and (nk) is Cauchy in N, for any two R-modules M and N), the left two maps are isomorphisms. Consequently, the rightmost arrow must be an isomorphism (see Exercise 1 on page 403 of Dummit and Foote).

Theorem 2.14 [4, Theorem 8.8] The I-adic completion of R is ﬂat over R.

Proof. For any ring R and R-module M, M is flat over R iff for every finitely generated ideal I of R the canonical map I ⊗R M → R ⊗R M is injective (this is [4, Theorem 7.7]). So, it’s enough to show that ∼ for every ideal J of R that J ⊗R Rb → Rb is injective. But we know from the previous theorem J ⊗ Rb = Jb, and by the corollary before that Jb → Rb is injective because J → R is.

Observation 2.15 For any ring R, the (X1, ..., Xn)-adic completion of R[X1, ..., Xn] can be identiﬁed with R[[X1, ..., Xn]]. Indeed, an element of this completion is just a sequence of polynomials (fk) ∈ (R[X1, ..., Xn])N. k with deg fk < k such that fk+1 mod (X1, ..., Xn) = fk, so at each step we can write fk as a sum of terms of the various degrees (from 0 up to degree k − 1) which matches with all fj (j ≤ k), meaning limk→∞ fk is a well-deﬁned power series. Thus, we have a natural one-to-one correspondence (which is a map of R[X1, ..., Xn]-algebras) between the (X1, ..., Xn)-adic completion of R[X1, ..., Xn] and R[[X1, ..., Xn]], so they’re isomorphic. Using this, we obtain a very nice characterization of the completion of a Noetherian ring.

Theorem 2.16 [4, Theorem 8.12] Let R be a Noetherian ring and I = (a1, ..., an) be an ideal of R. Then the I-adic completion Rb of R is isomorphic to R[[X1, ..., Xn]]/(X1 − a1, ..., Xn − an). In particular, Rb is a Noetherian ring.

0 Proof. Let B = R[X1, ..., Xn], I = (X1, ..., Xn) ⊆ B, and J = (X1 − a1, ..., Xn − an) ⊆ B. Then, B/J =∼ R, and the I0-adic topology on R considered as the B-module B/J coincides with the I-adic topology of R. Indeed, we can deﬁne an R-algebra map q : B → A by Xi 7→ ai, which makes A a ﬁnite B-module, and

6 the I-adic topology on A will coincide with the I0-adic topology as a ﬁnitely generated B-module naturally. 0 Now, if we write b for the I -adic completion of B-modules, we have by Theorem 2.12 [4, Theorem 8.11]

Rb = B/b Jb = B/Jb Bb = R[[X1, ..., Xn]]/(X1 − a1, ..., Xn − an).

Theorem 2.17 [4, Theorem 8.13] Let R be a Noetherian ring, I an ideal, M a ﬁnite R-module, and Mc be the I-adic completion of M. Then, the topology of Mc is the I-adic topology of Mc as an R-module and is the IRb-adic topology of Mc as an Rb-module.

∗ Proof. As shown earlier, the family {Mn := ker πn} form the fundamental system of neighborhoods of ∗ n n 0 in Mc that deﬁne the usual “inverse limit” topology of Mc. Thus, we wish to show Mn = I Mc. M/I M is discrete in the I-adic topology (as we noted in Observation 1.2), so we have M/I\nM = M/InM (discrete topological spaces are complete), and we know the kernel of Mc → M/I\nM is InMc by Theorem 2.12 [4, ∗ n n n Theorem 8.11]. Thus, we have Mn = I Mc, as desired. Moreover, I Mc can be written as (I Rb)Mc (since Mc is natural an Rb-module) and InRb = (IRb)n, so that the topology of Mc is also the IRb-adic topology.

Theorem 2.18 [4, Theorem 8.14] Let R be a Noetherian ring and I be an ideal of R. If we consider R with the I-adic topology, the following are equivalent: (i) I ⊆ rad(R), where rad(R) denotes the Jacobson radical of R, which is defined to be the intersection of all maximal ideals of R. (ii) every ideal of R is a closed set. (iii) The I-adic completion Rb of R is faithfully flat over R, meaning for every sequence S of R-modules, S is exact iff S ⊗R Rb is exact.

Proof. See Matsumura, page 62.

∞ k Proposition 2.19 [3, Proposition III.1.2]Let R be a Noetherian ring, I an ideal such that ∩k=0I = (0), and Rb be the completion of R in the I-adic topology. The following are equivalent: (i) Rb is compact. (ii) R/I is finite. (iii) Each prime ideal containing I is maximal with finite residue field. (iv) For every integer n, R/In is finite.

Proof. Recall we showed in Theorem 2.17 that R/b Icn =∼ R/In for each n. (i) =⇒ (ii): If R/I were infinite, then R/b Ib would be too. In this case, we’d cover Rb by infinitely many disjoint neighborhoods homeomorphic to Ib, which would not admit a finite subcover, contradicting our compactness hypothesis. (ii) =⇒ (iii): trivial. (iii) =⇒ (iv): (iii) implies that R/In is an Artinian ring with finite residue fields, which is always a finite ring, for each n. n ∼ n (iv) =⇒ (i): Suppose that R/b Ic = R/I is finite for each n. Let {xn} be a sequence in Rb. By our finiteness hypothesis, infinitely many of its terms are in the same residue class modulo Ib - say these terms form a subset X1. Exploiting our finiteness hypothesis again (for each i ∈ N), inductively, define from Xi−1 a i new infinite subset Xi with all terms in the same class modulo Ib , for each i ≥ 2. Thus, we have a decreasing n sequence {Xn} of subsets (the elements of which lie in the same class modulo Ic). If we let xkn be the first

term of Xn, then the subsequence {xkn } of {xn} converges (since it’s a Cauchy sequence) in Rb, meaning (since {xn} was arbitrary) that Rb is sequentially compact and hence compact since it’s a metric space.

In particular, if D is a DVR, its completion Db is compact if and only if its residue ﬁeld is ﬁnite.

7 3 Nice properties of complete rings

What’s the use of being complete?

Theorem 3.1 [4, Theorem 8.2] Let R be a ring, I an ideal of R, and M an R-module. Then, (i) If R is I-adically complete, then I ⊆ rad(R); (ii) If M is I-adically complete and a ∈ I, then multiplication by 1 + a is an automorphism of M.

Theorem 3.2 (Hensel’s Lemma) [4, Theorem 8.3] Let (A, m, k) be a local ring and suppose A is m-adically complete. Let F (X) ∈ A[X] be a monic polynomial and F ∈ k[X] be the polynomial obtained by reducing the coeﬃcients of F modulo m. If there are monic polynomials g, h ∈ k[X] such that (g, h) = 1 and F = gh, then there exist monic polynomials G and H in A[X] such that F = GH, G = g, and H = h.

Theorem 3.3 [4, Theorem 8.4] Let A be a ring, I an ideal, and M an A-module. Suppose A is I-adically complete and M is Hausdorﬀ in the I-adic topology. If M/IM is generated over A/I by ω1, ..., ωn, then for any collection {ωi}1≤i≤n where ωi ∈ M is an arbitrary inverse image of ωi in M for each i M is generated over A by ω1, ..., ωn.

4 Stone-Weierstrass for Integer-Valued Polynomials

Lemma 4.1 [3, Lemma I.3.19] Let R be a ring and f ∈ R[X]. Then, for each a, b ∈ R, b − a divides (f(b) − f(a)).

Pn i Pn i i Proof. If f = i=0 ciX , then (f(b) − f(a)) = i=1 ci(b − a ), and we know for i ≥ 1 b − a divides bi − ai, so the result follows immediately.

Proposition 4.2 [3, Proposition III.2.1] Let (D, m) be a one-dimensional Noetherian local domain with quotient ﬁeld K, and let f ∈ K[X]. Then, (i) f is uniformly continuous on D in the m-adic topology, meaning there is an integer h such that for all r ∈ N and all a, b ∈ D if (b − a) ∈ mr+h, then (f(b) − f(a)) ∈ mr. (ii) f is continuous on K.

Proof. (i) Let d ∈ D, d 6= 0, be such that df ∈ D[X]. Then, by the lemma, (b − a) divides d(f(b) − f(a)) in D. Since D is one-dimensional, Noetherian, and local, threre is an integer h such that mh ⊆ dD. Then, we have if (b − a) ∈ mr+h, then b − a ∈ dmr, meaning d(f(b) − f(a)) ∈ dmr. Now, cancel the d’s in the last relation for the desired conclusion. (ii) Let a ∈ K, d1 ∈ D be such that d1a ∈ D. Then, consider g(X) = f(X/d1). We know from the ﬁrst assertion that g is uniformly continuous on D, meaning there is an integer h such that for all r ∈ N and all r+h r r+h a1, b1 ∈ D if (b1 − a1) ∈ m , then (f(b1) − f(a1)) ∈ m . Let b be such that (b − a) ∈ m . Then, if we r+h r set a1 = d1a and b1 = d1b, we have (b1 − a1) ∈ m , so (f(b) − f(a)) = (g(b1) − g(a1)) ∈ m . Thus, f is continuous at a, and since a ∈ K was arbitrary, this means f is continuous on K. In fact, the Artin-Rees Lemma can get us a slightly more general result with the same lemma we just used.

Proposition 4.3 [3, Proposition III.2.3] Let I be an ideal of a Noetherian domain D. Then, each f ∈ Int(D) is uniformly continuous on D in the I-adic topology.

Of course, uniformly continuous functions from D to D can be uniquely extended to completions, so we have Int(D) ⊆ C(D,b Db). Our aim is to establish (in appropriate contexts) an analogue to the Stone-Weierstrass theorem for integer-valued polynomials, meaning we wish to establish whether Int(D) is dense in the space of continuous functions from Db to itself (endowed with the uniform convergence topology). The only way denseness really makes sense in this space is in the following way: for each continuous function ϕ : Db → Db and n ∈ N, there is f ∈ Int(D) such that for all x ∈ Db, f(x) − ϕ(x) ∈ Icn. As we showed earlier, if D is Noetherian, Icn =∼ Ibn.

8 Theorem 4.4 [3, Lemma III.3.3, Theorem III.3.4, Corollary III.3.5] If V is a discrete valuation domain with quotient field K, maximal ideal m generated by t, residue field of cardinality q, and corresponding valuation v, and we denote by Vˆ the completion of V in the m-adic topology, then the following hold: (i) The topology defined by its valuation is the m-adic topology, and Vˆ is also a discrete valuation domain with valuation vˆ an extension of v. (ii) Each continuous function from Vˆ to itself can be approximated modulo mˆ by some f ∈ Int(V ). (iii) Int(V ) is dense in the space of continuous functions from Vˆ to itself in the uniform convergence topology. (iv) If U1, ..., Ur are disjoint open subsets covering Vˆ and n1, ..., nr are arbitrary integers, then there exists a polynomial h ∈ K[X] such that vˆ(h(x)) = ni if x ∈ Ui, for each i ∈ {1, ..., r}.

Of course, the case of the discrete valuation domain and its maximal ideal seems very special, but in fact whenever we have a Dedekind domain (like the integers), the same holds. In fact, with any domain, we can reduce to the local one-dimensional case.

Proposition 4.5 [3, Lemma III.4.8] For a Noetherian domain D and ideal I, if Int(D) is dense in the space of continuous functions from Dˆ to itself in the I-adic topology, then (i) the completion Dˆ is a domain. (ii) the radical of I is a height-one maximal ideal m with ﬁnite residue ﬁeld. (iii) the I-adic topology is equivalent to the m-adic topology.

Proposition 4.6 [3, Proposition III.4.9] If m is a maximal ideal of a Noetherian domain D, then (i) the m-adic topology of D is induced by the mDm-adic topology of Dm, (ii) D is dense in Dm in the mDm-adic topology, (iii) Int(D) is dense in Int(Dm) in the mDm-adic topology.

So, what these last two results tell us is that we may restrict ourselves to the m-adic topology where m is a maximal ideal, Dˆ (the completion of D in the m-adic topology) is also the completion of Dm in the mDm-adic topology, and that Int(D) is dense in the space of continuous functions from Dˆ to itself if and only if Int(Dm) is dense in the space of continuous functions from Ddm to itself. Therefore, for all intents and purposes, when dealing with these kinds of Krull topology questions, we may as well restrict ourselves to the local case with the Krull topology with respect to a maximal ideal. So, in particular, when dealing with Dedekind domains, it suﬃces to consider them as discrete valuation domains.

5 A classification of the spectrum of Int(Z) and consequences In this section, we build up to a complete classification of the spectrum (that is, the set of all prime ideals) of Int(Z), which is [3, Proposition V.2.7]. To prove [3, Proposition V.2.7], we need a few preliminary results. To get these results, we start with a more general treatment of integer-valued polynomials, dealing with Int(D) = {f ∈ K[X] | f(a) ∈ D ∀ a ∈ D} for a domain D with quotient field K. As a consequence of this classification, it’s not much more work to show that Int(Z) is Prüferand has Krull dimension 2. At the conclusion of this section, we’ll show that Int(Z) satisfies another particularly nice property: the strong Skolem property.

Proposition 5.1 [3, Proposition V.1.1] Let p be a prime ideal of a domain D with infinite residue field (we say “field” by abuse of vocabulary; we may only have a domain). The prime ideals of D above p are of the following types: (i) the prime ideals pDp[X] ∩ Int(D), which is the set of polynomials of Int(D) with coeffficients in pDp, (ii) for each polynomial q of Dp[X] which is irreducible modulo pDp, the prime ideal (p, q)Dp[X]∩Int(D), which is the set of polynomials of Int(D) which are divisible by q modulo pDp.

Proof. Fix a prime ideal p of D with inﬁnite residue ﬁeld. Then, by [3, Proposition I.3.4], Int(D)p = Dp[X]. As a result, since all the prime ideals of Int(D) contracting to p in D lift to the prime ideals of

9 Int(D)p = Dp[X] contracting to p, we get the desired characterization by contracting the latter ideals back to Int(D). We have the following immediate (but important) corollary for the set of ideals in Int(D) which contract to (0):

Corollary 5.2 [3, Corollary V.1.2] The nonzero prime ideals of Int(D) above (0) are in one-to-one correspondence with the monic irreducible polynomials of K[X]. To the irreducible polynomial q corresponds the prime ideal Bq = qK[X] ∩ Int(D). Moreover, the localization of Int(D) with respect to the prime ideal Bq is the ring of q-adic valuation of K(X), K[X](q).

In particular, the above corollary holds for Int(Z). We now take a short pause in our classiﬁcation of the spectrum of Int(Z) to establish the Krull dimension of Int(Z). Proposition 5.3 [3, Proposition V.1.5] Let E be a subset of K. Then,

dim(Int(E,D)) ≥ dim(D).

If, moreover, E is a fractional subset of D, then

dim(Int(E,D)) ≥ dim(D) + 1.

Proof. Take a chain of prime ideals (0) = p0 ( p1 ( ··· ( pn and an element a of E. Then, as a consequence, we have a chain of prime ideals of Int(E,D) of the same length: Bp0,a ( Bp1,a ( ··· ( Bpn,a, the strict containments following from the fact that each Bpi,a = {f ∈ Int(E,D) | f(a) ∈ pi} lies above pi. So, we have the former result. To get the latter result, notice that if for some d 6= 0 in D we have dE ⊆ D,

Bp0,a contains the nonzero polynomial d(X − a) and hence the principal ideal generated by it. So, in particular, we get dim(Int(Z)) ≥ 2 as a consequence. In fact, since Z is Noetherian, we even get equality, thus showing dim(Int(Z)) = 2, as our next result shows. But, first, we need a few definitions, since we can attack this in an even more general setting: Jaffard domains.

Deﬁnition 5.4 The valuative dimension of a domain D, denoted dimv(D) is the supremum of the Krull dimensions of the valuation overrings of D.

Deﬁnition 5.5 If the Krull dimension of a domain is equal to its valuative dimension, then we say the domain is a Jaﬀard domain.

Some facts about Jaffard domains that will be used here but won’t be proved are (1) Noetherian domains are Jaffard domains, (2) if D is Jaffard then so is D[X], and (3) being Jaffard (according to the above definition) is equivalent to satisfying the property that dim(D[X]) = dim(D) + 1. We’ll make use of them in this next proof, which shows that Int(Z) has dimension 2. Proposition 5.6 [3, Proposition V.1.8] Let D be a Jaffard domain and E be a fractional subset of D. Then,

dim(Int(E,D)) = dim(D[X]) = dim(D) + 1.

Proof. Let d ∈ D − {0} be such that dE ⊆ D. Then, Int(E,D) contains D[dX] and so each overring of Int(E,D) is an overring of D[dX], and so we get dimv(Int(E,D)) ≤ dimv(D[dX]). Thus, we obtain

dim(Int(E,D)) ≤ dimv(Int(E,D)) ≤ dimv(D[dX]) = dim(D[dX]) = dim(D) + 1.

We obtained the reverse inequality from the previous proposition, [3, Proposition V.1.5], so we have the desired equality. Now, we are ready to continue our classification of the spectrum of Int(Z). Lemma 5.7 [3, Lemma V.1.9] Let D be a domain, a an ideal of D such that D/a is finite, and E a subset of K. Then each prime ideal of Int(E,D) containing Int(D, a) is maximal. If in particular a = m is maximal, these primes lie above m and have residue field isomorphic to D/m.

10 Proof. Let {u1, ..., uk} be a set of representatives of D modulo a and M be a prime ideal of Int(E,D) Qk containing Int(D, a). For each f ∈ Int(E,D), the product i=1(f −ui) belongs to Int(E, a) and hence to M. Therefore, one of its factors belongs to M, and {u1, ..., uk} is a set of representatives of Int(E,D) modulo M. Finite integral domains are ﬁelds, and thus M is maximal.

Deﬁnition 5.8 A maximal ideal m is pseudo-principal if there is a positive integer n and t ∈ m such that mn ⊆ Dt.

Lemma 5.9 [3, Lemma V.1.10] Let E be a subset of K. The prime ideals of Int(E,D) above a pseudo- principal maximal ideal m contain the ideal Int(E, m).

Proof. Let M be a prime ideal of Int(E,D) above m, and let f ∈ Int(E, m). By premise, there is an element t ∈ m such that mn ⊆ Dt. So, f n(E) ⊆ tD, meaning f n ∈ tInt(E,D), and as M contains m, t ∈ M. Hence, f n ∈ M and ﬁnally f ∈ M. Putting these two lemmas together we immediately get the following:

Proposition 5.10 [3, Proposition V.1.11] Let E be a subset of K. The prime ideals of Int(E,D) above a pseudo-principal maximal ideal m with finite residue field contain the ideal Int(E, m), they are maximal, and their residue fields are isomorphic to D/m.

Now, if D is Noetherian, we may as well localize and assume D is local, by [3, Proposition I.2.3], which states that when D is Noetherian and S is a multiplicative subset of D, S−1Int(D) = Int(S−1D). In the case where D = Z, we obtain a discrete valuation ring when we localize at a prime ideal. Proposition 5.11 [3, Proposition V.2.2] Let D be a one-dimensional local Noetherian domain with ﬁnite residue ﬁeld and E be a fractional subset of D. Then each prime ideal of Int(E,D) above the maximal ideal m of D is of the form Mm,α = {f ∈ Int(E,D) | f(α) ∈ mˆ }, where α is an element of the topological closure E¯ of E in the m-adic completion of K.

Proof. Let M be a prime ideal of Int(E,D) above m which is not Mm,α. Since D is Noetherian, local, and one-dimensional, m is necessarily the radical of every nonzero proper principal ideal, and so it is necesarily a pseudo-principal ideal. Therefore, by [3, Proposition V.1.11], we have that M is maximal. Suppose for contradiction that M is not of the form Mm,α, or equivalently (as they are maximal ideals) that M is not contained in any such ideals, meaning for each α ∈ E¯ there is f ∈ M such that f(α) ∈/ mˆ . Since f is continuous and the m-adic topology determines an ultrametric, there is a clopen neighborhood Uα of α such q−1 that f(t) ∈/ mˆ for each t ∈ Uα. If |D/m| = q, we notice that the values of f are always congruent to 0 or 1 modulo mˆ and definitely congruent to 1 on Uα. As D’s residue field is finite, the completion Dˆ of D is compact [3, Proposition III.1.2]. Also, E¯ is compact; this is clear if E ⊆ D, but if not, then for some d ∈ D (because E is fractional) dE¯ ⊆ Dˆ is necessarily compact and it is also homeomorphic to E¯, and therefore E¯ is compact regardless of scenario. Therefore, E¯ may be covered by finitely many open sets Ui, 1 ≤ i ≤ r, and we may find for each Ui a polynomial fi ∈ M such that the values of fi are congruent to 0 or 1 modulom ˆ and definitely congruent to Qr 1 on Ui (this was done in the first paragraph). Now, consider the polynomial g = i=1(1 − fi). Then g ∈ M and g(α) is congruent to 1 modulo mˆ for each α ∈ E¯, and in particular we have g(α) ≡ 1 mod m for each α ∈ E and (g − 1) ∈ Int(E, m). But since m is pseudo-principal, M contains Int(E, m)[3, Lemma V.1.10], so g − 1 ∈ M, which implies (since g ∈ M) that 1 ∈ M, a contradiction. So, each prime ideal of Int(E,D) must indeed be of the form Mm,α. If D is a discrete valuation domain with finite residue field, then Int(V ) is dense in the set of continuous functions from Vˆ to Vˆ [3, Theorem III.3.4]. If α 6= β in Vˆ , there is a continuous function ϕ such that ϕ(α) ∈ mˆ and ϕ(β) ∈/ mˆ . This function can be approximated by an integer-valued polynomial, and this means that the prime ideals Mm,α and Mm,β are distinct. Hence, we have the following:

Proposition 5.12 [3, Proposition V.2.3] Let V be a discrete valuation domain with ﬁnite residue ﬁeld and m be its maximal ideal. Then the prime ideals of Int(V ) above the maximal ideal m of V are maximal and in one-to-one correspondence with the elements of the completion Vˆ of V : to each element α ∈ Vˆ corresponds the maximal ideal Mm,α = {f ∈ Int(V ) | f(α) ∈ mˆ }.

11 From this proposition comes an easy and important corollary:

Corollary 5.13 [3, Corollary V.2.4] Let V be a discrete valuation domain with finite residue field. The prime ideals of Int(V ) above the maximal ideal m of V are not finitely generated.

Proof. Assume by way of contradiction that Mm,α is generated by polynomials f1, ..., fr. Then, fi(α) ∈ m for all i, and so by continuity if β is close enough to α in Vˆ , then fi(β) ∈ m for each i, meaning Mm,α = Mm,β (as both ideals are maximal), contradicting the previously established fact that these ideals are distinct. Next, we consider the prime ideals above (0) of Int(V ) and their possible containment relations with Mm,α.

Proposition 5.14 [3, Proposition V.2.5] Let V be a discrete valuation domain with finite residue field, q be a polynomial irreducible in the quotient field K of V , and α ∈ Vˆ . The prime ideal Bq = qK[X] ∩ Int(V ) is contained in the maximal ideal Mm,α if and only if q(α) = 0.

Proof. Clearly, if q(α) = 0, then Bq ⊂ Mm,α. For the converse, assume for contradiction q(α) 6= 0; multiplying q by a common denominator of its coefficients we may assume q ∈ V [X]. Lettingv ˆ be the valuation associated with Vˆ ( the completion of a DVR is a DVR; this is exercise 11 on page 68 of [3]), vˆ(q(α)) = n is finite. Since q is a continuous function, there is some clopen neighborhood U of α in Vˆ such thatv ˆ(q(x)) = n for each x ∈ U. From the Stone-Weirstrass Theorem, there is a polynomial h with coefficients in K such thatv ˆ(h(x)) = −n if x ∈ U andv ˆ(h(x)) ≥ 0 otherwise [3, Corollary III.3.5]. Dissecting these last two sentences, this latter bit means h is integer-valued outside of U meaning hq is there as well, and by choice of U,v ˆ(h(x)q(x)) = 0 when x ∈ U, so h(x)q(x) is a unit in V . As a result, hq is integer-valued, so hq ∈ Bq, whilev ˆ(h(α)q(α)) = 0, thus hq∈ / Mm,α, a contradiction.

Remark 5.15 From this proposition, we notice that each prime ideal of V [X] above (0) lifts uniquely in Int(V ): for each polynomial q irreducible in K[X], the prime ideal qK[X] ∩ V [X] lifts to Bq in Int(V ). Furthermore, we have the following (i) The prime ideal m[X] of V [X] is lost in Int(V ) (we can’t construct a prime that lies above it, as we see from the preceding propositions) (ii) In the case of the prime ideal (m, g) of V [X], where g is such that its image in V/m[X]g ¯ is irreducible, we have the following: (a) if deg(¯g) > 1, this prime is lost in Int(V ), (b) if g ≡ X − a mod m, then all the primes Mm,α of Int(V ) where α ≡ a mod mˆ are above (m, g).

Thus, we obtain from this remark and the preceding propositions from V.2 an explicit description of the spectrum of Int(Z). Proposition 5.16 [3, Proposition V.2.7] (i) The prime ideals of Int(Z) above a prime number p are in one-to-one correspondence with the elements ˆ ˆ of the p-adic completion Zp of Z: to each element α ∈ Zp corresponds the maximal ideal Mp,α = {f ∈ Int(Z) | ˆ f(α) ∈ pZp}. Moreover, these ideals are not ﬁnitely generated and hence Int(Z) is not Noetherian. (ii) The nonzero prime ideals of Int(Z) lying above (0) are in one-to-one correspondence with the monic polynomials irreducible in Q[X]. To the irreducible polynomial q corresponds the prime Bq = qQ[X]∩Int(Z). ˆ (iii) The ideal Mp,α is of height one if α ∈ Zp is transcendental over Q and is height two otherwise. In the case that it is height two, it contains the prime Bq where q is the minimal polynomial of α. From this, we get that the Krull dimension of Int(Z) is equal to two. (iv) The maximal ideals of Int(Z) are the ideals Mp,α. Indeed, (i) is just [3, Proposition V.2.2], [3, Proposition V.2.3], and [3, Corollary V.2.4], since although Z isn’t local, we may localize and the p-adic topology is the same whether we localize at pZ or not, by [3, Proposition III.4.9]. Also, (ii), (iii), and (iv) all follow from [3, Proposition V.2.5] and the remark following it.

12 6 Pr¨ufer

To show that Int(Z) is Prüfer,we’ll show that every finitely generated ideal is invertible. In fact, we’ll show this for more general domains D than Z. First, we need a definition and lemma. Definition 6.1 An ideal A of Int(D) is said to be unitary if it contains nonzero constants - in other words, A ∩ D 6= (0). Lemma 6.2 [3, Lemma VI.1.2] Let D be a domain and A be a finitely generated nonzero ideal of Int(D). Then there exists d ∈ D and g ∈ D[X] such that B = (d/g)A is a unitary ideal. Proof. Extending to the PID K[X], AK[X] = gK[X] for some g ∈ K[X]. WLOG, g ∈ Int(D) by clearing denominators. Since A is finitely generated, there is some d ∈ D such that dA ⊆ gInt(D), and so we get our desired B. If D is a Dedekind domain with finite residue fields (like the integers; note that Dedekind domains always have Krull dimension 1) and if M is a maximal ideal of Int(D) above a prime ideal p of D, then Int(D)M is actually a localization of Int(Dp) at a prime ideal. We can break down the result here into two cases of interest: (1) if p = (0), then Int(Dp) = K[X], which is a PID, so obviously its localizations at prime ideals are DVRs; (2) if p is maximal, then Dp is a DVR with finite residue field. So, in obtaining a proof that Int(D) is Prüferwhen D is Dedekind, it suffices to restrict ourselves to the local case, where we let V be a DVR with maximal ideal m, finite residue field, corresponding valuation v, Vˆ and Kˆ the completions of V and its quotient field K in the m-adic topology, andv ˆ the extension of v to Kˆ . Lemma 6.3 [3, Lemma VI.1.4] Let V be a discrete valuation domain as described in the previous paragraph. Then Int(V ) is Prüfer. Proof. Let A be a finitely generated nonzero ideal of Int(V ). By the preceding lemma, we may assume that A is unitary. Let A−1 = (Int(V ): A) = {f ∈ K[X] | fA ⊆ Int(V )}. Then, we have AA−1 ⊆ Int(V ), but we need to show the reverse containment. Since AA−1 contains A, it contains nonzero constants, so it lies above m. As a result, all, we need to do is show that AA−1 is not contained in any maximal ideals above m, which we know by the previous section are all of the form Mm,α = {f ∈ Int(V ) | vˆ(f(α)) > 0}, where α ∈ Vˆ . Let α ∈ Vˆ and let n = inff∈A{vˆ(f(α))}. Since A is unitary, n must be finite. Let f1, ..., fr be generators of A. Since these polynomials are continuous, there are neighborhoods Ui of α in Vˆ such thatv ˆ(fi(x)) ≥ n for r all x ∈ Ui. So, if we let U = ∩i=1Ui, then for each f ∈ A we havev ˆ(f(x)) ≥ n for each x ∈ U. Then, by a variant of the Stone-Weirstrass theorem [3, Corollary III.3.5], there is a polynomial h ∈ K[X] such that −1 vˆ(h(x)) = −n if x ∈ U andv ˆ(h(x)) ≥ 0 otherwise, meaning h ∈ A . So, there is a polynomial f0 ∈ A such −1 thatv ˆ(f0(α)) = n, andv ˆ(h(α)f0(α)) = 0. Therefore, hf0 ∈ AA , but hf0 ∈/ Mm,α. Thus, we have shown that Int(V ) is Prüfer. In particular, as a consequence, Int(Z) is a Prüferdomain by our remarks preceding this lemma.

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