NOTES ON FILTRATIONS, TOPOLOGIES, AND COMPLETIONS

J.P. MAY

Contents

1. Filtered groups 1 2. Filtered rings and modules 2 3. The Artin–Rees Lemma and Krull intersection theorem 3 4. I-adic completion 4 5. Faithfully flat R-modules 5 6. Zariski rings 7 7. The I-adic metric on R 7 8. Hensel’s lemma 8

1. Filtered groups

1. Let G be a group with a decreasing filtration by normal subgroups G_{i}. Then G is a topological group. The G_{i} form a fundamental system of neighborhoods of the identity. The open subsets are the arbitrary unions of finite intersections of cosets gG_{i}.

2. The following are equivalent.

(i) G is Hausdorff.

(ii) Points in G are closed.

(iii) The G_{i} intersect in {e}. Of course, (i) implies (ii) by general topology. For (ii) implies (i), the diagonal in G × G is μ^{−1}(e), where μ(g, h) = gh^{−1}. Since G − G_{i} is the union of the cosets gG_{i} with g ∉ G_{i}, G − G_{i} is open, hence G − G_{i} is both open and closed, hence so is G_{i}. Now (iii) clearly implies (ii). If x is in all G_{i} and is not e, then there are no open neighborhoods separating e and x, so G is not Hausdorff.

3. If H ⊃ G_{i}, then H − G_{i} is open since it is the union of the cosets hG_{i}, h ∈ H − G_{i} and, similarly, H is also closed.

4. G/ ∩ G_{i} is the associated Hausdorff group of G.

5. Consider the canonical map γ: G ⟶ lim G/G_{i}, obtained from the quotient homomorphisms γ_{i}: G ⟶ G/G_{i}. Give the target the inverse limit topology, where the G/G_{i} are discrete. Then γ is continuous since γ_{i}^{−1}(eG_{i}) = G_{i}. If γ is a bijection, then it is a homeomorphism. Indeed, the G_{i} then give a fundamental system of neighborhoods of the identity in both. We say that G is complete when this holds.

6. Define the completion of G to be Ĝ = lim G/G_{i}; it is more accurate to view the map γ: G ⟶ Ĝ as the completion of G. Let Ĝ_{i} be the kernel of Ĝ ⟶ G/G_{i}. This gives Ĝ a decreasing filtration, and the topology on Ĝ is the same as the topology associated to this filtration. Moreover, G/G_{n} ≅ Ĝ/Ĝ_{n}.

7. Two filtrations of G give the same topology if for each m and n there exist p and q such that G_{m} ⊂ G′_{p} and G′_{n} ⊂ G_{q}. By cofinality, the completions are isomorphic as topological groups (homeomorphic via an isomorphism of groups).

2. Filtered rings and modules

Let R be a commutative ring and M an R-module. We consider decreasing filtrations R = R_{0} ⊃ R_{1} ⊃ R_{2} ⊃ ⋯ by ideals such that R_{i} · R_{j} ⊂ R_{i+j}. Similarly, we consider decreasing filtrations M = M_{0} ⊃ M_{1} ⊃ M_{2} ⊃ ⋯ by sub R-modules such that R_{i} · M_{j} ⊂ M_{i+j}. In favorable cases ∩R_{i} = {0} and ∩M_{i} = {0}. The notations F_{i}R = R_{i} and F_{i}M = M_{i} are frequently used. In the most important example, we take an ideal I of R and define R_{i} = I^{i} and M_{i} = I^{i}M. These are called I-adic filtrations.

We can apply the constructions of the previous section to the underlying filtered Abelian groups of R and M. The completion R̂ inherits a multiplication from R, and the completion M̂ becomes an R̂-module. We are interested in understanding the exactness properties of these constructions. We start in the general case in this section and specialize to I-adic completions in the next.

We start over with a filtered R-module M, with no filtration given on R = R_{0}, so that each M_{i} is an R-module. Of course each M_{i} is open and closed in the resulting “linear topology” on M. Let N be a sub R-module of M and let P = M/N. Then N has the filtration given by N_{i} = N ∩ M_{i}, P has the filtration given by letting P_{i} be the image of M_{i}, and N and P have associated linear topologies.

1. The subspace topology on N coincides with the linear topology. Indeed, a subset X ⊂ N is open in the subspace topology iff X = N ∩ U for some open subset U of M, while X is open in the linear topology iff X is a union of finite intersections of subsets of the form x + N_{i} = x + N ∩ M_{i}, x ∈ N. Here U is a union of finite intersections of the form y + M_{i}, but if N ∩ (y + M_{i}) is non-empty, then y + M_{i} = x + M_{i} for some x ∈ N.

2. The closure N̄ of N in M is given by N̄ = ∩_{i}(N + M_{i}). Indeed, x ∈ N̄ iff (x + M_{i}) ∩ N ≠ ∅ for all i, and that holds if and only of x ∈ N + M_{i} for all i. Therefore N is closed in M iff ∩(N + M_{n}) = N, and this holds iff ∩P_{i} = 0, that is, iff P is Hausdorff in the linear topology.

3. The quotient topology on P coincides with the linear topology. Indeed, let X ⊂ P. Then X is open in the quotient topology iff the inverse image, Y say, of X in M is open. This means that if y ∈ Y then y + M_{i} ⊂ Y for some i. Reducing mod N, this means that if x ∈ X, then x + P_{i} ⊂ X for some i, which means that X is open in the linear topology on P.

4. Since P/P_{i} ≅ M/N + M_{i}, we have the short exact sequences

\xymatrix{0 \ar[r] & N/N ∩ M_{i} \ar[r] & M/M_{i} \ar[r] & P/P_{i} \ar[r] & 0.}

On passage to limits , there results a short exact sequence

\xymatrix{0 \ar[r] & N̂ \ar[r] & M̂ \ar[r] & P̂ \ar[r] & 0.}

That is, { \atop M/N} ≅ M̂/N̂. Here N̂ is the closure of the image of N in M̂. Exactness at the left is a general fact on inverse sequences. Exactness at the right uses that the maps in our inverse systems are epimorphisms. In detail, let (p_{j}) ∈ P̂. Inductively, suppose chosen m_{i}, i < j, such that m_{i} ⟶ p_{i} and m_{i} ⟶ m_{i−1} for each i. Choose m′_{j} that maps to p_{j}. Then m′_{j} − m_{j−1} is in N + M_{j−1}, say m′_{j} − m_{j−1} = n + m.

Let m_{j} = m′_{j} − n = m_{j−1} + m. Inductively, this gives an element (m_{j}) of M̂ that maps to (p_{j}).

3. The Artin–Rees Lemma and Krull intersection theorem

Let I be a proper ideal in R and N ⊂ M be R-modules. Filtering M by the I^{i}M, we obtain two filtrations on N, namely the I-adic filtration given by the I^{i}N and the submodule filtration given by the N ∩ I^{i}M. Clearly I^{i}N ⊂ N ∩ I^{i}M. The opposite inclusion fails, but the two linear topologies are sometimes the same. Let R be Noetherian and M be finitely generated throughout this section. Then we have the following result.

Lemma 3.1 (Artin–Rees). There is an m such that

N ∩ I^{n}M = I^{n−m}(N ∩ I^{m}M)

and therefore N ∩ I^{n}M ⊂ I^{n−m}N for all n > m.

Proof. Define a graded ring B_{I}(R) = R ⊕ I ⊕ I^{2} ⊕ ⋯; it is called the Rees ring of (R, I). Think of B_{I}(R) as R[It] ⊂ R[t] for an indeterminate t. Write E_{I}^{0}R (or gr_{I}(R)) for the associated graded ring of R with respect to the I-adic filtration, that is, ⊕_{i≥0}I^{i}/I^{i+1}. Observe that B_{I}(R)/IB_{I}(R) ≅ E_{I}^{0}R. Thus the construction replaces the associated graded by a simple quotient. Let M_{∗} = {M_{i}} be any decreasing I-filtration of M, meaning that IM_{i} ⊂ M_{i+1}. Say that the I-filtration is I-stable if IM_{n} = M_{n+1} for all sufficiently large n. The filtration {I^{i}M} is certainly I-stable, and the claim is that the filtration {N_{i} = N ∩ I^{i}M} is I-stable. Define B(M_{∗}) = M ⊕ M_{1} ⊕ M_{2} ⊕ ⋯ and observe that B(M_{∗}) is a graded B_{I}(R)-module. The second of the following two lemmas is a generalized version of the result we are after. □

Lemma 3.2. The B_{I}(R)-module B(M_{∗}) is finitely generated iff the I-filtration M_{∗} is I-stable.

Proof. Suppose that B(M_{∗}) is finitely generated. Its generators lie in the first m terms for some m. Replace the generators by their homogeneous components (or work homogeneously from the start). These components are still finite in number and still generate B(M_{∗}). Thus B(M_{∗}) is generated by the elements of the M_{i} for i ≤ m. This implies that M_{m} ⊕ M_{m+1} ⊕ ⋯ is generated as a B_{I}(R)-module by M_{m}. This means that M_{i+m} = I^{i}M_{m} for i ≥ 0 or, equivalently, that the filtration is I-stable. Conversely, if M_{i+m} = I^{i}M_{m} for some m and all i ≥ 0, then B(M_{∗}) is generated by the union of the sets of generators of the M_{i} for i ≤ m, which is a finite set. □

Lemma 3.3. Let M_{∗} be any I-stable filtration of M, such as {I^{i}M}, and let N_{i} = N ∩ M_{i}. Then N_{∗} is an I-stable filtration of N.

Proof. Clearly B(N_{∗}) is a sub B_{I}(R)-module of B(M_{∗}). Since M_{∗} is I-stable, B(M_{∗}) is finitely generated. Since I is finitely generated, B_{I}(R) is finitely generated as an R-algebra. Therefore, by the Hilbert basis theorem, B_{I}(R) is a Noetherian ring. But then B(N_{∗}) is finitely generated and therefore N_{∗} is I-stable. □

Corollary 3.4 (Krull intersection theorem). Let N = ∩I^{i}M. Then there exists r ∈ R such that 1 − r is in I and rN = 0. If I ⊂ \sqrt{R}, then N = 0.

Proof. By the Artin-Rees theorem and N ⊂ I^{n}M, there exists m such that

N = N ∩ I^{m+1}M = I(N ∩ I^{m}M) = IN.

The existence of r is now either a standard lemma in the proof of Nakayama’s lemma (which says that N = 0 if I ⊂ \sqrt{R}) or is sometimes itself referred to as Nakayama’s lemma. One proof is by induction on the number of generators of modules N such that IN = N. The last statement follows since r is a unit if I ⊂ \sqrt{R}. □

4. I-adic completion

Let I be a proper ideal of a commutative ring R and let M be an R-module. We have the completion γ: R ⟶ R̂_{I} = lim R/I^{i}, which is a continuous homomorphism of topological rings. We say that R is complete at I if γ is an isomorphism. We also have the completion γ: M ⟶ M̂_{I} = lim M/I^{i}M, which is a continuous homomorphism of topological R-modules. Let Î = {(r_{i})|r_{0} = 0}. Then Î is an ideal of R̂_{I} such that Î^{n} = {(r_{i})|r_{i} = 0 if i < n}. The associated graded rings E_{I}^{0}R and E0 ; ÎR̂ are the same. The Artin–Rees lemma gives the following fundamental result.

Lemma 4.1. If R is Noetherian, then completion is an exact functor on the category of finitely generated R-modules. Therefore R̂_{I} is a flat R-module.

Algebraic topologists must often work with rings that are not Noetherian and modules that are not finitely generated even when R is Noetherian. For them, the “right” notion of completion is not I-adic completion, but rather its zeroth left derived functor (and its first left derived functor is also relevant).

Lemma 4.2. Let 𝔪 be a maximal ideal of R. Then R̂_{𝔪} is a local ring with maximal ideal 𝔪̂, and R/𝔪 = R̂_{𝔪}/𝔪̂. The completion R ⟶ R̂_{𝔪} is the composite of the localization R ⟶ R_{𝔪} and the completion R_{𝔪} ⟶ R̂_{𝔪}.

Proof. We must show that an element (r_{i}) of R̂_{I} that is not in 𝔪̂ is a unit. Now (r_{i}) ∉ 𝔪̂ if and only if r_{0} ≠ 0 in R/𝔪. Since r_{i} maps to r_{0} under R/𝔪^{i} ⟶ R/𝔪, r_{i} is not in 𝔪R/𝔪^{i}, so is a unit in R/𝔪^{i}. The sequence (r_{i}^{−1}) is (r_{i})^{−1} in R̂_{𝔪}. □

A complete local ring R is a Noetherian local ring which is complete at its maximal ideal 𝔪. Such rings are central to number theory and algebraic geometry.

Example 4.3. The p-adic integers ℤ̂_{(p)} are usually denoted ℤ_{p} (or sometimes ℤ̂_{p}). They can be represented in terms of “infinite p-adic expansions” ∑ a_{i}p^{i}, where 0 ≤ a_{i} < p.

Example 4.4. The completion of the polynomial ring R[x_{1}, ⋯, x_{n}] at the ideal I = (x_{1}, ⋯, x_{n}) is isomorphic to the power series ring R[[x_{1}, ⋯, x_{n}]]. Explicitly, send a formal power series f to the element (f mod I^{i}) of the completion. For the inverse, consider an element (f_{i}) of the completion. Here f_{i} can be represented (mod I^{i}) as a polynomial of degree less than i in the x_{q}, and then f_{i} = f_{i+1} plus terms of degree i + 1. The formal power series f_{0} + (f_{1} − f_{0}) + (f_{2} − f_{1}) + ⋯ gives the corresponding element of the power series ring.

It is left as an exercise to prove that if R is I-adically complete, then I is contained in the radical of R. If M is I-adically complete, then it is an R̂_{I}-module and therefore multiplication by 1 + a, a ∈ I is an automorphism of M.

It is also left as an exercise to prove that if I = (a_{1}, ⋯, a_{n}) is an ideal in a Noetherian ring R, then R̂_{I} is isomorphic to

R̂_{I} ≅ R[[x_{1}, …, x_{n}]]/(x_{1} − a_{1}, ⋯, x_{n} − a_{n}).

This has the following basic consequence.

Corollary 4.5. If R is Noetherian, then R̂_{I} is Noetherian.

One way to work the excercise just cited is to use the following result, which shows that ideal theory passes nicely to completions.

Proposition 4.6. Let I and J be ideals in a commutative Noetherian ring R and let M be a finitely generated R-module. Then

{ \atop (JM)}_{I} = J · M̂_{I} and { \atop (M/JM)}_{I} ≅ M̂_{I}/JM̂_{I}.

Moreover, { \atop (JM)}_{I} is the closure of JM in M̂_{I}.

Proof. By Artin–Rees, the short exact sequence

\xymatrix{0 \ar[r] & JM \ar[r] & M \ar[r] & M/JM \ar[r] & 0}

gives an exact sequence

\xymatrix{0 \ar[r] & { \atop (JM)}_{I} \ar[r] & M̂_{I} \ar[r] & { \atop (M/JM)}_{I} \ar[r] & 0}

on passage to I-adic completion, and this implies that { \atop (JM)}_{I} is the closure of JM in M̂_{I}. Certainly J · M̂_{I} ⊂ { \atop (JM)}_{I}. Let J = (a_{1}, ⋯, a_{r}) and define ϕ: M^{r} ⟶ M by ϕ(m_{1}, ⋯, m_{r}) = ∑ a_{i}m_{i}. The image of ϕ is JM, so we have an exact sequence

\xymatrix{M^{r} \ar[r]^{ϕ} & M \ar[r]^{π} & M/JM \ar[r] & 0.}

On passage to limits , there results an exact sequence

\xymatrix{(M̂_{I})^{r} = { \atop (M^{r})}_{I} \ar[r]^{ϕ̂} & M̂_{I} \ar[r]^{π̂} & { \atop (M/JM)}_{I} \ar[r] & 0.}

Since the kernel of π̂ must be { \atop (JM)}_{I}, this gives { \atop (M/JM)}_{I} ≅ M̂_{I}/{ \atop (JM)}_{I}. Here again, ϕ̂(m_{1}, ⋯, m_{r}) = ∑ a_{i}m_{i}, where now m_{i} ∈ M̂_{I}. The image of ϕ̂ is J · M̂_{I}, and this is equal to the kernel, ĴM_{I}, of π̂. □

Corollary 4.7. Let M̂_{i} denote the kernal of the projection M̂_{I} ⟶ M/I^{i}M. Then M̂_{i} = I^{i} · M̂_{I}. Thus the linear topology of M̂_{I} coincides with its I-adic topology as an R-module, which in turn coincides with its I ·R̂_{I}-adic topology as an R̂_{I}-module.

Proof. M/I^{i}M = { \atop (M/I^{i}M)}_{I}, and the kernel of M̂_{I} ⟶ { \atop (M/I^{i}M)}_{I} is I^{i} · M̂_{I} by

the previous result. □

5. Faithfully flat R-modules

An R-module N is said to be faithfully flat if a sequence of R-modules is exact if and only if it becomes exact on tensoring with N. We shall relate this notion to completions. We record the following general result.

Proposition 5.1. An R-module N is flat if and only if the canonical map

I ⊗_{R} N ⟶ R ⊗_{R} N ≅ N

is a monomorphism for all finitely generated ideals I, so that I ⊗_{R} N ≅ IN.

Proof. The forward implication is clear. Assume the condition on ideals. Any ideal is the colimit of its finitely generated ideals, and tensoring with N commutes with colimits. We conclude that I ⊗_{R} N ⟶ N is a monomorphism for any ideal I. Let M′ ⟶ M be a monomorphism. We must show that M′ ⊗_{R} N ⟶ M ⊗_{R} N is a monomorphism. Clearly M is the colimit of the sums M′ + M′′, where M′′ ⊂ M is finitely generated. By induction on the number of generators and passage to colimits, it suffices to show the required monomorphism when M = M′ + Rx for some x ∈ M. Let I = {r|rx ∈ M′}. We then have a short exact sequence

\xymatrix{0 \ar[r] & M′ \ar[r] & M \ar[r] & R/I \ar[r] & 0.}

Since we have a short exact sequence

\xymatrix{0 \ar[r] & I \ar[r] & R \ar[r] & R/I \ar[r] & 0}

and since I ⊗_{R} N ⟶ R ⊗_{R} N is a monomorphism, we see that Tor_{1}^{R}(R/I, N) = 0. Therefore M′ ⊗_{R} N ⟶ M ⊗_{R} N is a monomorphism. □

The following result, together with Nakayama’s lemma, shows that flat modules are often necessarily faithfully flat.

Proposition 5.2. The following conditions on an R-module N are equivalent. (i) N is faithfully flat. (ii) N is flat and M ⊗_{R} N = 0 implies M = 0. (iii) N is flat and 𝔪N ≠ N if 𝔪 is a maximal ideal.

Proof. (i) ⟹ (ii): If M ⊗_{R} N = 0, then 0 ⟶ M ⟶ 0 becomes exact after tensoring with N, hence is exact, and M = 0. (ii) ⟹ (iii): N/𝔪N ≅ R/𝔪 ⊗_{R} N, so this is clear. (iii) ⟹ (ii): Assume M ⊗_{R} N = 0 and x ∈ M is non-zero. Then Rx ≅ R/I, where I is the annihilator of x. Embed I in a maximal ideal 𝔪. Since IN ⊂ 𝔪N ≠ N, Rx ⊗_{R} N ≅ N/IN ≠ 0. Since N is flat Rx ⊗_{R} N ⟶ M ⊗_{R} N is a monomorphism, which contradicts the assumption that M ⊗_{R} N = 0. (ii) ⟹ (i): Let M′ ⟶^{f} M ⟶^{g} M′′ be a sequence such that

M′ ⊗_{R} N ⟶^{f⊗id} M ⊗_{R} N ⟶^{g⊗id} M′′ ⊗_{R} N

is exact. Then, using that N is flat, Im(g ∘ f) ⊗_{R} N = 0. Therefore Im(g ∘ f) = 0 and g ∘ f = 0. Let H = Ker(g)/Im(f). Again, H ⊗_{R} N = 0, hence H = 0. □

A ring homomorphism f: R ⟶ S is said to be faithfully flat if S is faithfully flat as an R-module.

Proposition 5.3. Let f: R ⟶ S be a faithfully flat ring homomorphism. (i) For any R-module M, extension of scalars

id ⊗f: M = M ⊗_{R} R ⟶ M ⊗_{R} S

is a monomorphism. In particular, f is a monomorphism. (ii) Regard f as an inclusion. If I is an ideal in R, then IS ∩ R = I.

Proof. For (i), let x ∈ M be non-zero. The monomorphism Rx ⊗_{R} S ⟶ M ⊗_{R} S has image (x ⊗ 1)S, hence x ⊗ 1 is non-zero. To see (ii), apply (i) to M = R/I, noting that M ⊗_{R} S ≅ S/IS. If r ∈ R is not in I, then it is also not in IS. □

6. Zariski rings

A pair (R, I) consisting of a commutative Noetherian ring R and an ideal I ⊂ \sqrt{R} is called a Zariski ring. The interest of this notion comes from the following result.

Theorem 6.1. The following conditions are equivalent for an ideal I in a commutative Noetherian ring R.

(i) I ⊂ \sqrt{R}.

(ii) Every ideal of R is closed in the I-adic topology.

(iii) R̂_{I} is faithfully flat over R.

Proof. (i) ⟹ (ii): More generally, N ⊂ M is closed for any submodule of a finitely generated R-module M since M/N is Hausdorff by the Krull intersection theorem and therefore {0} is closed in M/N. (ii) ⟹ (iii): It suffices to show that 𝔪R̂_{I} ≠ R̂_{I} for every maximal ideal 𝔪. Since {0} is closed in R, ∩I^{i} = 0 and the completion γ: R ⟶ R̂_{I} is a monomorphism. Since 𝔪 is closed in R and 𝔪R̂_{I} is the closure of 𝔪 in R̂_{I}, 𝔪R̂_{I} ∩R = 𝔪 and therefore 𝔪R̂_{I} ≠ R̂_{I}. (iii) ⟹ (ii): 𝔪R̂_{I} ≠ R̂_{I} for any maximal ideal 𝔪. Since R̂_{I} is I-adically complete, Î_{I} is contained in the radical of R̂_{I}. As in (i) ⟹ (ii), if N is a submodule of a finitely generated R̂_{I}-module M, then N is closed in M. Since γ is continuous, 𝔪 = 𝔪R̂_{I} ∩ R is closed in R. If I is not contained in 𝔪, then I^{i} + 𝔪 = R for all i > 0, contradicting that 𝔪 is closed in R. Therefore I ⊂ \sqrt{R}. □

Consider a Noetherian local ring R with maximal ideal 𝔪. Obviously, 𝔪 = \sqrt{R}. We have proven the following results. Recall that R is said to be complete if it is 𝔪-adically complete.

1. ∩𝔪^{i} = 0.

2. If N is a submodule of a finitely generated R-module M, then N is closed in the 𝔪-adic topology. That is, N = ∩(N + 𝔪^{i}N).

3. Let R̂ = R̂_{𝔪}. Then R̂ is faithfully flat over R, R ⊂ R̂, and I = IR̂ ∩ R for any ideal I.

4. R̂ is a Noetherian local ring with maximal ideal 𝔪R̂, and R̂/𝔪^{i}R̂ ≅ R/𝔪^{i} for i > 0. In particular, R and R̂ have the same residue field.

5. If R is a complete local ring and I is a proper ideal, then R/I is a complete local ring.

7. The I-adic metric on R

Let I be an ideal in R and define d(x, y) to be 1/n if x − y is in I^{n} and not in I^{n+1} and to be 0 if x − y ∈ ∩I^{n}. Then d(x, y) = d(y, x) and

d(x, z) ≤ max(d(x, y), d(y, z)) ≤ d(x, y) + d(y, z).

Thus d is a pseudo-metric on R, and it is a metric if ∩I^{n} = 0.

Any pseudo-metric space X is normal. If A and B are disjoint closed subset of X, let U = {x|d(A, x) < d(B, x)} and V = {x|d(B, x) < d(A, x)}. Then U and V are disjoint open subsets that contain A and B. Of course, X need not be Hausdorff since points need not be closed.

We assume the reader knows what a Cauchy sequence is and what it means for two Cauchy sequences to be equivalent. We say that X is complete if every Cauchy sequence converges, and every Cauchy sequence then converges to a unique point if X is Hausdorff. We define the completion of X to be the set of equivalence classes of Cauchy sequences with the induced metric topology, where d((x_{n}), (y_{n})) is the limit of the d(x_{n}, y_{n}). The completion γ: X ⟶ X̂ sends x to the constant sequence at x, and it is a continuous map with dense image.

Proposition 7.1. The completion of R at I is canonically homeomorphic to its completion in the I-adic metric.

Indeed, the metric topology is the same as the I-adic topology on R. More explicitly, an element (r_{i}) of limR/I^{i} can be viewed as an equivalence class of Cauchy sequences in R.

8. Hensel’s lemma

Here is one fundamental and beautiful reason to care about complete rings, and especially complete local rings.

Lemma 8.1 (Hensel’s lemma). Let R be an I-adically complete Noetherian ring and let k = R/I be the residue ring. Use small letters for polynomials in k[x] and capital letters for polynomials in R[x]. Let F be a polynomial in R[x] that reduces mod I to a polynomial f = gh in k[x], where g and h are relatively prime and g is monic. Then there is a factorization F = GH in R[x] such that G and H reduce mod I to g and h and G is monic. If h is also monic, then H can be chosen to be monic and the resulting factorization is unique.

Sketch Proof. Choose any polynomials G_{1} and H_{1} that reduce mod I to g and h, taking G_{1} to be monic and taking deg(G_{1}) = deg(g) and deg(H_{1}) = deg(h). Proceeding inductively, suppose given G_{n} and H_{n} that reduce mod I^{n} to g and h, where G_{n} is monic, deg(G_{n}) = deg(g) and deg(H_{n}) = deg(h). Write F −G_{n}H_{n} = ∑ a_{i}J_{i}, where a_{i} ∈ I^{n} and deg(J_{i}) < deg(F). Since (g, h) = 1, there are polynomials u_{i} and v_{i} such that j_{i} = gu_{i} + hv_{i}, and we can arrange that deg(u_{i}) < deg(h) by replacing u_{i} by its remainder after division by h and adjusting v_{i} accordingly. Then deg(hv_{i}) = deg(j_{i} − gu_{i}) < deg(f) and therefore deg(v_{i}) < deg(g). Choose U_{i} and V_{i} that reduce mod I to u_{i} and v_{i}, with deg(U_{i}) = deg(u_{i}) and deg(V_{i}) = deg(v_{i}). Set G_{n+1} = G_{n} + ∑ a_{i}V_{i} and H_{n+1} = H_{n} + ∑ a_{i}U_{i}. A quick check shows that F ≡ G_{n+1}H_{n+1} mod I^{n+1}, G_{n+1} is monic, deg(G_{n+1}) = deg(g) and deg(H_{n+1}) = deg(h). Then (G_{n}) and (H_{n}) are Cauchy sequences (coefficientwise) and we can pass to limits to obtain polynomials G and H as required. When h is monic, we can choose the H_{n} to be monic, and comparison shows that different choices of the sequences G_{n} and H_{n} give equivalent Cauchy sequences. □

Corollary 8.2. If F ∈ R[x] and a ∈ R are such that F′(a) is a unit in R and F (a) ≡ 0 mod I, then there exists b ∈ R such that F (b) = 0 and b ≡ a mod I.

Proof. Reducing mod I, f(x) = (x − ā)g(x). Since

f′(x) = g(x) + (x − ā)g′(x),
g(x) ≡ f′(x) ≡ f′(ā) mod (x − ā).

Since f′(ā) is a unit in k, g(x) and x−ā generate k[x] and thus are relatively prime. Hensel’s lemma gives F (x) = (x − b)G(x), where G reduces to g and x − b reduces to x − ā mod I. □