Properties of Completions

We now examine a few additional properties of completion.

Completion and Exact Sequences

Theorem 1 Fix a Noetherian ring \(A\) and an ideal \(\mathfrak{a}\), and let \(\widehat{A}\) be the completion of \(A\) with respect to \(\mathfrak{a}\). Then the following hold.

1. \(\widehat{A}\) is Noetherian.
2. \(\widehat{A}/\mathfrak{a}^i\widehat{A}=A/\mathfrak{a}^i\) holds for all \(i\).

Proof

First, since \(A\) is Noetherian, \(A/\mathfrak{a}\) is also Noetherian and \(\mathfrak{a}/\mathfrak{a}^2\) is a finitely generated \(A/\mathfrak{a}\)-module. Thus \(\gr_\mathfrak{a}A\) is an \(A/\mathfrak{a}\)-algebra generated by \(\mathfrak{a}/\mathfrak{a}^2\), and therefore by §Basic Notions, ⁋Theorem 12 (Hilbert basis theorem), \(\gr_\mathfrak{a}A\) is Noetherian. Now, following the notation of §Completion, §§Definition of Completion, we write the kernel of the canonical morphism \(\rho_i:\widehat{A} \rightarrow A/\mathfrak{a}^i\) as \(\widehat{\mathfrak{a}}_i\), and let the filtration of \(\widehat{A}\) formed by these be \(\widehat{\mathcal{J}}\). Then by the isomorphism \(\gr_\mathfrak{a}A\cong \gr_{\widehat{\mathcal{J}}}\widehat{A}\) obtained in the same place, we see that \(\gr_{\widehat{\mathcal{J}}}\widehat{A}\) is also Noetherian. On the other hand, for any ideal \(\mathfrak{b}\subseteq \widehat{A}\), the initial ideal \(\initial(\mathfrak{b})\) is generated by finitely many elements by the argument above, so we obtain the first result by §Completion, ⁋Proposition 7.

For the second result, since \(A\) is Noetherian, we can choose finitely many elements such that \(\mathfrak{a}^i=(a_1,\ldots, a_n)\). First, since \(\rho_i\) sends \(\mathfrak{a}^i\widehat{A}\) to \(0\), we have \(\mathfrak{a}^i\widehat{A}\subseteq \widehat{\mathfrak{a}}_i\). On the other hand, an element of \(\widehat{\mathfrak{a}}_i\) has an initial form of degree at least \(i\), and conversely, a homogeneous element of degree \(d\geq i\) belongs, via \(\mathfrak{a}^d=\mathfrak{a}^{d-i}\mathfrak{a}^i\), to the ideal generated by \(\initial(a_1),\ldots, \initial(a_n)\); hence inside \(\gr_{\widehat{\mathcal{J}}}\widehat{A}=\gr_\mathfrak{a}A\), \(\initial(\widehat{\mathfrak{a}}_i)\) is generated by \(\initial(a_1),\ldots, \initial(a_n)\). Therefore, by §Completion, ⁋Proposition 7, \(\widehat{\mathfrak{a}}_i\) is generated by \(a_1,\ldots, a_n\), and thus \(\widehat{\mathfrak{a}}_i=\mathfrak{a}^i\widehat{A}\). Now since \(\widehat{A}/\widehat{\mathfrak{a}}_i\cong A/\mathfrak{a}^i\), we obtain the desired result.

The following lemma can be proved easily by using the topological structure and base of the completion discussed in §Completion, §§The \(\mathfrak{a}\)-adic Topology, together with [Topology] §Bases of a Topological Space, ⁋Proposition 2, to examine when two bases of a topological space define the same topology.

Lemma 2 Suppose we are given two filtrations of an \(A\)-module \(M\) by submodules,

\[\mathcal{J}:\qquad M=M_0\supseteq M_1\supseteq M_2\supseteq\cdots\]

and

\[\mathcal{J}': \qquad M=M_0'\supseteq M_1'\supseteq M_2'\supseteq\cdots\]

If for each \(M_i\) there exists some \(M_j'\) such that \(M_j'\subseteq M_i\), and for each \(M_i'\) there exists some \(M_j\) such that \(M_j\subseteq M_i'\), then \(\widehat{M}_\mathcal{J}\cong \widehat{M}_{\mathcal{J}'}\) holds.

Meanwhile, by [Category Theory] §Limits, ⁋Proposition 10, taking inverse limits is left exact. The following lemma shows that, under a suitable finiteness assumption, taking completion is exact.

Lemma 3 Fix a Noetherian ring \(A\) and an ideal \(\mathfrak{a}\). Then for a short exact sequence of finitely generated \(A\)-modules

\[0 \rightarrow M' \xrightarrow{\alpha} M \rightarrow M'' \rightarrow 0\]

the following sequence

\[0 \rightarrow \varprojlim M'/\mathfrak{a}^i M' \rightarrow \varprojlim M/\mathfrak{a}^i M \rightarrow \varprojlim M''/\mathfrak{a}^i M'' \rightarrow 0\]

is an exact sequence.

Proof

Since \(\alpha\) is injective, regarding \(M'\) as a submodule of \(M\), since \(\mathfrak{a}^iM''\) is the image of \(\mathfrak{a}^iM\), for each \(i\) we obtain the following short exact sequence

\[0 \rightarrow M'/(M'\cap \mathfrak{a}^iM) \rightarrow M/\mathfrak{a}^iM \rightarrow M''/\mathfrak{a}^iM'' \rightarrow 0\]

Applying the left exactness of inverse limits discussed earlier to this, we see that \(\varprojlim M'/(M'\cap \mathfrak{a}^iM) \rightarrow \varprojlim M/\mathfrak{a}^iM\) is injective and that its image equals the kernel of \(\varprojlim M/\mathfrak{a}^iM \rightarrow \varprojlim M''/\mathfrak{a}^iM''\). Meanwhile, since \(\{\mathfrak{a}^iM\}\) is an \(\mathfrak{a}\)-stable filtration, by §Blowup Algebras, ⁋Lemma 7 (Artin-Rees), \(\{M'\cap \mathfrak{a}^iM\}\) is also \(\mathfrak{a}\)-stable, so there exists some \(l\) such that \(M'\cap \mathfrak{a}^iM=\mathfrak{a}^{i-l}(M'\cap \mathfrak{a}^lM)\) for all \(i\geq l\). From this we obtain

\[\mathfrak{a}^iM'\subseteq M'\cap \mathfrak{a}^iM,\qquad M'\cap \mathfrak{a}^{i+l}M\subseteq \mathfrak{a}^iM'\]

so the two filtrations \(\{\mathfrak{a}^iM'\}\) and \(\{M'\cap \mathfrak{a}^iM\}\) satisfy the conditions of Lemma 2, and the resulting isomorphism \(\varprojlim M'/\mathfrak{a}^iM'\cong \varprojlim M'/(M'\cap \mathfrak{a}^iM)\) is compatible with the maps into \(\varprojlim M/\mathfrak{a}^iM\) induced by \(M'\subseteq M\). Therefore, it suffices to show that \(\varprojlim M/\mathfrak{a}^i M \rightarrow \varprojlim M''/\mathfrak{a}^i M''\) is surjective.

Given an element \((c_i+\mathfrak{a}^i M'')\) of \(\varprojlim M''/\mathfrak{a}^i M''\), let us find \((b_i+\mathfrak{a}^iM)\) satisfying the following two conditions:

• \(b_i\mapsto c_i\pmod{\mathfrak{a}^i M''}\),
• \(b_i\equiv b_j\pmod{\mathfrak{a}^iM}\) for \(i<j\)

Then it is clear that for a fixed \(i\) we can find \(b_i\) satisfying the first condition alone, and that for the second condition it suffices to show only the case \(j=i+1\). Therefore, suppose inductively that we have found \(b_1,\ldots, b_k\) satisfying these two conditions, and let us find \(b_{k+1}\). First, choose \(b_{k+1}'\) satisfying \(b_{k+1}'\mapsto c_{k+1}\pmod{\mathfrak{a}^{k+1} M''}\). Then \(b_{k+1}'\) and \(b_k\) map to the same element in \(M''/\mathfrak{a}^k M''\), so from the following exact sequence

\[M'/\mathfrak{a}^{k}M' \rightarrow M/ \mathfrak{a}^{k}M \rightarrow M''/ \mathfrak{a}^{k}M'' \rightarrow 0\]

we can find some \(m'\in M'\) such that \(m'\) maps to \(b_k-b_{k+1}'\) in \(M/\mathfrak{a}^kM\). Now setting \(b_{k+1}=b_{k+1}'+\alpha(m')\), we obtain the desired result.

From this we obtain the following theorem.

Theorem 4 Fix a Noetherian ring \(A\) and an ideal \(\mathfrak{a}\), and let \(\widehat{A}\) be the completion with respect to \(\mathfrak{a}\) of \(A\). Then the following hold.

1. For any finitely generated \(A\)-module \(M\),

   \[\widehat{A}\otimes_A M \rightarrow\varprojlim_i M/\mathfrak{a}^iM\]

   is an isomorphism.

2. \(\widehat{A}\) is a flat \(A\)-module.

Proof

Since both \(\varprojlim\) and \(\otimes\) commute with finite direct sums, the first result holds trivially when \(M\) is a finitely generated free module. Now, for an arbitrary finitely generated \(A\)-module \(M\), if we take a free presentation of \(M\)

\[F \rightarrow G \rightarrow M \rightarrow 0\]

and apply \(\widehat{A}\otimes_A-\), we obtain the desired result by Lemma 3 and [Homological Algebra] §Diagram chasing, ⁋Proposition 1 (The four lemma).

The second result, by §Flatness, ⁋Proposition 1, is equivalent to showing that for any finitely generated ideal \(\mathfrak{a}\), \(\widehat{\mathfrak{a}} \rightarrow \widehat{A}\) is injective, and this in turn is immediate from the left exactness of completion discussed in Lemma 3.

Hensel’s Lemma

A typical example of a complete ring is the ring of formal power series \(A[[\x_i]]_{i\in I}\) discussed in §Completion, ⁋Example 4. Meanwhile, we saw in [Algebraic Structures] §Algebras, ⁋Proposition 8 that the polynomial ring \(A[\x_i]_{i\in I}\) plays the role of a free functor \(\Set \rightarrow \cAlg{A}\); a similar kind of universal property holds for \(A[[\x_i]]_{i\in I}\) as well.

Theorem 5 Fix a ring \(A\) and an \(A\)-algebra \(E\), and suppose that \(E\) is complete with respect to some ideal \(\mathfrak{a}\subseteq E\). If \(\alpha_1,\ldots,\alpha_n\in \mathfrak{a}\), then the following hold.

1. Sending each \(\x_i\) to \(\alpha_i\), there exists a unique \(A\)-algebra homomorphism \(\phi:A[[\x_1,\ldots, \x_n]]\rightarrow E\).
2. If \(A \rightarrow E/\mathfrak{a}\) is surjective and the \(\alpha_i\) generate \(\mathfrak{a}\), then \(\phi\) is also surjective.
3. If \(\gr\phi: A[\x_1,\ldots, \x_n]\cong \gr_{(\x_1,\ldots, \x_n)}A[[\x_1,\ldots, \x_n]] \rightarrow \gr_\mathfrak{a}E\) is a monomorphism, then so is \(\phi\).

Proof

The first claim is almost identical to [Algebraic Structures] §Algebras, ⁋Proposition 8, and like that proposition, it holds for infinitely many variables as well.

For the second claim, from the given hypothesis we know that

\[(\x_1,\ldots, \x_n)/(\x_1,\ldots, \x_n)^2 \rightarrow \mathfrak{a}/\mathfrak{a}^2\]

is surjective, and since \(\mathfrak{a}/\mathfrak{a}^2\) generates \(\gr_\mathfrak{a}E\), we know that \(\gr\phi\) is surjective. Now choose an arbitrary \(y\in E\). Since the case \(y=0\) is trivial, assume \(y\neq 0\). Because \(E\) is complete, \(E \rightarrow \widehat{E}\) is injective and hence \(\bigcap_i \mathfrak{a}^i=0\), so, satisfying \(y\in \mathfrak{a}^i\), there exists a largest \(i\). Therefore, there exists \(x_1\in (\x_1,\ldots, \x_n)^i\) such that the initial form of \(x_1\) can be mapped by \(\gr\phi\) to the initial form of \(y\), and then \(y-\phi(x_1)\in \mathfrak{a}^{i+1}\). Repeating this process, we can have an infinite family of elements of \(A[\x_1,\ldots, \x_n]\), \((x_1,x_2,\ldots)\), satisfy the following sum

\[y=\sum_{j=1}^\infty \phi(x_j)=\phi\left(\sum_{j=1}^\infty x_j\right)\]

Here, since the degree of \(x_j\) increases with \(j\), \(\sum_j x_j\) defines an element of \(A[[\x_1,\ldots, \x_n]]\), and because \(\phi\) sends \((\x_1,\ldots, \x_n)^i\) into \(\mathfrak{a}^i\), we can interchange \(\phi\) with the infinite sum as in the second equality above.

As for the last claim, if \(x\) is an element of \(A[[\x_1,\ldots, \x_n]]\) not equal to \(0\), then \(\initial(x)\) is also not \(0\), and thus \((\gr\phi)(\initial(x))\) is not \(0\) either. Now, letting the degree of \(\initial(x)\) be \(d\), since

\[x\equiv \initial(x)\pmod{(\x_1,\ldots, \x_n)^{d+1}}\]

we have

\[\phi(x)\equiv(\gr\phi)(\initial(x))\pmod{\mathfrak{a}^{d+1}}\]

so that \(\phi(x)\neq 0\).

Then the following holds.

Corollary 6 Fix a power series \(f\in \x A[[\x]]\), and define \(\phi: A[[\x]] \rightarrow A[[\x]]\) by the formula

\[\phi: A[[\x]] \rightarrow A[[\x]];\qquad \x\mapsto f\]

Then \(\phi\) is an isomorphism if and only if \(f'(0)\) is a unit in \(A\).

Proof

First, the elements not belonging to \((\x)\) in \(A[[\x]]\) are precisely those with constant term not equal to \(0\), and \(\phi\) of the form given in the condition preserves such elements. Now if \(\phi\) is an isomorphism, it follows from this that \(\phi((\x))=(\x)\). Moreover, since \(\phi\) must send a generator of \((\x)\) again to a generator of \((\x)\), we know from this that \(f+(\x^2)\) must generate \((\x)/(\x^2)\), and now since

\[f\equiv f'(0)\x\pmod{\x^2}\]

we know from this that \(f'(0)\) must be a unit of \(A\).

Conversely, suppose \(f'(0)\) is a unit of \(A\). Then first, \(\gr_{(\x)}A[[\x]]\cong \gr_{(\x)}A[\x]=A[\x]\), and setting \(u=f'(0)\), by definition \(\gr\phi: A[\x] \rightarrow A[\x]\) sends \(\x\) to \(u\x\). Now by the third result of Theorem 5, \(\phi\) is injective; also, for some \(h\in A[[\x]]\), writing \(f=u\x+h\x^2=(u+h\x)\x\), we know that \(f\) generates \((\x)\). Therefore, by the second result of Theorem 5 again, we obtain the desired conclusion.

Then we obtain the following theorem, which is the central result of this section.

Theorem 7 (Hensel) Suppose a ring \(A\) is complete with respect to an ideal \(\mathfrak{a}\), and let \(f(\x)\in A[\x]\) and \(a\in A\). If

\[f(a)\equiv 0\pmod{f'(a)^2 \mathfrak{a}}\]

then there exists some \(b\in A\) such that

\[f(b)=0,\qquad b\equiv a\pmod{f'(a)\mathfrak{a}}\]

holds. Moreover, if \(f'(a)\) is a non-zerodivisor, then such a \(b\) is uniquely determined.

Proof

For notational convenience, let \(f'(a)=e\). Then from

\[f(a+e\x)=f(a)+f'(a)e\x+\cdots\]

we can choose \(h\) such that \(f(a+e\x)=f(a)+f'(a)e\x+h(\x)(e\x)^2\). Then

\[f(a+e\x)=f(a)+e^2(\x+\x^2h(\x))\]

and by the first result of Theorem 5, there exists a unique \(A\)-algebra homomorphism \(\phi:A[[\x]] \rightarrow A[[\x]]\) sending \(\x\) to \(\x+\x^2h(\x)\). Meanwhile, by Corollary 6, \(\phi\) is an isomorphism, and therefore its inverse \(\phi^{-1}\) exists. Now applying \(\phi^{-1}\) to the above equation, we obtain

\[f(a+e\phi^{-1}(\x))=f(a)+e^2\x\]

and by the given assumption there exists \(\alpha\in \mathfrak{a}\) such that \(f(a)=e^2\alpha\). Using the first result of Theorem 5 again, if we use the \(A\)-algebra homomorphism \(\psi: A[[\x]] \rightarrow A\) sending \(\x\) to \(-\alpha\), from the above equation we obtain

\[f(a+e\psi(\phi^{-1}(\x)))=0\]

Therefore, setting \(b=a+e\psi(\phi^{-1}(\x))\), we obtain the desired result. Here, since \(\phi^{-1}(\x)\in (\x)\) and \(\psi((\x))\subseteq \mathfrak{a}\), we have \(b-a\in e\mathfrak{a}\), that is, \(b\equiv a\pmod{f'(a)\mathfrak{a}}\) also holds.

For uniqueness, suppose \(e\) is not a zero divisor and let \(b,b'\) be two elements satisfying the given conditions. Then by definition they must be of the form \(a+er\) and \(a+er'\), respectively. Now, applying the first result of Theorem 5, let us choose \(\beta,\beta': A[[\x]] \rightarrow A\) sending \(\x\) to \(r\) and \(r'\), respectively. Applying these, since

\[f(a)+e^2(r+r^2h(r))=f(a+er)=0=f(a+er')=f(a)+e^2(r'+(r')^2h(r'))\]

we obtain \(\beta(\phi(\x))=\beta'(\phi(\x))\). Now, from the fact that \(\phi\) is an isomorphism and the uniqueness in Theorem 5, we obtain the desired result.

Finally, we mention the following theorem and conclude.

Theorem 8 (Cohen structure theorem) For a complete local Noetherian ring \((A, \mathfrak{m})\) and residue field \(\kappa\), if \(A\) contains a field, then there exist some \(n\) and an ideal \(\mathfrak{a}\) such that \(A\cong\kappa[[\x_1,\ldots, \x_n]]/\mathfrak{a}\).

---

References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

---