Properties of Completion

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We now examine some 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 \(\mathfrak{a}\)-adic completion of \(A\). Then the following hold.

1. \(\widehat{A}\) is Noetherian.
2. \(\widehat{A}/\mathfrak{a}^i\widehat{A}=A/\mathfrak{a}^i\) 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. Hence \(\gr_\mathfrak{a}A\) is generated by \(\mathfrak{a}/\mathfrak{a}^2\) as an \(A/\mathfrak{a}\)-algebra, and therefore \(\gr_\mathfrak{a}A\) is Noetherian by §Basic Notions, ⁋Theorem 12 (Hilbert basis theorem). Now since \(\gr_{\widehat{\mathfrak{a}}}\widehat{A}=\gr_\mathfrak{a}A\), we see that \(\gr_{\widehat{\mathfrak{a}}}\widehat{A}\) is also Noetherian. On the other hand, for any ideal \(\widehat{\mathfrak{a}}\subseteq \widehat{A}\), the initial ideal \(\initial(\widehat{\mathfrak{a}})\) is generated by finitely many elements by the argument above, so the first result follows from §Completion, ⁋Proposition 7.

For the second result, we again use §Completion, ⁋Proposition 7 to see that the equality \(\widehat{\mathfrak{a}}^i=\mathfrak{a}^i \widehat{A}\) is equivalent to their initial ideals being equal, and thus we obtain the desired result.

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

Lemma 2 Let two filtrations of a ring \(A\)

\[\mathcal{J}:\qquad A=\mathfrak{a}_0\supseteq \mathfrak{a}_1\supseteq \mathfrak{a}_2\supseteq\cdots\]

and

\[\mathcal{J}': \qquad A=\mathfrak{a}_0'\supseteq \mathfrak{a}_1'\supseteq \mathfrak{a}_2'\supseteq\cdots\]

be given. If for each \(\mathfrak{a}_i\) there exists some \(\mathfrak{a}_j'\) with \(\mathfrak{a}_j'\subseteq \mathfrak{a}_i\), and for each \(\mathfrak{a}_i'\) there exists some \(\mathfrak{a}_j\) with \(\mathfrak{a}_j\subseteq \mathfrak{a}_i'\), then \(\widehat{A}_\mathcal{J}\cong \widehat{A}_{\mathcal{J}'}\).

On the other hand, by §Limits, ⁋Proposition 10, taking completion is left exact. The following lemma shows that under an appropriate finiteness hypothesis, taking completion is also right exact.

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

\[0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0\]

the sequence

\[0 \rightarrow \varprojlim A/\mathfrak{a}^i A \rightarrow \varprojlim B/\mathfrak{a}^i B \rightarrow \varprojlim C/\mathfrak{a}^i C \rightarrow 0\]

is exact.

Proof

By the preceding argument, we only need to show that \(\varprojlim B/\mathfrak{a}^i B \rightarrow \varprojlim C/\mathfrak{a}^i C\) is surjective.

Let an element \((c_i+\mathfrak{a}^i C)\) of \(\varprojlim C/\mathfrak{a}^i C\) be given, and let us find \((b_i+\mathfrak{a}^iB)\) satisfying the following two conditions:

• \(b_i\mapsto c_i\pmod{\mathfrak{a}^i C}\),
• \(b_i\equiv b_j\pmod{\mathfrak{a}^iB}\) for \(i<j\).

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

\[A/\mathfrak{a}^{k}A \rightarrow B/ \mathfrak{a}^{k}B \rightarrow C/ \mathfrak{a}^{k}C \rightarrow 0\]

we can find suitable \(a\in A\) mapping to \(b_k-b_{k+1}'\) in \(B/\mathfrak{a}^kB\). Now setting \(b_{k+1}=b_{k+1}'+\alpha_{k+1}(a_{k+1})\) gives 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 \(\mathfrak{a}\)-adic completion 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 is obvious when \(M\) is a finitely generated free module. Now for any finitely generated \(A\)-module \(M\), applying \(\widehat{A}\otimes_A-\) to a free presentation

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

yields the desired result by Lemma 3 and §Diagram chasing, ⁋Proposition 1 (The four lemma).

For the second result, by §Flatness, ⁋Proposition 1 it suffices to show that \(\widehat{\mathfrak{a}} \rightarrow \widehat{A}\) is injective for any finitely generated ideal \(\mathfrak{a}\), and this is clear from the left exactness of completion examined 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. We saw in §Algebras, ⁋Proposition 8 that the ring of power series \(A[\x_i]_{i\in I}\) plays the role of the free functor \(\Set \rightarrow \cAlg{A}\), and a similar universal property holds for \(A[[\x_i]]_{i\in I}\).

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

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

Proof

The first assertion is almost identical to §Algebras, ⁋Proposition 8, and as in that proposition it holds for infinitely many variables as well.

For the second assertion, 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 see that \(\gr\phi\) is surjective. Now for any \(y\in E\), there exists a largest \(i\) such that \(y\in \mathfrak{a}^i\). Hence there exists \(x_1\in (\x_1,\ldots, \x_n)^i\) whose initial form maps under \(\gr\phi\) to the initial form of \(y\), and then \(y-\phi(x_1)\in \mathfrak{a}^{i+1}\). Repeating this process, we can find an infinite family \((x_1,x_2,\ldots)\) of elements of \(A[\x_1,\ldots, \x_n]\) such that

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

For the last assertion, if \(x\) is a nonzero element of \(A[[\x_1,\ldots, \x_n]]\), then \(\initial(x)\) is also nonzero, and hence \((\gr\phi)(\initial(x))\) is nonzero. Letting \(d\) be the degree of \(\initial(x)\),

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

so

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

and thus \(\phi(x)\neq 0\).

Then the following holds.

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

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

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

Proof

First, the elements of \(A[[\x]]\) not belonging to \((\x)\) are exactly those with nonzero constant term, and any \(\phi\) of the given form preserves such elements. Now if \(\phi\) is an isomorphism, then \(\phi((\x))=(\x)\). Moreover \(\phi\) must send a generator of \((\x)\) to a generator of \((\x)\), so we see that \(f+(\x^2)\) must generate \((\x)/(\x^2)\), and since

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

we conclude that \(f'(0)\) must be a unit in \(A\).

Conversely, suppose \(f'(0)\) is a unit in \(A\). Then \(\gr_{(\x)}A[[\x]]\cong \gr_{(\x)}A[\x]=A[\x]\), and \(\gr\phi: A[\x] \rightarrow A[\x]\) sends \(x\) to \(ux\) by definition. By the third result of Theorem 5, \(\phi\) is injective, and writing \(f=u\x+h\x^2=(u+h\x)\x\) for suitable \(h\in A[[\x]]\), we see that \(f\) generates \((\x)\). Hence by the second result of Theorem 5 we obtain the desired conclusion.

We thus obtain the following theorem, which is the central result of this section.

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

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

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

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

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

Proof

For convenience set \(f'(a)=e\). Then from

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

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

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

and by the first result of Theorem 5, there exists a unique \(A\)-algebra homomorphism \(\phi:A[[\x]] \rightarrow A[[\x]]\) sending \(\x\) to \(\x+\x^2h(\x)\). On the other hand, by Corollary 6, \(\phi\) is an isomorphism, and hence its inverse \(\phi^{-1}\) exists. Applying \(\phi^{-1}\) to the above equation yields

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

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

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

Thus setting \(b=e\psi\phi^{-1}\) gives the desired result.

For uniqueness, assume \(e\) 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 \(a+er\), \(a+er'\). Applying the first result of Theorem 5, choose \(\beta,\beta': A[[\x]] \rightarrow A[[\x]]\) sending \(\x\) to \(r\) and \(r'\) respectively. Applying these gives

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

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

Finally, we mention the following theorem and conclude.

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

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References

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

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