This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Now we further examine some 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.
- \(\widehat{A}\) is noetherian.
- \(\widehat{A}/\mathfrak{a}^i\widehat{A}=A/\mathfrak{a}^i\) holds for all \(i\).
Proof
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 generated by \(\mathfrak{a}/\mathfrak{a}^2\) as an \(A/\mathfrak{a}\)-algebra, and therefore \(\gr_\mathfrak{a}A\) is Noetherian by ##ref##. Now since \(\gr_{\widehat{\mathfrak{a}}}\widehat{A}=\gr_\mathfrak{a}A\), we know 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 above argument, so we obtain the first result by §Completion, ⁋Proposition 7.
For the second result, we again know by §Completion, ⁋Proposition 7 that the equality of \(\widehat{\mathfrak{a}}^i\) and \(\mathfrak{a}^i \widehat{A}\) is equivalent to the equality of their initial ideals, and thus we obtain the desired result.
The following lemma can be easily proved by examining the topological structure and base of completion studied in §Completion, §§\(\mathfrak{a}\)-adic topology and using [Topology] §Bases of Topological Spaces, ⁋Proposition 2 to determine when two bases of a topological space 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 a suitable \(\mathfrak{a}_j'\) such that \(\mathfrak{a}_j'\subseteq \mathfrak{a}_i\), and for each \(\mathfrak{a}_i'\) there exists a suitable \(\mathfrak{a}_j\) such that \(\mathfrak{a}_j\subseteq \mathfrak{a}_i'\), then \(\widehat{A}_\mathcal{J}\cong \widehat{A}_{\mathcal{J}'}\) holds.
On the other hand, by [Category Theory] §Limits, ⁋Proposition 10, taking completion is left exact. The following lemma shows that if an appropriate finiteness condition is assumed, then taking completion is also right exact.
Lemma 3 Fix a Noetherian ring \(A\) and an ideal \(\mathfrak{a}\). Then for a short exact sequence of finitely generated \(A\)-modules
\[0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0\]the following 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 an exact sequence.
Proof
By the preceding argument, it suffices 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 \(i<j\).
Then it is obvious that for a fixed \(i\) we can find \(b_i\) satisfying only the first condition, and that for the second condition it suffices to show the case \(j=i+1\). Thus, assuming inductively that we have found \(b_1,\ldots, b_k\) satisfying these two conditions, let us find \(b_{k+1}\). First choose \(b_{k+1}'\) satisfying \(b_{k+1}'\mapsto c_{k+1}\pmod{\mathfrak{a}^{k+1} C}\). 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 a suitable \(a\in A\) such that \(a\) maps to \(b_k-b_{k+1}'\) in \(B/\mathfrak{a}^kB\). Now setting \(b_{k+1}=b_{k+1}'+\alpha_{k+1}(a_{k+1})\) 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 of \(A\) with respect to \(\mathfrak{a}\). Then the following hold.
-
For any finitely generated \(A\)-module \(M\),
\[\widehat{A}\otimes_A M \rightarrow\varprojlim_i M/\mathfrak{a}^iM\]is an isomorphism.
-
\(\widehat{A}\) is a flat \(A\)-module.
Proof
Since \(\varprojlim\) and \(\otimes\) both commute with finite direct sums, the first result obviously holds when \(M\) is a finitely generated free module. Now for any finitely generated \(A\)-module \(M\), taking \(\widehat{A}\otimes_A-\) on a free presentation
\[F \rightarrow G \rightarrow M \rightarrow 0\]of \(M\) yields the desired result by Lemma 3 and [Homological Algebra] §Diagram chasing, ⁋Proposition 1.
The second result reduces, by §Flatness, ⁋Proposition 1, to showing that \(\widehat{a} \rightarrow \widehat{A}\) is injective for any finitely generated ideal \(\mathfrak{a}\), and this is again obvious from the left exactness of completion examined in Lemma 3.
Hensel’s Lemma
A representative example of a complete ring is the ring of formal power series \(A[[\x_i]] _{i\in I}\), which we examined in §Completion, ⁋Example 4. On the other hand, we saw in [Algebraic Structures] §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 kind of universal property also holds for \(A[[\x_i]]_{i\in I}\).
Theorem 5 Fix a ring \(A\) and an \(A\)-algebra \(E\), and suppose that \(E\) is complete with respect to some ideal \(\mathfrak{a}\subseteq E\). Let \(\alpha_1,\ldots,\alpha_n\in \mathfrak{a}\). Then the following hold.
- There exists a unique \(A\)-algebra homomorphism \(\phi:A[[\x_1,\ldots, \x_n]]\rightarrow E\) sending each \(x_i\) to \(\alpha_i\).
- If \(A \rightarrow E/\mathfrak{a}\) is an epimorphism and the \(\alpha_i\) generate \(\mathfrak{a}\), then \(\phi\) is also an epimorphism.
- 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 [Algebraic Structures] §Algebras, ⁋Proposition 8, and just like that proposition it also holds for infinitely many variables.
For the second assertion, from the given assumption 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 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 is 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 arrange that an infinite family \((x_1,x_2,\ldots)\) of elements of \(A[\x_1,\ldots, \x_n]\) satisfies the sum
\(y=\sum_{j=1}^\infty \phi(x_j)=\phi\left(\sum_{j=1}^\infty x_j\right)\).
For the last assertion, if \(x\) is a nonzero element of \(A[[\x_1,\ldots, \x_n]]\), then \(\initial(x)\) is also nonzero and therefore \((\gr\phi)(\initial(x))\) is nonzero as well. But if the degree of \(\initial(x)\) is \(d\), then
\[x\equiv \initial(x)\pmod{(\x_1,\ldots, \x_n)^{d+1}}\]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 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 of \(A[[\x]]\) not belonging to \((\x)\) are exactly those with nonzero constant term, and a \(\phi\) of the form given in the condition preserves such elements. Now if \(\phi\) is an isomorphism, then \(\phi((\x))=(\x)\). Moreover, since \(\phi\) must send a generator of \((\x)\) to a generator of \((\x)\), we know that \(f+(\x^2)\) must generate \((\x)/(\x^2)\), and now since
\[f\equiv f'(0)\x\pmod{x^2}\]we know that \(f'(0)\) must be a unit of \(A\).
Conversely, suppose that \(f'(0)\) is a unit of \(A\). Then first \(\gr_{(\x)}A[[\x]]\cong \gr_{(\x)}A[\x]=A[\x]\), and \(\gr\phi: A[\x] \rightarrow A[\x]\) sends \(x\) to \(ux\) by definition. Now by the third result of Theorem 5, \(\phi\) is injective, and writing \(f=u\x+h\x^2=(u+h\x)\x\) for a suitable \(h\in A[[\x]]\), we see that \(f\) generates \((\x)\). Hence again by the second result of Theorem 5 we obtain the desired conclusion.
Then we obtain the following theorem, which is the main 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 a suitable \(b\in A\) 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 set \(f'(a)=e\). 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 \(\x+\x^2h(\x)\). On the other hand, by Corollary 6, \(\phi\) is an isomorphism, and hence its inverse \(\phi^{-1}\) exists. Now applying \(\phi^{-1}\) to the above equation we obtain
\[f(a+e\phi^{-1}(x))=f(a)+e^2x\]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, let \(\psi: A[[\x]] \rightarrow A[[\x]]\) be the \(A\)-algebra homomorphism sending \(\x\) to \(-\alpha\); then from the above equation we obtain
\[f(a+e\psi\phi^{-1}(x))=0\]Therefore setting \(b=e\psi\phi^{-1}\) we obtain the desired result.
For uniqueness, assume that \(e\) 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 apply the first result of Theorem 5 to 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 \(\beta(\phi(\x))=\beta'(\phi(\x))\). 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 noethherian ring \((A, \mathfrak{m})\) and residue field \(\kappa\), if \(A\) contains some field then there exist a suitable \(n\) and an ideal \(\mathfrak{a}\) satisfying \(A\cong\kappa[[\x_1,\ldots, \x_n]]/\mathfrak{a}\).
References
[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.
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