Completion

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Definition of Completion

Given an abelian group \(G\) and a decreasing sequence of its subgroups

\[\mathcal{J}:\qquad G=H_0\supseteq H_1\supseteq\cdots\]

the quotient maps \(G/ H_{i+1} \rightarrow G/H_{i}\) are well-defined, and more generally, by composing them appropriately, we obtain maps \(\rho_{ji}:G/H_j \rightarrow G/H_i\) for all \(j>i\). From these data we form the inverse limit

\[\widehat{G}_\mathcal{J}=\varprojlim_i G/H_i=\left\{(g_1,g_2,\ldots)\in \prod G/H_i\,\middle\vert\,\text{$\rho_{ji}(g_j)=g_i$ for all $j>i$}\right\}\]

together with the canonical morphisms \(\rho_i:\widehat{G}_{\mathcal{J}} \rightarrow G/ H_i\), which satisfy \(\rho_{ji}\circ\rho_j=\rho_i\) for all \(j>i\). For notational convenience, when \(\mathcal{J}\) is clear from context we simply write \(\widehat{G}\).

These can be viewed as categorical limits, as we saw in §Limits, ⁋Example 5 (Inverse limit), and therefore they also satisfy the following universal property.

Whenever maps \(K \rightarrow G/H_i\) satisfying \(\rho_{ji}\circ\pi_j=\pi_i\) are given, there exists a unique \(\pi:K \rightarrow \widehat{G}\) making the following diagram

commute.

If \(G\) is equipped with a ring structure and the \(H_i\) are ideals, then \(\widehat{G}\) naturally inherits a ring structure as well. The situation we shall study is the following.

Definition 1 Fix a ring \(A\) and an ideal \(\mathfrak{a}\). For an \(\mathfrak{a}\)-filtration of ideals of \(A\)

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

we call

\[\widehat{A}=\varprojlim_i A/\mathfrak{a}_i\]

the completion of \(A\) with respect to this filtration. If the natural map \(A \rightarrow \widehat{A}\) is an isomorphism, we call \(A\) a complete ring with respect to this filtration.

In particular, if the above filtration is given by

\[A\supseteq\mathfrak{a}\supseteq \mathfrak{a}^2\cdots\]

we call it the \(\mathfrak{a}\)-adic completion of \(A\). In this case, if \(\mathfrak{a}\) is a maximal ideal then \(\widehat{A}\) becomes a local ring with unique maximal ideal \(\widehat{\mathfrak{a}}\), so we call \(\widehat{A}\) a complete local ring.

First, the natural map \(\rho:A \rightarrow \widehat{A}\) is obtained by applying the universal property to the canonical morphisms \(\pr_i: A \rightarrow A/\mathfrak{a}_i\). Then by definition

\[x\in\ker\rho\iff\rho(x)=0\iff \rho_i(\rho(x))=0\text{ for all $i$}\iff \pr_i(x)=0\text{ for all $i$}\iff x\in \mathfrak{a}_i\text{ for all $i$}\]

so \(\rho\) is injective if and only if \(\bigcap \mathfrak{a}_i=0\).

Now let us write \(\widehat{\mathfrak{a}}_i\) for the kernel of the canonical morphism \(\rho_i:\widehat{A}\rightarrow A/\mathfrak{a}_i\). Then by definition \(\mathfrak{a}_i=\rho^{-1}(\widehat{\mathfrak{a}}_i)\), and since the \(\pr_i\) are surjective and \(\pr_i=\rho_i\circ\rho\), the \(\rho_i\) are all surjective; hence by the first isomorphism theorem

\[\widehat{A}/\widehat{\mathfrak{a}}_i\cong A/\mathfrak{a}_i\]

holds. Therefore the descending chain of ideals of \(\widehat{A}\)

\[\widehat{A}=\widehat{\mathfrak{a}}_0\supseteq \widehat{\mathfrak{a}}_1\supseteq\cdots\tag{1}\]

is an \(\mathfrak{a}\)-filtration, and from the above isomorphism we have

\[\widehat{A}=\varprojlim_i A/\mathfrak{a}_i\cong\varprojlim_i \widehat{A}/\widehat{\mathfrak{a}}_i\]

so \(\widehat{A}\) is complete with respect to the given filtration. Moreover, this isomorphism also yields

\[\gr_\mathcal{J}A\cong\gr_{\widehat{\mathcal{J}}}\widehat{A}.\]

The \(\mathfrak{a}\)-adic Topology

On the other hand, the process of constructing \(\widehat{A}\) from \(A\) can also be understood by endowing \(A\) with a special kind of topology. First, suppose a topological abelian group \(G\) is given. Fixing an element \(g\in G\), the translation map \(T_g\) defined by it is continuous, so the neighborhood filter at each point of \(G\) is completely determined by the neighborhood filter at \(0\in G\). This process can, of course, be reversed.

As in the previous section, suppose a decreasing sequence of subgroups of \(G\)

\[G=H_0\supseteq H_1\supseteq\cdots\]

is given. Then defining

\[\mathcal{N}(0)=\{U\subseteq G\mid\text{$G_n\subseteq U$ for some $n$}\}\]

we know that this satisfies all the conditions of §Open Sets, ⁋Proposition 6. Now for arbitrary \(g\in G\) and \(U\in \mathcal{N}(0)\), declaring \(g+U\in \mathcal{N}(g)\) gives a topology on \(G\).

In particular, applying this to the situation of Definition 1, the topology defined by the above process is called the \(\mathfrak{a}\)-adic topology. In this case, \(0\in A\) has the countable local base

\[\mathfrak{a}\supseteq \mathfrak{a}^2\supseteq\cdots\tag{2}\]

so the topology on \(A\) thus defined is first countable.

Returning to a general topological abelian group \(G\), we can weaken the condition defining a convergent sequence to define the following.

Definition 2 For a topological group \((G, +, 0)\), a sequence \((x_i)_{i\in \mathbb{N}}\) of elements of \(G\) is called a Cauchy sequence if for any neighborhood \(U\) of \(0\) there exists a natural number \(N\) such that

\[m,n>N \implies x_m-x_n\in U\]

holds.

Then, just as the completion of a general topological group can be defined as the set of equivalence classes of Cauchy filters, we can define when two Cauchy sequences \((x_m)\) and \((y_n)\) are regarded as the same, and through that define the (topological) completion. However, the object of our interest is the first countable topological group \(A\) defined by the filtration (2) above, and since a first countable space is sequential, for convenience in the following definition we assume \(G\) is a first countable space and use Cauchy sequences in place of Cauchy filters.

Definition 3 Two Cauchy sequences \((x_m)\), \((y_n)\) of a topological group \((G, +, 0)\) are called equivalent if for any neighborhood \(U\) of \(0\) there exists a natural number \(N\) such that

\[m,n>N \implies x_m-y_n\in U\]

holds. We call the set of equivalence classes of all Cauchy sequences of a first countable topological group \(G\), equipped with this equivalence relation, the completion of \(G\), and denote it by \(\widehat{G}\).

Now for an open neighborhood \(U\) of \(0\in G\), define

\[\widehat{U}=\{[(x_n)]\in \widehat{G}:\text{for any $(y_n)\in [(x_n)]$, $y_n\in U$ for all but finitely many $n$}\}.\]

Then by a short calculation, one can verify that the collection \(\mathcal{N}(0)\) of subsets of \(\widehat{G}\) having the \(\widehat{H}_i\) as a coinitial subset satisfies all the conditions of §Open Sets, ⁋Proposition 6, and therefore we can define a topology on \(\widehat{G}\). By definition \(\widehat{G}\) is also first countable, and the function \(G \rightarrow \widehat{G}\) sending \(x\in G\) to the constant sequence \((x_i=x)\) is continuous. Moreover, this function coincides exactly with the \(G \rightarrow \widehat{G}\) defined in the previous section.

Basic Properties of Completion

Now let us examine the basic properties of completion. By Definition 3 above, any element of \(\widehat{A}\) can be viewed as a Cauchy sequence in the \(\mathfrak{a}\)-adic topology on \(A\). Then for \(b_j\in \mathfrak{a}^j\), writing

\[a_i=\sum_{j=1}^i b_j\tag{3}\]

the sequence \((a_i)\) is a Cauchy sequence in \(\widehat{A}\), and therefore the limit of this sequence

\[\sum_{j=1}^\infty b_j\]

defines an element of \(\widehat{A}\). Conversely, given any element \((a_n')\) of \(\widehat{A}\), using the local base (2) of \(0\) we can find a Cauchy sequence equivalent to this element and having the form (3).

Example 4 If \(A=\mathbb{K}[\x]\) and \(\mathfrak{a}=(\x)\), then \(\widehat{A}\) is the ring of formal power series \(\mathbb{K}[[\x]]\).

The ring \(\mathbb{K}[[\x]]\) is a discrete valuation ring with unique nonzero prime ideal \(\mathfrak{m}=(\x)\). That is, any element not in \((\x)\) is a unit, which essentially follows from the identity

\[\frac{1}{1+\x}=1-\x+\x^2-\cdots\]

This equality, or equivalently

\[(1+\x)(1-\x+\x^2-\cdots)=1\]

is obtained, as in the discussion above, by observing that for the partial sum of \(1-\x+\x^2-\cdots\) up to degree \(i\)

\[1-\x+\x^2-\cdots+(-1)^i\x^i\]

we have

\[(1+\x)(1-\x+\x^2-\cdots+(-1)^i\x^i)=1+(-1)^i\x^i\in \mathfrak{m}^i\]

so this product is equivalent to the constant sequence \((1)\).

Generalizing this calculation, we obtain the following two results.

Proposition 5 Suppose \(A\) is complete with respect to an ideal \(\mathfrak{a}\). Then the set

\[U=\{1+a\mid a\in \mathfrak{a}\}\]

consists of units of \(A\), and \(U\) is multiplicatively closed.

Proof

Replace \(\x\) by \(a\) in the above argument.

Corollary 6 For a local ring \((A, \mathfrak{m})\), the ring \(A[[\x_1,\ldots, \x_n]]\) is also a local ring, and its unique maximal ideal is \(\mathfrak{m}+(\x_1,\ldots, \x_n)\).

Proof

Any element outside \(\mathfrak{m}+(\x_1,\ldots,\x_n)\) has a nonzero constant term, so by Proposition 5 it is a unit.

Moreover, the following holds.

Proposition 7 Fix a filtration of ideals of \(A\)

\[A=\mathfrak{a}_0\supseteq \mathfrak{a}_1\supseteq\cdots\]

and the associated graded ring \(\gr A\) with respect to this filtration. If \(A\) is complete with respect to this filtration, then for an ideal \(\mathfrak{a}\) of \(A\) and its elements \(a_1,\ldots, a_n\), if \(\initial(\mathfrak{a})\) is generated by \(\initial(a_1),\ldots, \initial(a_n)\), then \(\mathfrak{a}\) is also generated by \(a_1,\ldots, a_n\).

Proof

Let \(\mathfrak{a}'\) be the ideal generated by \(a_1,\ldots, a_n\) and let us show \(\mathfrak{a}=\mathfrak{a}'\). Without loss of generality we may assume all these elements are nonzero. Also, if \(a_k\in \mathfrak{a}_i\) held for all \(i\), then by the canonical morphism \(A \rightarrow \widehat{A}\) the element \(a_k\) is sent to \(0\in \widehat{A}\), and since \(A\) is complete this means \(a_k=0\); thus we can choose \(d\) so that \(a_k\not\in \mathfrak{a}_i\) holds for all \(k\).

On the other hand, from the assumption that \(\initial(\mathfrak{a})\) is generated by the \(\initial(a_k)\), for any \(a\in \mathfrak{a}\) there exist \(\beta_k\in \gr_\mathfrak{a}A\) satisfying

\[\initial(a)=\sum_{k=1}^n \beta_k\initial(a_k)\tag{4}\]

and considering degrees in the above formula, we know that the \(\beta_k\) are homogeneous and their degrees must satisfy

\[\degree(\beta_k)=\degree (\initial(a))-\degree(\initial(a_k))>\degree(\initial(a))-d.\]

Therefore for \(b_k\in A\) satisfying \(\initial(b_k)=\beta_k\), the element \(a-\sum_k b_k a_k\) lies in \(\mathfrak{m}_{\degree(\initial(a))+1}\). Repeating this process, we can choose \(a'\in \mathfrak{a}'\) such that

\[a-\underbrace{\sum_k b_k a_k-\cdots}_{=a'} \in \mathfrak{a}_{d+1}.\]

Now, since \(a'\) is generated by the \(a_k\) anyway, showing that \(a\) is generated by the \(a_k\) is the same as showing that \(a-a'\) is generated by the \(a_k\). That is, without loss of generality we may assume \(a\) lies in \(\mathfrak{a}_{d+1}\).

Now let us revisit formula (4) under this assumption. Setting \(\degree(\initial(a))=e\), we saw above that the degree of \(\beta_k\) must be at least \(e-d\). Therefore we can choose the \(b_k\) from \(\mathfrak{a}_{e-d}\), and now by the same logic as above

\[a-\sum_{k=1}^n b_ka_k\]

lies in \(\mathfrak{a}_{e-d+1}\). Repeating this, we can choose \(b_k^{(l)}\in \mathfrak{a}_{e-d+l}\) such that

\[a-\sum_{k=1}^n\sum_{l=0}^j b_k^{(l)}a_k\in \mathfrak{a}_{e+j+1}.\]

Now since \(A\) is complete, the infinite sum

\[\sum_{l=0}^\infty b_k^{(l)}\]

can be regarded as an element \(c_k\) of \(A\). Then

\[a-\sum_{k=1}^n c_k a_k\in \bigcap \mathfrak{a}_i=0\]

so we obtain the desired result.

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References

[AM] M.F. Atiyah and I.G. Macdonald, Introduction to commutative algebra, Basic Books, 1969.
[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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