the maps \(G/ H_{i+1} \rightarrow G/H_{i}\) are well-defined, and more generally, composing these appropriately defines \(\rho_{ji}:G/H_j \rightarrow G/H_i\) whenever \(j>i\). From these data we obtain 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 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 also simply write \(\widehat{G}\).
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 \(G\) carries a ring structure and the \(H_i\) are ideals, then \(\widehat{G}\) also carries a natural ring structure. The situation we will consider is the following.
Definition 1 Fix a ring \(A\) and an ideal \(\mathfrak{a}\). Then for an \(\mathfrak{a}\)-filtration of ideals of \(A\) (§Blowup Algebras, ⁋Definition 3)
the completion of \(A\) defined by 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 in the form
we call it the \(\mathfrak{a}\)-adic completion of \(A\). In this case, if \(\mathfrak{a}\) is a maximal ideal, then \(\widehat{A}\) is a local ring with the 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 \(\rho\) being injective is equivalent to \(\bigcap \mathfrak{a}_i=0\).
Now let us denote the kernels of the canonical morphisms \(\rho_i:\widehat{A}\rightarrow A/\mathfrak{a}_i\) by \(\widehat{\mathfrak{a}}_i\). 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, and therefore by the first isomorphism theorem
Meanwhile, the process of passing from \(A\) to \(\widehat{A}\) can also be viewed by endowing a special kind of topology. First, let a topological abelian group \(G\) be given. Then, after fixing in \(G\) an element \(g\), the translation map \(T_g\) defined using it is continuous, so the neighborhood filter at each point of \(G\) is entirely determined by the neighborhood filter at \(0\in G\). Of course, this process can be carried out in reverse as well.
As in the previous section, suppose we are given a decreasing sequence of subgroups of \(G\)
\[G=H_0\supseteq H_1\supseteq\cdots\]
Then defining
\[\mathcal{N}(0)=\{U\subseteq G\mid\text{$H_n\subseteq U$ for some $n$}\}\]
we know that this satisfies all the conditions of [Topology] §Open Sets, ⁋Proposition 6. Now, for arbitrary \(g\in G\) and \(U\in \mathcal{N}(0)\), setting \(g+U\in \mathcal{N}(g)\) gives a topology on \(G\).
In particular, applying this to the situation of Definition 1, we call the topology defined through the above process the \(\mathfrak{a}\)-adic topology. In this case, since \(0\in A\) has a countable local base
the topology on \(A\) defined in this way is first countable.
Returning to a general topological abelian group \(G\), we can weaken the condition for a convergent sequence and define the following.
Definition 2 For a topological group \((G, +, 0)\), a sequence of elements of \(G\), \((x_i)_{i\in \mathbb{N}}\), is called a Cauchy sequence if, whenever an arbitrary neighborhood of \(0\), \(U\), is given, there exists a natural number \(N\) such that the following statement
\[m,n>N \implies x_m-x_n\in U\]
is true.
Then, just as one defines the completion in a general topological group as the collection of equivalence classes of Cauchy filters, given two Cauchy sequences \((x_m)\), \((y_n)\) we can decide when to regard them as the same and thereby define the (topological) completion. However, what we are interested in is the first countable topological group \(A\) defined by the filtration (2) above, and since a first countable space is sequential, in the following definition we assume for convenience that \(G\) is a first countable space and use Cauchy sequences instead of Cauchy filters.
Definition 3 For a topological group \((G, +, 0)\), two Cauchy sequences \((x_m)\), \((y_n)\) are said to be equivalent if, whenever an arbitrary neighborhood of \(0\), \(U\), is given, there exists a natural number \(N\) such that the following statement
\[m,n>N \implies x_m-y_n\in U\]
is true. The set obtained by imposing this equivalence relation on the set of all Cauchy sequences of a first countable topological group \(G\) is called the completion of \(G\), and we write it as \(\widehat{G}\).
Now, for \(0\in G\) and an open neighborhood \(U\) of it, define
\[\widehat{U}=\{[(x_n)]\in \widehat{G}\mid\text{for any $(y_n)\in [(x_n)]$, $y_n\in U$ for all but finitely many $n$}\}\]
Then a short computation shows that, with the \(\widehat{H}_i\) as a coinitial subset, the collection of subsets of \(\widehat{G}\) denoted by \(\mathcal{N}(0)\) satisfies all the conditions of [Topology] §Open Sets, ⁋Proposition 6, and therefore we can define a topology on \(\widehat{G}\). By definition, \(\widehat{G}\) is also first countable, and one can see that the function taking \(x\in G\) to the constant sequence \((x_i=x)\), \(G \rightarrow \widehat{G}\), is continuous. Moreover, this function is exactly the same as the map \(G \rightarrow \widehat{G}\) defined in the previous section.
Basic Properties of Completion
Let us now examine the basic properties of completion. By Definition 3 examined above, an arbitrary element of \(\widehat{A}\) can be thought of as a Cauchy sequence in \(A\) with respect to the \(\mathfrak{a}\)-adic topology. Then, for elements satisfying \(b_j\in \mathfrak{a}^j\), if for these \(b_j\) we write
\[a_i=\sum_{j=0}^i b_j\tag{3}\]
then \((a_i)\) is a Cauchy sequence in \(\widehat{A}\), and therefore the limit of this sequence
\[\sum_{j=0}^\infty b_j\]
defines an element of \(\widehat{A}\). Conversely, given an arbitrary element of \(\widehat{A}\), \((a_n')\), one can use the local base (2) of \(0\) to 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 \(\mathbb{K}[[\x]]\) of formal power series.
The ring \(\mathbb{K}[[\x]]\) is a local ring with the unique nonzero prime ideal \(\mathfrak{m}=(\x)\). That is, any element not belonging to \((\x)\) is a unit, and this essentially follows from the identity
\[\frac{1}{1+\x}=1-\x+\x^2-\cdots\]
The above identity, or equivalently the identity
\[(1+\x)(1-\x+\x^2-\cdots)=1\]
is obtained, as in the discussion above, from the fact that for the partial sum of \(1-\x+\x^2-\cdots\) up to degree \(i\)
and since the difference between this product and \(1\) belongs to \(\mathfrak{m}^{i+1}\), this product is equivalent to the constant sequence \((1)\).
Generalizing this computation, we obtain the following two results.
Proposition 5 Suppose \(A\) is complete with respect to an ideal \(\mathfrak{a}\). Then every element of the set
\[U=\{1+a\mid a\in \mathfrak{a}\}\]
is a unit of \(A\), and \(U\) is multiplicatively closed.
Proof
It suffices to replace \(\x\) with \(a\) in the above argument.
Corollary 6 For a local ring \((A, \mathfrak{m})\), \(A[[\x_1,\ldots, \x_n]]\) is also a local ring, and its unique maximal ideal is \(\mathfrak{m}+(\x_1,\ldots, \x_n)\).
Proof
For an element outside \(\mathfrak{m}+(\x_1,\ldots,\x_n)\), since its constant term does not belong to \(\mathfrak{m}\), the constant term is a unit of \(A\); thus by Proposition 5 one can show that it is a unit.
Returning to the situation of Definition 1, let us consider the case where \(\mathfrak{a}\) is a maximal ideal and \(\widehat{A}\) is the \(\mathfrak{a}\)-adic completion with respect to it. First, since \(\widehat{A}/\widehat{\mathfrak{a}}_1\cong A/\mathfrak{a}\) is a field, \(\widehat{\mathfrak{a}}_1\) is a maximal ideal of \(\widehat{A}\). Also, for \(k>1\), since \(\rho_1=\rho_{k1}\circ\rho_k\) and \(\widehat{\mathfrak{a}}_1=\ker\rho_1\), we have \(\rho_k(\widehat{\mathfrak{a}}_1)\subseteq \ker\rho_{k1}=\mathfrak{a}/\mathfrak{a}^k\), and therefore \(\rho_k(\widehat{\mathfrak{a}}_1^k)\subseteq (\mathfrak{a}/\mathfrak{a}^k)^k=0\), that is, \(\widehat{\mathfrak{a}}_1^k\subseteq \ker \rho_k=\widehat{\mathfrak{a}}_k\). Then for an arbitrary \(a\in \widehat{\mathfrak{a}}_1\), since \((-a)^k\in \widehat{\mathfrak{a}}_k\), the partial sums of the series \(\sum_{k=0}^\infty (-a)^k\) form a Cauchy sequence in \(\widehat{A}\), and since \(\widehat{A}\) is complete with respect to the filtration (1), the argument of Proposition 5 applies directly, so \(1+a\) is a unit. Now, if \(x\in \widehat{A}\) does not belong to \(\widehat{\mathfrak{a}}_1\), then since \(\widehat{A}/\widehat{\mathfrak{a}}_1\) is a field there exists \(y\) such that \(xy-1\in \widehat{\mathfrak{a}}_1\), and since \(xy\) is a unit by the previous result, \(x\) is also a unit. That is, every element of \(\widehat{A}\) not belonging to \(\widehat{\mathfrak{a}}_1\) is a unit, and therefore \(\widehat{A}\) is a local ring with \(\widehat{\mathfrak{a}}_1\) as its unique maximal ideal.
and the associated graded ring \(\gr A\) with respect to the filtration. Suppose \(A\) is complete with respect to this filtration. Then, for an ideal of \(A\), \(\mathfrak{a}\), 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 the ideal generated by the elements \(a_1,\ldots, a_n\) be \(\mathfrak{a}'\), and let us show that \(\mathfrak{a}=\mathfrak{a}'\). Without loss of generality, we may assume that these elements are all not \(0\). Also, if \(a_k\in \mathfrak{a}_i\) held for all \(i\), then under the canonical morphism \(A \rightarrow \widehat{A}\), \(a_k\) is mapped to \(0\in \widehat{A}\), and since \(A\) is complete, this would mean that \(a_k=0\); therefore we can choose a suitable \(d\) so that \(a_k\not\in \mathfrak{a}_d\) holds for all \(k\).
Meanwhile, from the assumption that \(\initial(\mathfrak{a})\) is generated by the \(\initial(a_k)\), for any \(a\in \mathfrak{a}\) the equation
holds for some \(\beta_k\in \gr A\), and considering degrees in the above equation, we know that the \(\beta_k\) are homogeneous and that their degree must satisfy
Therefore, for elements satisfying \(\initial(b_k)=\beta_k\), with \(b_k\in A\), the element \(a-\sum_k b_k a_k\) belongs to \(\mathfrak{a}_{\degree(\initial(a))+1}\). Repeating this process, so that
holds, we can choose \(a'\in \mathfrak{a}'\). Here, 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 that \(a\) belongs to \(\mathfrak{a}_{d+1}\).
Now let us examine equation (4) again under this assumption. Setting \(\degree(\initial(a))=e\), we saw earlier 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 reasoning as above, we know that
\[a-\sum_{k=1}^n b_ka_k\]
belongs to \(\mathfrak{a}_{e+1}\). Repeating this, so that
[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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