formal power series as inverse limits

1 Motivation and Overview

The ring of formal power series can be described as an inverselimit.¹¹It is worth pointing out that, since we aredealing with formal series, the concept of limit used here hasnothing to do with convergence but is purely algebraic. Thefundamental idea behind this approach is that of truncation —given a formal power series $a_{0}+a_{1}t+a_{2}t^{2}+a_{3}t^{3}+\\cdots$ and an integer $n\\geq 0$ , we can truncate the series toorder $n$ to obtain $a_{0}+a_{1}t+\\cdots+a_{n}t^{n}+O(t^{n+1})$ .²²Here, the symbol “ $O(t^{n+1})$ ” is notused in the sense of Landau notation but merely as an indicatorthat the power series has been truncated to order $n$ .(Indeed, we must do this in practical computation since it isonly possible to write down a finite number of terms at a time;thus the approach taken here has the advantage of being closeto actual practice.) Furthermore, this procedure of truncationcommutes with ring operations — given two formal power series,the truncation of their sum is the sum of their truncations andthe truncation of their product is the product of theirtruncations. Thus, for every integer $n$ the set of powerseries truncated to order $n$ forms a ring and truncation is amorphism from the ring of formal power series to this ring.

To obtain our definition, we will proceed in the oppositedirection. We will begin by defining rings of truncatedpower series and exhibiting truncation morphisms betweendifferent truncations. Then we will show that these ringsand morphisms form an inverse system which has a limit whichwe will take as the definition of the ring of formal powerseries. Finally, we will complete the circle by demonstratingthat the object so constructed is isomorphic with the usualdefinition for ring of formal power series.

2 Formal Development

In this section, we will carry out the develpment outlinedabove in rigorous detail. We begin by formalizing thisnotion of truncation.

Theorem 1.

Let $A$ be a commutative ring and let $n$ be a positiveinteger. Then $A[[x]]/\\langle x^{n}\\rangle$ is isomorphicto $A[x]/\\langle x^{n}\\rangle$ .

Proof.

We may identify $A[x]$ with the subring of $A[[x]]$ consistingof series which have all but a finite number of coeficientsequal to zero. Consider an element $f=\\sum_{k=0}^{\\infty}c_{n}x^{n}$ of $A[[x]]$ . We may write $f=\\sum_{k=0}^{n-1}c_{n}x^{k}+x^{n}\\sum_{k=0}^{\\infty}c_{k+n}x^{k}$ . Thus, every element of $A[[x]]$ is equivalent to an element of the subring $A[x]$ modulo $x^{k}$ . Hence, if follows rather immediately from thedefinition of quotient ring that $A[[x]]\\langle x^{n}\\rangle$ is isomorphic to $A[x]/\\langle x^{n}\\rangle$ .∎

Let us call the isomorphism between $A[[x]]/\\langle x^{n}\\rangle$ and $A[x]/\\langle x^{n}\\rangle$ which is described above $I_{n}$ . We now define a few more morphisms.

Definition 1.

Suppose $m, n$ are integers satifying the inequalities $m>n\\geq 0$ . Then define the morphisms $t_{mn},T_{mn},p_{n},P_{n}$ as follows:

•
Define $t_{nm}\\colon A[x]/\\langle x^{m}\\rangle\\to A[x]/\\langle x^{n}\\rangle$ as the map which sends eachequivalence class $a$ modulo $x^{n}$ to the unique equivalenceclass $b$ modulo $x^{m}$ such that $a\\subset b$ .
•
Define $T_{nm}\\colon A[[x]]/\\langle x^{m}\\rangle\\to A[[x]]/\\langle x^{n+1}\\rangle$ as the map which sends eachequivalence class $a$ modulo $x^{n}$ to the unique equivalenceclass $b$ modulo $x^{m}$ such that $a\\subset b$ .
•
For every integer $n>0$ , let $Q_{n}$ be the quotientmap from $A[[x]]$ to $A[[x]]/\\langle x^{n}\\rangle$ .
•
For every integer $n>0$ , let $q_{n}$ be the quotientmap from $A[x]$ to $A[x]/\\langle x^{n}\\rangle$ .

These morphisms commute with each other in ways which aredescribed by the next theorem:

Theorem 2.

Suppose $m, n, k$ are integers satifying the inequalities $m>n>k\\geq 0$ . Then we have the following relations:

1.
$t_{nk}\\circ t_{mn}=t_{mk}$
2.
$I_{n}\\circ T_{mn}=t_{mn}\\circ I_{m}$
3.
$T_{nk}\\circ T_{mn}=T_{mk}$
4.
$t_{mn}\\circ q_{m}=q_{n}$
5.
$T_{mn}\\circ Q_{m}=Q_{n}$

[More to come]

单词	FormalPowerSeriesAsInverseLimits
释义	formal power series as inverse limits 1 Motivation and Overview The ring of formal power series can be described as an inverselimit.¹¹It is worth pointing out that, since we aredealing with formal series, the concept of limit used here hasnothing to do with convergence but is purely algebraic. Thefundamental idea behind this approach is that of truncation —given a formal power series $a_{0}+a_{1}t+a_{2}t^{2}+a_{3}t^{3}+\\cdots$ and an integer $n\\geq 0$ , we can truncate the series toorder $n$ to obtain $a_{0}+a_{1}t+\\cdots+a_{n}t^{n}+O(t^{n+1})$ .²²Here, the symbol “ $O(t^{n+1})$ ” is notused in the sense of Landau notation but merely as an indicatorthat the power series has been truncated to order $n$ .(Indeed, we must do this in practical computation since it isonly possible to write down a finite number of terms at a time;thus the approach taken here has the advantage of being closeto actual practice.) Furthermore, this procedure of truncationcommutes with ring operations — given two formal power series,the truncation of their sum is the sum of their truncations andthe truncation of their product is the product of theirtruncations. Thus, for every integer $n$ the set of powerseries truncated to order $n$ forms a ring and truncation is amorphism from the ring of formal power series to this ring. To obtain our definition, we will proceed in the oppositedirection. We will begin by defining rings of truncatedpower series and exhibiting truncation morphisms betweendifferent truncations. Then we will show that these ringsand morphisms form an inverse system which has a limit whichwe will take as the definition of the ring of formal powerseries. Finally, we will complete the circle by demonstratingthat the object so constructed is isomorphic with the usualdefinition for ring of formal power series. 2 Formal Development In this section, we will carry out the develpment outlinedabove in rigorous detail. We begin by formalizing thisnotion of truncation. Theorem 1. Let $A$ be a commutative ring and let $n$ be a positiveinteger. Then $A[[x]]/\\langle x^{n}\\rangle$ is isomorphicto $A[x]/\\langle x^{n}\\rangle$ . Proof. We may identify $A[x]$ with the subring of $A[[x]]$ consistingof series which have all but a finite number of coeficientsequal to zero. Consider an element $f=\\sum_{k=0}^{\\infty}c_{n}x^{n}$ of $A[[x]]$ . We may write $f=\\sum_{k=0}^{n-1}c_{n}x^{k}+x^{n}\\sum_{k=0}^{\\infty}c_{k+n}x^{k}$ . Thus, every element of $A[[x]]$ is equivalent to an element of the subring $A[x]$ modulo $x^{k}$ . Hence, if follows rather immediately from thedefinition of quotient ring that $A[[x]]\\langle x^{n}\\rangle$ is isomorphic to $A[x]/\\langle x^{n}\\rangle$ .∎ Let us call the isomorphism between $A[[x]]/\\langle x^{n}\\rangle$ and $A[x]/\\langle x^{n}\\rangle$ which is described above $I_{n}$ . We now define a few more morphisms. Definition 1. Suppose $m, n$ are integers satifying the inequalities $m>n\\geq 0$ . Then define the morphisms $t_{mn},T_{mn},p_{n},P_{n}$ as follows: • Define $t_{nm}\\colon A[x]/\\langle x^{m}\\rangle\\to A[x]/\\langle x^{n}\\rangle$ as the map which sends eachequivalence class $a$ modulo $x^{n}$ to the unique equivalenceclass $b$ modulo $x^{m}$ such that $a\\subset b$ . • Define $T_{nm}\\colon A[[x]]/\\langle x^{m}\\rangle\\to A[[x]]/\\langle x^{n+1}\\rangle$ as the map which sends eachequivalence class $a$ modulo $x^{n}$ to the unique equivalenceclass $b$ modulo $x^{m}$ such that $a\\subset b$ . • For every integer $n>0$ , let $Q_{n}$ be the quotientmap from $A[[x]]$ to $A[[x]]/\\langle x^{n}\\rangle$ . • For every integer $n>0$ , let $q_{n}$ be the quotientmap from $A[x]$ to $A[x]/\\langle x^{n}\\rangle$ . These morphisms commute with each other in ways which aredescribed by the next theorem: Theorem 2. Suppose $m, n, k$ are integers satifying the inequalities $m>n>k\\geq 0$ . Then we have the following relations: 1. $t_{nk}\\circ t_{mn}=t_{mk}$ 2. $I_{n}\\circ T_{mn}=t_{mn}\\circ I_{m}$ 3. $T_{nk}\\circ T_{mn}=T_{mk}$ 4. $t_{mn}\\circ q_{m}=q_{n}$ 5. $T_{mn}\\circ Q_{m}=Q_{n}$ [More to come]
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