代数章节翻译.doc

2
Integers and rational numbers
2.1 The integers
The integers are familiar to us from elementary arithmetic, but here we shall want to express that familiarity in precise terms. We do this by writing down a list of properties which the integers posses and on which we shall base all our deductions. Later, in Vol.2, we shall see that all the properties listed here can actually be deduced from quite a brief list of axioms, but this is immaterial at present.
We denote the set of positive integers (also called natural numbers) 1,2,3...by N and write Z for the set of all integers, positive, negative, and zero. Here N stands for number and Z for Zahl, the German for number; both abbreviations are generally used in mathematics (see Appendix 2 ).
The set z admits three operations: addition, x+y, subtraction, x-y,and multiplication, x.y, or xy.Often it is convenient to express subtraction by adding the negative:x-y=x+(-y). These operations are connected by the following laws.
Z.1 Associative law: (x+y)+z=x+(y+z), (xy)z=x(yz).
Z.2 Commutative law: x+y=y+x, xy=yx.
Z.3 Existence of neutral element:x+0=x, x1=x.
Z.4 Existence of (additive) inverse: x+(-x)=0.
The number 0 is said to be neutral for addition because adding it to any number x leaves x unchanged; likewise 1 is neutral for multiplication. Every number x has the additive inverse -x (which undoes the effort of adding x), but apart from 1 and -1, no integer has a multiplicative inverse. However we shall find such inverse once we come to consider rational numbers in 2.4.
In addition to the above laws, there is a further law, relating addition and multiplication:
Z.5 Distributive law: x(y+z)=xy+xz.
A set R with two operations x+y,xy and a negative -x, satisfying Z.1-5 is called a ring;more precisely it is commutative ring (because the multiplication is commutative,cf.Ex.(1) ,6.1). Thus the set Z of all integers is a commutative ring.However, these laws are not yet sufficient to determine Z; in Ch.6 we shall give a general definition of a ring and we shall find that there are many different types.
We now look at some consequence of the above laws. It follows from the distributive law that x0=0 for all x. By the associative law, the sum of any number of terms is independent of the way in which brackets are placed, and by the commutative law the order of the terms is immaterial. A similar remark applies to multiplication; for the present we shall accept this without proof and return to this point in Ch.3 to give a general proof.
Thus the sum of numbers a1,...,an may be written a1+...+an. Often one abbreviates this expression by writing down the general term an with a capital sigma, S,to show that the sum is to be taken, with some indiction of the range over which the terms are to be summed (unless this is clear from the context). So instead of a1+...+an we may write
or or or simply ,
where in each case v is a dummy variable(cf. 1.1). When n=0, the sum written here is empty and ,by convention, this is taken to be 0. This notation is not only briefer; it can also help to make our formula more perspicuous as well as more accurate. For instance,in the expression
1+2+...+n,
the reader is expected to guess that he is dealing with an arithmetic progression; the expression

removes all doubt. Thus the formula for the sum of the first n natural numbers may be written =½n(n+1). We observe that for n=0 the right-hand side reduces to 0, so with our convention about empty sums,this formula still holds for n=0.
For another example consider the distributive law. This has a generalized version which reads (cf. Ex.(2))
(a1+...+ar)(b1+...+bs)=a1b1+a1b2+...+arbs,
or in abbreviated from Sau .Sbv=Su,vaubv. Here we have not indicated the precise range of summation u,v.
A similar abbreviation exists for repeated products, using capital pi,P, in place of S..Thus instead of a1a2...an we write
or or or simply .
Integers and rational numbers
For example, the factorial function may be defined as n!=. An empty product is taken to be 1; thus empty sums and products are neutral for addition and multiplication respectively.
It is an important property of the integers that the product of two non-zero integers is never zero:
Z.6 For any integers a, b, if a≠0 and b≠0,then a*b≠0; morever 1≠0.
This has the following useful consequence:
Cancellation law For a, b, c∈Z, if ca=cb and c≠0,then a=b.
This asserts that multiplication by a non-zero integer is an injective mapping of Z into itself. To prove it, suppose that a≠b, then a-b≠0 and hence(by Z.6) c(a-b)≠0,therefore ca-cb=c(a-b)≠0.
Besides the operations on Z We have an order relation, i.e. an ordering on Z: x≤y or y≥x. If x≤y but x≠y, We write x＜y or also y>x. This relation satisfies the requirements for a total ordering (see 1.5) and is related to the operations of Z by the following rules:
Z.7 If x1≤x2 and y1≤y2, then x1+y1≤x2+y2.
Z.8 If x≤y and z>0, then zx≤zy.
The presence of these rules means that Z is a totally ordered ring. Using the ordering we can describe the set N of positive integers as
N={x∈Z∣x>0}.
Later we shall see how to reconstruct Z from N; for the moment we note that, for every x∈Z, either x=0 or x∈N or –x∈N and that these three possibilities are mutually exclusive. In fact, this is true in any totally ordered ring, taking N to be defined by (1). For we know that just one of the following holds (because we have a total order): x=0 or x>0 or x<0. Now x+ (-x) =0, hence, if x<0, then 0<-x by Z.7. Thus either x=0 or x>0 or –x>0, as asserted.
In order to fix Z completely, we use the following condition on the set N of positive integers:
I (principle of induction): Let S be a subset of N such that 1∈S and n+1∈S whenever n∈S. Then S=N.
This principle forms the basis of the familiar method of proof by induction. Let P(n) be an assertion about a positive integer n, e.g. ,P(n) might be'the sum of the first n positive integers is n(n+1)/2 ’.Suppose we wish to prove P(n) for all n, i.e. ("n)P(n). Then by I it will be enough to prove (ⅰ) P(1) and (ⅱ)("n)(P(n)=>P(n+1)). For this means that the set S of all n for which P(n) holds contains 1 and contains n+1 whenever it contains n.

2.1 The integers
Hence by I,S=N, i.e. P(n) holds for all nÎN.
There are two alternative forms of I that are often useful.
I' Let S be a subset of N such that 1ÎS and nÎS whenever mÎS for all m<n; then S=N.
I" (Principle of the least element): Every non-empty set of positive integers has a least element.
To prove I,I' and I" equivalent we shall establish the implications I=>I'=>I"=>I,
I=>I'. Let S be such that 1ÎS and nÎS whenever mÎS for all m<n. Define T={xÎN|yÎS for all y≤x}, thus xÎT precisely when all the numbers from 1 to x lie in S.Clearly TÍS,so it will be enough to show that T=N.Since 1ÎS, we have 1ÎT and, if nÎT,then yÎS for all y≤n,hence n+1ÎS and so yÎS for all y≤n+1; but this means that n+1ÎT. Applying I, we see that T=N.
I'=>I". Let S be a set of positive integers without a least element;we shall show that S is empty. Denoting the complement of S by S', we must show that S'=N.Now, since S has no element,1ÏS,so 1ÎS', moreover,if mÎS' for all m<n,then nÎS',for otherwise n would be the least element in S. Thus by I',S=N and S must be empty.
I"=>I. 1 is the least element of N(for if the least element r satisfied r<1,then r2<r,a contradiction ).Let S be a subset of N such that 1ÎS and n+1ÎS whenever nÎS,then the complement S' of S in N has no least element.For 1ÏS' and if nÎS',then n-1ÎS',hence, by I",S'=Æ and so S=N as we wished to show.
We end this section with a practical remark on proofs induction.Generally a theorem is easier to prove,the stronger the hypothesis and the weaker conclusion.But in an induction proof the conclusion at the nth step becomes the hypothesis at the (n+1)th step and the theorem may actually become easier to prove if the conclusion is strengthened; for an instance of this see Lemma 4,4.3.We also remark that an induction frequently starts at 0(instead of 1); clearly this does not affect the validity.

2
整数和有理数
2.1 整数
我们较为熟悉整数的基本算法，但现在要用明确的术语来表示其精确性。通过写下一些整数所拥有的性质及由这些性质所得出的推导来实现以上操作。随后，在卷2中可看出列出来的所有性质事实上都可以由一些极简单的原理推导得来，但这在目前不重要。
我们用N表示正整数（也叫自然数）集合。用Z表示所有整数集合，包括正的，负的和零。此处N代表数且Z也代表数，德语中的数。二者的缩写通常用在数学中（见附录2）。集合Z允许三种运算：加法，x+ y,减法，x-y，和乘法，x.y或xy。通常用加上一个数的负数来表示减法是比较简便的：x-y=x+(-y)
这些运算遵循以下规律：
Z.1 结合律：（x + y)+z=x+(y +z )， (xy)z=x(yz)。
Z.2 交换律： x+y=y+x， xy=yx。
Z.3存在中性元素：x+0=x， x1=x。
Z.4存在加法逆： x+(-x)=0。
对于加法来说0是中性的，因为0加上任意x后x不改变。同样在乘法中1也是中性的。每个数x都有一个加法逆-x（加x的反效果）但除了1 和-1，其他整数都没有乘法逆。然而在2.4中我们考虑有理数时则会有这样的逆。
除了以上规律，还有一条关于加法和乘法的更进一步的规律：
Z.5 分配率：x(y+z)=xy+xz。
满足Z.1-5且带有两种运算x+y，xy，和一个负数 -x的集合R叫做环。更巧，它是一个交换环（因为乘法是可交换的，如之前（1），6.1）。这样含所有整数的集合Z是一个交换环。但是，这些规律对于测定Z来说还是不够的。第六章将给出环的常规定义以及许多不同类型。
现在来研究以上规律的一些推论。对所有x遵循分配律x0=0。由结合律的各项数相加所得的和与括弧所在的位置无关，由交换律得各项的顺序是无关紧要的。类似的结论应用于乘法中。现在没有这样结论的证明，在第3章中会给出针对这点的一般证明。
a1,¼an的和可写作a1+¼+an。通常这个表达式可由一般项an和一个用来表示要进行求和运算的大写字母西格玛缩写，以及所需求和项的范围（除了上下文很明显的）。因此a1+¼+an可写作或或 或简单写作，其中每种情况下n是一个哑变量（如1.1）。当n=0时和为空，约定俗成，计算结果为0。这种表述不仅更为简洁，还使公式更容易理解更精确。例如，表达式1+2+......+n，其实是要处理一个算术级数。表达式 消除所有疑虑。如此，这个公式是对前n个自然数求和，写作=½n(n+1)。观察得，当n=0时等式右边推出0，由空的和的协定得：当n=0时该公式任然成立。
对于另外一个例子可利用分配律。这样得出广义版本，写作（如之前（2））（a1+.....ar）(b1+.....bs)=a1b1+a1b2+......arbs，或缩写为 Sau .Sbv=Su,vaubv。此时没有必要显示总和的精确范围，仅出现总和的上下指标。
累乘有相似的缩写形式，将S用大写的Pi，即P来替代，这样，论及a1a2.....an计算，可写为 或或或简单写作。
整数和有理数
例如，阶乘函数可以定义为n!= 。空的乘积运算为1，从而对于空分别在加法和乘法中的和和乘积的运算是无确定性质的。
“两个非零整数的乘积不为零”是整数的一个重要性质：
Z．6 对于任意整数a、b,如果a≠0，b≠0,那么ab≠0;此外1≠0。这样得出以下有用的结论：
消除法：对于a、b、c∈Z,若ca=cb且c≠0，则a=b。这表明非零整数的乘法是这个整数本身的一个单射。证明：假设a≠b,则a-b≠0且因为c（a-b）≠0(由Z.6得)，所以ca-cb≠c(a-b)≠0。
对整数除了这些操作还有一种序关系，也就是对整数进行一个排序：x≤y或y≥x。若x≤y但x≠y，可得x＜y或y＞x。这种关系满足总体排序要求（参考1.5）并且与整数遵循以下原则的操作有关：
Z.7如果x1≤x2且y1≤y2，那么x1+y1≤x2+y2。
Z.8如果x≤y且z＞0，那么zx≤zy。
这些规则的存在意味着整数是一个完全有序环。利用这种序可将正整数集合N表述为N={x∈Z︳x＞0}。
随即可明白怎样由N推想Z,暂时记作：对于任意x∈Z,-x=0或x∈N或-x∈N这三种可能是相互排斥的。事实上，依据（1）定义N在任意一个完全有序环中都为真。对于以下结论中只有一个成立（因为存在全序）：x=0或x＞0或x＜0。已知x+(-x)=0，因此，若x＜0，则由Z.7得0＜-x。所以x=0或x＞0或-x＞0，同样皆可得到。
为了完全确定Z，可以对正整数集合N使用以下条件：
I（归纳法则）：设S为集合N的一个字集，这样有1∈S且n+1∈S,n∈S恒成立。那么S=N。
这项基本原则可通过类似归纳法来证明。令P(n)是由n推断而来，举例说明:P(n)可以是‘前n个正整数之和为n(n+1)/2’。假使要证明对所有n的P(n)，例如：("n)P(n)。那么通过I将足以证明(ⅰ)P(1)和(ⅱ) ("n){P(n)=>P(n+1)}。这意味着对所有n构成的集合S,当P(n)包含1和n+1成立时，P(n)包含n同样成立。

2.1 整数
因为通过I,S=N，举例来说，对于所有nÎN，P（n)成立。
I有两个可供选择的形式通常很有用。
I'令S是N的一个子集，这样1ÎS且nÎS，当m<n时mÎS成立，则有S=N。
I''（最小元素原理）：每个非空正整数集合有一个最小元素。
为了证明I，I'和I''是等价的，可以建立这样的联系I=>I'=>I''=>I。
I=>I'。令S满足1ÎS且nÎS，则当m<n时mÎS成立；定义T={xÎN|yÎS 所有y≤x}，从1到x的所有数都位于S中时，xÎT。显然TÍ S，这样足以证明T=N。因为1ÎS，1ÎT，如果nÎT，则当y≤n时，yÎS成立。因此n+1ÎS，这样当y≤n+1时yÎS；这意味着n+1ÎT。应用I，可得T=N。
I'=>I''。令S为一个无最小元素的正整数集合，可看出S为空。用S'表示S的补集，我们必须证明S'=N。已知S没有最小元素，1ÏS，此外，当m<n时mÎS'成立，则nÎS'，否则n是S中的最小元素。这样通过I'，S'=N得S必为空。
I''=>I。1是N中的最小元素，（因为如果最小元素满足r<1，那么r2<r，矛盾）。令S是N的一个子集，这样有1ÎS且n+1ÎS，当nÎS则N中集合S的补集S'没有最小元素。对于1ÏS’且如果nÎS'，则n-1ÎS'，因此，通过I"，S'=Æ，S=N得证。
用关于通过推导证明的实际性的评论来结束本节内容。通常定理比较容易证明，假设越强结论越弱。但是在一项推导证明中，如果结论被强化，第n步的结论变成等n+1步的假设，定理将更容易得证。以4.3中引理4为例。我们习以为常从0（取代1）开始推导，显然这并不影响其正确性。