现代高级代数：群与环的互动探索

需积分: 41 55 浏览量更新于2024-07-24 收藏 5.64MB PDF 举报

"《高级现代代数》是Joseph J. Rotman撰写的一本关于代数的教科书，涵盖了从基础到进阶的代数概念。全书共有1040页，分为第一版（2002年）和第二版印刷（2003年）。书中以群论和环论之间的相互作用为组织原则，同时引入了模的概念。内容包括基本定义、完整的定理及其简明证明，并涉及代数几何、计算机科学、同调理论和表示论等主题。作者旨在将结果和思想置于上下文中，让学生理解为何要研究特定话题以及定义的来源。章节涵盖群论、交换环论、模、主理想域、代数、上同调与表示论以及同调代数。这本书适合对自学高级代数及相关领域感兴趣的人士使用。" 本书详细介绍了代数的基础和高级概念。在"群论I"部分，读者将学习群的基本概念，如置换、群的性质、拉格朗日定理、同态和商群，以及群的作用。这些内容为理解群的结构和性质奠定了基础。接着，"交换环论I"章节探讨了环的基本属性、多项式环、最大公因数、环同态以及欧几里得环。这部分还引入了线性代数的概念，如向量空间和线性变换，以及商环和有限域。 "域论"章节深入研究了五次方程的不可解性，通过群论的视角解释了为什么某些代数方程无法用根式求解。并讨论了伽罗瓦理论的基本定理，这是理解域扩张和群论之间关系的关键。在"群论II"中，Rotman介绍了有限阿贝尔群，特别是直和、基定理和阿贝尔群的基本定理。他还讨论了西洛定理，这些定理在研究群的子群结构时非常关键。此外，还有约当-霍尔德定理，它涉及到群的链条件，以及项目模态群和表示。这些概念在群论的深入研究中扮演着重要角色。 "交换环论II"涵盖了素理想和极大理想，唯一因子分解域以及诺特环。此外，还利用佐恩引理解决了环论中的某些应用问题。《高级现代代数》是一部综合性的教材，适合希望深入了解代数结构、理论及其应用的学生和学者。书中的内容不仅限于理论，还包括实际背景，使得学习过程更加生动和有意义。

Things Past Ch. 1

Proposition 1.1. Every integer n ≥ 2 is either a prime or a product of primes.

Proof. Let C be the subset of N consisting of all those n ≥ 2 for which the proposition

is false; we must prove that C = ∅. If, on the contrary, C is nonempty, then it contains a

smallest integer, say, m. Since m ∈ C, it is not a prime, and so there are natural numbers

a and b with m = ab, a < m, and b < m. Neither a nor b lies in C, for each of them is

smaller than m, which is the smallest integer in C, and so each of them is either prime or a

product of primes. Therefore, m = ab is a product of (at least two) primes, contradicting

the proposition being false for m. •

There are two versions of induction.

Theorem 1.2 (Mathematical Induction). Let S(n) be a family of statements, one for

each integer n ≥ m, where m is some ﬁxed integer. If

(i) S(m) is true, and

(ii) S(n) is true implies S(n + 1) is true,

then S(n) is true for all integers n ≥ m.

Proof. Let C be the set of all integers n ≥ m for which S(n) is false. If C is empty, we

are done. Otherwise, there is a smallest integer k in C.By(i),wehavek > m, and so there

is a statement S(k − 1).Butk − 1 < k implies k − 1 /∈ C,fork is the smallest integer in

C. Thus, S(k − 1) is true. But now (ii) says that S(k) = S([k − 1] + 1) is true, and this

contradicts k ∈ C [which says that S(k) is false]. •

Theorem 1.3 (Second Form of Induction). Let S(n) be a family of statements, one for

each integer n ≥ m, where m is some ﬁxed integer. If

(i) S(m) is true, and

(ii) if S(k) is true for all k with m ≤ k < n, then S(n) is itself true,

then S(n) is true for all integers n ≥ m.

Sketch of Proof. The proof is similar to the proof of the ﬁrst form. •

We now recall some elementary number theory.

Theorem 1.4 (Division Algorithm). Given integers a and b with a = 0, there exist

unique integers q and r with

b = qa +rand0 ≤ r < |a|.

Sketch of Proof. Consider all nonnegative integers of the form b − na, where n ∈ Z.

Deﬁne r to be the smallest nonnegative integer of the form b − na, and deﬁne q to be the

integer n occurring in the expression r = b − na.

If qa + r = q



a + r



, where 0 ≤ r



< |a|, then |(q − q



)a|=|r



− r |.Now0≤



− r | < |a| and, if |q − q



| = 0, then |(q − q



)a|≥|a|. We conclude that both sides

are 0; that is, q = q



and r = r



. •

Sec. 1.1 Some Number Theory

Deﬁnition. If a and b are integers with a = 0, then the integers q and r occurring in the

division algorithm are called the quotient and the remainder after dividing b by a.

Warning! The division algorithm makes sense, in particular, when b is negative. A

careless person may assume that b and −b leave the same remainder after dividing by a,

and this is usually false. For example, let us divide 60 and −60 by 7.

60 = 7 ·8 +4 and − 60 = 7 · (−9) +3

Thus, the remainders after dividing 60 and −60 by 7 are different.

Corollary 1.5. There are inﬁnitely many primes.

Proof. (Euclid) Suppose, on the contrary, that there are only ﬁnitely many primes. If

, p

,..., p

is the complete list of all the primes, deﬁne M = (p

···p

) + 1. By

Proposition 1.1, M is either a prime or a product of primes. But M is neither a prime

(M > p

for every i ) nor does it have any prime divisor p

, for dividing M by p

gives

remainder 1 and not 0. For example, dividing M by p

gives M = p

( p

···p

) + 1, so

that the quotient and remainder are q = p

···p

and r = 1; dividing M by p

gives M =

( p

···p

) + 1, so that q = p

···p

and r = 1; and so forth. This contradiction

proves that there cannot be only ﬁnitely many primes, and so there must be an inﬁnite

number of them. •

Deﬁnition. If a and b are integers, then a is a divisor of b if there is an integer d with

b = ad. We also say that a divides b or that b is a multiple of a, and we denote this by

a | b.

There is going to be a shift in viewpoint. When we ﬁrst learned long division, we

emphasized the quotient q; the remainder r was merely the fragment left over. Here, we

are interested in whether or not a given number b is a multiple of a number a, but we are

less interested in which multiple it may be. Hence, from now on, we will emphasize the

remainder. Thus, a | b if and only if b has remainder r = 0 after dividing by a.

Deﬁnition. A common divisor of integers a and b is an integer c with c | a and c | b.

The greatest common divisoror gcd of a and b, denoted by (a, b), is deﬁned by

(a, b) =



0ifa = 0 = b

the largest common divisor of a and b otherwise.

Proposition 1.6. If p is a prime and b is any integer, then

( p, b) =



p if p | b

1 otherwise.

Sketch of Proof. A positive common divisor is, in particular, a divisor of the prime p, and

hence it is p or 1. •

Things Past Ch. 1

Theorem 1.7. If a and b are integers, then (a, b) = d is a linear combination of a and

b; that is, there are integers s and t with d = sa + tb.

Sketch of Proof. Let

I ={sa + tb : s, t ∈ Z}

(the set of all integers, positive and negative, is denoted by Z). If I ={0},letd be the

smallest positive integer in I ; as any element of I ,wehaved = sa + tb for some integers

s and t. We claim that I = (d), the set of all multiples of d. Clearly, (d) ⊆ I .Forthe

reverse inclusion, take c ∈ I . By the division algorithm, c = qd + r, where 0 ≤ r < d.

Now r = c − qd ∈ I , so that the minimality of d is contradicted if r = 0. Hence, d | c,

c ∈ (d), and I = (d). It follows that d is a common divisor of a and b, and it is the largest

such. •

Proposition 1.8. Let a and b be integers. A nonnegative common divisor d is their gcd if

and only if c | d for every common divisor c.

Sketch of Proof. If d is the gcd, then d = sa +tb. Hence, if c | a and c | b, then c divides

sa +tb = d. Conversely, if d is a common divisor with c | d for every common divisor c,

then c ≤ d for all c, and so d is the largest. •

Corollary 1.9. Let I be a subset of Z such that

(i) 0 ∈ I;

(ii) if a, b ∈ I , then a − b ∈ I;

(iii) if a ∈ I and q ∈ Z, then qa ∈ I.

Then there is a natural number d ∈ I with I consisting precisely of all the multiples of d.

Sketch of Proof. These are the only properties of the subset I in Theorem 1.7 that were

used in the proof. •

Theorem 1.10 (Euclid’s Lemma). If p is a prime and p | ab, then p | aorp| b. More

generally, if a prime p divides a product a

···a

, then it must divide at least one of the

factors a

Sketch of Proof. If p  a, then ( p, a) = 1 and 1 = sp + ta. Hence, b = spb +tab is a

multiple of p. The second statement is proved by induction on n ≥ 2. •

Deﬁnition. Call integers a and b relatively prime if their gcd (a, b) = 1.

Corollary 1.11. Let a, b, and c be integers. If c and a are relatively prime and if c | ab,

then c | b.

Sketch of Proof. Since 1 = sc + ta,wehaveb = scb + tab. •

剩余1055页未读，继续阅读

月夜星空10

粉丝: 0
资源: 2

现代高级代数：群与环的互动探索

Advanced Modern Algebra, Part 1, Joseph Rotman.pdf

Advanced Modern Algebra（抽象代数英文版）

advanced algebra is so difficult

gilberts- elements of modern algebra

advanced linear algebra

modern computer algebra

data.loc[ : ,'algebra'].apply(lambda x:algebra(x))

line algebra done right 云盘

labview解方程Linear Algebra

matrix algebra james pdf

最新资源