• 0 Votes - 0 Average
• 1
• 2
• 3
• 4
• 5
 Ash:【Abstract Algebra The Basic Graduate Year】
12-30-2010, 02:37 PM
Post: #1
 elim Moderator Posts: 580 Joined: Feb 2010 Reputation: 0
Ash:【Abstract Algebra The Basic Graduate Year】

Chapter 0

Prerequisites

0.1 Elementary Number Theory
$\quad\quad \gcd(a,b) =: \min\{d \mid d\in \mathbb{N}^+,\; (d \mid a)\wedge (d\mid b)\}\left(=\min\{sa+tb \mid s,t \in \mathbb{Z},\; sa+tb > 0\} \right )$

$\quad\quad \textbf{0.1.1} \quad d=\gcd(a,b) \Rightarrow \exists s,t \in \mathbb{Z}\; (d=sa+tb)$

$\quad\quad \textbf{0.1.2} \quad p\in \mathbb{Z}$ is called a prime if $(|p| > 1) \wedge (\forall d\in \mathbb{N}^+\;(d \mid p \Rightarrow |d|\in \{1,|p|\}) )$
$\quad\quad\quad\quad$We have that $(a_1,\cdots,a_n \in \mathbb{Z},\; p \text{ is prime},\; p \mid a_1\cdots a_n)\Rightarrow (\exists k\in \{1,\cdots ,n\}\;\; (p\mid a_k))$
$\quad\quad\quad\quad \textbf{0.1.2.1}\quad (p \nmid c)\Rightarrow \gcd(p,c)=1$ $\left(\gcd(p,c)=d \Rightarrow (d \mid p \nmid c) \Rightarrow (d \in \{1,p\}) \Rightarrow (d=1)\right)$
$\quad\quad\quad\quad \textbf{0.1.2.2}\quad (c \mid ab)\wedge (\gcd(c,a)=1)\Rightarrow (c \mid b)$ $c \mid scb+tab =b\gcd(c,a)=b$

$\quad\quad \textbf{0.1.3 Unique Factorization }\forall a\in \mathbb{N}\setminus \{1\} \,\exists ! m\in \mathbb{N},\; \text{primes} \{p_k\}\; p_i\le p_{i+1}\; (a = \prod_1^m p_k)$

$\quad\quad \textbf{0.1.4 The Integers Modulo }m\;$ If $a,b\in \mathbb{Z}\ni m\ge 2$, then $a\equiv b\mod m\;$ if $m\mid (a-b)$
$\quad\quad\quad\quad\;$ (i.e. $\;a$ is congruent to $b$ modulo $m$). Congruence modulo $m$ is an equivalence relation in $\mathbb{Z}$
$\quad\quad\quad\quad\;$ and the resulting equivalence classes (called residue classes mod $m$) form a set $\mathbb{Z}_m$
$\quad\quad\quad\quad\;\;\, [a]+[b]:=[a+b],\;[a][b]:=[ab]$ are well defined operations on $\mathbb{Z}_m$ where
$\quad\quad\quad\quad\;\;\, [n] = \{k\in\mathbb{Z}: k\equiv n\mod m\}\in\mathbb{Z}_m$ has the property $([x] = [y])\Longleftrightarrow (x\equiv y\mod m)$
$\quad\quad\quad\quad\;\;\, \mathbb{Z}_m$ is a commutative ring under $+$ and $\cdot$(usually omitted); it's a field iff $m$ is prime.(see 2.4)

$\quad\quad \textbf{0.1.5}$
$\quad\quad\quad\quad$ (1) $(a,m) = 1 \Longleftrightarrow \exists b\; ([a][b] =_m [1])$
$\quad\quad\quad\quad$ (2) $(c\mid ab)\wedge ((a,c) = 1) \implies (c\mid b)$
$\quad\quad\quad\quad$ (3) $(a,m) = (b,m) = 1\implies (ab,m) = 1$
$\quad\quad\quad\quad$ (4) $(ax\equiv ay\mod m)\wedge ((a,m)=1) \implies (x\equiv y\mod m)$
$\quad\quad\quad\quad$ (5) $d = \gcd(a,b) \implies (a/d,b/d) = 1$
$\quad\quad\quad\quad$ (6) $(ax\equiv ay\mod m)\wedge (d=\gcd(a,m))\implies (x\equiv y\mod (m/d))$
$\quad\quad\quad\quad$ (7) $(a_i\mid b\,(i=\overline{1,r})) \wedge ((a_i,a_j)=1\,(i\ne j))\implies (a_1\cdots a_r \mid b)$
$\quad\quad\quad\quad$ (8) $ab = \gcd(a,b)\cdot\text{lcm}(a,b)$

$\quad\quad \textbf{0.1.6 Chinese Remainder Theorem}$
$\quad\quad\quad\quad$ $\left((m_i,m_j) = 1\;(1\le i < j\le r),\, m=m_1\cdots m_r\right)$
$\quad\quad\quad\quad\quad\quad \implies \left(\forall \{b_j\}\in \mathbb{Z}^{\{1,\cdots,r\}}\,\exists ! [x_0]\in\mathbb{Z}_m\,(x\equiv b_j\mod m_j\,(j=\overline{1,r}))\Longleftrightarrow x\in [x_0]\right)$
$\quad\quad\quad\quad$ This can be derived from the abstract form of the Chinese remainder theorem. (see 2.3)

$\quad\quad \textbf{0.1.7 Eular's Theorem}$
$\quad\quad\quad\quad$ The Eular phi function is defined by $\varphi(n) = |\{k\in\{1,\cdots,n\}: (k,n)=1\}|$
$\quad\quad\quad\quad\;$ (For an explicit formula of $\varphi(n)$, see Ex.1.1.13)
$\quad\quad\quad\quad$ Euler's theorem states that $\displaystyle{((n\ge 2)\wedge (a,n)=1)\implies (a^{\varphi(n)}\equiv 1\mod n)}$

$\quad\quad \textbf{0.1.8 Permat's Little Theorem}$
$\quad\quad\quad\quad$ If $a\in\mathbb{Z}$ and $p (\nmid a)$ is a prime, then $a^{p-1}\equiv 1\mod p$. Thus $\forall a\in\mathbb{Z}\,\forall \text{ prime } p\,(a^p \equiv a\mod p)$

$\quad\quad$ See 1.3.4 for proofs of (0.1.7) and (0.1.8)
09-17-2012, 01:31 PM
Post: #2
 elim Moderator Posts: 580 Joined: Feb 2010 Reputation: 0
RE: Ash:Abstract Algebra The Basic Graduate Year
...Space...
09-17-2012, 01:58 PM
Post: #3
 elim Moderator Posts: 580 Joined: Feb 2010 Reputation: 0
0.2 Set Theory -- Ash:【Abstract Algebra The Basic Graduate Year】

0.2 Set Theory
$\quad\quad$ 0.2.1
$\quad\quad\quad\quad$ A partial ordering on a set $S$ is a relation on $S$ that is
$\quad\quad\quad\quad\quad$ reflexive $\quad\quad\;\, (\forall x\in S\,(x\le x))$,
$\quad\quad\quad\quad\quad$ antisymmetric $((x\le y) \wedge(y\le x))\implies (x=y)$,
$\quad\quad\quad\quad\quad$ and transitive $\;((x\le y)\wedge (y\le z))\implies (x\le z)$

$\quad\quad\quad\quad$ If $\forall x,y\in S\,(x\le y)\vee (y\le x)$, the ordering is total.

$\quad\quad$ 0.2.2
$\quad\quad\quad\quad$ A well-ordering on $S$ is a partial ordering such that $\forall A\in 2^S\setminus \{\varnothing\}\,\exists a\in A\,\forall b\in A\,(a\le b)$
$\quad\quad\quad\quad$ (nonempty subset has the smallest element)
$\quad\quad\quad\quad\quad\quad$ (If $A,\,B$ are sets, then $(A\times\{0\})\cap (B\times \{1\}) = \varnothing. And$|A\times \{0\}|= |A|,\, |B\times \{0\}|= |B|$)$\quad\quad$0.2.3 Well-Ordering Principle$\quad\quad\quad\quad$Every set can be well-ordered.$\quad\quad$0.2.4 Maximum Principle$\quad\quad\quad\quad$If$T$is any chain (totally ordered subset) of a partially ordered set$S$, then$T$is contained$\quad\quad\quad\quad$in a maximal chain$M$. (Maximal means that$M$is not properly contained in a larger chain.)$\quad\quad$0.2.5 Zorn's Lemma$\quad\quad\quad\quad$If$S(\ne \varnothing)$is a partially ordered set such that every chain of$S$has an upper bound in$S$,$\quad\quad\quad\quad$then$S$has a maximal element.$\quad\quad\quad\quad\quad\quad$(An element$x$is an upper bound of a set$A$if$\forall a\in A\,(a\le x)$. Note that$x$need not$\quad\quad\quad\quad\; \in A$, but in the sataement of Zorn's lemma, we require that if$A$is a chain of$S$, then A has$\quad\quad\quad\quad$an upper bound that actually belongs to$S$.)$\quad\quad$0.2.6 Axiom of Choice$\quad\quad\quad\quad$Given a family of nonempty sets$\{S_i\}_{i\in I}$,$\displaystyle{\exists f:I\to \bigcup_{i\in I} S_i}$such that$f(i)\in S_i\,(\forall i\in I)\quad\quad$The well-ordering principle, the maximum principle, Zorn's lemma, and the axiom of choice are equivalent in the sense that if any one of these statements is added to the basic axioms of set theory, all the others can be proved. The statements themselves cannot be proved from the basic axioms. Constructivist mathematics rejects the axiom of choice and its equivalents. In this philosophy, an assertion that we can choose an element from each$S_i$must be accompanied by an explicit algorithm. The idea is appealing, but its acceptance results in large areas of interesting and useful mathematics being tossed onto the scrap heap. So at present, the methematicall mainstream embraces the axiom of choice, Zorn's lemma et al.$\quad\quad$0.2.7 Proof by Transfinite Induction$\quad\quad\quad\quad$To prove that statement$P_i$holds for all$i$in the well-ordered set$I$, do the following:$\quad\quad\quad\quad\quad$1. Prove$P_0$is true, where$0$is the smallest element of$I$.$\quad\quad\quad\quad\quad$2. Assume$P_j$holds for all$j<i$with$i>0$and prove$P_i$.$\quad\quad$0.2.8 If there is an injective map from$A$to$B$, we say the size of$A$is less than or equal$\quad\quad\quad\quad$to the size of$B$($A\le_s B$). We say that$A,\;B$have the same size ($A =_s B$) if there$\quad\quad\quad\quad$is a bijection between$A$and$B$.$\quad\quad$0.2.9 Schröder-Bernstein Theorem$\quad\quad\quad\quad(A\le_s B)\wedge (B\le_s A)\implies (A=_s B)$. (This can be proved without the axiom of choice.)$\quad\quad$0.2.10$\quad\quad\quad\quad$If sets of the same size are called equivalent (0.2.9), then$\le_s$among equivalence classes$\quad\quad\quad\quad$is a partial ordering. It follows with the aid of Zorn's lemma that the ordering is total. The$\quad\quad\quad\quad$equivalence class of a set$A$, written$|A|$, is called the cardinal number or cardinality of$A$.$\quad\quad\quad\quad$In practice,$|A|$may be identified with a member of it. For example$\quad\quad\quad\quad\quad\quad 0=\varnothing, 1=\{0\},\cdots,n=\{0,1,\cdots,n-1\}$and we identify$|n|$with$n$.$\quad\quad$0.2.11$\quad\quad\quad\quad$For any set$A$,$|A| <_s 2^A$where$2^A = \{S\mid S\subset A\}$is the power set (of$A$).$\quad\quad$0.2.12$\quad\quad\quad\quad$Define$|A|+|B| = |A\cup B|\;(A\cap B=\varnothing),\; |A||B| = |A\times B|$(addition& multiplication)$\quad\quad\quad\quad$Note that ($S_j = S\times \{j\}\implies |S_j|=|S|,\quad A_i\cap B_j = \varnothing\,(i\ne j)$)$\quad\quad$0.2.13$\quad\quad\quad\quad$If$\aleph_0$is the cardinal number of a countable infinite set, then$\aleph_0+\aleph_0=\aleph_0\aleph_0=\aleph_0\quad\quad\quad\quad\quad$More generally, if$\alpha,\,\beta$are cardinals,$\quad\quad\quad\quad\quad$(a)$(\alpha\le\beta \ge\aleph_0)\implies \alpha+\beta=\beta,\quad$(b)$(|\varnothing| = 0 < \alpha\le\beta \ge\aleph_0)\implies \alpha\beta = \beta$.$\quad\quad$0.2.14$\quad\quad\quad\quad$If$|A|\ge \aleph_0$, then$A$and the set of all finite subsets of$A$have the same cardinality. 09-18-2012, 03:42 PM Post: #4  elim Moderator Posts: 580 Joined: Feb 2010 Reputation: 0 RE: Ash:Abstract Algebra The Basic Graduate Year ...Comments... 09-18-2012, 05:13 PM Post: #5  elim Moderator Posts: 580 Joined: Feb 2010 Reputation: 0 0.3 Linear Algebra --Ash:【Abstract Algebra The Basic Graduate Year】 It is not feasible to list all results presented in an undergraduate course in linear algebra. Instead, here is a list of topics that are covered in a typical course.$\;$1. Sums, products, transposes, inverses of matrices; symmetric matrices.$\;$2. Elementary row and column operations; reduction to echelon form.$\;$3. Determinants; evaluation by Laplace expansion and Cramer's rule.$\;$4. Vector spaces over a field; subspaces, linear independence and bases.$\;$5. Rank of a matrix, homogeneous and nonhomogeneous linear equations.$\;$6. Null space and range of a matrix; the dimension theorem.$\;$7. Linear transformations and their representation by matrices.$\;$8. Coordinates and matrices under change of basis.$\;$9. Inner product spaces and the projection theorem. 10. Eigenvalues and eigenvectors; diaonalization of matrices with distinct eigenvalues, symmetric$\quad\;$and Hermitian matrices. 11. Quadratic forms. A more advanced course might cover the following topics: 12. Generalized eigenvectors and the Jordan canonical form. 13. The minimal and characteristic polynomials of a matrix; Cayley-Hamilton theorem. 14. The adjoint of a linear operator. 15. Projection operators. 16. Normal operators and the spectral theorem. 09-19-2012, 05:32 PM Post: #6  elim Moderator Posts: 580 Joined: Feb 2010 Reputation: 0 Chapter 1 Group Fundamentals --Ash:【Abstract Algebra The Basic Graduate Year】 Chapter 1 Group Fundamentals 1.1 Groups and Subgroups$\quad\quad$1.1.1 Definition A group is a nonempty set$G$with its binary operation$(a,b)\to ab$such that$\quad\quad\quad\quad$Closure:$\quad\quad\quad a,b\in G\implies ab\in G$;$\quad\quad\quad\quad$Associativity:$\;\, \forall a,b,c\in G\,(a(bc) = (ab)c)$;$\quad\quad\quad\quad$Identity:$\quad\quad\;\;\, \exists 1\in G\,\forall a\in G\,(a1 = 1a = a)$;$\quad\quad\quad\quad$Inverse:$\quad\quad\;\;\; \forall a\in G\,\exists a^{-1}\in G\,(a a^{-1} = a^{-1} a = 1).\quad\quad\quad\quad$A group$G$is abelian if the operation is commutative (i.e.$ab = ba\,(\forall a,b\in G)$)$\quad\quad\quad\quad$In this case the binary operation is often written additively ($(a,b)\to a+b$), and the$\quad\quad\quad\quad$identity denoted as$0$rather than$1\quad\quad$Under addition,$\mathbb{Z},\,\mathbb{Q},\,\mathbb{R},\,\mathbb{C}$and the integers$\mathbb{Z}_m\,$modulo$m$are abelian groups.$\quad\quad$The associative law generalizes to products of any finite number of elements:$a_1\cdots a_n$is thus$\quad\quad\$ well defined.
 « Next Oldest | Next Newest »

Forum Jump: