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+~~META:title=Daily topics~~
+<WRAP centeralign>
+===== Math 401 - 01 Daily Topics - part 3 (Fall 2018)=====
+</WRAP>
+----
+{{page>people:fer:401ws:defs&nofooter&noeditbtn}}
+[[people:fer:401ws:fall2018:home| Home]]
+^Week 12^Topics^
+|11/05/2018|Sylow Theorems |
+| |Examples: (1)  $|G|=35$  <html>&nbsp;&nbsp;</html> (2)  $|G|=455$  <html>&nbsp;&nbsp;</html> (3)  $|G|=21$  <html>&nbsp;&nbsp;</html> (4)  $|G|=256$ |
+|11/06/2018|Test 2|
+|11/07/2018|Rings. Definitions: ring, unity, ring with unity (unitary ring), commutative ring, units of a unitary ring|
+| |Examples|
+| |Prop: The units of a ring,  $U(R)$  form a multiplicative group.|
+|11/09/2018|No class.|
+^Week 13^Topics^
+|11/12/2018|Thm. 12.1|
+| |Thm. 12.2|
+| |Subrings, definition, examples|
+| |Direct Products (Sums), definition, examples|
+| |Ring homomorphisms, definition|
+| |kernel, Ideal|
+| |Homo, mono, epi, iso, endo, auto|
+|11/13/2018|Test 2 returned|
+| |R/I definition|
+| |Thm. 12.3|
+| |Integral Domains, zero-divisors|
+| |Prop. Let  $R$  be a commutative ring.  TFAE|
+| |(1)  $R$  has no zero-divisors|
+| |(2)  $R$  satisfies the cancellation law:  $ab=ac$  and  $a\neq 0 \imp b=c$ .|
+| |(3)  $R$  satisfies:  $ab=0 \imp a=0\ \text{or}\ b=0$ |
+| |Definition: integral domain|
+| |Examples:  $\Z$ ,  $\Q$ ,  $\R$ ,  $\C$ ,  $\Q(\sqrt{2})$ ,  $\Z_p$ .|
+| |Thm. (1) Any field is an integral domain.|
+| |(2) Any finite ID is a field.|
+| |Cor:  $\Z_n$  is a field iff it is an ID iff  $n$  is a prime.|
+|11/14/2018|Examples:  $\Q(\sqrt{2})$  is a field.|
+| | $\Z_3[i]$  is a field.|
+| | $\Z_5[i]$  is not a field.|
+| |Prop: If  $R$  is an ID, then  $R[x]$  is an ID, and for any  $f,g\in R[x]$  we have  $\deg(fg)=\deg(f)+\deg(g)$ .|
+| |Example:  $\Z_6[x]$  is not an ID and the degree formula does not hold.|
+|11/16/2018|Snow day.  Class cancelled.|
+^Week 14^Topics^
+|11/19/2018|Prop:Let  $R$  be a commutative ring with unity, and  $a\in R$ .
+| |(1)  $\pbr{a}:​=aR=\{ar|r\in R\}$  is an ideal of  $R$ .|
+| |(2)  $a\in\pbr{a}$ .|
+| |(3) If  $I\normaleq R$  and  $a\in I$  then  $\pbr{a} \leq I$ .|
+| |Def:  $\pbr{a}$  is called the ideal generated by  $a$ . It is the smallest ideal of  $R$  that contains  $a$ .|
+| |Example: In the ring  $\Q[x]/​\pbr{x^2-2}$  the element  $u=x+I$  where  $I=\pbr{x^2-2}$ , satisfies  $u^2=2$ , i.e. it is a root of the polynomial  $x^2-2$ .|
+| |Characteristic of a ring. Thms. 13.3 and 13.4.|
+|11/20/2018|Comparison of  $\Q[\sqrt{2}]$  and  $\Q[x]/​\pbr{x^2-2}$ . Intuitive motivation for the construction  $\Q[x]/​\pbr{x^2-2}$ . |
+| |Given a commutative ring with unity  $R$ , and an ideal  $I\normaleq R$ , |
+| |(Q1) when is  $R/I$  an I.D.?|
+| |(Q2) when is  $R/I$  a field?|
+| |Def: prime ideal|
+| |Thm 14.3.   $R/I$  is an ID iff  $I$  is a prime ideal.|
+^Week 15^Topics^
+|11/26/2018|Lemma: Let  $R$  be a commutative ring with unity, and  $I, J\normaleq R$ . |
+| |(1) *  $I\intersection J \normaleq R$ |
+| | *  $I\intersection J \leq I, J$ |
+| | * if  $K\normaleq R$  and  $K\leq I, J$  then  $K\leq I\intersection J$ .|
+| |(2) *  $I+J := \{x+y|x\in I, y\in J\} \normaleq R$ |
+| | *  $I, J \leq I+J$  |
+| | * if  $K\normaleq R$  and  $I, J\leq K$  then  $I+J\leq K$ .|
+| |Prop: The set  $\idl(R):​=\{I|I\normaleq R\}$  of ideals of  $R$  is a lattice, i.e. a partially ordered set, in which any two elements have a  $\glb$  and a  $\lub$ .|
+| |Thm 14.4   $R/I$  is a field iff  $I$  is a maximal proper ideal.
+| |Cor: Let  $R$  be a commutative ring with unity and  $I\normaleq R$ . If  $I$  is maximal then it is prime.|
+|11/27/2018|Board presentation, PS 11.|
+| |Example:  $R=\Z[x]$ ,  $I=\pbr{x}$  is a prime ideal but it is not maximal.|
+| |Fact: In the ring  $\Z$  every ideal is a principal ideal, and every prime ideal is maximal.|
+|11/28/2018|Def: Principal ideal domain (PID).|
+| |Prop:  $\Z$  is a principal ideal domain.|
+| |Thm: If  $R$  is a PID and  $I\normaleq R$  is prime then  $I$  is a maximal proper ideal.|
+| |Cor:  $\Z[x]$  is not a PID.|
+| |Example: in  $\Z[x]$  the ideal  $K=\pbr{2}+\pbr{x}$  is not a principal ideal.|
+| |Chapter 15. Divisibility by  $9$  criterion.|
+| |Divisibility by  $7$  criterion:  $n=10m+d_0$  is divisible by  $7$  iff  $m-2d_0$  is divisible by  $7$ |
+|11/30/2018|Thm 15.1|
+| |Thm. 15.3|
+| |Lemma: Let  $R$  be a ring with unity. For  $a,b\in R$ ,  $n,m\in \Z$ ,  $(m\cdot a)(n\cdot b)=(mn)\cdot(ab)$ |
+| |Thm. 15.5|
+^Week 16^Topics^
+|12/03/2018|Corollaries 1, 2, and 3.|
+| |Field of fractions(quotients)|
+| |Alternative construction of number fields, by constructing  $\Q$  from  $\Z$ .
+|12/04/2018|Thm. 15.6 Moreover,  $F$  is minimal. If  $E$  is a field that contains a copy of  $D$ , then  $E$  contains a copy of  $F$ .|
+| |Examples. 1) The field of fractions of  $\Z$  is  $\Q$ .|
+| |2) Let  $D$  be an integral domain, and  $D[x]$  the ring of polynomials over  $D$ . The field of fractions of  $D[x]$  is denoted by  $D(x)$ , and its elements are called //rational functions// over  $D$ . A //rational function// is a quotient of two polynomials  $f(x)/​g(x)$ , with  $g(x)\neq 0$ .|
+| |External and internal direct product of groups.|
+| |Def: Internal semi direct product of groups. Given a group  $G$ ,  $N\normaleq G$ ,  $H\leq G$  such that  $N\intersection H=1$  and  $NH=G$ , we say that  $G$  is the (internal) semi direct product of  $N$  and  $H$ .|
+| |Example: The dihedral group  $D_n$  is the semi direct product of the subgroup of rotations  $\pbr{R}$  and the subgroup  $\pbr{F}$  generated by one reflection  $F$ .
+|12/05/2018|Def: External semi direct product. Given two groups  $N$  and  $H$  and a homomorphism  $\alpha:​H\to\aut(N)$ , write  $\alpha(h)$  as  $\alpha_h$ . Consider the cartesian product   $N\times H$  with the following operation:  $(n_1,​h_1)(n_2,​h_2)=(n_1\alpha_{h_1}(n_2),​h_1h_2)$ |
+| |Thm: 1) The operation just defined makes  $N\times H$  into a group. We denote it by  $N\rtimes_\alpha H$ .  We omit the subscript  $\alpha$  is it is understood from the context.|
+| |2)  $\bar{N}=\{(n,​1)\mid n\in N\}$  is a normal subgroup of  $N\rtimes_\alpha H$ , isomorphic to  $N$ , via the map  $N\to N\rtimes_\alpha H\\ n\mapsto \bar{n}=(n,​1)$ .|
+| |3)  $\bar{H}=\{(1,​h)\mid h\in H\}$  is a subgroup of  $N\rtimes_\alpha H$ , isomorphic to  $H$ , via the map  $H\to N\rtimes_\alpha H\\ h\mapsto \bar{h}=(1,​h)$ .|
+| |4)  $\bar{N}\intersection \bar{H}=1$  and  $\bar{N}\bar{H}=N\rtimes_\alpha H$ .|
+| |5)  $N\rtimes_alpha H$  is the internal semi direct product of  $\bar{N}$  and  $\bar{H}$ .|
+| |6) Given  $h\in H$  and  $n\in N$ , conjugation of  $\bar{n}$  by  $\bar{h}$  is given by  $\varphi_{\bar{h}}(\bar{n}) =\overline{\alpha_h(n)}.$ |
+| |Cor: When  $\alpha$  is the trivial homomorphism, i.e.  $\alpha_h=1$  for all  $h\in H$ , then the semi direct product is equal to the direct product,  $N\rtimes_\alpha H=N\oplus H$ . |
+| |Cor: The operation in  $N\rtimes_\alpha H$  is completely determined by the operations in  $N$  and  $H$ , and the relation  $\bar{h}\ \bar{n}=\overline{\alpha_h(n)}\ \bar{h}.$ |
+| |Example: Let  $N=\pbr{a}$  be cyclic of order  $7$ , and  $H=\pbr{b}$  cyclic of order  $3$ .  $\aut(N)\isom U_7$  is abelian of order  $6$ , hence cyclic. |
+| | $s:N\to N, a\mapsto a^2$  is an automorphism of  $N$  of order  $3$  since  $a^{2^3}=a^8=a$ . |
+| | $\aut(N)$  is generated by  $c:N\to N, a\mapsto a^3$ , and  $s=c^2$ , since  $a^{3^2}=a^9=a^2$ .|
+| |Any homomorphism  $\alpha:​H\to\aut(N)$  has to map  $b$ , which has order  $3$ , to an element of  $\aut(N)$  of order a divisor of  $3$ .  The only such elements are  $1$ ,  $s$  and  $s^{-1}=c^4=s^2$ .|
+|12/07/2018|Board presentation PS 12|
+| |Continuation of example. There are three different semi direct product of  $N$  and  $H$ , given by the three automorphisms  $\alpha(b)=1$ ,  $\beta(b)=s$ , and  $\gamma(b)=s^2$ . Let's write down the three.|
+| |Case 1:  $\alpha(b)=1$  is trivial. In this case  $N\rtimes_\alpha H = N\oplus H\isom C_{21}$ .|
+| |Case 2:  $\beta(b)=s$ .  $ba = \beta_b(a)b = s(a)b = a^2b$ |
+| |so  $N\rtimes_\beta H$  is not abelian, and not isomorphic to case 1.|
+| |Case 3: To distinguish from case 2, let's write  $N=\pbr{u}$  cyclic of order  $7$ , and  $H=\pbr{v}$  cyclic of order  $3$ ,  $\gamma(v)=s^2$ .  $vu = \gamma_v(u)v = s^2(u)v = u^4v$ |
+| |Again  $N\rtimes_\gamma H$  is not abelian, not isomorphic to case 1.|
+| |Claim:  $N\rtimes_\beta H$  is isomorphic to  $N\rtimes_\gamma H$  via the map  $a\mapsto u, b\mapsto v^{-1}$ .|
+| |Cor: there are only two non-isomorphic semi direct products of a cyclic group of order 7 and a cyclic group of order 3, namely, the direct product, and the non-abelian semi direct product of case 2.|
+| |Example: Let  $G$  be a group of order  $21$ .  By Sylow's theorem we have  $n_7=1$ . Let  $N$  be the Sylow 7-subgroup of  $G$ .  We also know that  $n_3$  is either 1 or 7.  Let  $H$  be a Sylow 3-subgroup of  $G$ . When  $n_3=1$   $H$  is a normal subgroup of  $G$  and  $G$  is the direct product of  $N$  and  $H$ . When  $n_3=7$ , then  $H$  is not normal, and  $G$  is the non-abelian semi direct product of  $N$  and  $H$ .|
+| |Therefore, there are exactly two non-isomorphic groups of order 21. |
+[[people:fer:401ws:fall2018:daily_topics|Daily topics (2)]]
+[[people:fer:401ws:fall2018:home| Home]]

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