people:fer:401ws:fall2018:daily_topics_3
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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}} | ||
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| + | [[people:fer:401ws:fall2018:home| Home]] | ||
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| + | ^Week 12^Topics^ | ||
| + | |11/05/2018|Sylow Theorems | | ||
| + | | |Examples: (1) $|G|=35$ <html>  </html> (2) $|G|=455$ <html>  </html> (3) $|G|=21$ <html>  </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| | ||
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| + | ^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. | | ||
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| + | [[people:fer:401ws:fall2018:daily_topics|Daily topics (2)]] | ||
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| + | [[people:fer:401ws:fall2018:home| Home]] | ||