Algebra (Lecture)

I collect here some of the notes of the lecture on Algebra I followed in the3rd_Semester at theLMU held by A. Rosenschon.

Recap

Elementary Group Theory

Most of theorems of this section were present already in Linear Algebra II (Lecture).

• Normal Divisor
  • for $H\subseteq G$ subgr., $aH:=\{ah:h\in H\}$, $Ha:=\{ha:h\in H\}$
    • $|aH|=|bH|$
    • $G={⨆}_{a\in R\subseteq G}aH$
    • $aH=bH\iff aH\cap bH=\varnothing \iff a\in bH\iff b^{-1}a\in H$
  • $G/H:=\{aH:a\in G\}$, $H\setminus G:=\{Ha:a\in G\}$, $|G:H|:=|G/H|$, $|G/H|=H\setminus G$
  • $|G|=|G:H|\cdot |H|$ (Lagrange)
  • for $H\subseteq G$ subgr., $H⊴G:{\forall }_{a\in G}aH=Ha$
  • Ways to prove $U⊴G$:
    • for $\phi :G\to G^{\prime }$, $ker(\phi )⊴G$
    • $|G:H|=2\Rightarrow H⊴G$, link.
  • $G\to G,x\mapsto gx{g}^{-1}$ is a gr.autom.
• Homomorphism and Isomorphism Theorems 1.12
  • for (i) $\phi :G\to G^{\prime }$ hom., (ii) $N⊴G$ s.t. $N\subseteq ker(\phi )$, (iii) $\pi :H\to G/N,a\mapsto aH$ then: ex. uniq. $\overline{\phi }:G/N\to G^{\prime }$ s.t. G \to^\pi G/N \to^\overline{\varphi} G', $G{\to }^{\phi }{G}^{\prime }$ groups_hom.png
    • $\phi$ surj, then $\overline{\phi }:G/ker(\phi )\to G^{\prime }$ is isom.
  • $H\subseteq G$ subgr. $N⊴G$, then (i) $HN\subseteq G$ subgr., (ii) $N⊴HN$, (iii) $H\cap N⊴H$, (iv) $H/H\cap N\cong HN/N$ (1st Isomorphism Theorem)
    • $HN:=\{hn|h\in H\wedge n\in N\}$
  • for $N,H⊴G$, $N\subseteq H\subseteq G$, then (i) $N⊴H$, (ii) $H/N⊴G/N$, (iii) $(G/N)(H/N)\cong G/H$ (2nd)
• Grothendieck
• Cyclical Groups
  • ...

Rings and Polynomials

Most of these theorems are present in Linear Algebra II (Lecture) already, though I list some of them here again:

• for $\phi$ hom, $\phi$ inj. $\iff \phi$ surj. No, find the right version
  • for $|R|<{\aleph }_0$ comm, ${\forall }_{a\in R}a|0\vee a\in R^{\times }$ (B.9.1.a)
• Rings and Polynomial Rings in one Variable
  • def. of Ring: for a group $(R,+)$ with $e=0$ and a monoid $(R,\cdot )$ with $e=1$ it holds $(a+b)c=ac+ab$ and $c(a+b)=ca+cb$. It is comm. if $ab=ba$.
    • if $0=1$, then $R=\{0\}$
  • $R^{\times }=\{a\in R|{\exists }_{b\in R}ab=ba=1\}$, $(R^{\times },{\cdot }_R)$ is a group.
  • For $(R_i{)}_{i\in I}$ a family of rings, ${\prod }_{i\in I}{R}_i$ is ring with component-wise $+$ and $\cdot$.
  • $R[X]\simeq (R^{(\mathbb{N})},{+}_{R^{\omega }},{\cdot }_{R^{\omega }})$
    • $R^{(\mathbb{N})}=\{(a_i{)}_{i\in I}:{\forall }_{i\in I}{a}_i\in R\wedge |\{a_i\in (a_i{)}_{i\in I}:a_i\ne 0\}|<{\aleph }_0\}$ with $+$ and $\cdot$ from $R^{\omega }$. 2.2-2.3
  • $grad(f)=max\{i|:a_i\ne 0\}$
    • $grad(f+g)\le max\{grad(f),grad(g)\}$, $grad(fg)\le grad(f)+grad(g)$.
• Ideal
  • def. ideal $I\subset R$ s.t. $(I,{+}_R)$ subgroup of $(R,+)$ and ${\forall }_{r\in R}{\forall }_{i\in I}ra=ar\in I$
  • def. generator $(a_j{)}_{j\in J}$, if $I={\sum }_{x\in X}R{a}_i$, fin.gen. if $|(a_j{)}_{j\in J}|<{\aleph }_0$, principle if $(a)$.
  • 2.14: $p$ prime Ideal if ${\forall }_{a,b\in R}ab\in p\to (a\in p\vee b\in p)$, $m$ max. ideal if all ideals $I$, ${\forall }_{I\subset R}m\subseteq I\subseteq R\to (I=m\vee I=R)$.
  • 2.15: $R$ comm., $I⊊R$ prime iff $R/I$ is ID. $I⊊R$ max. ideal $\iff$ $R/m$ is a field.
    • for $I$ max., I prime. $p\mathbb{Z}$ max. $\iff$ $p$ prime.
• Rings-homomorphisms and Factor-rings
  • def hom. $\phi :R\to S$ s.t. (i) $\phi (a{+}_{R}b)=\phi (a){+}_S\phi (b)$, (ii) $\phi (ab)=\phi (a)\phi (b)$, (iii) $\phi (1_R)=\phi (1_S)$.
    • $ker(\phi )$ is an ideal, $im(\phi )$ is a subr., $\phi$ determines a gr. hom. $R^{\times }\to S^{\times }$. $ker(\phi )=\{0\}\Rightarrow \phi$ inj.
  • Homomorphiesatz
  • Chinese Restsatz
• Prime Factorisation
  • def. let $\delta :R\setminus \{0\}\to \mathbb{N}$ s.t. $a,b\in R$, $b\ne 0$ there is $q,r\in R$ s.t. $a=qb+r$ where $\delta (r)>\delta (g)\vee r=0$, then $R$ is eucl. ring and $\delta$ a norm. 2.17
  • $R$ eucl $\Rightarrow R$ PID
  • $p\in R\setminus R^{\times }$ irr. if ${\forall }_{x,y\in R}p=xy\to x\in R^{\times }\vee y\in R^{\times }$
  • $p\in R\setminus R^{\times }$ prime if ${\forall }_{x,y\in R}p|xy\to p|x\vee p|y$
  • (i) $(p)$ max. $\Rightarrow p$ prime, (ii) $p$ prime $\Rightarrow p$ irr. (iii) $R$ PID: $p$ irr. $\iff p$ prime $\iff (p)$ max. 2.20
  • $R$ PID, $0\ne a\in R\setminus R^{\times }$, then $a=p_1\cdot p_2\cdot ...\cdot p_n$. unique up to order and mult. with units. 2.22
  • def Fact. Ring: ${\forall }_{0\ne a\in R\setminus R^{\times }}$ if (i) $a=p_1\cdot p_2\cdot ...\cdot p_n$ or (ii) $a=i_1\cdot i_2\cdot ...\cdot i_m$ irr elem. 2.23
    • $R$ Fact. Ring $p$ prime $\iff p$ irr.
    • $R$ eucl. $\Rightarrow R$ PID $\Rightarrow$ $R$ Fact Ring
    • $ggT(x_1,...,x_n)={\prod }_{p\in \mathbb{P}}{p}^{min\{v_p(x_1),...,v_p(x_n)\}}$
    • $(x_1,...,x_n)=(d)\Rightarrow d=ggT(x_1,...,x_n)$, $(x_1)\cap ...\cap (x_n)=(v)\Rightarrow v=kgV(x_1,...,x_n)$
      • in an $ID$ there could be no $gcd$ bc not every ideal is principal. 2.26
  • for $R$ is PID, $|R|<{\aleph }_0\Rightarrow R$ is a field, link.
  • $R[X]$ PID $\Rightarrow R$ is a filed, link.
  • $\mathfrak{p}$ pr. ideal $\Rightarrow R/\mathfrak{p}$ ID (B.9.1.c)
  • $\mathfrak{m}$ max. ideal $\Rightarrow R/\mathfrak{m}$ is a field. (B.7.1.a)
• Roots of Polynomials
  • $K$ a field $grad(f)=n\ge 0$ then $f$ has max. $n$ roots, exactly $n$ if $f={\prod }_{i=1}^n(X-a_i)\in K[X]$.
  • $f\in K[X]$, $f(\alpha )=0$, then $\alpha$ mult. root $\iff f^{\prime }(\alpha )=0\iff ggT(f,f^{\prime })(\alpha )=0$.
  • $M=R\times R\setminus \{0\}=\{(a,b):a,b\in R\wedge b\ne 0\}$, $(a,b)\sim (a^{\prime },b^{\prime })\iff a{b}^{\prime }=a^{\prime }b$, with the fraction operation define the the quotient field, called $Q(R)$. 2.28 Also $\frac{a}{b}=ϵ{\prod }_{p\in P}{p}^{v_p(x)}$.
  • $v_p(f):=min\{v_p(c_i):i\in \mathbb{N}\}$, (Gauss-lemma) $R$ fact. Ring $v_p(fg)=v_p(f)+v_p(g)$.
  • (Gauss) $R$ fact. $\Rightarrow R[X]$ fact.: $q\in R[X]$ pr. $\iff q\in R$ pr. $\vee (q$ primitive in $R[X]\wedge$ prime in $Q(R)[X]$.
    • $R$ fact. $q$ primitive in $R[X]$: $q$ irr in $R[X]\iff q$ irr. in $Q(R)[X]$.
  • for $f,g\in \mathbb{Q}[X]$, $cont(fg)=cont(f)\cdot cont(g)$, link.
    • $cont(f)$ is the $gcd(a_1,...,a_n)$ for all coefficients of $f$., wiki.
  • Irreducibility Criteria for $R$ fac. ring with quotientfield and $f$ primitive.
    • (Eisenstein) $f={\sum }_{i=0}^n{a}_i{X}^i$ primitive, $grad(f)>0$ and there is $p$ prime in $R$, s.t. $\neg p|a_n$, ${\forall }_{0\le i<n}p|a_i$, $\neg p^2|a_0$, then $f$ irr. in $R[X]$. 2.34
    • (Red. mod. $p$) $p\in R$, $0\ne f={\sum }_{i=0}^n{a}_i{X}^i\in R[X]$ s.t. $\neg p|a_n$ and $\pi :R[X]\to (R/(p))[X],{\sum }_{i=0}^n{a}_i{X}^i\mapsto {\sum }_{i=0}^n\overline{a_i}{X}^i$ for $\overline{a_i}:=a+(p)$. If $\pi (f)$ irr. then $f$ irr in Q(R)[X], if $f$ also primitive, then $f$ irr. in $R[X]$. 2.35
  • Methods (Sol. 9.2): for $R$ fac., $f={\sum }_{k=0}^d{a}_k{X}^k\in R[X]$, $d\ge 2$., check red. of $f$ in $Q(R)[X]$
    1. Eisenstein: check for all $p\in \mathbb{P}$ s.t. $p|a_0$, if it works, it is irr., else continue
    2. Roots: check poss. roots in $Q(R)$, i.e. $\frac{a}{b}\in Q(R)$ s.t. $gcd(a,b)=1$, $a|a_0$, $b|a_n$, if none, continue
    3. Degree: if $d\le 3$, $f$ has no roots $\iff$ $f$ irr. (if no roots, conclude)
    4. Reduction: if $R=\mathbb{Z}$, $\neg 2|a_d$, then $\overline{f}\in {\mathbb{F}}_2[X]$ irr. $\Rightarrow f\in \mathbb{Q}[X]$ irr., try further red.., else continue
       • recall irr. pol. of $d\le 4$ in ${\mathbb{F}}_2[X]$ (link): for $d\le 3$, try $\overline{f}(0),\overline{f}(1)$, else
         • $x^4+x^3+x^2+x+1$ and $x^4+x^n+1$ for $n\in \{1,2,3\}$
    5. Absurdum: if $d=4$, $f$ norm., no roots, try $f=(X^2+aX+b)(X^2+cX+d)$ get cont., else
    6. $R=\mathbb{Z}$, $d\le 5$ with roots ${\xi }_i\in \mathbb{C}\setminus \mathbb{Q}$, if ${\forall }_{i,j}{\xi }_i\cdot {\xi }_j\notin \mathbb{Q}$, $f$ irr.
    7. $f$ irr. $\iff f^{♭}$ irr. (Wiederholungsblatt 6.a, Tut.B.9-10)
       • $f^{♭}$ is $f$ with the order of the coefficients inverted, this can help to apply Eisenstein.

Algebraic Field Extensions

• Characteristic
  • def. for $R$ ID $\phi :\mathbb{Z}\to R,n\mapsto n\cdot 1_R$, for $p\in \mathbb{N}$ s.t. $(p)=ker(\phi )$ then $char(R)=p$.
    • $R\subseteq S$ subr. $\Rightarrow char(R)=char(S)$
  • def. prime field $P=\bigcap \{L:L\subseteq K$ subfield$\}\subseteq K$, smallest subfield of $K$.
    • (i) $char(K)=p>0\iff P\simeq {\mathbb{F}}_p$, (ii) $char(K)=0\iff P\simeq \mathbb{Q}$.
      • (iii) any field contains some ${\mathbb{F}}_p$ or a $\mathbb{Q}$ up to isomorphism.
• Field-Extentions
  • $L/K$ ext. if $K$ is a subfield of $L$. Note that $L$ is a $K$-VS.
    • $[L:K]=di{m}_K(L)$
      • $[L:K]=[L:E][E:K]$
        • $[L:K]\in \mathbb{P}\Rightarrow K=E\vee E=L$
  • $L/K$, $\alpha \in L$ algebraic on $K$ if ex. $f\in K[X]$ s.t. $f(\alpha )=0$. $L/K$ alg. if all $\alpha \in L$ is alg.
    • ${\phi }_{\alpha }:K[X]\to L,g\mapsto g(\alpha )$, $\alpha$ alg. on $K$ iff $ker({\phi }_{\alpha })\ne \{0\}$.
    • $\alpha \in L$ alg., ex! $f\in K[X]$ s.t. (i) norm., (ii) $ker({\phi }_{\alpha })=(f)$, then $(f)$ prime, $f$ min. of $\alpha$
      • $f$ is the min. poly. of $\alpha$ if it is normed, irr. and has $\alpha$ as a root.
    • All finite extensions are algebraic. 3.7
    • $L/K$ alg. $\iff$ each subring $R$, s.t. $K\subseteq R\subseteq L$ is a field B.6.4
  • $K(({\alpha }_i{)}_{i\in I}):=\bigcap \{E$ fi.$:{\forall }_{i\in I}({\alpha }_i\in E)\wedge K\subseteq E\subseteq L\}$ for $({\alpha }_i{)}_{i\in I}\subseteq L$
    • $K({\alpha }_1,...,{\alpha }_n)=Q(K[{\alpha }_1,...,{\alpha }_n])$ 3.9
    • Method: find $(\alpha +1{)}^{-1}\in \mathbb{Q}(a)$, for $f(\alpha )=0$, $f$ min.pol. B.6.2.b
      1. $f(x-1)(\alpha +1)=0$, rewrite the equation in order to get $\frac{1}{\alpha +1}=...$
  • $K[\alpha ]=im({\phi }_{\alpha })$, $f$ min of $\alpha$ then $K[X]/ker({\phi }_{\alpha })\cong K[\alpha ]$, hence $[K[\alpha ]:K]=deg(f)$. 3.8
  • $L/K$ fin.gen. if $L=K({\alpha }_1,...,{\alpha }_n)$, $L/K$ simple if $L=K(\alpha )$ 3.10
    • if $\alpha$s alg. in $L$: (i) $K[{\alpha }_1,...,{\alpha }_n]=K({\alpha }_1,...,{\alpha }_n)$, (ii) $L/K$ finite (alg. + fin. gen = finite) 3.11
    • $K\subseteq E\subseteq L$, $\alpha$ alg. on $E$, $E/K$ alg.: (i) $\alpha$ alg. on $K$, (ii) $L/K$ alg. $\iff L/E$, $E/K$ alg. 3.12
  • Method: $\mathbb{Q}({\alpha }_1,...,{\alpha }_n)$, find deg. and basis, B.6.1
    1. guess the min.pol., it is normed, irr. and has ${\alpha }_1$ as a root, repeat for each ${\alpha }_n$
       • multiply the degrees and compute the basis: $\{1,a\},\{1,b\}⇝\{1,a,b,a\cdot b\}$.
  • $[\mathbb{Q}(\sqrt{2},\sqrt[2]{2},...,\sqrt[n]{2}):\mathbb{Q})]=lcm(2,...,n)$, $\mathbb{Q}(\sqrt{2},\sqrt[2]{2},...,\sqrt[n]{2})=\mathbb{Q}(\sqrt[lcd(2,..,n)]{2})$
• Algebraic Closure
  • (Kronecker) for $K$ field and $f\in K[X]$, there is $L/K$ s.t. ${\exists }_{\alpha \in L}f(\alpha )=0$.
    • $f=g(X-\alpha )\in L[X]$, then after less than $deg(f)$ steps $f$ is red in lin. fact. $L_n$ 3.14
  • $K$ alg. cl. if all non-const $f\in K[X]$ have a root, i.e. $f=c{\prod }_{i=1}^n(X-\alpha )\in K[X]$, $deg(f)=n$, 3.15
    • $K$ alg. cl. $\Rightarrow |K|\ge {\aleph }_0$
    • $K$ alg. cl. $\Rightarrow$ ($L$ alg. cl $\to$ $L=K$)
    • each $K$ has an alg. closure $\overline{K}$. 3.17
  • Let $L$ field, $\sigma :K\to L$ hom., $f\in K[X]$ min of $\alpha$, ${\sigma }^{\prime }:K(\alpha )\to L$ s.t. ${\sigma }^{\prime }{|}_K=\sigma$:
    • (a): $f^{\sigma }\in L[X]\wedge f^{\sigma }({\sigma }^{\prime }(\alpha )=0$
    • (b): for $\beta \in L$, $f^{\sigma }(\beta )=0$, then ${\sigma }^{\prime }$ s.t. ${\sigma }^{\prime }(\alpha )=\beta$ is unique.
    • $|\{{\sigma }^{\prime }:\}|=|\{\alpha :f^{\sigma }(\alpha =0\}|\le deg(f^{\sigma })=deg(f)$ 3.18
  • For $K^{\prime }$ alg. ext. of $K$, $\sigma :K\to \overline{K}$:
    • (a): ${\sigma }^{\prime }$ s.t. ${\sigma }^{\prime }:K^{\prime }\to \overline{K}$ and ${\sigma }^{\prime }{|}_K=\sigma$ exists
    • (b): if $K^{\prime }$ alg. cl. and $\overline{K}/\sigma (K)$ alg. then ${\sigma }^{\prime }$ is an isom.
  • If $\overline{K}$, $\overline{K^{\prime }}$ alg. cl. and ext. of $K$ and $\sigma =i{d}_K$, then ${\sigma }^{\prime }:\overline{K}\to \overline{K^{\prime }}$ ex. and is isom. 3.20
  • $|K|<{\aleph }_0\Rightarrow |\overline{K}|={\aleph }_0$ also $|K|\ge {\aleph }_0\Rightarrow |\overline{K}|={\aleph }_0$, wiki.
• Splitting Fields
  • $K$ field, $(f_i{)}_{i\in I}\subset K[X]$, $L$ is called spl.fi. if (i) any $f_i$ splits in lin. factors, (ii) $L=K[{\alpha }_1,...,{\alpha }_n]$ roots of $f_i$s. $L/K$ is then called normal.
  • spl.fi. of $f\in K[X]$ is $K[{\alpha }_1,...,{\alpha }_n]$ for $\alpha$s the roots fo $f$. For more polynomials, let $f=f_1\cdot ...\cdot f_m$.
  • (Compactness) let $L$ be spl.fi. of $(f_i{)}_{i\in I}$ for $I={\aleph }_0$, let $\mathcal{J}:=\{J:J⊊I\wedge |J|<{\aleph }_0\}$ and $A_J=\{\alpha :{\exists }_{f_j\in J}{f}_j(\alpha )=0\}$, $L={\bigcup }_{J\in \mathcal{J}}K(A_J)$. 3.31.iii
  • Every two spl.fi. of the same polynomials are isomorphic.
  • $K\subseteq L$ alg. ex. tfae:
    • $L/K$ norm.
    • irr. $f\in K[X]$ s.t. ${\exists }_{\alpha \in L}f(\alpha )=0$ splits in lin. fact.
    • $\sigma :L\to \overline{L}$ defines a $K$-autom. of $L$.
  • $[L:K]=2\Rightarrow L/K$ norm.
  • $N$ normal closure of alg. ext. $L/K$ if (i) $L\subseteq N$ alg., (ii) $N/K$ norm., (iii) $N$ minimal. 3.25
    • $L/K$ alg. then:
      • unique up to isom normal closure $N/K$
      • $L/K$ finite $\Rightarrow N/K$ finite
      • if $M/L$ alg., $M/K$ norm. then ex. norm. cl. $N$ of $L/K$ s.t. $L\subseteq N\subseteq M$, $K=(\{{\sigma }_i(L)i\in I\})$ for $({\sigma }_i{)}_{i\in I}$ $K$-hom. ${\sigma }_i:L\to M$. $N$ is unique as a subfi. of $M$. 3.25
  • $|L|<{\aleph }_0\Rightarrow L/K$ norm. (tut.13.5)
  • Method: find the spl.fi. of $f$:
    1. Roots: find the roots of $f$, call them ${\alpha }_1,...,{\alpha }_n$.
    2. Minimise: try to find the smallest set of roots s.t. $K({\alpha }_i,...,{\alpha }_j)$ contains all roots
  • Methods: for $L$ spl.fi. for $f=X^n+a\in \mathbb{Q}[X]$, find $L$ and compute $[L:\mathbb{Q}]$ B.9.3.b
    1. Roots: those are $({\zeta }_n^j{)}_{j\in [0,n-1]}$, for ${\zeta }_n=exp(\frac{2\pi i}{n})\in \mathbb{C}$.
    2. Splitting field: then $L=\mathbb{Q}(({\zeta }_n^j\sqrt[n]{a}{)}_{j\in [0,n-1]})=\mathbb{Q}(\sqrt[n]{a},{\zeta }_n)$
    3. Degree: must divide $n\cdot (n-1)$ if $n$ prime check
       • Min. Pol.: $f$ irr., $[\mathbb{Q}(\sqrt[n]{a}):\mathbb{Q}]=deg(f)=n$.
       • compute $[\mathbb{Q}(\sqrt[n]{a},{\zeta }_n):\mathbb{Q}(\sqrt[n]{a})]$, recall $[\mathbb{Q}({\zeta }_p):\mathbb{Q}]=p-1$ (for $p\in \mathbb{P}$)
• Separable Field Extensions
  • $f\in K[X]$ sep. if $f$ has no mult. root in $\overline{K}$
    • $\alpha \in \overline{K}$ is mult. root of $f\in K[X]\iff ggt(f,f^{\prime })(\alpha )=0$
    • $f\in K[X]$ irr. then $f$ not sep. in $\overline{K}\iff f^{\prime }=0$ 3.26
  • $char(K)=0\wedge f\in K[X]$ irr. $\Rightarrow f$ sep.
    • $|K|<{\aleph }_0\wedge f\in K[X]$ irr. $\Rightarrow f$ sep., link, 3.27
  • $char(K)=p\in \mathbb{P}\setminus \{0\}$:
    • $f$ not sep. $\iff f^{\prime }=0\iff {\exists }_{g\in K[X]}f=g(X^p)$ ($deg(g)=\frac{deg(f)}{p}\Rightarrow K(a^p)⊊K(a)$)
    • (Frobenius) ${\forall }_{a,b\in K}(a+b{)}^p=a^p+b^p$, $F:K\to K,a\mapsto a^p$ inj., $K$ perf. $\Rightarrow F$ aut.
  • Defs: 3.28
    • $L/K$ alg., $\alpha \in L$ sep. over $K$ if for $f\in K[X]$ sep. $f(\alpha )=0$
    • $L/K$ alg. is sep. if any $\alpha \in L$ sep. over $K$.
    • $K$ perfect if all $L/K$ alg. are sep., wiki
      • (i) $char(K)=0\Rightarrow K$ perf., (ii) $|K|<{\aleph }_0\Rightarrow K$ perf. (iii) $\overline{K}$ per. (iv) alg.ext. of per. is per.
  • $L/K$ alg. sep.deg. is $[L:K{]}_s:=|Ho{m}_K(L,\overline{K})|$, for $Ho{m}_K(L,\overline{K})$ set of $K$-hom. 3.30
    • $L/K$ norm. $\Rightarrow [L:K]=|Au{t}_K(L)|=|Ho{m}_K(L,\overline{K})=[L,K{]}_s$, see Galois Theory.
    • $\phi :L\to \overline{K}$ is $K$-hom. if $\phi {|}_K=i{d}_K$.
  • $K(\alpha )/K$, $f\in K[X]$ min of $\alpha$ then 3.31
    • $[L:K{]}_s=|\{\beta \in \overline{K}:f(\beta )=0\}|$
    • $\alpha$ sep. over $K\iff [L:K{]}_s=[L:K]$
    • if $char(K)=p\in \mathbb{P}\setminus \{0\}$ and mult. of $\alpha$ in $f$ is $p^r$, then $[L:K]=p^r[L:K{]}_s$
  • $K\subseteq E\subseteq L$ alg. ext., $[L:K{]}_s=[L:E{]}_s[E:K{]}_s$ 3.32
    • $char(K)=0\Rightarrow [L:K]=[L:K{]}_s\Rightarrow L/K$ sep.
    • $char(K)\in \mathbb{P}\setminus \{0\}\Rightarrow {\exists }_{r\in \mathbb{N}}[L:K]=p^r[L:K{]}_s$
    • always: $[L:K{]}_s|[L:K]$
  • $L/K$ fin., then $L/K$ sep. $\iff$ $L=K({\alpha }_1,...,{\alpha }_n)$ for ${\alpha }_1,...,{\alpha }_n\in L$ sep. over $K\iff [L:K{]}_s=[L:K]$
  • $L/K$ alg., let $\mathcal{A}\subseteq L$ s.t. $K(\mathcal{A})=L$, then $L/K$ sep. $\iff$ any $\alpha \in \mathcal{A}$ sep. over $K\Rightarrow [L:K{]}_s=[L:K]$.
  • if $E/K$, $L/E$ alg. then $L/K$ sep. $\iff L/E$, $E/K$ sep. 3.35
  • $(H,{\cdot }_K)\subseteq (K^{\times },{\cdot }_K)$ subgr. $\Rightarrow H$ cycl.
• Theorem of the Primitive Element: $L/K$ fin. sep. ext. $\Rightarrow {\exists }_{\alpha \in L}L=K(\alpha )$. Also $\alpha$ is primitive to $L/K$.
• Finite Fields
  • For $p^n$, $p\in \mathbb{P}$, ex. unique up to isom. $\mathbb{F}$ s.t. $|\mathbb{F}|=p^n$.
    • For ${\mathbb{F}}_{p^n}$ is spli.fi. of $X^{p^n}-X\in {\mathbb{F}}_p[X]$
    • ${\mathbb{F}}_p\subseteq {\mathbb{F}}_{p^n}\subseteq \overline{{\mathbb{F}}_p}$ holds, hence ${\mathbb{F}}_p\subseteq {\mathbb{F}}_{p^n}\subseteq {\mathbb{F}}_{p^m}\subseteq \overline{{\mathbb{F}}_p}\iff n|m$
      • extensions ${\mathbb{F}}_{p^n}\subseteq {\mathbb{F}}_{p^m}$ are the only extensions of finite fields with char. $p$.
  • $|K|<{\aleph }_0\Rightarrow {\exists }_{p\in \mathbb{P}}{\exists }_{n\in \mathbb{N}}K\cong {\mathbb{F}}_{p^k}$ wiki.

Galois Theory

True love makes non-Galois couples: not normal or inseparable. And if you claim your lover to be perfect, be aware that they are separable to anyone.jokes

• $L/K$ is gal. $:=L/K$ norm. and sep., a gal. gr. $Gal(L/K)=Au{t}_K(L)$, Galois Examples.
  • $L/K$ normal, then:
    • recall: $L/K$ norm. $\Rightarrow Ho{m}_K(L,\overline{K})=Au{t}_K(L)$ 3.23.c
    • $|Au{t}_K(L)|=|Ho{m}_K(L,\overline{K})|=[L:K{]}_s\le [L:K]$
  • $L/K$ separable, then:
    • ($\iff$) $|Au{t}_K(L)|=[L:K]$
  • $L/K$ gal. $\iff |Gal(L/K)|=[L:K]$
  • $L/K$ gal. fin. $\Rightarrow L=K({\alpha }_1,...,{\alpha }_n)=K[{\alpha }_1,...,{\alpha }_n]$, for $\sigma \in Gal(L/K)$ is uniq. def. by $\sigma ({\alpha }_1),...,\sigma ({\alpha }_n)$.
  • $f\in K[X]$, root $\alpha \in L$, $\sigma \in Gal(L/K)\Rightarrow f(\sigma (\alpha ))=0$
    • $0=f(\alpha ){\sum }_{i=0}^n{a}_i{\alpha }^i\in L$, $0=\sigma (0)=\sigma (f(\alpha ))=\sigma ({\sum }_{i=0}^n{a}_i{\alpha }^i)={\sum }_{i=0}^n{a}_i\sigma ({\alpha }^i)=f(\sigma (\alpha ))$.
  • $L/K$ gal., $K\subseteq E\subseteq L$:
    • (i) $L/E$ gal., $Gal(L/E)\subseteq Gal(L/K)$ subgr.
    • (ii) if $E/K$ gal. $\Rightarrow Gal(L/K)\to Gal(E/K),\sigma \mapsto \sigma {|}_E$ is surj. hom.
  • $L$ field, $G\subseteq Aut(L)$, then $L^G:=\{a\in L|{\forall }_{\sigma \in G}\sigma (a)=a\}$ (fixkö. of $G$): 4.4
    • (i) $|G|<{\aleph }_0\Rightarrow (L/L^G$ gal. $\wedge G=Gal(L/L^G)$
    • (ii) $|G|\ge {\aleph }_0\wedge L/L^G$ alg. $\Rightarrow L/L^G$ gal. $\wedge G\subseteq Gal(L/L^G)$ subgr.
  • $L/K$ norm.: 4.5
    • (i) $L/L^{Au{t}_K(L)}$ gal. $\wedge Gal(L/L^{Au{t}_K(L)})=Au{t}_K(L)$
    • (ii) if $L/K$ sep. then $K=L^{Au{t}_K(L)}$
  • for $\psi :E\mapsto Gal(L/E)$, $\varphi :H\mapsto L^H$, then $\varphi \circ \psi =id$
  • (Gal.-Cor.) $L/K$ gal., $L^H/K$ norm. ($\Rightarrow$ gal.) $\iff H⊴G$ $\Rightarrow H=ker(Gal(L/K)\to Gal(L^H/K),\sigma \mapsto \sigma {|}_{L^H})$ 4.6
    • if $L/K$ fin. sep. then $|\{K^{\prime }:K\subseteq K^{\prime }\subseteq L\}|<{\aleph }_0$
      • if also $N/K$ norm.cl. then $N/K$ fin. gal. and $[N:K]=|Gal(N/K)|<{\aleph }_0$.
  • Method: find $Gal(f)$, $L$ spl.fi. of $f$ in $K$, and all $E$ s.t. $K\subseteq E\subseteq L$
    1. Galois Ext. $L$ is a spl.fi., hence $L/K$ is norm., prove it to be sep.
       • $char(K)=0\Rightarrow L/K$ sep.
       • $L/K$ fin., then $L/K$ sep. $\iff$ $L=K({\alpha }_1,...,{\alpha }_n)$ for ${\alpha }_1,...,{\alpha }_n\in L$ sep. over $K\iff [L:K{]}_s=[L:K]$
       • $L/K$ alg., let $\mathcal{A}\subseteq L$ s.t. $K(\mathcal{A})=L$, then $L/K$ sep. $\iff$ any $\alpha \in \mathcal{A}$ sep. over $K
    2. Degree: find $[L:K]$, conclude $[L:K]=|Gal(f)|$
    3. Elements:
       1. let $\sigma ({\alpha }_i)={\alpha }_j$
       2. determine $\sigma ({\alpha }_j)$ using that $\sigma$ is a $K$-automorphism.
       3. combine a fixed ${\alpha }_i$ with any ${\alpha }_j$, and check wether they are automorphism
       4. Stop when ${\alpha }_j$ are finished or you already got $[L:K]$ many automorphism.
    4. Isomorphic Class: check isomorphisms with the groups in the table using the listed methods
    5. Galois Correspondence: subfields of $L$ correspond to subgroups of $Gal(f)$
       1. for $\sigma \in Gal(f)$, $L^{⟨\sigma ⟩}$ are exactly the subfields
       2. to compute $L^{⟨\sigma ⟩}$ consider $\sigma (a_1{\beta }_1+...+a_n{\beta }_n)$ for ${\beta }_1,...,{\beta }_n$ the basis of $L/K$ and recall $L^G:=\{a\in L|{\forall }_{\sigma \in G}\sigma (a)=a\}$, for instance if \sigma(a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6})$$= a - b\sqrt{2} + c \sqrt{3} + d\sqrt{6}, then $L^{⟨\sigma ⟩}=\mathbb{Q}(\sqrt{3})$.
• Roots of the Unit
  • for $f_n:=X^n-1\in K[X]$ for $n\in \mathbb{N}\setminus \{0\}$
    • $U_n:=\{\alpha \in \overline{K}:{\alpha }^n=1\}$ subgr. of $({\overline{K}}^{\times },\cdot )$, $U_n$ cycl. 4.9
    • $\neg char(K)|n\Rightarrow |U_n|=n\wedge (U_n,{\cdot }_{\overline{K}})\cong (Z/(n),+)$.
      • if $\neg char(K)|n$ then $X^n-1\ne 0\wedge (X^n-1{)}^{\prime }=n{X}^{n-1}\ne 0$ hence $f_n$ sep. and $|U_n|=n$.
      • $\zeta \in U_n$ pr. if $⟨\zeta ⟩=U_n$ or equivalently $ord(\zeta )=n$.
      • for $\zeta$ pr. $U_n=\{1,\zeta ,...,{\zeta }^{n-1}\}$, $K[\zeta ]/K$ is spl.fi. of $x^n-1$
    • if $p=char(K)|n$, $n=p^{r}m$, $\neg p|m$ then $x^m-1$ sep., since $x^n-1=(x^n-1{)}^{p^r}$, $x^n-1$ and $x^m-1$ share all roots. From now on assume $\neg char(K)|n$.
    • $a\in G$, $ord(a)=n\in \mathbb{N}\Rightarrow {\forall }_{m\in \mathbb{Z}}ord(a^m)=\frac{m}{gcd(m,n)}$
      • if $G=⟨a⟩$ then $G=⟨a^m⟩\iff gcd(m,n)=1$
    • $\phi :\mathbb{N}\setminus \{0\}\to \mathbb{N},n\mapsto |\{m:1\le m\le n\wedge gcd(m,n)=1\}|$ $=|\{$gen., of $\mathbb{Z}/(n)\}|=$ $|\mathbb{Z}/(n{)}^{\times }|=|\{$gen., of $U_n\}|=|\{\zeta \in U_n:\zeta$ pr. $\}|$ (Euler-$\phi$-function)
      • $\phi (1)=1,\phi (2)=1,\phi (3)=2$
      • $gcd(m,n)=1\Rightarrow \phi (m\cdot n)=\phi (m)\cdot \phi (n)$
      • $n=p_1^{v_1}\cdot ...\cdot p_r^{v_r}$ for $v_i>0$ $\Rightarrow \phi (n)={\prod }_{i=1}^r{p}_i^{v_1-1}$
      • $n={\sum }_{d|n\wedge d>0}\phi (d)$
    • for $\neg char(K)|n$ and $({\zeta }_i{)}_{i\in [1,...,\phi (n)]}$ pr. ${\varphi }_n:={\prod }_{i=1}^{\phi (n)}(x-{\zeta }_i)={\prod }_{d|n}{\varphi }_d$ 4.11 (circ. div. pol.)
      • ${\varphi }_4=(x-i)(x+i)$
      • for all $n\in \mathbb{N}$, ${\varphi }_n\in \mathbb{Q}[X]$ is irr.
    • for $d|n$, $P_d:=\{{\zeta }_d:{\zeta }_d$ pr.$\}\subseteq U_n$, $U_n:={⨆}_{d|n}{P}_d$, $x^n-1={\prod }_{d|n}{\prod }_{\zeta \in P_d}(x-\zeta )={\prod }_{d|n}{\varphi }_d$
      • $M_d:=\{g\in \mathbb{Z}/(n):ord(g)=d\}$, ${⨆}_{d|n}{M}_d=\mathbb{Z}/(n)$
    • $\neg char(K)|n$, ${\zeta }_n$ pr., then $X^n-1$ sep., $K({\zeta }_n)$ spl.fi of $X^n-1$, i.e. fin.gal. check note 4.13.
    • for $n\ge 1$, $\zeta \in \overline{\mathbb{Q}}$, $\zeta$ pr. then $f={\varphi }_n$, also ${\varphi }_n\in \mathbb{Q}[X]$ irr. 4.15 ?
    • $\zeta \in \overline{\mathbb{Q}}$, $\zeta =\sqrt[n]{u}$ primitive for $u\in {\mathbb{Q}}^{\times }$, then (i) $\mathbb{Q}(\zeta )/\mathbb{Q}$ gal., (ii) $[\mathbb{Q}(\zeta )/\mathbb{Q}]=\phi (n)$, (iii) $Gal(\mathbb{Q}(\zeta )/\mathbb{Q})\cong \mathbb{Z}/(n{)}^{\times }$.
      • call gal. $L/K$ abelian if $Gal(L/K)$ is abelian. [[Algebra (Lecture)#example-with—zeta|Example with $zeta$]].
• Roots Extensions
  • $char(K)=0$, $L/K$ is rad. if there are $K_1\subseteq ...\subseteq K_r=L$ s.t. $K_{i+1}=K_i(\sqrt[n_i]{a_i})$ 4.18
    • $L/K$ rad. $\Rightarrow L/K$ fin.
    • $char(K)=0$, $L/K$ rad. then ex. $M/K$ s.t. $K\subseteq L\subseteq M$ s.t. $M/K$ rad. gal. 4.20
    • $L/K$ gal. rad., $char(K)=0$, then $Gal(L/K)$ is quot. of a gr. $G$ s.t. (i). 4.21
      • (i): exists $G_i\subseteq G$ subgr. s.t. $⟨e⟩=G_r⊴G_{r-1}⊴...⊴G_0=G$, $G_i/G_{i+1}$ ab. ($G$ solvable)
    • $char(K)=0$, $f\in K[X]$ is sol. with rad. if ex. $M/K$ rad. with all roots of $f$. 4.22
      • $f\in K[X]$ sol.w.rad. $\Rightarrow Gal(f)$ quot.gr. of $G$ s.t. (i)
        • $f\in K[X]$ solv.w.rad. $\iff Gal(f)$ is solv. (not proved)

Advanced Group Theory

• $X$ a set, $S(X)$ the set of bij. on $X$, $S(X)$ is a gr. on $\circ$, $G$ operates on $X$ if ex. $\Phi :G\to S(X)$ hom.
  • for $g\in G$, $x\in X$ write $g\cdot x=\Phi (g)(x)$, also $\Phi (1)=i{d}_X=1\cdot x=i{d}_X(x)=x$
  • $g_1,g_2\in G$, write $g_1{g}_2\cdot x=\Phi (g_1{g}_2)(x)=\Phi (g_1)(\Phi (g_2)(x))=g_1\cdot (g_2\cdot x)$.
  • Within these two rules, $\cdot$ can be freely chosen (similarly to the $\lambda \cdot v$ for a VS)
  • $Gal(f)$ oper. on roots of $f$.
• $G$ oper. on $X$, for $x\in X$ def:
  • $O_x=\{g\cdot x:g\in G\}\subseteq X$ (orbit of $x$)
    • $X={⨆}_{x\in R\subseteq X}{O}_x$
  • $G_x=\{g\in G:{\forall }_{x\in X}g\cdot x=x\}\subseteq G$ (stabilisator of $x$)
    • $G_x$ subgr of $G$, 5.3
  • for $|G|<{\aleph }_0$, $G$ oper. on $G$ with $g\cdot x=gx{g}^{-1}$ 5.6.a
    • $O_x=\{gx{g}^{-1}:g\in G\}$
    • $G_x=\{g\in G:gx=xg\}$
    • $Z(G):=\{g\in G:{\forall }_{x\in G}gx=xg\}$
      • (i) $Z(X)⊴G$, (ii) $G$ ab. $\iff Z(G)=G$, (iii) $x\in Z(G)\iff G_x=G\iff |O_x|=1$, (iv) $|G|=|Z|+{\sum }_{|O_x|>1}|O_x|$
  • $G/G_x\cong O_x$ ($\iff |G:G_x|=|O_x|$), since $g{G}_x\mapsto g\cdot x$ bij. 5.5
    • $|X|={\sum }_{O_x\text{ disj.}}|O_x|={\sum }_{O_x\text{ disj.}}|G/G_x|$
    • $|G|=|G_X|+{\sum }_{|O_x|>1}|O_x|$ Tut.12.1.a
    • obv. $G_x=G\iff |O_x|=1$
    • $|G|=p^k$, for $p\in \mathbb{P}$, $k\ge 1$ then $Z\ne \{1\}$
• $|G|=p^2\Rightarrow G$ abel. 5.6.c
• $|G|=p^r\cdot m$, $\neg p|m$, a syl.-$p$-subgr of $G$ is $U\subseteq G$ s.t. $|U|=p^r$, i.e. max. $p$-subgr.
• $|G|=p^r\cdot m$, $\neg p|m$, then for any $1\le s\le r$ $G$ has a subgr $U_s$ s.t. $|U_s|=p^s$ (Syl.-I)
  • for $p\in \mathbb{P}$, $p||G|\Rightarrow G$ has an elem. of order $p$. (Special Theorem of Cauchy)
• $|G|=p^r\cdot m$, $\neg p|m$, ${\forall }_{s\in [1,r]}{\exists }_{g\in G}g{U}_s{g}^{-1}\subseteq U_r$ (Syl.-II)
  • for $P,P^{\prime }$ syl.-$p$-subgr. then ${\exists }_{g\in G}g{P}^{\prime }{g}^{-1}=P$
  • $P$ the only syl.-$p$-subgr $\iff P⊴G$, (i.e. ${\exists }_{g\in G}gP{g}^{-1}=P$).
• $s_p|m$ and $s_p\equiv {\overline{1}}^p$ (mod. $p$) (Syl. III)
  • $|G|=p^r\cdot m$, $\neg p|m$, $s_p=|\{P:$ syl.-$p$- subgr. of $G\}|$
• Method: prove there is no simple group of order $n$, tut.B.13.1
  1. Sylow Subgruops: consider the factorisations of the from $|G|=p^r\cdot m$, $\neg p|m$,
  2. Sylow III: for each $p$, $m$ found above determine possible $s_p$ s.t. $s_p|m$ and $s_p\equiv {\overline{1}}^p$.
  3. Absurdum: assume for each $p$ found in 1., assume $s_p\ne 1$ and get contradiction
  4. Sylow II: since there is one $p$ s.t. $s_p=1$, then use $P$ the only syl.-$p$-subgr $\iff P⊴G$.
• $N⊴G$, $H⊴N$ and $H$ syl.subgr. of $N$, $\Rightarrow H⊴G$, tut.B.13.2 (weaker transitivity)
• Classification of Groups of Small Order
  • Abelian Groups
    • (a) for $A$ an ab.gr. s.t. $|A|=p^k$ then ex.uniq. a partition $l_0,...,l_s$, ${\sum }_{i=0}^s{l}_i=k$, $1\le l_0\le ...\le l_s\le k$ s.t. $A\cong \mathbb{Z}/p^{l_0}\mathbb{Z}\oplus ...\oplus \mathbb{Z}/p^{l_s}\mathbb{Z}$.
    • (b) for $A$ an ab.gr. s.t. $|A|=p\cdot q$, $p\ne q$ then $A\cong \mathbb{Z}/p\mathbb{Z}\oplus \mathbb{Z}/q\mathbb{Z}\cong \mathbb{Z}/p\cdot q\mathbb{Z}$.
  • Groups through Generators and Relations
    • $G=⟨\alpha ,\beta |{\alpha }^3=1={\beta }^2,\beta \alpha ={\alpha }^2\beta ⟩$ is the gr. generated by $\alpha ,\beta$ s.t. $\beta \alpha ={\alpha }^2\beta$.
      • Questions are: (i) is one of the conditions redundant? (ii) is it a trivial gr.?
      • $G\cong S_3$ for $\alpha =(123),\beta =(12)$
    • $D_n=⟨\alpha ,\beta |{\alpha }^n=1={\beta }^2,\beta \alpha ={\alpha }^{-1}\beta ⟩$ (gr. of symm. of an $n$-sided reg. polygon)
      • $D_3=⟨\alpha ,\beta |{\alpha }^3=1={\beta }^2,\beta \alpha ={\alpha }^{-1}\beta ={\alpha }^2\beta ⟩\cong S_3$
      • max. ord. of the first rotation $\alpha$ is $\frac{ord(D_n)}{2}$
        • for $ord(G)$ equal $6$ or $10$ this is enough, for $8$ it might still be $\mathbb{Z}/(2)\oplus \mathbb{Z}/(4)$.
      • the reflection $\beta$ has ord. $2$.
        • this information is not very helpful since there is such a $\beta$ in each possible group.
      • $\beta \alpha ={\alpha }^{-1}\beta$
        • this shall be checked for $ord(G)=8$.
        • recall $D_n$ is not abelian
    • $Q=⟨\alpha ,\beta |{\alpha }^4=1{\beta }^4,{\alpha }^2={\beta }^2,\beta \alpha ={\alpha }^3\beta ⟩$, $|Q|=8$
      • $Q=⟨\begin{array}{cc}0& 1\\ -1& 0\end{array},\begin{array}{cc}0& i\\ i& 0\end{array}⟩\subseteq G{L}_2(\mathbb{C})$
      • $Q=\{\pm 1,\pm i,\pm j,\pm k\}$, $i^2=j^2=k^2=-1$, yt,
        • $i⇝j⇝k⇝i$ to get $+$ (i.e. $ij=k$ snd inverse for $-$, i.e. $ik=-j$)
        • for $\alpha \in \{\pm i,\pm j,\pm k\}$, $⟨\alpha ⟩=\{1,\alpha ,-1,-\alpha \}$.
    • $p>2$ prime, $|G|=2p\Rightarrow G\cong \mathbb{Z}/2p\mathbb{Z}$ or $G\cong D_p$. 5.19
    • $G$ not ab., $|G|=8$ $\Rightarrow G\cong Q\vee F\cong D_4$ 5.20
    • 
• Solvable Groups
  • $G$ sol. if ex. $(G_i{)}_{i\in [0,r]}$ s.t. (i) $G_i\subseteq G$ subgr., (ii) $\{e\}=G_r⊴...⊴G_1⊴G_0=G$, (iii) $G_i/G_{i+1}$ ab.
    • $G$ ab. $\Rightarrow$ $G$ sol. ($G_0=\{e\}$, $G_1=G$, $G_0/G_1\cong G$ ab.)
  • For $U\subseteq G$ subgr., $N⊴G$, then: 5.23
    • $G$ sol. $\Rightarrow G/N$ sol.
    • $G$ sol. $\Rightarrow U$ sol.
    • $N,N/G$ sol. $\Rightarrow G$ sol.
  • $|G|=p^k\cdot q^l\Rightarrow G$ sol. (Burnside)
    • we only proved: $|G|=p^k\Rightarrow G$ sol. 5.24
  • $|G|=2n+1\Rightarrow G$ sol. (Feit-Thompson) (nor proved)
• Simple Groups
  • $G$ sim. if $G\ne H\ne \{e\}\Rightarrow H/⊴G$ i.e. there’s no non-trivial norm.div. 5.25
    • $\mathbb{Z}/p\mathbb{Z}$ sim.
  • $A_5$ sim. 5.26
    • $n\ge 5\Rightarrow S_n$ not sol
    • $A_5$ is smallest s.t. (i) not ab., (ii) sim.

Polynomials with $S_n$ as Galois Group

• $Gal(f)\cong S_n$ for $n\ge 5$ $\Rightarrow f$ not sol. with rad.
  • since $Gal(f)$ permutes roots of $f$, $Gal(f)\subseteq S_n$ subrgr.
  • since $n\ge 5$, $S_n$ not sol.
• for $n\ge 2$: 6.1
  • (i) $S_n=⟨(12),(\mathrm{123...}n)⟩$
  • (ii) if ex. $H$ s.t. (iia) $H\subseteq S_n$ trans.subgr. and contains a transp. and $(n-1)$-cycle, then $H=S_n$
  • for $1\ge i,j\ge n$, $i\ne j$ ex. $\sigma \in H$ s.t. $\sigma (i)=j$ then $H$ trans subgr.
• $p\in \mathbb{P}$, $f\in \mathbb{Q}[X]$ irr. s.t. $deg(f)=p$ and has exactly 2 not-real roots, then $Gal(f)\cong S_p$ 6.2
• $Gal(X^{p^n}-X)=Gal({\mathbb{F}}_{p^n}/{\mathbb{F}}_p)$ cycl.,
  • if $m||Gal(f)|$, ex. an elem. of order $m$.
    • for $f\in {\mathbb{F}}_p[X]$ irr., ${\exists }_{\sigma \in Gal(f)}ord(\sigma )=n$
  • $|Gal({\mathbb{F}}_{p^n})|=n$, ${\mathbb{F}}_{p^n}/{\mathbb{F}}_p$ is simple and separable (since ${\mathbb{F}}_p$ is finite)
• ${\mathbb{F}}_{p^n}/{\mathbb{F}}_p$ fin. sep., if ${\mathbb{F}}_{p^n}={\mathbb{F}}_p[\alpha ]$, $f\in {\mathbb{F}}_p[X]$ is min. of $\alpha$, so $f$ irr. of deg. $n$.
  • ${\forall }_{p\in \mathbb{P}}{\forall }_{n\ge 1}{\exists }_{f\in {\mathbb{F}}_p[X]}f$ irr.$\wedge deg(f)=n$.
• $A$ fac.ring., $\mathfrak{p}\subseteq A$ pr.ideal, $\overline{A}=A/\mathfrak{p}$, $K=Q(A)$, $k=Q(\overline{A})$. For $f\in A[X]$ primitive, let $\overline{f}\in \overline{A}$ be $f$ $mod(p)$, then: (i) $\overline{f}$ sep $\Rightarrow f$ sep., (ii) $Gal(\overline{f})\subseteq Gal(f)$ subgr. 6.4 (not proved)
• ${\forall }_{n\ge 1}{\exists }_{f\in \mathbb{Q}[X]}deg(f)=n\wedge Gal(f)\cong S_n$

Examples

For Galois Theory I decided to write down some of the examples too, I collect them here:

Galois Examples

• (a). $L=\mathbb{Q}(\sqrt{p})$, $K=\mathbb{Q}$, since $f=x^2-p\in \mathbb{Q}[X]$ hence norm., $char(\mathbb{Q})=0\Rightarrow L/K$ sep. Also $[L:K]=2\Rightarrow |Gal(L/K)|=2\Rightarrow Gal(L/K)\cong {\mathbb{Z}}_2$, elem. in $Gal(L/K)$ premute roots ($\pm \sqrt{p}$) of $f$, i.e. $\sqrt{p}\mapsto \sqrt{p}$ and $\sqrt{p}\mapsto -\sqrt{p}$ (uniq. def. by $\alpha$s)
• (b). $f=(x^2-2)(x^2-3)\in \mathbb{Q}[X]$ has zer. kö. $\mathbb{Q}[\sqrt{2},\sqrt{3}]=L$, $[L:\mathbb{Q}]=4$. $L/K$ is gal., $\Rightarrow |Gal(L/\mathbb{Q})|=[L:\mathbb{Q}]=4$. Elements perm. roots, (22:33, 2-2:33, 22:3-3, 2-2:3-3), $Gal(L/K)=\{id,\sigma ,\tau ,\sigma \tau \}\cong {\mathbb{Z}}_2\times {\mathbb{Z}}_2$.
• (c) $f=x^3-2\in \mathbb{Q}[X]$, let $\omega$ s.t. ${\omega }^2+\omega +1=0$, roots of $f$ is $\{\sqrt[3]{2},\sqrt[3]{2}\omega ,\sqrt[3]{2}{\omega }^2\}$, $L=\mathbb{Q}(\sqrt[3]{2},\omega )$ is zer.kö. of $f$, $[L:\mathbb{Q}]=6$, $Gal(L/\mathbb{Q})$ prem. the roots, hence $Gal(L/\mathbb{Q})=S_3$ (smallest non-abelian), gal. gr. must not be abelian. The extension between, $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$, $\mathbb{Q}(\sqrt[3]{2}\subset \mathbb{R}$, $\sigma (\sqrt[3]{2})\in \mathbb{R}$ hence $\sigma (\sqrt[3]{2})=\sqrt[3]{2}$, hence $|Gal(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})|=1\ne 3=[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]$, hence it is not a gal. ext.
• $p\in \mathbb{P}$, $r,n\in \mathbb{N}\setminus \{0\}$, $q=p^r$, $q=p^{rn}$, ${\mathbb{F}}_p\subseteq {\mathbb{F}}_q\subseteq {\mathbb{F}}_q^{\prime }\subseteq \overline{{\mathbb{F}}_p}$ and ${\mathbb{F}}_q\subseteq {\mathbb{F}}_{q^{\prime }}$ sep. ($\Leftarrow$ finite), (norm. $\Leftarrow {\mathbb{F}}_q^{\prime }$ zer.kö. for $X^q-X\in {\mathbb{F}}_q[X]$) $\Rightarrow |Gal({\mathbb{F}}_{q^{\prime }}/{\mathbb{F}}_q)|=n$, see script for conclusion.

Galois Correspondence Examples

• (a) $L=\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$ is gal. and $Gal(L/\mathbb{Q})=\{id,\sigma ,\tau ,\sigma \tau \}$ as in Galois Examples (b). For the other examples, see the script.

Example with $\zeta$

• $n=12$, $\zeta =\sqrt[12]{u}$ prim. for $u\in {\mathbb{Q}}^{\times }$, $\phi (12)=\phi (3\cdot 2^2)=\phi (3)\phi (2^2)=2\cdot 2=4$. ${\mathbb{Z}}^{\times }=\{1,5,7,11\}$, $Gal(\mathbb{Q}(\zeta )/\mathbb{Q})=\{{\sigma }_m:m\in {\mathbb{Z}}^{\times }\}=:G$ for ${\sigma }_m:\zeta \mapsto {\zeta }^m$. ${\sigma }_{11}={\sigma }_3\circ {\sigma }_7$. Also note $G\cong \mathbb{Z}/(2)\times \mathbb{Z}/(2)$. Also $\mathbb{Q}(\zeta )/\mathbb{Q}$ has three non trivial fields in between: $\mathbb{Q}(\zeta {)}^{⟨{\sigma }_5⟩}$, ${\zeta }^3\sqrt[4]{u}=\pm i$, hence $\mathbb{Q}(\zeta {)}^{⟨{\sigma }_5⟩}=\mathbb{Q}(i)$. The second field is: ${\mathbb{Q}}^{⟨{\sigma }_7⟩}$, ${\zeta }^2=\sqrt[6]{u}=\frac{-1\pm \sqrt{-3}}{2}$, then $\mathbb{Q}(\zeta {)}^{⟨{\sigma }_7⟩}=\sqrt{-3}$. $\mathbb{Q}(\zeta {)}^{⟨{\sigma }_{11}⟩}$, similarly ${\mathbb{Q}}^{⟨{\sigma }_{1}1⟩}=\mathbb{Q}(\sqrt{3})$.