# Abstract Algebra, Judson

> Source: https://ollybritton.com/notes/textbooks/abstract-algebra-theory-and-applications/ · Updated: 2024-11-09

![abstract-algebra.jpg](https://ollybritton.com/assets/attachments/img/abstract-algebra.jpg)

A [free-to-read online](http://abstract.ups.edu/aata/aata-toc.html) textbook about abstract algebra, written by Thomas W. Judson.

- Relevant university courses:
	- I read to help with:
	- [Course - Galois Theory HT25](https://ollybritton.com/notes/uni/part-b/ht25/galois-theory/)
	- But also relevant for:
	- [Course - Linear Algebra MT22](https://ollybritton.com/notes/uni/prelims/mt22/linear-algebra/)
	- [Course - Linear Algebra II HT23](https://ollybritton.com/notes/uni/prelims/ht23/linear-algebra/)
	- [Course - Groups and Group Actions HT23](https://ollybritton.com/notes/uni/prelims/ht23/groups/)
	- [Course - Groups and Group Actions TT23](https://ollybritton.com/notes/uni/prelims/tt23/groups/)
	- [Course - Linear Algebra MT23](https://ollybritton.com/notes/uni/part-a/mt23/linear-algebra/)
	- [Course - Rings and Modules HT24](https://ollybritton.com/notes/uni/part-a/ht24/rings-and-modules/)

## Table of Contents

### Chapter 10: Normal Subgroups and Factor Groups
- [Factor Groups and Normal Subgroups](http://abstract.ups.edu/aata/normal-section-factor-groups.html)
	- Definition: $H$ is a normal subgroup of $G$ if $gH = Hg$ for all $g \in G$.
	- Example: Every subgroup of an abelian group is normal.
	- Example: The subgroup $\{(1), (12)\}$ of $S_3$ is not normal since $(123)H \ne H(123)$
	- Theorem: The following are equivalent
		- $N$ is a normal subgroup of $G$
		- For all $g \in G$, $g N g^{-1} \subset N$
		- For all $g \in G$, $g N g^{-1} = N$
	- Theorem: If $N \trianglelefteq G$, then the cosets of $N$ in $G$ form a group $G/N$ of order $[G:N]$.
	- Example: $S_3 / A_3 \cong C_2$
	- Example: $\mathbb Z / n \mathbb Z$
	- Example: $D_n / R_n \cong C_2$
- [The Simplicity of the Alternating Group](http://abstract.ups.edu/aata/normal-section-simplicity-of-an.html)
	- F: Definition: Groups with no nontrivial normal subgroups are called simple groups.
	- Example: $\mathbb Z / p\mathbb Z$ is simple for $p$ prime, trivially simple since no nontrivial subgroups, let alone nontrivial *normal* subgroups
	- Lemma: $A_n$ is generated by $3$-cycles for $n \ge 3$
	- F: Lemma: If $N \trianglelefteq A_n$ for $n \ge 3$ and $N$ contains a three cycle, then $N = A_n$.
	- F: Lemma: For $n \ge 5$, every nontrivial normal subgroup $N$ of $A_n$ contains a 3-cycle
	- F: Theorem: The alternating group $A_n$ is simple for $n \ge 5$

### Chapter 13: The Structure of Groups
- [Finite Abelian Groups](http://abstract.ups.edu/aata/struct-section-finite-abelian-groups.html)
	- Definition: The group generated by $\{g_i : i \in I\}$ is the smallest group containing all elements, and the $g_i$s are said to be the generators.
	- Definition: If $G = \langle g_1, \ldots, g_n \rangle$ finite, then $G$ is finitely generated
	- Example: All finite groups are finitely generated, but e.g. $\mathbb Z \times \mathbb Z_n$ is an infinite group but finitely generated
	- Example: $\mathbb Q$ is not finitely generated
	- Proposition: If $G = \langle g_i : i \in I\rangle$, then every element can be written as a product of powers of group elements
	- Definition: A group $G$ is a $p$-group if the order of every element of $G$ is a power of $p$.
	- Theorem: Fundamental Theorem of Finite Abelian Groups: Every finite abelian group $G$ is isomorphic to a direct product of cyclic groups of prime-power order.
	- Lemma: If $G$ is a finite abelian group of order $n$, and $p$ is a prime such that $p \mid n$, then $G$ contains an element of order $p$
	- Lemma: A finite abelian group is a $p$-group iff its order is a power of $p$
	- Lemma: Let $G$ be a finite abelian group of order $n = \prod p_i^{\alpha_i}$ where $p_i$ are prime and $\alpha_i$ positive integers. Then $G$ is the internal direct product of subgroups $G_1, \ldots, G_k$ where $G_i$ is the subgroup of $G$ consisting of all elements of order $p_i^r$ for some integer $r$.
	- Lemma: Let $G$ be a finite abelian $p$-group and suppose $g \in G$ has maximal order. Then $G$ is isomorphic to $\langle g \rangle \times H$ for some subgroup $H$ of $G$.
	- Theorem: Fundamental Theorem of Finitely Generated Abelian Groups: Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups of the form $\mathbb Z_{p_1}^{\alpha_1} \times \mathbb Z_{p_2}^{\alpha_2} \times \cdots \times \mathbb Z_{p_n}^{\alpha_n} \times \mathbb Z \times \cdots \times \mathbb Z$
- [Solvable Groups](http://abstract.ups.edu/aata/struct-section-solvable-groups.html)
	- F: Definition: A subnormal series of a group $G$ is a finite sequence of subgroups $G = H_n \supset H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{e\}$ where $H_i$ is a normal subgroup of $H_{i+1}$.
	- F: Definition: The series is normal if each subgroup  $H_i$ is normal in $G$
	- Definition: The length of a normal series is the number of proper inclusions
	- F: Example: Any series of subgroups of an abelian group is a normal series, e.g:
		- $\mathbb Z \supset 9\mathbb Z \supset 45 \mathbb Z \supset 180 \mathbb Z \supset \{0\}$
		- $\mathbb Z_{24} \supset \langle 2 \rangle \supset \langle 6 \rangle \supset \langle 12 \rangle \supset \{0\}$
	- Example: $D_4 \supset \{(1), (12)(34), (13)(24), (14)(23)\} \supset \{(1),(12)(34)\} \supset \{e\}$ is a subnormal series that is not normal
	- Definition: A subnormal series $\{K_j\}$ is a refinement of a subnormal series $\{H_i\}$ if $\{H_i\} \subset \{K_j\}$, i.e. each $H_i$ appears in $\{K_j\}$
	- Example: $\mathbb Z \supset 3 \mathbb Z \supset 9 \mathbb Z \supset \mathbb 45 \supset 90 \mathbb Z \supset 180 \mathbb Z \supset \{0\}$ is a refinement of $\mathbb Z \supset 9 \mathbb Z \supset 45 \mathbb Z \supset 180\mathbb Z \supset \{0\}$
	- Remark: It's useful to study the factor groups $H_{i+1} / H_i$
	- F: Definition: Two subnormal series $\{H_i\}$ and $\{K_j\}$ of a group $G$ are isomorphic if there is a one-to-one correspondence between $\{H_{i+1} / H_i\}$ and $\{K_{j+1} / K_j\}$.
	- Example:
		- $\mathbb Z_{60} \supset \langle 3 \rangle \supset \langle 15 \rangle \supset \{0\}$
		- $\mathbb Z_{60} \supset \langle 4 \rangle \supset \langle 20 \rangle \supset \{0\}$
		- These are isomorphic since:
		- $\mathbb Z_{60} / \langle 3 \rangle \cong \langle 20 \rangle / \langle 0 \rangle \cong \mathbb Z_3$
		- $\langle 3 \rangle / \langle 15 \rangle \cong \langle 4 \rangle / \langle 20 \rangle \cong \mathbb Z_5$
		- $\langle 15 \rangle / \{0\} \cong \mathbb Z_{60} / \langle 4 \rangle \cong \mathbb Z_4$
	- F: Definition: A subnormal series $\{H_i\}$ of a group $G$ is a composition series if all the factor groups are simple, equivalently none of the factor groups contains a normal subgroup
	- F: Definition: A normal series $\{H_i\}$ of $G$ is called a principal series if all the factor groups are simple
	- Example:
		- $\mathbb Z_{60}$ has a composition series $\mathbb Z_{60} \supset \langle 3 \rangle \supset \langle 15 \rangle \supset \langle 30 \rangle \supset \{0\}$
		- The factor groups are:
		- $\mathbb Z_{60} / \langle 3 \rangle \cong \mathbb Z_3$
		- $\langle 3 \rangle / \langle 15 \rangle \cong \mathbb Z_5$
		- $\langle 15 \rangle / \langle 30 \rangle \cong \mathbb Z_2$
		- $\langle 30 \rangle / \{0\} \cong \mathbb Z_2$
		- Since $\mathbb Z_{60}$ is abelian, this is automatically a principal series.
	- F: Example: For $n \ge 5$, $S_n \supset A_n \supset \{(1)\}$ is a composition series for $S_n$ as $S_n / A_n \cong \mathbb Z_2$ and $A_n$ is simple.
	- Example: Not every group has a principal series. Suppose $\mathbb Z = H_n \supset H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{0\}$ is a subnormal series for the integers under addition. Then $H_1$ is of the form $k \mathbb Z$ for $k \in \mathbb N$. But then $H_1 / H_0$ is an infinite cyclic group with many nontrivial proper normal subgroups
	- F: Theorem: Jordan-Hölder: Any two composition series of $G$ are isomorphic.
	- F: Definition: A group $G$ is solvable if it has a subnormal series $\{H_i\}$ such that all the factor groups $H_{i+1}/H_i$ are abelian.
	- Example:
		- F: $S_4$ is solvable since $S_4 \supset A_4 \supset \{(1), (12)(34), (13)(24), (14)(23)\} \supset \{(1)\}$ has abelian factor groups
		- F: $S_5$ is not solvable since for $n \ge 5$, $S_n \supset A_n \supset \{(1)\}$ is a composition series for $S_n$ but $A_n$ is not abelian
	

### Chapter 21: Fields
- [Extension Fields](http://abstract.ups.edu/aata/fields-section-extension-fields.html)
	- Definition of an extension field and a base field
	- Example of $\mathbb Q[\sqrt{2} + \sqrt{3}]$ being extension field of $\mathbb Q[\sqrt 2]$
	- Example of extension of $\mathbb Z_2$ to add a root of $1 + x + x^2$
	- "Fundamental theorem of Field Theory": For any nonconstant polynomial over a field $F$, there exists an extension field $E$ such that there's some $a \in E$ where $p(\alpha) = 0$.
	- Example of another extension of $\mathbb Z_2$ to add roots for $1 + x^4 + x^5$
	- Definition of algebraic element: $\alpha$ is algebraic over $F$ if you can find a polynomial over $F$ that annihiliates
	- Definition of transcendental: not algebraic
	- Definition of algebraic extension $E / F$: Every element of $E$ is algebraic over $F$
	- Definition of $F(\alpha_1, \ldots, \alpha_n)$ as smallest field containing all $\alpha_i$
	- F: Definition of $E/F$ being a simple extension of $F$: $E = F(\alpha)$ for some $\alpha$
	- Example that $\pi$ and $e$ are algebraic over the reals, but nontrivial to show they are transcendental over the rationals
	- Definition of an "algebraic number": algebraic over $\mathbb Q$
	- Definition of "transcendental": not algebraic over $\mathbb Q$
	- Example showing $\sqrt{2 + \sqrt 3}$ is algebraic over $\mathbb Q$
	- Theorem: If $E/F$ is a field extension and $\alpha \in E$, then $\alpha$ is transcendental over $F$ iff $F(\alpha)$ is isomorphic to $F(x)$, the field of fractions for $F[x]$.
	- Theorem: existence of a unique minimal polynomial for arbitrary extension fields
	- Example: $x^2 - 2$ is the minimal polynomial of $\sqrt 2$
	- Proposition: $F(\alpha) \cong F[x]/\langle m_\alpha(x) \rangle$
	- Theorem: In a simple extension $F(\alpha) : F$ where $\alpha$ is algebraic, every element is a linear combination of powers of $\alpha$
	- Example: construction of real numbers as $\mathbb R[x] / \langle x^2 + 1\rangle$
	- Definition of a finite extension: $[E : F] = n$
	- Theorem: every finite extension $E/F$ is an algebraic extension
	- Remark: Converse is not true, i.e. not every algebraic extension is a finite extension (take algebraic completion of the rationals)
	- Theorem: tower law
	- Corollary: you can find the degree of a sequence of field extensions by multiplying up the intermediate ones
	- Corollary: Suppose $E/F$ is an extension and $\alpha \in E$ is algebraic with minimal polynomial $m_\alpha$. If $\beta \in F(\alpha)$ and has minimal polynomial $m_\beta$, then $\deg m_\beta \mid \deg m_\alpha$.
	- Example: $\mathbb Q(\sqrt 3, \sqrt 5) = \mathbb Q(\sqrt 3 + \sqrt 5)$ 
	- Example: $\mathbb Q(\sqrt[6]{5} i) = \mathbb Q(\sqrt[3]{5}, \sqrt 5 i)$ 
	- Theorem: several equivalent statements
		- $E$ is a finite extension of $F$
		- There exists a finite number of algebraic elements $\alpha_1, \ldots, \alpha_n \in E$ such that $E = F(\alpha_1, \ldots, \alpha_n)$.
		- There exists a sequence of fields $E = F(\alpha_1, \ldots, \alpha_n) \supset F(\alpha_1, \ldots, \alpha_{n-1}) \supset \cdots \supset F(\alpha_1) \supset F$
	- Theorem: If $E/F$, then the elements of $E$ that are algebraic over $F$ form a field
	- Corollary: The set of all algebraic numbers forms a field
	- Definition: Algebraic closure of a field $F$ contains all elements that are algebraic over $F$
	- Theorem: $F$ is algebraically closed iff every nonconstant polynomial in $F$ factors into linear factors over $F[x]$
	- Definition: A field $F$ is algebraically closed if every nonconstant polynomial over $F$ has a root in $F$
	- Corollary: An algebraically closed field $F$ has no proper algebraic extension $E$
	- Theorem: Every $F$ has a unique algebraic closure (difficult to prove)
- [Splitting Fields](http://abstract.ups.edu/aata/fields-section-splitting-fields.html)
	- Definition: $E/F$ is a splitting field of $p(x)$ if there exist elements $\alpha_1, \ldots, \alpha_n$ in $E$ such that $E = F(\alpha_1, \ldots, \alpha_n)$ and $p(x) = (x - \alpha_1) \cdots (x - \alpha_n)$.
	- Definition: A polynomial $p(x) \in F[x]$ splits in $E$ if it is the product of linear factors in $E[x]$.
	- Example: $p(x) = x^4 + 2x^2 - 8$ has splitting field $\mathbb Q[\sqrt 2, i]$
	- Example: $p(x) = x^3 - 2$ has a root in the field $\mathbb Q[\sqrt[3]{3}]$, but this is not a splitting field since it doesn't contain the complex roots
	- F: Theorem: Every $p(x) \in F[x]$ nonconstant has a splitting field $E$
	- Lemma: If $\phi : E \to F$ is a field isomorphism, $K$ is an extension field of $E$, $\alpha \in K$ is algebraic over $E$ with minimal polynomial $m_\alpha$, $L$ is an extension field of $F$ such that $\beta$ is the root of the polynomial in $F[x]$ obtained from $p(x)$ under the image of $\phi$. Then $\phi$ extends to a unique isomorphism $\overline \phi : E(\alpha) \to F(\beta)$ such that $\overline \phi(\alpha) = \beta$ and $\overline \phi$ agrees with $\phi$ on $E$
	- Theorem: Field extensions are unique, sort of: Let $\phi : E \to F$ be an isomorphism of fields, $p(x)$ is a nonconstant polynomial in $E[x]$ and $q(x)$ is corresponding polynomial in $F[x]$ under the isomoprhism. If $k$ is a splitting field of $p(x)$ and $L$ is a splitting field of $q(x)$, then $\phi$ extends to an isomorphism $\psi : K \to L$.
- [Geometric Constructions](http://abstract.ups.edu/aata/fields-section-constructions.html)
- [Reading Questions](http://abstract.ups.edu/aata/reading-questions-21.html)
- [Exercises](http://abstract.ups.edu/aata/fields-exercises.html)
- [References and Suggested Readings](http://abstract.ups.edu/aata/fields-references.html)
- [Sage](http://abstract.ups.edu/aata/fields-sage.html)
- [Sage Exercises](http://abstract.ups.edu/aata/fields-sage-exercises.html)

### Chapter 22: Finite Fields
- [Structure of a Finite Field](http://abstract.ups.edu/aata/finite-section-field.html)
	- Proposition: Finite fields have prime characteristic
	- Proposition: If $F$ is a finite field of characteristic $p$, then order of $F$ is $p^n$ for some $n \in \mathbb N$.
	- Lemma: Freshman's dream
	- Definition: A polynomial $f(x) \in F[x]$ of degree $n$ is separable if it has $n$ distnct roots in splitting field of $f(x)$
		- ⚠️: This is not the same definition [Course - Galois Theory HT25](https://ollybritton.com/notes/uni/part-b/ht25/galois-theory/) gives.
	- Definition: An extension $E$ of $F$ is separable extension of $F$ if every element in $E$ is the root of a separable polynomial in $F[x]$
	- Example: $x^2 - 2$ is separable over $\mathbb Q$
	- Example: $\mathbb Q[\sqrt 2]$ is a separable extension of $\mathbb Q$ (check every element is the root of a separable polynomial)
	- Definition: Derivative of a polynomial
	- F: Lemma: If $F$ is a field and $f(x) \in F[x]$, then $f$ is a separable iff $f(x)$ and $f'(x)$ are relatively prime
	- F: Theorem: For every prime $p$ and every positive integer $n$, there exists a finite field with $p^n$ elements. Furthermore, any field of order $p^n$ is isomorphic to the splitting field of $x^{p^n}  - x$ over $\mathbb Z_p$
	- Definition: The (unique up to isomorphism) field with $p^n$ elements is called the Galois field of order $p^n$, denoted $\text{GF}(p^n)$.
	- Theorem: Every subfield of $\text{GF}(p^n)$ has $p^m$ elements where $m \mid n$, and if $m \mid n$ for $m > 0$, there exists a unique subfield of $GF(p^n)$ isomorphic to $\text{GF}(p^m)$.
	- Example: subfields of $\text{GF}(p^{24})$
	- Theorem: If $F$ is a field and $F^\ast$ is the multiplicative group of nonzero elements then if $G$ is a finite subgroup of $F^\ast$, $G$ is cyclic
	- Corollary: Multiplicative group of all nonzero elements of a finite field is cyclic
	- Corollary: Every finite extension $E$ of a finite field $F$ is a simple extension of $F$
	- Example: $\text{GF}(2^4)$ is isomorphic $\mathbb Z_2 / \langle 1 + x + x^4 \rangle$
- [Polynomial Codes](http://abstract.ups.edu/aata/finite-section-poly-codes.html)
- [Reading Questions](http://abstract.ups.edu/aata/reading-questions-22.html)
- [Exercises](http://abstract.ups.edu/aata/finite-exercises.html)
- [Additional Exercises: Error Correction for BCH Codes](http://abstract.ups.edu/aata/finite-exercises-bch-codes.html)
- [References and Suggested Readings](http://abstract.ups.edu/aata/finite-references.html)
- [Sage](http://abstract.ups.edu/aata/finite-sage.html)
- [Sage Exercises](http://abstract.ups.edu/aata/finite-sage-exercises.html)

### Chapter 23: Galois Theory
- [Field Automorphisms](http://abstract.ups.edu/aata/galois-section-field-automorphisms.html)
	- Proposition: The set of all automorphisms of a field $F$ is a group under composition.
	- Proposition: The set of all $F$-linear automorphisms of a field extension $E/F$ of a group under composition.
	- (the $F$-linear automorphisms of a field extension $E/F$ are the automorphisms of $E$ which fix $F$ elementwise)
	- Definition: If $E/F$ is a field extension, define $\text{Gal}(E/F)$ as the set of all $F$-linear automorphisms of the field
	- Definition: If $f(x)$ is a polynomial in $F[x]$ and $E$ is the splitting field of $f(x)$ over $F$, then the Galois group of $f$ is defined as $\text{Gal}(E/F)$.
	- Example: complex conjugation is in $\text{Gal}(\mathbb C / \mathbb R)$.
	- Example: $\mathbb Q$-linear automorphisms of $\mathbb Q[\sqrt 3, \sqrt 5]$ give a group of size $4$
	- F: Proposition: If $E/F$ is a field of extension of $f(x) \in F[x]$, then any element of $\text{Gal}(E/F)$ defines a permutation of the roots of $f$ that lie in $E$
	- F: Proposition: Any permutation of the roots of $f$ that lie in $E$ is also an element of $\text{Gal}(E/F)$
	- Definition: If $E/F$ is an algebraic extension, two elements $\alpha, \beta \in E$ are said to be conjugate if they have the same minimal polynomial.
	- Example: $\sqrt 2$ and $-\sqrt 2$ are conjugate over $\mathbb Q$ since they're both roots of $x^2 - 2$.
	- Proposition: If $\alpha$ and $\beta$ are conjugate over $F$, then there exists an isomorphism $\sigma : F(\alpha) \to F(\beta)$ such that $\sigma$ is the identity when restricted to $F$
	- F: Theorem: Let $f(x)$ be a polynomial in $F[x]$ and suppose that $E$ is the splitting field for $f(x)$ over $F$. If $f(x)$ has no repeated roots, then $|\text{Gal}(E/F)| = [E : F]$.
	- F: Corollary: Let $F$ be a **finite field** with a finite extension $E$ such that $[E : F] = k$. Then $\text{Gal}(E/F)$ is cyclic of order $k$.
	- Example: $\text{Gal}(\mathbb Q[\sqrt 3, \sqrt 5] / \mathbb Q) \cong \mathbb Z_2 \times \mathbb Z_2$
	- Example: $\text{Gal}(\mathbb Q[\omega] : \mathbb Q) = \mathbb Z_4$ where $\omega$ is a root of $f(x) = x^4 + x^3 + x^2 + x + 1$.
	- Definition: Multiplicity of a root
	- Definition: Simple root is a root with multiplicity 1
	- Proposition: Let $f(x)$ be an irreducible polynomial over $F$. If the characteristic of $F$ is $0$, then $f(x)$ is separable. If the characteristic of $F$ is $p$ and $f(x) \ne g(x^p)$ for some $g(x)$ in $F[x]$, then $f(x)$ is also separable
	- Definition: $\alpha \in E$ is a primitive element of the field extension $E/F$ if $E = F(\alpha)$.
	- F: Theorem: Primitive element theorem: Let $E$ be a finite separable extension of a field $F$, then there exists an $\alpha \in E$ such that $E = F(\alpha)$.
- [The Fundamental Theorem](http://abstract.ups.edu/aata/galois-section-fund-theorem-galois-theory.html)
	- Proposition: Let $S = \{\sigma_i \mid i \in I\}$ be a collection of automorphisms of $F$. Then $F^{S} = \{a \in F \mid \sigma_i(a) = a \text{ for all } \sigma_i \in S\}$ is a subfield of $F$
	- Corollary: If $F$ is a field and $G$ is a subgroup of $\text{Aut}(G)$, then $F^G = \{\alpha \in F \mid \sigma(a) = a \text{ for all } \sigma \in G\}$ is a subfield of $F$.
	- Definition: The subfield $F^S$ is called the fixed field of $S$.
	- Example: If $\sigma : \mathbb Q(\sqrt 3, \sqrt 5) \to \mathbb Q(\sqrt 3, \sqrt 5)$ is the automorphism that maps $\sqrt 3$ to $-\sqrt 3$, then $\mathbb Q(\sqrt 3, \sqrt 5)^{\langle\sigma\rangle} = \mathbb Q[\sqrt 5]$, i.e. the subfield left alone
	- Proposition: Let $E$ be a splitting field of $F$ of a separable polynomial. Then $E^{\text{Gal}(E/F)} = F$.
	- Lemma: $E$ is a field. Let $G$ be a finite group of automorphisms of $E$ and let $F = E^G$. Then $[E : F] \le |G|$. (Why important?)
	- Definition: If $E/F$ is an algebraic extension and every irreducible polynomial in $F[x]$ with a root in $E$ has all of its roots in $E$, then $E$ is called a normal extension of $F$. In other words, every irreducible polynomial in $F[x]$ containing a root in $E$ is the product of linear factors in $E[x]$.
	- Theorem: $E/F$ is a finite extension. The following are equivalent:
		- $E$ is a finite, normal, separable extension
		- $E$ is a splitting field over $F$ of a separable polynomial
		- $F = E^G$ for some finite group $G$ of automorphisms of $E$
	- Corollary: If $K/F$ is a field extension such that $F = K^G$ for some finite group of automorphisms $G$ of $K$, then $G = \text{Gal}(K/F)$.
	- Example: Lattice of subgroups of $\text{Gal}(\mathbb Q(\sqrt 3, \sqrt 5)/\mathbb Q)$ and the lattice of subfields of $\mathbb Q(\sqrt 3, \sqrt 5)$.
	- Theorem: **Fundamental Theorem of Galois Theory**: Let $F$ be a finite field or a field of characteristic zero. If $E$ is a finite normal extension of $F$ with Galois group $\text{Gal}(E/F)$, then the following holds:
		- The map $K \mapsto \text{Gal}(E/K)$ is a bijection of subfields $K$ of $E$ containing $F$ with the subgroups of $\text{Gal}(E/F)$.
		- If $F \subset K \subset E$, then $[E : K] = |G(E/K)|$ and $[K : F] = [G(E/F) : G(E/K)]$
		- $F \subset K \subset L \subset E$ iff $\{\text{id}\} \subset \text{Gal}(E/L) \subset \text{Gal}(E/K) \subset \text{Gal}(E/F)$.
		- $K$ is a normal extension of $F$ iff $\text{Gal}(E/K)$ is a normal subgroup of $\text{Gal}(E/F)$. In this case, $\text{Gal}(K/F) \cong \text{Gal}(E/F) / \text{Gal}(E/K)$.
	- Example: Lattice of subgroups of Galois group of $f(x) = x^4 - 2$ compared to the lattice of field extensions of the splitting field of $x^4 - 2$.
- [Applications](http://abstract.ups.edu/aata/galois-section-applications.html)
	- Definition: An extension $E/F$ is called an extension by radicals if there is a chain of subfields $F = F_0 \subset F_1 \cdots F_r = E$ for $i = 1, 2, \ldots, r$ and $F_i = F_{i-1}(\alpha_i)$ and $\alpha_i^n \in F_{i-1}$.
	- Definition: A polynomial $f(x)$ is solvable by radicals over $F$ if the splitting field $K$ of $f(x)$ over $F$ is contained in an extension of $F$ by radicals
	- Example: $x^n - 1$ is solvable by radicals over $\mathbb Q$ for any $n$
	- Lemma: Let $F$ be a field of characteristic zero and $E$ be the splitting field of $x^n - a$ over $F$ with $a \in F$. Then $\text{Gal}(E/F)$ is a solvable group.
	- Lemma: Let $F$ be a field of characteristic zero and let $F = F_0 \subset F_1 \subset \cdots \subset F_r = E$ be a radical extension of $F$. Then there exists a normal radical extension $F = K_0 \subset K_1 \subset \cdots \subset K_r = K$ such that $K$ contains $E$ and $K_i$ and $K_i$ is a normal extension of $K_{i-1}$.
	- Theorem: Let $f(x)$ be in $F[x]$ where the characteristic of $F$ is zero. Then $f(x)$ is solvable by radicals iff the Galois group of $f(x)$ over $F$ is solvable.
	- Lemma: If $p$ is a prime, then any subgroup of $S_p$ that contains a transposition and a cycle of length $p$ must be all of $S_p$.
	- Example: $f(x) = x^5 - 6x^3 - 27x - 3 \in \mathbb Q[x]$ has Galois group $S_5$. But $S_5$ is not solvable, so this polynomial is not solvable by radicals.
	- Theorem: **Fundamental Theorem of Algebra**: The field of complex numbers is algebraically closed, i.e. every polynomial $\mathbb C[x]$ has a root in $\mathbb C$.
- [Reading Questions](http://abstract.ups.edu/aata/reading-questions-23.html)
- [Exercises](http://abstract.ups.edu/aata/galois-exercises.html)
- [References and Suggested Readings](http://abstract.ups.edu/aata/galois-references.html)
- [Sage](http://abstract.ups.edu/aata/galois-sage.html)
- [Sage Exercises](http://abstract.ups.edu/aata/galois-sage-exercises.html)

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Olly Britton — https://ollybritton.com. Machine-readable index: https://ollybritton.com/llms.txt