Algebra Math Notes • Study Guide Abstract Algebra

Table of Contents Rings ...... 3 Rings ...... 3 Subrings and Ideals ...... 4 Homomorphisms ...... 4 Fraction Fields ...... 5 Integers ...... 5 Factoring ...... 7 Factoring ...... 7 UFDs, PIDs, and Euclidean Domains ...... 7 Algebraic Integers ...... 8 Quadratic Rings ...... 8 Gaussian Integers ...... 9 Factoring Ideals ...... 9 Ideal Classes ...... 10 Real Quadratic Rings ...... 11 Polynomials ...... 12 Polynomials ...... 12 ℞[푋] ...... 12 Irreducible Polynomials ...... 13 Cyclotomic Polynomials ...... 14 Varieties ...... 14 Nullstellensatz ...... 15 Fields...... 17 Fundamental Theorem of Algebra ...... 17 Algebraic Elements ...... 17 Degree of a Field Extension ...... 17 Finite Fields ...... 19 Modules ...... 21 Modules ...... 21 Structure Theorem ...... 22 Noetherian and Artinian Rings ...... 23 Application 1: Abelian Groups ...... 23 Application 2: Linear Operators ...... 24 Polynomial Rings in Several Variables ...... 25 Tensor Products ...... 25 Galois Theory ...... 27 Symmetric Polynomials ...... 27 Discriminant ...... 27 Galois Group ...... 27 Fixed Fields ...... 28 Galois Extensions and Splitting Fields ...... 28 Fundamental Theorem ...... 29 Roots of Unity ...... 29 Cubic Equations ...... 30 Quartic Equations ...... 30 Quintic Equations and the Impossibility Theorem ...... 31 Transcendence Theory ...... 32 Algebras ...... 33 Division Algebras ...... 33 References ...... 34

1 Rings 1-1 Rings

A ring 푅 is a set with the operations of addition (+) and multiplication (×) satisfying: 1. R is an abelian group 푅+ under addition. 2. Multiplication is associative. In a commutative ring, multiplication is commutative. A ring with identity has a multiplicative identity 1. 3. Distributive law: 푎 푏 + 푐 = 푎푏 + 푎푐, 푎 + 푏 푐 = 푎푐 + 푏푐 A left/ right zero divisor is an element 푎 ∈ 푅 so that there exists 푏 ≠ 0 with 푎푏 = 0, 푏푎 = 0, respectively. A commutative ring without zero divisors is an integral domain. A ring is a division ring if 푅 − {0} is a multiplicative group under multiplication (inverses exist). A division ring is a field if the multiplicative group is abelian. Ex. of Rings: ℞, 푅[푥] (the ring of polynomials in x, where R is a ring), 푅[푥1, … , 푥푛 ]

Group G

Abelian group

Ring R

Ring with Commutative identity ring

Division ring Integral domain

Unique Field F factorization domain (UFD)

Principal ideal domain (PID)

Euclidean domain

Note: A field is a UFD, though not a very interesting one.

Basic properties: 1. 푎0 = 0푎 = 0 2. 푎 −푏 = −푎 푏 = −(푎푏) 3. The multiplicative identity 1 is unique (if it exists). 4. A multiplicative inverse is unique if it exists. 5. If R is not the zero ring {0}, then 1 ≠ 0. 6. If R is an integral domain, the cancellation law holds. Conversely, if the left (right) cancellation law holds, then R has no left (right) zero divisors. 7. A field is a division ring, and any finite integral domain is a field.

From here on rings are assumed to be commutative with identity unless otherwise specified.

A unit is an element with a multiplicative inverse. The characteristic of a ring is the smallest 푛 > 0 so that 1 + ⋯ + 1 = 0. If this is never true, 푛 the characteristic is 0.

1-2 Subrings and Ideals

A subring is a subset of a ring that is closed under addition and multiplication. An ideal I is a subset of a ring satisfying: 1. I is a subgroup of 푅+. 2. If 푠 ∈ 퐼, 푟 ∈ 푅 then 푟푠 ∈ 퐼. Every linear combination of elements 푠푖 ∈ 퐼 with coefficients 푟푖 ∈ 푅 is in I.

The principal ideal 푎 = 푎푅 generated by 푎 is the ideal of multiples of 푎. The smallest ideal generated by 푎1, … , 푎푛 is 푎1, 푎2, … , 푎푛 = 푟1푎1 + ⋯ + 푟푛 푎푛 푟푖 ∈ 푅

Ideals and fields: 1. The only ideals of a field are the zero ideal and the unit ideal. 2. A ring with exactly two ideals is a field. 3. Every homomorphism from a field to a nonzero ring is injective.

A principal ideal domain (PID) is an integral domain where every ideal is principal. Ex. ℞ is a PID. If F is a field, 퐹[푥] (polynomials in x with coefficients in F) is a PID: all polynomials in the ideal are a multiple of the unique monic polynomial of lowest degree in the ideal.

A maximal ideal of R is an ideal strictly contained in R that is not contained in any other ideal.

1-3 Homomorphisms

A ring homomorphism 휑: 푅 → 푅′ is compatible with addition and multiplication: 1. 휑 푎 + 푏 = 휑 푎 + 휑(푏) 2. 휑 푎푏 = 휑 푎 휑(푏) 3. 휑 1 = 1 An isomorphism is a bijective homomorphism and an automorphism is an isomorphism from R to itself. The kernel is {푠 ∈ 푅|휑 푠 = 0}, and it is an ideal.

1-4 Quotient and Product Rings (11.4,6)

If I is an ideal 푅/퐼 is the quotient ring. The canonical map 휋: 푅 → 푅/퐼 sending 푎 → 푎 + 퐼 is a ring homomorphism that sends each element to its residue. If 퐼 = (푎1, … , 푎푛 ), the quotient ring is obtained by “killing” the 푎푖, i.e. imposing the relations 푎1, … , 푎푛 = 0.

Mapping Property: Let 푓: 푅 → 푅′ be a ring homomorphism with kernel K and let 퐼 ⊆ 퐾 be an ideal. Let 휋: 푅 → 푅/퐼 = 푅 be the canonical map. There is a unique homomorphism 푓: 푅 → 푅′ so that 푓휋 = 푓. 푓 푅 푅′ 휋 푓 푅 = 푅/퐼 First Isomorphism Theorem: If f is surjective and 퐼 = 퐾 then 푓 is an automorphism. Correspondence Theorem: If f is surjective with kernel K,  There is a bijective correspondence between ideals of 푅 and ideals of 푅′ that contain 퐾. A subgroup of G containing K is associated with its image.  If 퐼 corresponds to 퐼′, 푅/퐼 ≅ 푅′/퐼′.  Corollary: Introducing relations one at a time (killing 푎, then 푏 ) or together (killing 푎, 푏) gives the same ring. Useful in polynomial rings.

The product ring 푅 × 푅′ is the product set with componentwise addition and multiplication. 1. The additive identity is (0,0) and the multiplicative identity is (1,1). 2. The projections 휋 푥, 푥′ = 푥, 휋′ 푥, 푥′ = 푥′ are ring homomorphisms to 푅, 푅′. 3. 1,0 , (0,1) are idempotent elements- elements such that 푒2 = 1. Let e be an idempotent element of a ring S. 1. 푒′ = 1 − 푒 is idempotent, and 푒푒′ = 0. 2. 푒푆 is a ring (but not a subring of S unless e=1) with identity e, and multiplication by e is a ring homomorphism 푆 → 푒푆. 3. 푆 ≅ 푒푆 × 푒′푆.

1-5 Fraction Fields

Every integral domain can be embedded as a subring of its fraction field F. Its elements 푎 푎1 푎2 are 푎, 푏 ∈ 푅, 푏 ≠ 0 , where = ⇔ 푎1푏2 = 푎2푏1. Addition and multiplication are defined 푏 푏1 푏2 푎 푐 푎푑 +푏푐 푎 푐 푎푐 푎 as in arithmetic: + = , = , and = 푎. The field of fractions of the polynomial 푏 푑 푏푑 푏 푑 푏푑 1 ring 퐾[푥], K a field, is the field of rational functions 퐾(푥).

Mapping Property: Let R be an integral domain with field of fractions F, and let 휑: 푅 → 퐹′ be an injective homomorphism from 푅 to a field 퐹’. 휑 can be extended uniquely to an injective 푎 homomorphism Φ: 퐹 → 퐹′ by letting Φ = 휑 푎 휑 푏 −1. 푏

1-6 Integers

Peano’s Axioms for ℕ: 1. 1 ∈ ℕ 2. Successor function: The map 휎: ℕ → ℕ sends an integer to the next integer 푥′ = 휎(푥); 휎 1 = 2, 휎 2 = 3 … 휎 is injective, and 휎 푛 ≠ 1 for any 푛 (i.e. 1 is the first element). 3. Induction axiom: Let S be a subset of ℕ having the following properties: a. 1 ∈ 푆 b. If 푛 ∈ 푆 then 푛′ ∈ 푆. Then 푆 = ℕ. (Counting runs through all the natural numbers.)

Inductive/ Recursive Definition: By induction, if an object 퐶1 is defined, and a rule for determining 퐶푛′ = 퐶푛+1 from 퐶푛 is given, then the sequence 퐶푘 is determined uniquely.

Recursive definition of… 1. Addition: 푚 + 1 = 푚′, 푚 + 푛′ = (푚 + 푛)′ (i.e. 푚 + 푛 + 1 = 푚 + 푛 + 1) 2. Multiplication: 푚 × 1 = 푚, 푚 × 푛′ = 푚 × 푛 + 푚 (i.e. 푚 × 푛 + 1 = 푚 × 푛 + 푚) From these, the associative, distributive, and distributive laws can be verified. (“Peano playing”) The integers can be developed from ℕ by introducing an additive inverse for every element.

2 Factoring 2-1 Factoring

Vocabulary u is a unit if 푢 has a multiplicative inverse 푢 = (1) in R a divides b if 푏 = 푎푞 for some 푞 ∈ 푅 푏 ⊆ (푎) a properly divides b if 푏 = 푎푞 for some 푞 ∈ 푅 and 푏 ⊂ 푎 ⊂ 1 neither 푎 nor 푞 are units a and b are relatively 푎 and 푏 have no common divisor 푎 ∩ 푏 = 1 prime except units. a and b are associates if each divides the other, or if 푎 = 푏 푏 = 푢푎, u a unit a is irreducible if a is not a unit and has no 푎 ⊂ 1 , and there is no proper divisor (its only divisors principal ideal (푐) such are units and associates) that 푎 ⊂ 푐 ⊂ (1). p is a prime element if 푝 is not a unit, and whenever 푝 푎푏 ∈ 푝 ⇒ 푎 ∈ 푝 divides 푎푏, 푝 divides 푎 or 푝 or 푏 ∈ (푝) divides 푏.

Ex. Nonunique factorization: 2 ⋅ 3 = 6 = 1 + −5 1 − −5 in ℞ −5 .

2-2 UFDs, PIDs, and Euclidean Domains

A size function is a function 휎: 푅 → ℕ. An integral domain is an Euclidean domain if there is a norm 휎 such that division with remainder is possible:  For any 푎, 푏 ∈ 푅, 푎 ≠ 0, there exist 푞, 푟 ∈ 푅 so that 푏 = 푎푞 + 푟, and 푟 = 0 (in which case 푎 divides 푏) or 휎 푟 < 휎(푎). Examples: 1. For ℞, 휎 푎 = 푎 . 2. For 퐹[푥], 휎 푓 = deg 푓. 3. For ℞[푖], 휎 푎 = 푎 2. A characterization of the GCD: Let R be a UFD. A greatest common divisor 푑 of 푎, 푏 ∈ 푅 is 푑 such that If 푒|푎, 푏 then 푒|푑. If R is a PID, this is equivalent to 푅푑 = 푅푎 + 푅푏, and there exist 푝, 푞 ∈ 푅 so 푑 = 푝푎 + 푞푏 (Bezout). 푎, 푏 are relatively prime iff 푑 is a unit.

An integral domain is a unique factorization domain (UFD) if factoring terminates (stopping when all elements are irreducible), and the factorization is unique up to order and multiplication by units. The following are equivalent: 1. Factoring terminates. 2. R is Noetherian: it does not contain an infinite strictly increasing chain of principal ideals 푎1 ⊂ 푎2 ⊂ ⋯.

Class (Each Definition Reasons Properties includes the next) Ring Integral Ring with no zero Prime element domain divisors irreducible Unique Integral domain where For equivalence: If every Irreducible element factorization  Factoring terminates irreducible element prime, prime and there are two GCD exists (factor domain (UFD)  Factorization unique factorizations, an element on or equivalently, and look at common the left divides an element on prime divisors) every irreducible the right; they are associates element is prime. and can be canceled. Principal Ideal All ideals generated A PID is a UFD: Maximal ideals Domain (PID) by one element 1- Irreducible element prime: generated by 푅푑 = 푅푎 + 푅푏 ⇒ Bezout’s irreducible elements identity. Irreducible element prime since 푝 푎푏, 푝 ∤ 푎 ⇒ 1 = 푠푝 + 푡푎 ⇒ 푝 푏 = 푠푝푏 + 푡푎푏 2- Factoring terminates: If 푎1 ⊆ 푎2 ⊆ ⋯ then ⋃ 푎푖 = (푏) is a principal ideal; 푏 ∈ (푎푗 ) for some j. Euclidean Integral domain with A Euclidean Domain is a Euclidean algorithm domain norm compatible with UFD: works. (Given a, b, division with The element of smallest write 푎 = 푞푏 + 푟, remainder size in an ideal replace 푎 with 푏, 푏 generates it. with 푟.)

Ex. ℞, ℞ 푖 , 퐹[푥] are UFDs.

2-3 Algebraic Integers

An algebraic number is the root of an integer polynomial equation. It is an algebraic integer if it is the root of a monic integer polynomial equation. Equivalently, its irreducible polynomial over ℞ is monic. A rational number is an algebraic integer iff it is an integer.

2-4 Quadratic Rings

ℚ 푑 is a quadratic number field when 푑 ∈ ℞ is not a square. The algebraic integers in ℚ 훿 , 훿 = 푑, d squarefree, have the form 훼 = 푎 + 푏훿, where  If 푑 ≡ 2,3(mod 4) then a and b are integers. 1  If 푑 ≡ 1(mod 4) then 푎 and b are both integers or half-integers (푛 + ). In other words, 2 1 they have the form 푎 + 푏휂, 휂 = 1 + 푑 ; 푎, 푏 ∈ ℞. 2 The algebraic integers in ℚ 푑 form the ring of integers R in the field, ℞[훿] or ℞[휂]. Complex quadratic rings can be represented by a lattice in the complex plane. The norm is defined by 푁 푧 = 푧 푧.  The norm is a multiplicative function, and 푁 푧 = 푁(푧 ).  Hence 푥 푦 ⇒ 푁 푥 푁 푦 , 푥|푦 ⇔ 푁 푥 |푦푥 .  An element is an unit iff its norm is 1. If 푑 = −3, −2, −1,2,3,5 then R is an Euclidean domain, and hence a UFD. The only other values of d where R is a complex UFD are -7, -11, -19, -43, -67, and -163. d Units Is 2 prime? -3 1 3 Yes ±1, ± ± 2 2 -2 1, −1 No -1 ±1, ±푖 No 2 Infinitely many No 3 Infinitely many No 5 Infinitely many Yes All primes in R have norm p or 푝2. If the integer prime 푝 is odd and…  푑 is a perfect square modulo p, then p is the product of 2 conjugate primes of norm p.  푑 is not a perfect square modulo p, then p is prime in R. Some Theorems 1. Chinese Remainder: Given pairwise coprime 푏푖 and any 푎푖, there exists a unique number 푥 ∈ 푅 modulo ∏푏푖 such that 푥푖 ≡ 푎푖(mod 푏푖). 2. 푅/휋푅 has 푁(휋) elements. It is a field iff π is prime. (Pf. If N(π)=p then 1,…p are the distinct residues. If 푁(휋) = 푝2 then 휋 = 푝; take 푎 + 푏푖, 1 ≤ 푎, 푏 ≤ 푝. Use induction on the number of prime factors of π.) 3. Fermat’s Little Theorem: 푎푁 휋 ≡ 푎(mod 휋). 4. Euler’s Theorem: 푎휑 휋 ≡ 1(mod 휋). 푚 푎푖 푚 푁 휋푖 −1 5. Totient: 휑 ∏푖=1 휋푖 = 푁 휋 ∏푖=1 . 푁 휋푖 6. Wilson’s Theorem: If π is prime, the product of all nonzero residues modulo π is congruent to -1 modulo π. 7. There are finitely many pairwise non-associated numbers with given norm.

2-5 Gaussian Integers

℞[푖] is the ring of Gaussian integers. 1. If 휋 is a Gauss prime, then 휋휋 is an integer prime or the square of an integer prime. 2. Each integer prime 푝 is a Gauss prime or the product 휋휋 where 휋 is a Gauss prime. 3. Primes congruent to 3 modulo 4 are Gauss primes. (They are not the sum of two squares.) 4. Primes congruent to 1 modulo 4, and 2, are the product of complex conjugate Gauss primes.

Pf. (3-4) 2 p is a Gauss prime iff (푝) is a maximal ideal in ℞[푖], iff 푅 = ℞ 푖 (푝) = 픽푝 푥 (푥 + 1) is a 2 field, iff 푥 + 1 is irreducible (doesn’t have a zero) in 픽푝 [푥]. -1 is a square modulo 푝 iff 푝 = 2 or 푝 ≡ 1 (mod 4).

2-6 Factoring Ideals

If A and B are ideals, then 퐴퐵 = 푖 푎푖 푏푖 푎푖 ∈ 퐴, 푏푖 ∈ 퐵 . 푅 is the unit ideal since 퐴푅 = 퐴 for any R.  Let R be a complex quadratic ring. Then 퐴퐴 = 푛 = 푛푅 for some n; i.e. the product is a principal ideal. (Pf. Let 푎, 푏 , (푎 , 푏 ) be lattice bases for A, B. Let 푛 = gcd(푎 푎, 푏 푏, 푏 푎 + 푎 푏); then (푛) ⊆ 퐴 퐴. Show n divides the four generators 푏 푎 푎 푏 푎 푎, 푏 푏, 푏 푎, 푎 푏, by showing , are algebraic integers, so 퐴 퐴 ⊆ 푛 .) 푛 푛  Cancellation Law: Let A, B, C be nonzero ideals. 퐴퐵 = 퐴퐶 iff 퐵 = 퐶. 퐴퐵 ⊂ 퐴퐶 ⇔ 퐵 ⊂ 퐶.  퐴|퐵 ⇔ 퐴 ⊃ 퐵. Note that if A divides B, then A is “larger” than B- it contains more elements; upon multiplication, many of the elements disappear.

A prime ideal P satisfies any of the following equivalent conditions: 1. 푅/푃 is an integral domain. 2. 푃 ≠ 푅, and 푎푏 ∈ 푃 ⇒ 푎 ∈ 푃 or 푏 ∈ 푃. 3. 푃 ≠ 푅, and if A, B are ideals of R, 퐴퐵 ⊆ 푃 ⇒ 퐴 ⊆ 푃 or 퐵 ⊆ 푃. Any maximal ideal is prime. If P is prime and not zero, and 푃|퐴퐵, then 푃|퐴 or 푃|퐵. Letting R be a complex quadratic ring, and B be a nonzero ideal, 1. B has finite index in R. 2. Finitely many ideals in R contain B. (Pf. Look at lattice) 3. B is in a maximal ideal. 4. B is prime iff it is maximal. (Pf. 푅/푃 is a field.) Every proper ideal of a complex quadratic ring R factors uniquely into a product of prime ideals (up to ordering).1 (Pf. Use 2 and cancellation law.) Warning: Most rings do not have unique ideal factorization since 푃 ⊇ 퐵 ⇏ 푃|퐵. The complex quadratic ring R is a UFD iff it is a PID. (Pf. If P is prime, P contains a prime divisor π of some element in P. Then 푃 = (휋).) Below, P is nonzero and prime, p is an integer prime, and π is a Gauss prime. 1. If 푃 푃 = (푛) then 푛 = 푝 or 푝2 for some p. 2. (푝) is a prime ideal (p “remains prime”), or 푝 = 푃 푃 (p splits; if 푃 = 푃 then p ramifies). 3. If 푑 ≡ 2,3(mod 4), then p generates a prime ideal iff d is not a square modulo p, iff 2 푥 − 푑 is irreducible in 픽푝 [푥]. [Pf. by diagram] 1−푑 4. If 푑 ≡ 1 mod 4 , then p generates a prime ideal iff 푥2 − 푥 + is irreducible in 픽 [푥]. 4 푝 Random: For nonzero ideals A, B, C, 퐵 ⊃ 퐶 ⇒ 퐵: 퐶 = 퐴퐵: 퐴퐶 . (Show for prime ideal A, split into 2 cases based on (2).)

2-7 Ideal Classes

Ideals A and B are similar if 퐵 = 휆퐴 for some 휆 ∈ ℂ. (The lattices are similar and oriented the same.) Similarity classes of ideals are ideal classes; the ideal class of A is denoted by ⌌퐴⌍. The class of the unit ideal ⌌푅⌍ consists of the principal ideals. The ideal classes form the (abelian) class group C of R, with ⌌퐴⌍⌌퐵⌍ = ⌌퐴퐵⌍. (Note ⌌퐴⌍−1 = ⌌퐴 ⌍.) The class number |퐶| tells how “badly” unique factorization of elements fails. Measuring the size of an ideal:  Norm: 푁 퐴 = 푛, where (푛) = 퐴 퐴. Multiplicative function.  Index [푅: 퐴] of A in R.  Δ(퐴), the area of the parallelogram spanned by a lattice basis.  Minimal norm of nonzero elements of A. Relationships: Δ 퐴  푁 퐴 = 푅: 퐴 = Δ 푅 푑 2 , if 푑 ≡ 2,3 (mod 4) 3  If 푎 ∈ 퐴 has minimal nonzero norm, 푁 푎 ≤ 푁 퐴 휇, 휇 = . (Pf. 푑 , if 푑 ≡ 1 (mod 4) 3 2 푁 푎 ≤ Δ(퐴)) 3 1. Every ideal class contains an ideal A with norm 푁 퐴 ≤ 휇. (Pf. Choose nonzero

1 A Dedekind domain is a Noetherian, integrally closed integral domain with every nonzero prime ideal maximal (such as the complex quadratic rings). “Integrally closed” means that the ring contains all algebraic integers that are in its ring of fractions. Prime factorization of ideals holds in any Dedekind domain. See Algebraic Number Theory. element with minimal norm, 푎 = 퐴퐶 for some C, 푁 퐶 ≤ 휇, ⌌퐶⌍ = ⌌퐴 ⌍,⌌퐴⌍ = ⌌퐶 ⌍.) 2. The class group C is generated by ⌌푃⌍, for prime ideals P with prime norm 푝 ≤ 휇. (Pf. Factor. Either 푁(푃) = 푝 or 푃 = (푝).) 3. The class group is finite. (Pf. There are at most 2 prime ideals with norms a given integer since 퐴퐴 = (푝) has unique factorization; use multiplicativity of norm.) Computing the Class Group of 푅 = ℞[ 푑] . 1. List the primes 푝 ≤ ⌊휇⌋. 2. For each p, determine whether p splits in R by checking whether 푥2 − 푑 푑 ≡ 2,3mod4, 푥2−푥+1−푑4 푑≡1mod4 are reducible. 3. If 푝 = 퐴 퐴 splits in R, include ⌌퐴⌍ in the list of generators. 4. Compute the norm of some small elements (with prime divisors in the list found above), like 푘 + 훿. 푘 ∈ ℞. Factor 푁(푎) to factor 푎 푎 = 푁 푎 ; match factors using unique factorization. Note ⌌ 푎 ⌍ = ⌌ 푎 ⌍ = ⌌푅⌍ = 1. As long as 푎 is not divisible by one of the prime factors of 푁(푎), this amounts to replacing each prime in the factorization of 푁 푎 by one of its corresponding ideals in (3) and setting equal to 1. Repeat until 푎푖 there are enough relations to determine the group. [This works since if ∏푖⌌푃푖⌍ = 푎푖 1, 푁 푃푖 = 푝푖 then there is an element 푎, 푁 푎 = ∏푖 푝푖 .] 5. For the prime 2, a. If 푑 ≡ 2,3(mod 4), 2 ramifies: 2 = 푃푃 . 푃 has order 2 for 푑 ≠ −1, −2. b. If 푑 ≡ 2(mod 4), 푃 = 2, 훿 . c. If 푑 ≡ 3 mod 4 , 푃 = 2,1 + 훿 .

2-8 Real Quadratic Rings

Represent ℞[ 푑] as a lattice in the plane by associating 훼 = 푎 + 푏 푑 with 푢, 푣 = (푎 + 푏 푑, 푎 − 푏 푑). (The coordinates represent the two ways that 퐹[훿] can be embedded in the real numbers, where δ is the abstract square root of d: 훿2 = 푑.) The norm is 푁 푎 = 푎2 − 푏2푑 (sometimes the absolute value is used).

The units satisfy 푁 푎 = 1; they lie on the hyperbola 푢푣 = ±1. The units form an infinite group in R. 2 proofs: 1. The Pell’s equation has infinitely many solutions. 푢2 2 2. Let Δ be the determinant for the lattice of ℞[ 푑] and 퐷 = {(푢, 푣)| + 푠2푣2 ≤ Δ}. For 푠 푠2 3 any lattice with determinant Δ, 퐷1 contains a nonzero lattice point (take the point nearest 푦 the origin and use geometry). 휑 푥, 푦 = (푠푥, ) is an area-preserving map; by applying 휑 푠 to the lattice, we get that every 퐷푠 has a nonzero lattice point. These points have bounded norm; one norm is hit an infinite number of times. The ratios of these quadratic integers are units.

3 Polynomials 3-1 Polynomials

A polynomial with coefficients in R is a finite combination of nonnegative powers of the variable x. The polynomial ring over R is 푛 푛 푅 푋 = 푎푖푥 |푎푖 ∈ 푅 푖=0 푖 with X a formal variable and addition and multiplication defined by: (푓 푥 = 푖 푎푖푥 , 푔 푥 = 푖 푖 푏푖푥 ) 푖 1. 푓 푥 + 푔 푥 = 푖 푎푖 + 푏푖 푥 푖+푗 2. 푓 푥 푔 푥 = 푖 푎푖푏푗 푥 R is a subring of R[X] when identified with the constant polynomials in R. The ring of formal series 푅[ 푋 ] is defined similarly but combinations need not be finite. Iterating, the ring of polynomials with variables 푥1, 푥2, … , 푥푛 is 푅[푥1, 푥2, … , 푥푛 ]. (Formally, use the substitution principle below to show we can identify 푅 푥 푦 with 푅[푥, 푦].)

Basic vocab: monomial, degree, constant, leading coefficient, monic

Division with Remainder: Let 푓, 푔 ∈ 푅[푋] with the leading coefficient of 푓 a unit. There are unique polynomials 푞, 푟 ∈ 푅[푋] (the quotient and remainder) so that 푔 푥 = 푓 푥 푞 푥 + 푟 푥 , deg 푟 < deg 푓

Substitution Principle: Let 휑: 푅 → 푅′ be a ring homomorphism. Given 푎1, … , 푎푛 ∈ 푅′, there is a unique homomorphism Φ: 푅 푥1, … , 푥푛 → 푅′ which agrees with the map 휑 on constant polynomials, and sends 푥푖 → 푎푖. For 푅 = 푅′, this map is just substituting 푎푖 for the variable 푥푖 and evaluating the polynomial.

A characterization of the GCD: The greatest common divisor 푑 of 푓, 푔 ∈ 푅 = 퐹[푥] is the monic polynomial defined by any of the following equivalent conditions: 1. 푅푑 = 푅푓 + 푅푔 2. Of all polynomials dividing f and g, 푑 has the greatest degree. 3. If 푒|푓, 푔 then 푒|푑. From (1), there exist 푝, 푞 ∈ 푅 so 푑 = 푝푓 + 푞푔.

Adjoining Elements Suppose 훼 is an element satisfying no relation other than that implied by 푓 훼 = 0 where 푓 ∈ 푅[푥] and has degree n. Then 푅 훼 ≅ 푅 푥 /(푓) is the ring extension. If f is monic, its distinct elements are the polynomials in 훼 (with coefficients in R) of degree less than n, with (1, 훼, … , 훼푛−1) a basis. A polynomial in 훼 is equivalent to its remainder upon division by f. 푅 can be identified with a subring of 푅[훼] as long as no constant polynomial is identified with 0.

3-2 ℞[푋]

Tools for factoring in ℞[푋]: 1. Inclusion in ℚ[푋]. 2. Reduction modulo prime p:휓푝 : ℞ 푋 → 픽푝 [푋]. A polynomial in ℞[푋] is primitive if the greatest common divisor of its coefficients is 1, it is not constant, and the leading coefficient is positive.

Gauss’s lemma: The product of primitive polynomials is primitive. Pf. Any prime 푝 in ℞ is prime in ℞[푥], and a prime divides a polynomial iff it divides every coefficient.

 Every nonconstant polynomial 푓 푥 ∈ ℚ[푥] can be written uniquely as 푓 푥 = 푐푓0 푥 , 푐 ∈ ℚ and 푓0(푥) primitive.  If 푓0 is primitive and 푓0|푔 ∈ ℞[푥] in ℚ[푥], then 푓0|푔 in ℞[푥]. If two polynomials have a common nonconstant factor in ℚ[푥] , they have a common nonconstant factor in ℞[푥].  An element of ℞[푥] is irreducible iff it is a prime integer or a primitive irreducible polynomial in ℚ[푥], so every irreducible element of ℞[푥] is prime.  Hence ℞[푥] is a UFD, and each nonzero polynomial can be written uniquely (up to ordering) as (푝푖 primes, 푞푖 primitive irreducible polynomials) 푓 푥 = ±푝1 ⋯ 푝푚 푞1 푥 ⋯ 푞푛 (푥)  This technique can be generalized: replace ℞ with a UFD R, and ℚ with the field of fractions of R.  By induction, if R is a UFD, then 푅[푥1, … , 푥푛 ] is a UFD.

푛 푏0 Rational Roots Theorem: Let 푓 푥 = 푎푛 푥 + ⋯ + 푎0, 푎푛 ≠ 0. – (in reduced form) is a root 푏1 iff 푏1푥 + 푏0 divides f; if 푏1푥 + 푏0 divides f then 푏1|푎푛 and 푏0|푎0.

3-3 Irreducible Polynomials

The derivative of a polynomial is formally defined (without calculus) as 푛 ′ 푛 푖 푖−1 푎푖 푥 = 푖푎푖 푥 . 푖=0 푖=1  훼 is a multiple root of 푓 푥 ∈ 퐹 푥 iff 훼 is a root of both 푓(푥) and 푓′ 푥 . 훼 appears as a root exactly 푛 times if 푓 푖 훼 = 0 for 0 ≤ 푖 < 푛 but 푓 푛 훼 ≠ 0.  푓, 푓′ have a common factor other than 1 iff there is a field extension where f has a multiple root.  If f is irreducible, and 푓′ ≠ 0, then f has no multiple root (in any field extension). If F has characteristic 0, then f has no multiple root (in any field extension).

푛 If 푓 푥 = 푎푛 푥 + ⋯ + 푎0 ∈ ℞[푥], 푝 ∤ 푎푛 , and the residue of f modulo p is irreducible in 픽푝 [푥],

then f is irreducible in ℚ[푥]. Irreducible polynomials in 픽푝 [푥] can be found using the sieve method.

푛 2 Eisenstein Criterion: If 푓 푥 = 푎푛 푥 + ⋯ + 푎0 ∈ ℞ 푥 , 푝 ∤ 푎푛 , 푝|푎푛−1, … , 푎0, 푝 ∤ 푎0, then f is irreducible in ℚ[푥]. 푟 푠 2 Pf. Factor and take it modulo p; get 푓 = 푏푟 푥 (푐푠푥 ). But if 푟, 푠 ≠ 0, 푝 |푎0. Extended Eisenstein’s Criterion

Schönemann’s Criterion: Suppose  푘 = 푓푛 + 푝푔, 푛 ≥ 1, 푝 prime, 푓, 푔 ∈ ℞[푥]  deg 푓푛 > deg⁡(푔)  푘 primitive

 푓 is irreducible in 픽푝 [푥]  푓 ∤ 푔 . Then k is irreducible in ℚ[푥].

푛 Cohn’s Criterion: Let 푏 ≥ 2 and let p be a prime number. Write 푝 = 푎푛 푏 + ⋯ + 푎1푏 + 푎0 in 푛 base b. Then 푓 푋 = 푎푛 푋 + ⋯ + 푎1푋 + 푎0 is irreducible in ℚ[푋].

Capelli’s Theorem: Let K be a subfield of ℂ and 푓, 푔 ∈ 퐾[푋]. Let 푎 be a complex root of f and assume that f is irreducible in 퐾[푋] and 푔 푋 − 푎 is irreducible in 퐾 푎 [푋]. Then 푓 푔 푋 is irreducible in 퐾[푋].

[Add proof sketches.]

3-4 Cyclotomic Polynomials

A primitive nth root of unity satisfies 휔푛 = 1, but 휔푚 ≠ 1 for 푛 ∈ ℕ, 0 < 푚 < 푛. The nth cyclotomic polynomial is

Φ푛 푋 = (푋 − 휔). 휔 primitive nth root The cyclotomic polynomial is an irreducible polynomial in ℞[푋] of degree 휑(푛). Each polynomial 푋푛 − 1 is a product of cyclotomic polynomials: 푛 푛 푋 − 1 푋 = Φ푑 (푋) ⇒ Φ푛 푋 = ∏푑|푛,푑<푛 Φ푑 푋 푑|푛 so Φ푛 (푋) has integer coefficients. Pf. of irreducibility: Lemma- if 휔 is a primitive nth root of unity and a zero of 푓 ∈ ℞[푋] then 푝 휔 is a root for 푝 ∤ 푛. Proof- Suppose Φ푛 = 푔푕, 푕 휔 = 0, h irreducible (the minimal 푝 polynomial of 휔). 휔 is also a zero of Φ푛 . Suppose it is a zero of 푔. Then 휔 is a zero of 푔(푥푝 ), so 푔 푥푝 = 푕푘. Mod p, 푔 푥 푝 = 푕푘, so g and h have a root in cogmmon, contradicting 푛 푛 that 푋 − 1 ∈ ℞푝 [푋] has no multiple root (derivative has no common factors with 푋 − 1). Hence 휔푝 is a zero of h. ∎ Take powers to different primes to show that all primitive nth roots of unity divide h. Then 푕 = Φ푛 is irreducible.

3-5 Varieties

1. The set 퐴푛 of n-tuples in a field K is the affine n-space. 2. If S is a set of polynomials in 퐾 푋1, … , 푋푛 then 푉 푆 = 푥 ∈ 퐴푛 푓 푥 = 0 for all 푓 ∈ 푆 is a (affine) variety. 3. For 푋 ⊆ 퐴푛 , define the ideal of X to be 퐼 푋 = 푓 ∈ 퐾 푋1, … , 푋푛 푓 vanishes on 푋 푛 푛 Note 푆 ⊆ 퐾 푋1, … , 푋푛 , 푋 ⊆ 퐴 , 푉 푆 ⊆ 퐴 , 퐼 푋 ⊆ 퐾 푋1, … , 푋푛 . 4. The radical of the ideal I in a ring T is the ideal 퐼 = {푓 ∈ 푅|푓푟 ∈ 퐼 for some 푟 ∈ ℕ}

퐴푛 can be made into a topology (the Zariski topology) by taking varieties as closed sets (1- 4 below). Properties: 1. 푉 푆 = 푉 퐼 where I is the ideal generated by S. 2. ∩ 푉 퐼푗 = 푉(∪ 퐼푗 ) 푟 3. If 푉푗 = 푉(퐼푗 ) then ⋃푗=1 푉푗 = 푉( 푓1 ⋯ 푓푟 푓푗 ∈ 퐼푗 , 1 ≤ 푗 ≤ 푟 ). 4. 퐴푛 = 푉 0 , 휙 = 푉(1) 5. 푋 ⊆ 푌 ⇒ 퐼 푌 ⊆ 퐼(푋), 푆 ⊆ 푇 ⇒ 푉 푇 ⊆ 푉(푆) 6. 푆 ⊆ 퐼푉 푆 , 푋 ⊆ 푉퐼(푋) 7. 푉퐼푉 푆 = 푉 푆 , 퐼푉퐼 푋 = 퐼(푋) 8. 퐼 0 = 퐾[푋1, … , 푋푛 ] 9. If K is an infinite field, 퐼 퐴푛 = {0}. a. For n=1, a nonzero polynomial only has finitely many zeros. Extend by induction. 10. 퐼 푎1, … , 푎푛 = (푋1 − 푎1, … , 푋푛 − 푎푛 ) (the ideal generated by 푋푖 − 푎푖 ) a. Use division with remainder. 푛 11. If 푎1, … 푎푛 ∈ 퐴 then 퐼 = (푋1 − 푎1, … , 푋푛 − 푎푛 ) is a maximal ideal. a. Apply the division algorithm to 푓 ∈ 퐽\퐼. 12. 퐼 ⊆ 퐼푉(퐼)

3-6 Nullstellensatz

Hilbert Basis Theorem: If R is Noetherian, then 푅[푥] is Noetherian. Thus so is 푅[푥1, … , 푥푛 ]. In particular, this holds for 푅 = ℞ or a field F. Pf. For 퐼 ⊆ 푅[푥] an ideal, the set whose elements are the leading coefficients of the polynomials in I (including 0) forms the ideal of leading coefficients in R. Take generators for this ideal and polynomials with those leading coefficients, multiplying by x as necessary so they have the same degree n. The polynomials with degree less than n form a free, Noetherian R-module P with basis (1, 푥, … 푥푛−1). 푃 ∩ 퐼 has a finite generating set. Put the two generating sets together. The ring of formal power series 푅[ 푋 ] is also Noetherian.

Noether Normalization Lemma: Let A be a finitely generated K-algebra. There exists a subset 푦1, … , 푦푟 of A such that the 푦푖 are algebraically independent over K and A is integral over 퐾[푦1, … , 푦푟] (all elements of A are algebraic integers over 퐾[푦1, … , 푦푟]). Pf. Induct on n; n=1 trivial. Take a maximally algebraically independent subset 푥1, … , 푥푟 ⊆ {푥1, … , 푥푛 }; assume 푛 > 푟. By algebraic dependency, there exists 푓 ∈ 퐾 푋1, … , 푋푛 so that 푓 푥1, … , 푥푛 = 0. Lexicographically order the monomials, and choose weights 푤푖 to match 푤푖 the lexicographic order. Set 푥푖 = 푧푖 + 푥푛 . Then we get a polynomial in 푥푛 , where term with the highest power of 푥푛 is uncancelled. 푥푛 is integral over 퐾[푧1, … , 푧푛−1]; finish by induction. Cor. If B is a finitely generated K-algebra, and I is a maximal ideal, then 퐵 퐼 is a finite extension of K (K is embedded via 푐 ⇝ 푐 + 퐼.): In the above, we must have 푟 = 0; B/I is algebraic over K.

Hilbert’s Nullstellensatz: For any field K and 푛 ∈ ℕ, the following are equivalent: 1. K is algebraically closed. 2. (Maximal Ideal Theorem) The maximal ideals of 퐾[푋1, … , 푋푛 ] are the ideals of the form (푋1 − 푎1, … , 푋푛 − 푎푛 ). 3. (Weak Nullstellensatz) If I is a proper ideal of 퐾[푋1, … , 푋푛 ], then 푉(퐼) is not empty. 4. (Strong Nullstellensatz) If I is an ideal of 퐾[푋1, … , 푋푛 ] then 퐼푉 퐼 = 퐼. (2)⇒(3): For I a proper ideal, let J be a maximal ideal containing it. Then 푉 퐽 ⊆ 푉(퐼) so 퐽 = (푋1 − 푎1, … , 푋푛 − 푎푛 ), 푎 ∈ 푉(퐽). (3)⇒(4): Rabinowitsch Trick: ∗ 1. Let 푓 ∈ 퐼푉 퐼 . Let 푓1, … , 푓푚 be a generating set for 퐾[푋1, … , 푋푛 , 푌]. Let 퐼 be the ideal generated by 푓1, … , 푓푚 , 1 − 푌푓. 2. 푉 퐼∗ = 휙 푛+1 ∗ a. If 푎1, … , 푎푛 , 푎푛+1 ∈ 퐴 and 푎1, … , 푎푛 ∈ 푉 퐼 , then 푎1, … , 푎푛 , 푎푛+1 ∉ 푉(퐼 ) since it is not a zero of 1 − 푌푓. ∗ b. If 푎1, … , 푎푛 ∉ 푉 퐼 then 푎1, … , 푎푛 , 푎푛+1 ∉ 푉(퐼 ) (an even weaker statement). 3. Using the weak Nullstellensatz (see below), we can write 푚

1 = 푔푖푓푖 + 푕 1 − 푌푓 푖=1 ∗ ∗ since 푉 퐼 = 휙 ⇒ 1 ∈ 퐼 = 퐾[푋1, … , 푋푛 , 푌]. 4. Set 푌 = 1/푓, and multiply to clear denominators 푚 푟 푓 = 푕푖 푋1, … , 푋푛 푓푖 푋1, … 푋푛 ∈ 퐼 푖=1 (4)⇒(3): The radical of an ideal is the intersection of all prime ideals containing I. I is in a maximal, prime ideal P; so is 퐼, so 퐼 is proper. 퐼푉(퐼) is proper by (4), so 푉 퐼 ≠ 휙. (3)⇒(2): For I maximal, there is 푎 ∈ 푉(퐼). Since I is maximal, it must be in (푋1 − 푎1, … , 푥푛 − 푎푛 ). (1)⇒(2): Let I be a maximal ideal. K can be imbedded via 푐 ⇝ 푐 + 퐼 in 퐾 푋1, … , 푋푛 퐼; by the corollary to Noether Normalization Lemma, this is a finite extension of K so must be K (since it is algebraically closed). Then 푋푖 + 퐼 = 푎푖 + 퐼 ⇒ 푋푖 − 푎푖 ∈ 퐼, 푋1 − 푎1, … , 푋푛 − 푎푛 ⊆ 퐼, with equality since the LHS is maximal. (2)⇒(1): If f is a nonconstant polynomial in 퐾[푋1] with no root in K, regard it as a polynomial 푛 in 퐾[푋1, … , 푋푛 ] with no root in 퐴 . Then I is in a maximal ideal (푋1 − 푎1, … , 푋푛 − 푎푛 ), so has root 푋1 = 푎1.

Combinatorial Nullstellensatz: Let F be a field, 푓 ∈ 퐹[푋1, … , 푋푛 ] and let 푆1, … , 푆푛 be nonempty subsets of F. 1. If 푓 푠1, … , 푠푛 = 0 for all 푠1, … , 푠푛 ∈ 푆1 × ⋯ × 푆푛 (i.e. 푓 ∈ 푉(푆1 × ⋯ × 푆푛 )) then 푓 lies ∏ in the ideal generated by the polynomials 푔푖 푋푖 = 푠∈푆푖(푋푖 − 푠). The polynomials 푕1, … , 푕푛 satisfying 푓 = 푔1푕1 + ⋯ + 푔푛 푕푛 can be chosen so that deg 푕푖 ≤ deg 푓 − deg⁡(푔푖) for all i. If 푔1, … , 푔푛 ∈ 푅[푋1, … , 푋푛 ] for some subring 푅 ⊆ 퐹, we can choose 푕푖 ∈ 푅[푋1, … , 푋푛 ]. a. The 푠 ∈ 푆푖 are all zeros of 푔푖 of degree |푆푖| so by subtracting multiples of 푔푖 푘 푆푖 −1 every 푋푖 in f can be replaced with a linear combination of 1, … , 푥푖 . Then by counting zeros the result is actually 0; i.e. f is in the form above. 2. If deg 푓 = 푡1 + ⋯ + 푡푛 where 푡푖 are nonnegative integers with 푡푖 < |푆푖|, and if the 푡1 푡푛 coefficient of 푋1 ⋯ 푋푛 is not 0, then there exist 푠푖 ∈ 푆푖 so that 푓 푠1, … , 푠푛 ≠ 0, i.e. 푓 ∉ 푓 ∈ 푉(푆1 × ⋯ × 푆푛 ).

4 Fields

Examples of Fields An extension K of a field F (also denoted 퐾/퐹) is a field containing F.  Number field: subfield of ℂ  Finite field: finitely many elements  Function field: Extensions of ℂ(푡) of rational functions

4-1 Fundamental Theorem of Algebra

Fundamental Theorem of Algebra: Every nonconstant polynomial with complex coefficients has a complex zero. Thus, the field of complex numbers is algebraically closed, and every polynomial in ℂ[푥] splits completely. 푛 푛−1 휃푖 Pf. Let 푓 푥 = 푥 + 푎푛−1푥 + ⋯ + 푎1푥 + 푎0. Let 푥 = 푟푒 - the parameterization of a circle about the origin. 푓 푥 − 푥푛 is small for large r, so for large r, 푓(푥) winds around the origin n times as 휃 goes from 0 to 2휋푛.2 (f is a person walking a dog on a circular path n times around the origin. The dog will also walk around the origin n times provided the leash is shorter than the radius.) For small r, 푓 푥 makes a small loop around 푎0, and winds around the origin 0 times. For some intermediate value, 푓 푥 will pass through the origin.

4-2 Algebraic Elements

Let K be an extension, and 훼 be an element of K. 훼 is algebraic over F if it is the zero of a polynomial in 퐹[푥], and transcendental otherwise. 훼 is transcendental iff the substitution homomorphism 휑: 퐹 푥 → 퐾 is injective.

퐹 훼 is the smallest subfield of K that contains F and 훼. The irreducible polynomial 푓 of 훼 over F is the monic polynomial of lowest degree in 퐹[푥] having 훼 as a zero.  Any polynomial in 퐹[푥] having 훼 as a zero divides 푓.  퐹 푥 /(푓) is an extension field of F with 푥 a root of 푓 푥 = 0. The substitution map 퐹 푥 /(푓) → 퐹[훼] is an isomorphism, so 퐹 푥 /(푓) ≅ 퐹(훼). [*]  For every polynomial in F[x], there is an extension field in which it splits (factors) completely, i.e. every polynomial has a splitting field.  If f has degree 푛 then 퐹(훼) is a vector space with dimension 푛 and basis (1, 훼, … , 훼푛−1).  Let 훼, 훽 be elements of 퐾/퐹, 퐿/퐹 algebraic over F. There is an isomorphism of fields 퐹 훼 → 퐹(훽) sending 훼 ⇝ 훽 iff 훼, 훽 have the same irreducible polynomial. Let 퐾, 퐿 be extensions of F. A F-isomorphism is an isomorphism 휑: 퐾 → 퐿 that restricts to the identity on F. Then K and L are isomorphic as field extensions. If 훼 is a zero of 푓 in 퐾, then 휑(훼) is a zero of 푓.

4-3 Degree of a Field Extension

The degree [퐾: 퐹] is the dimension of K as an F-vector space.  If 훼 is algebraic over F, then [퐹 훼 : 퐹] is the degree of the irreducible polynomial of 훼 over F, and if 훼 is transcendental, 퐹 훼 : 퐹 = ∞.  퐾: 퐹 = 1 iff 퐹 = 퐾.

2 i.e. it is homotopic to the loop going around the origin n times in ℂ − 0 .  If 퐾: 퐹 = 2 then K can be obtained by adjoining a square root 훿 of an element in F: 훿2 ∈ 퐹  Multiplicative property: If 퐹 ⊆ 퐾 ⊆ 퐿, then 퐿: 퐹 = 퐿: 퐾 [퐾: 퐹]. o If K is a finite extension of F, and 훼 ∈ 퐾, then the degree of 훼 divides [퐾: 퐹]. o If 훼 ∈ 퐿 is algebraic over F, it is algebraic over 퐾 with degree less than or equal to its degree over 퐹. o A field extension generated by finitely many algebraic elements is a finite extension. o The set of elements of K that are algebraic over F is a subfield of K.  Let L be an extension field of F, and let 퐾, 퐹′ be subfields of L that are finite extensions of F. Let 퐾′ : 퐹 = 푁, 퐾: 퐹 = 푚, 퐹′ : 퐹 = 푛. Then m and n divide N, and 푁 ≤ 푚푛.

≤ 푛 퐾′ ≤ 푚

퐾 푁 ≤ 푚푛 퐹′

푚 푛 퐹

Finding the Irreducible Polynomial of 훾 (The dum way) Compute powers of 훾, and find a relation between them. (The smart way) 1. If 훾 = 푎1 + ⋯ + 푎푛 , 푎1 ⋯ 푎푛 where each 푎1 is algebraic (for example a nth root), then its conjugates (other zeros of the irreducible polynomial) are in the form 푏1 + ⋯ + 푏푛 , 푏1 ⋯ 푏푛 , respectively where 푏푖 is a conjugate of 푎푖. (Not all these may be conjugates.) 2. The irreducible polynomial is ∏훾′ conjugate of 훾(푥 − 훾′). [Note: This gives an elementary proof of the fact that the algebraic numbers/ integers form a field/ ring. For 푎, 푏 algebraic, expand the product (푥 − 푎′ − 푏′ ), 푎′ , 푏′ running over the conjugates of 푎, 푏, and use the Fundamental Theorem on Symmetric Polynomials.]

4-4 Application: Constructions!

Rules: 1. Two points (0,0) and (1,0) are given (constructed). 2. If P, Q have been constructed, we can draw (construct) a. the line through them b. a circle with center at P and passing through Q. 3. Points of intersection of constructed lines and circles are constructed. 4. A number is constructible if (a,0) is constructible. Finding all constructible numbers: 1. If 푃 = 푎0, 푎1 , 푄 = (푏0, 푏1) have coordinates in F, then the line in (2a) is defined by a linear equation with coefficients in F, while the circle in (2b) is defined by a quadratic equation with coefficients in F. 2. The point of intersection of two lines/ circles whose equations have coefficients in F has coordinates in a real quadratic extension of F. 3. Let P be a constructible point. There is a chain of fields ℚ = 퐹0 ⊂ 퐹1 ⊂ ⋯ ⊂ 퐹푛 = 퐾 such that a. K is a subfield of ℝ. b. The coordinates of P are in 퐹푛 . 푛 c. 퐹푖+1: 퐹푖 = 2, and 퐾: ℚ = 2 . Thus a constructible number has degree over ℚ a power of 2. 4. Conversely, if a chain of fields satisfies the conditions above, then every element of K is constructible. Examples: 1. Trisecting the angle with compass and straightedge is impossible. cos 20° is not constructible. 2. For p prime, a regular p-gon can be constructed iff 푝 = 2푟 + 1.

4-5 Finite Fields

푟 A finite field is a vector space over 픽푝 for some prime p, so has order 푞 = 푝 . The (unique) field of order q is denoted by 픽푞 . 1. The elements in a field of order q are roots of 푥푞 − 푥 = 0 (everything is modulo p). × a. The multiplicative group 픽푞 of nonzero elements has order 푞 − 1. The order of 푞−1 any element divides 푞 − 1 so 훼 = 0 for any 훼 ∈ 픽푞. × 2. 픽푞 is a cyclic group of order 푞 − 1. × a. By the Structure Theorem for Abelian Groups, 픽푞 is a direct product of cyclic 푑푘 subgroups of orders 푑1| ⋯ |푑푘, and the group has exponent 푑푘. 푥 − 1 = 0 has at most 푑푘 roots, so 푘 = 1, 푑푘 = 푞 − 1. 3. There exists a unique field of order q (up to isomorphism). a. Existence: Take a field extension where 푥푞 − 푥 splits completely. If 훼, 훽 are roots of 푥푞 − 푥 = 0 then 훼 + 훽 푞 = 훼 + 훽. Since −1 is a root, −훼 is a root. The roots form a field. b. Uniqueness: Suppose 퐾, 퐾′ have order 푞. Let 훼 be a generator of 퐾×; 퐾 = 퐹(훼). The irreducible polynomial 푓 ∈ 퐾[푥] with root 훼 divides 푥푞 − 푥. 푥푞 − 푥 splits completely in both 퐾, 퐾′, so f has a root 훼′ ∈ 퐾′. Then 퐹 훼 ≅ 퐹 푥 /(푓) ≅ 퐹(훼′ ) = 퐾′. 4. A field of order 푝푟 contains a subfield of order 푝푘 iff 푘|푟. (Note this is a relation between the exponents, not the orders.) a. 픽푝푘 ⊆ 픽푝푟 ⇒ 푘|푟: Multiplicative property of the degree. 푟 푘 × 푘 b. 픽푝푘 ⊆ 픽푝푟 ⇐ 푘|푟: 푝 − 1|푝 − 1. Cyclic 픽푝푟 contains a cyclic group of order 푝 . 푘 Including 0, they are the roots of 푥푝 − 푥 = 0 and thus form a field by 3a. 푞 5. The irreducible factors of 푥 − 푥 over 픽푝 are the irreducible polynomials 푔 in 퐹[푥] whose degrees divide r. a. ⇒: Multiplicative property b. ⇐: Let 훽 be a root of g. If 푘|푟, by (4), 픽푞 contains a subfield isomorphic to 푞 퐹(훽). 푔 has a root in 픽푞 so divides 푥 − 푥. 6. For every r there is an irreducible polynomial of degree r over 픽푝. 푟 a. 픽푞 (푞 = 푝 ) has degree r over 픽푝, and has a cyclic multiplicative group generated by an element 훼. 픽푝 (훼) has degree r over 픽푝.

푞 푟−1 To compute in 픽푞 , take a root 훽 of the irreducible factor of 푥 − 푥 of degree r; (1, 훽, … , 훽 ) is a basis.

Let 푊푝 (푑) be the number of irreducible monic polynomials of degree d in 픽푝. Then by (2), 푛 푝 = 푑푊푝 (푑). 푑|푛 By Möbius inversion, 1 푛 푊 푛 = 휇 푝푑 . 푝 푛 푑 푑|푛

Primitive elements Let K be a field extension of F. An element 훼 ∈ 퐾 such that 퐾 = 퐹(훼) is a primitive element for the extension. Primitive Element Theorem: Every finite extension of a field F contains a primitive element. Proof when F has characteristic 0: Show that if 퐾 = 퐹(훼, 훽) with 훼, 훽 ∈ 퐾 then 훾 = 훽 + 푐훼 is a primitive element for K over F. Let 푓, 푔 be the irreducible polynomials of 훼, 훽, and 훼푖, 푏푗 be the conjugates of 훼, 훽. For all but finitely many 푐, the numbers 훽푗 + 푐훼푖 are all distinct. For such a value of 푐, to show that 훾 = 훽 + 푐훼 is a primitive element, consider 푕 푥 = 푔 훾 − 푐푥 ∈ 퐹(훾). gcd 푓, 푕 = 푥 − 훼 by choice of c, so 훼 ∈ 퐹(훾) as desired.

5 Modules

More No division Division structure→ More complex Ring Field ↓ Module Vector Space Algebra Division Algebra

5-1 Modules

A left/right R-module 푅푀/푀푅 over the ring R is an abelian group (M,+) with addition and scalar multiplication (푅 × 푀 → 푀 or 푀 × 푅 → 푀) defined so that for all 푟, 푠 ∈ 푅 and 푥, 푦 ∈ 푀, Left Right 1. Distributive 푟 푥 + 푦 = 푟푥 + 푟푦 푥 + 푦 푟 = 푥푟 + 푦푟 2. Distributive 푟 + 푠 푥 = 푟푥 + 푠푥 푥 푟 + 푠 = 푥푟 + 푥푠 3. Associative 푟 푠푥 = 푟푠 푥 푥푟 푠 = 푥(푟푠) 4. Identity 1푥 = 푥 푥1 = 푥

A (S,R)-bimodule 푆푀푅has both the structure of a left S-module and right R-modules. Modules are generalizations of vector spaces. All results for vector spaces hold except ones depending on division (existence of inverse in R). 푆 ⊆ 푀 is linearly dependent if there exist 푣1, … , 푣푛 ∈ 푆 such that 푟푖푣푖 ≠ 0 and 푟1푣1 + ⋯ + 푟푛 푣푛 = 0. Again, a basis is a linearly independent set that generates the module. Note that if elements are linearly independent, it is not necessary that one element is a linear combination of the others, and bases do not always exist. Every basis for V (if it exists) contains the same number of elements. V is finitely generated if it contains a finite subset spanning V. The rank is the size of the smallest generating set.

A submodule of a R-module is a nonempty subset closed under addition and scalar multiplication. Viewing R as an R-module, the submodules of R are the ideals in R. In ℞푛 , submodules correspond to lattices, with the area/volume of a fundamental parallelogram/parallelepiped equal to the determinant.

A free module with n generators has a basis with n elements. It is isomorphic to 푅푛 . An isomorphism preserves addition and scalar multiplication. Unlike vector spaces, not all finitely generated modules are isomorphic to some 푅푛 .

Basic Theorems: 1. If W is a submodule of V, the quotient module 푉/푊 is a R-module, and the canonical map 휋: 푉 → 푉/푊 is a homomorphism. 2. Mapping Property: Let 푓: 푉 → 푉′ be a homomorphism of R-modules with kernel containing W. There is a unique homomorphism 푓 with 푓 = 푓 ∘ 휋.

V' 푓 휋 푓 V G 3. First Isomorphism Theorem: If 푓 is surjective with kernel W, 푓 is an isomorphism. 4. Correspondence Theorem: Let 푓: 푉 → 푉′ be a surjective homomorphism. There is a bijective correspondence between submodules of V' and submodules of V that containing ker 푓 = 푊: 푆 with 푊 ⊆ 푆 ⊆ 푉 is associated with 푓(푆); 푉/푆 ≅ 푉′ /푓(푆). 5. Second Isomorphism Theorem: Let 푆 and 푇 be submodules of 푀, and let 푆 + 푇 = 푥 + 푦 ∶ 푥 ∈ 푆, 푦 ∈ 푇 . Then 푆 + 푇 and 푆 ∩ 푇 are submodules of 푀 and 푆 + 푇 푆 ≅ 푇 푆 ∩ 푇 6. Third Isomorphism Theorem: Let 푁 ⊆ 퐿 ⊆ 푀 be modules. Then 푀/퐿 ≅ (푀/푁)/(퐿/푁).

5-2 Structure Theorem

Matrices, invertible matrices, the general linear group, the determinant, and change of bases matrices all generalize to a ring R. However, a R-matrix A is invertible iff its determinant is a unit. Properties of matrices in a field such as det(퐴퐵) = det 퐴 det 퐵 continue to hold in a ring, because they are polynomial identities.

Smith Normal Form For a matrix over an Euclidean domain R [*] (such as ℞ or 퐹[푡]), elementary row/ column operations correspond to left and right multiplication by elementary matrices and include: (1) Interchanging 2 rows/ columns (2) Multiplying any row/ column by a unit (3) Adding any multiple of a row/ column to another row/ column However, note arbitrary division in R is illegal.

A 푚 × 푛 matrix is in Smith (or Hermite) normal form if 1. It is diagonal. 2. The entries 푑1, … , 푑푛 on the main diagonal satisfy 푑푘|푑푘+1, 1 ≤ 푘 < min⁡(푚, 푛). (Ones at the end may be 0.) Every matrix is equivalent to a unique matrix N in normal form. For a 푚 × 푛 matrix A, follow this algorithm to find it: 푝 0 1. Make the first column ⋮ . 0 a. Choose the nonzero entry 푓 in the first column that has the least norm. b. For each other nonzero entry 푝, use division to write 푝 = 푓푞 + 푟, where 푟 is the remainder upon division. Subtract 푞 times the row with 푓 from the row with 푝. c. Repeat a and b until there is (at most) one nonzero entry. Switch the first row with that row if necessary. 2. Put the first row in the form 푝 0 ⋯ 0 by following the steps above but exchanging the words “rows” and “columns”. 3. Repeat 1 and 2 until the first entry 푔 is the only nonzero entry in its row and column. (This process terminates because the least degree decreases at each step.) 4. If 푔 does not divide every entry of A, find the first column with an entry not divisible by g and add it to column 1, and repeat 1-4; the degree of “g” will decrease. Else, go to the next step. 5. Repeat with the 푚 − 1 × (푛 − 1) matrix obtained by removing the first row and column. Solving 퐴푋 = 퐵 in R: 1. Write 퐴 = 푄퐴′푃−1, where 퐴′ is in normal form. Suppose the nonzero diagonal entries are 푑1, 푑2, … , 푑푘. 2. The solutions 푋′ of the homogeneous system 퐴′ 푋′ = 0 are the vectors whose first k coordinates are 0. 3. The solutions of 퐴푋 = 0 are in the form 푋푕 = 푃푋′. 푟1푑1 ⋮ ′ ′ 4. The equation has a solution iff 퐵 is in the form 푄푌 , 푌 = 푟푘 푑푘 . Use linear algebra to 0 ⋮ find a particular solution 푋푝 . (The condition guarantees that the entries are in R.) Then the solutions are 푋푕 + 푋푝, for 푋푕 a homogeneous solution. Structure Theorem: (a.k.a. Fundamental Decomposition Theorem) Let M be a finitely generated module over the PID R (such as ℞ or 퐹[푡]). Then M is a direct sum of cyclic modules and a free module 퐶1 ⊕ ⋯ ⊕ 퐶푘 ⊕ 퐿, where 퐶푖 ≅ 푅/(푑푖). The cyclic modules can be chosen to satisfy either of the following conditions: 1. 푑1|푑2 ⋯ |푑푘 2. Each 푑푖 is the power of an irreducible element.

5-3 Noetherian and Artinian Rings

The following conditions on a R-module V are equivalent: 1. Every submodule is finitely generated. 2. Ascending chain condition: There is no infinite strictly increasing chain 푊1 ⊂ 푊2 ⊂ ⋯ of submodules. (Pf. of ⇒: Union of chain finitely generated.) A ring is Noetherian if every ideal of R is finitely generated. In a Noetherian ring, every proper ideal is contained in a maximal ideal. Let 휑 be a homomorphism of R-modules. 1. If V is finitely generated and 휑 is surjective, then 푉′ is finitely generated. 2. If ker 휑 , im 휑 are finitely generated, so is V. (Pf. Take a generating set for the kernel and some preimage of a generating set for the image.) 3. In particular, if 푉 is finitely generated, so is 푉/푊. If 푉/푊 and 푊 are finitely generated, so is 푉. If R is Noetherian, then every submodule of a finitely generated R-module is finitely generated. Pf. Using a surjective map 휑: 푅푚 → 푉, it suffices to prove it for 푅푚 . Induct on m. For the projection 휋: 푅푚 → 푅푚−1, the image and kernel are finitely generated.

A ring is Artinian if it satisfies the descending chain condition on ideals: There is no infinite strictly decreasing chain 퐼1 ⊃ 퐼2 ⊃ ⋯ of ideals.

5-4 Application 1: Abelian Groups

An abelian group corresponds to a ℞-module with integer multiplication defined by 푛푣 = 푣 + 푣 + ⋯ + 푣. Abelian groups and ℞-modules are equivalent concepts, so 푛 times generalizing linear algebra for modules helps us study abelian groups.

If W is a free abelian group of rank m, and U is a subgroup, then U is a free abelian group of rank at most 푚. Pf. Choose a (finite) set of generators 훽 = (푢1, … , 푢푛 ) for U and a basis 훾 = (푤1, … , 푤푚 ) for 푛 W. Write 푢푗 = 푖 푤푖푎푖푗 . Then left multiplication by A is an surjective homomorphism ℞ → ℞푚 . Diagonalizing A, we can find an explicit basis for U with at most 푛 ≤ 푚 elements. 푛 푚 ℞ 퐴 ℞ 퐵 퐶 푖 U W L Generators and Relations Let A be a matrix, and let 퐴푅푛 denote the image of 푅푛 under left multiplication by 퐴. The module 푉 = 푅푚 /퐴푅푛 is presented by the matrix A. An abelian group G with n generators 푣1, … , 푣푛 can be identified with a quotient module of 푛 푇 푛 ℞ . The set of 푎1, … , 푎푛 ∈ ℞ such that 푛

푎푖 푣푖 = 0 푖=1 forms a submodule of ℞푛 . They are the relations in G, and form the kernel of the map 푛 푇 푛 ℞ → 퐺. If 푎푖1, … , 푎푖푛 ∈ ℞ generate this submodule, letting A be the matrix with these rows, 푛 푏1 푛 푏푖푣푖 = 0 ⇔ ⋮ = 퐴푣 for some 푣 ∈ ℞ 푖=1 푏푛 Then G corresponds to the module ℞푛 /퐴℞푛 ; G is presented by A. An abelian group that is presented by a matrix needs only satisfy the relations implied (through linearity) by the relations given by the columns of A. To determine a group from its presentation: 1. Use elementary row and column operations to write A in normal form. 2. Delete any columns of 0’s (trivial relations). 3. If 1 is the only nonzero entry in its column, delete that row and column. (This is a relation of the form 푣 = 0.) 4. If the diagonal entries are 푑1, … , 푑푘, and there are 푙 zero rows, then the group is 푙 isomorphic to ℞푑1 ⊕ ⋯ ⊕ ℞푑푘 ⊕ ℞ ≅ 퐶푑1 ⊕ ⋯ ⊕ 퐶푑푘 ⊕ 푙퐶∞.

Structure Theorem for Finitely Generated Abelian Groups: a.k.a Unicity Theorem for Abelian Group Decomposition, Basis Theorem A finitely generated abelian group V is the direct sum of cyclic subgroups and a free abelian

group: 푉 = 퐶푑1 ⊕ ⋯ ⊕ 퐶푑푘 ⊕ 퐿. The orders can be chosen uniquely to satisfy either of the following two conditions: 1. 1 < 푑1|푑2 ⋯ |푑푘 2. All the 푑푖 are prime power orders. (Uniqueness follows from counting orders in p- groups.)

5-5 Application 2: Linear Operators

A linear operator T on a F-vector space corresponds to a 퐹[푡]-module with multiplication by a polynomial defined by 푡푣 = 푇 푣 , 푓 푡 푣 = 푓 푇 (푣). A submodule corresponds to a T-invariant subspace. The structure theorem gives:

Rational Canonical Form: Every linear operator T on finite-dimensional V has a rational canonical form. 퐶1 푂 ⋯ 푂 푂 퐶 ⋯ 푂 푇 = 2 훽 ⋮ ⋮ ⋮ 푂 푂 ⋯ 퐶푟 where each 퐶푖 is the companion matrix of an invariant factor 푝푖. The rational canonical form is unique under the condition 푝푖+1|푝푖 for each 1 ≤ 푖 < 푟. 푘−1 푘 The companion matrix of the monic polynomial 푝 푡 = 푎0 + 푎1푡 + ⋯ + 푎푘−1푡 + 푡 is 0 0 ⋯ 0 −푎 0 1 0 ⋯ 0 −푎1 푘 퐶(푝) = 0 1 ⋯ 0 −푎2 because the characteristic polynomial of C(p) is −1 푝(푡). ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 ⋯ 1 −푎푘−1 The product of the invariant factors is the characteristic polynomial of T.

5-6 Polynomial Rings in Several Variables

푘 Let 푅 = ℂ 푥1, … , 푥푘 = ℂ 푋 푋 ∈ ℂ , and V be a finitely generated R-module with presentation matrix 퐴(푋). V is a free module of rank 푟 iff 퐴(푐) has rank 푚 − 푟 at every point 푐 ∈ ℂ푘 . The subspace 푊(푐) spanned by the columns varies continuously as c if the dimension does not “jump around.” Continuous families of vector spaces are vector bundles. V is free iff 푊(푐) forms a vector bundle over ℂ푛 .

5-7 Tensor Products

Commutative Rings: The tensor product 푀 ⊗푅 푁 of R-modules M and N is the (unique) R-module T (along with a bilinear map 푕: 푀 × 푁 → 푇) satisfying the Universal Mapping Property: for any bilinear map 푓: 푀 × 푁 → 푃, there is a unique R-homomorphism 푔: 푇 → 푃 such that 푓 = 푔푕. 푇 푕 푔 푀 × 푁 푃 푓 One way to construct it is as follows: Let F be the free module with basis 푀 × 푁, and let 퐺 be the submodule generated by 푥 + 푥′ , 푦 − 푥, 푦 − (푥′ , 푦), 푥, 푦 + 푦′ − 푥, 푦 − (푥, 푦′ ), ′ ′ 푟푥, 푦 − 푟(푥, 푦), 푥, 푟푦 − 푟(푥, 푦) for all 푥, 푥 ∈ 푀; 푦, 푦 ∈ 푁, 푟 ∈ 푅. Then 푀 ⊗푅 푁 = 퐹/퐺; 푥 ⊗ 푦 denotes 푥, 푦 + 퐺. The relations make 푥 ⊗ 푦 linear: 1. 푥 + 푥′ ⊗ 푦 = 푥 ⊗ 푦 + 푥′ ⊗ 푦 2. 푥 ⊗ 푦 + 푦′ = 푥 ⊗ 푦 + 푥 ⊗ 푦′ 3. 푟 푥 ⊗ 푦 = 푟푥 ⊗ 푦 = 푥 ⊗ 푟푦 Note that in general, an element of 푀 ⊗ 푁 is a sum (linear combination) of elements in the form 푥 ⊗ 푦.

Basic properties (prove using UMP) 1. 푀 ⊗ 푁 ≅ 푁 ⊗ 푀 2. 푀 ⊗ 푁 ⊗ 푃 ≅ 푀 ⊗ (푁 ⊗ 푃) 3. 푀 ⊗ 푁 ⊕ 푃 ≅ 푀 ⊗ 푁 ⊕ (푀 ⊗ 푃) 푚 푚 푚 4. 푅 ⊗푅 푀 ≅ 푀, 푅 ⊗ 푀 ≅ 푀 where 퐴 means the direct sum of m copies of A. 5. 푅푚 ⊗ 푅푛 = 푅푚푛 .

The tensor product 푓1 ⊗ 푓2 of homomorphisms 푓1: 푀1 → 푁1, 푓2: 푀2 → 푁2 is the map 푓: 푀1 ⊗ 푀2 → 푁1 ⊗ 푁2 such that 푓 푥1 ⊗ 푥2 = 푓1 푥1 ⊗ 푓2(푥2). Note that 푔1 ⊗ 푔2 ∘ 푓1 ⊗ 푓2 = 푔1 ∘ 푓1 ⊗ (푔2 ∘ 푓2). When applied to the linear transformations corresponding to the homomorphisms, the tensor (Kronecker) product of 푝 × 푞 matrix A and 푟 × 푠 matrix B is 푎11퐵 ⋯ 푎1푞 퐵 퐴⨂퐵 = ⋮ ⋱ ⋮ . The ordering of the basis is 푣1 ⊗ 푤1, … , 푣1 ⊗ 푤푞 , … , 푣푝 ⊗ 푤푞 . 푎푝1퐵 ⋯ 푎푝푞 퐵

For the tensor product of algebras, multiplication is given by 푎 ⊗ 푏 푎′ ⊗ 푏′ = 푎푎′ ⊗ 푏푏′.

Noncommutative rings:

The tensor product 푀 ⊗푅 푁 of right R-module 푀푅 and left R-module 푅푁 is an abelian group T (along with a bilinear map 푕: 푀 × 푁 → 푇) satisfying the Universal Mapping Property: for any biadditive, R-balanced map 푓: 푀 × 푁 → 푃 1. 푓 푥 + 푥′ , 푦 = 푓 푥, 푦 + 푓 푥′ , 푦 , 푓 푥, 푦 + 푦′ = 푓 푥, 푦 + 푓 푥, 푦′ 2. 푓 푥푟, 푦 = 푓(푥, 푟푦) there is a unique abelian group homomorphism 푔: 푇 → 푃 such that 푓 = 푔푕. 푇 푕 푔 푀 × 푁 푃 푓 One way to construct it is as follows: Let F be the free module with basis 푀 × 푁, and let 퐺 be the submodule generated by 푥 + 푥′ , 푦 − 푥, 푦 − (푥′ , 푦), 푥, 푦 + 푦′ − 푥, 푦 − (푥, 푦′ ), ′ ′ 푥푟, 푦 − (푥, 푟푦) for all 푥, 푥 ∈ 푀; 푦, 푦 ∈ 푁, 푟 ∈ 푅. Then 푀 ⊗푅 푁 = 퐹/퐺; 푥 ⊗ 푦 denotes 푥, 푦 + 퐺. The relations make 푥 ⊗ 푦 biadditive and R-balanced: 1. 푥 + 푥′ ⊗ 푦 = 푥 ⊗ 푦 + 푥′ ⊗ 푦 2. 푥 ⊗ 푦 + 푦′ = 푥 ⊗ 푦 + 푥 ⊗ 푦′ 3. 푥푟 ⊗ 푦 = 푥 ⊗ 푟푦

If M is a (S,R) bimodule and N is a (R,T) module, then 푀 ⊗ 푁 is a S-T bimodule. Definitions generalize to more modules with multiadditive balanced maps. The above properties all hold except for commutativity; (2) is replaced by 푀 ⊗푅 푁 ⊗푆 푃 ≅ 푀 ⊗푅 푁 ⊗푆 푃 ≅ 푀 ⊗푅 (푁 ⊗푆 푃).

6 Galois Theory 6-1 Symmetric Polynomials

Symmetric Polynomials A symmetric polynomial is fixed by every permutation of the variables. The elementary symmetric polynomial in n variables 푥1, … , 푥푛 of degree k is

푠푘 = 푥푖 1 ⋯ 푥푖 푘 . 1≤푖(1)<⋯<푖(푘)≤푛 Vieta’s Theorem: If 푛 푛−1 푥 − 훼1 ⋯ 푥 − 훼푛 = 푥 + 푎1푥 + ⋯ + 푎푛 푖 Then 훼푖 = −1 푎푖.

푘 푘 Newton’s identities: Let 푤푘 = 훼1 + ⋯ + 훼푛 . Then 푘 푘 푤푘 − 푠1푤푘−1 + ⋯ + −1 푠푘푤1 + −1 푘푠푘 = 0

Fundamental Theorem of Symmetric Polynomials: Every symmetric polynomial in 푅[푥] can be written in a unique way as a polynomial in the elementary symmetric polynomials. The polynomial will be in 푅[푥]. Pf. Introduce a lexicographic ordering of the monomials and use induction. Corollaries: 1. If 푓 푥 ∈ 퐹[푥] has roots 훼1, … , 훼푛 in 퐾 ⊇ 퐹, and 푔 푥1, … , 푥푛 ∈ 퐹[푥1, … , 푥푛 ] is a symmetric polynomial, then 푔 훼1, … , 훼푛 ∈ 퐹. 2. If 푝1, … , 푝푘 is the orbit of 푝1 for the operation of the symmetric group on the variables, and 푕(푥1, … , 푥푘 ) is a symmetric polynomial, then 푕(푝1, … , 푝푘 ) is a symmetric polynomial.

6-2 Discriminant

푛 푛−1 If 푃 푥 = 푥 + 푎1푥 + ⋯ + 푎푛 has roots 훼1, … , 훼푛 , then the discriminant of P is 2 퐷 = 훼푖 − 훼푗 . 1≤푖<푗≤푛 Since D is a symmetric polynomial in the 푢푖, it can be written in terms of the elementary symmetric polynomials. (It can always be defined in terms of Δ whether or not P splits completely.) 푛 퐷 훼1, … , 훼푛 = Δ 푠1, … , 푠푛 = Δ(−푎1, … , −1 푎푛 ) Ex. 1. 퐷 푥2 + 푏푥 + 푐 = 푏2 − 4푐. 2. 퐷 푥3 + 푝푥 + 푞 = −4푝3 − 27푞2.

6-3 Galois Group

Assume fields have characteristic 0. The F-automorphisms of an extension K form the Galois group 퐺(퐾/퐹) of K over F. 퐾/퐹 is a Galois extension3 if |퐺(퐾/퐹)| = [퐾: 퐹]. Ex. 퐺(ℂ/ℝ) = {퐼, conjugation}. Any two splitting fields of 푓 over F are isomorphic.

3 In general (when K is not necessarily a finite extension) a Galois extension is defined as a normal, separable field extension. A separable polynomial has no repeated roots in a splitting field; n element is separable over F if its minimal polynomial is separable; a separable field extension has every element separable over F. Pf. (a) An extension field contains at most one splitting field of 푓 over F. (b) Suppose 퐾1, 퐾2 are 2 splitting fields. Take a primitive element 훾 ∈ 퐾1 with irreducible polynomial 푔. Choose an extension L of 퐾2 so that g has a root 훾′. Then use (a).

6-4 Fixed Fields

Let H be a group of automorphisms of a field K. The fixed field of H, 퐾퐻, is the set of elements of K fixed by every group element. 퐾퐻 = 훼 ∈ 퐾 휎 훼 = 훼 ∀휎 ∈ 퐻 Fixed Field Theorem: 1. 퐾: 퐾퐻 = 퐻 : The degree of K over 퐾퐻 is equal to the order of the group. 2. 퐻 = 퐺(퐾/퐾퐻): K is a Galois extension of 퐾퐻 with Galois group H. Pf. 1. Let 훽1 ∈ 퐾 with H-orbit 훽1, … , 훽푟 . For any 푖, there is 휎 ∈ 퐻 with 휎 훽1 = 훽푖. Thus 퐻 푥 − 훽푖 |푕. Since 푔 푥 = 푥 − 훽1 ⋯ 푥 − 훽푟 ∈ 퐾 [푥] by symmetry, 푔 is the irreducible 퐻 polynomial for 훽1 over 퐾 . r divides the order of H. 2. If 퐾: 퐹 = ∞, there exist elements in K whose degrees over F are arbitrarily large. 3. By (1), 퐾/퐾퐻 is algebraic, so by (2), [퐾: 퐾퐻] is finite. The stabilizer of a primitive element is trivial, so the orbit has order 푛 = |퐻|. By (1), 훾 has degree 푛 over 퐾퐻. Then 퐾: 퐾퐻 = 푛. 4. Let 퐺 = 퐺(퐾/퐾퐻). Then 퐻 ⊆ 퐺 ⇒ 퐾퐺 ⊆ 퐾퐻. By definition, every element of G acts as the identity on 퐾퐻 so 퐾퐻 ⊆ 퐾퐺 .

Lüroth’s Theorem: Let 퐹 ⊃ ℂ be a subfield of the field ℂ(푡) of rational functions. Then 퐹 is a field ℂ(푢) of rational functions.

6-5 Galois Extensions and Splitting Fields

Splitting Fields A splitting field of 푓 ∈ 퐹[푥] over 퐹 is an extension 퐾/퐹 such that 1. 푓 splits completely in K: 푓 푥 = 푥 − 훼1 ⋯ 푥 − 훼푛 , 훼푖 ∈ 퐾. 2. 퐾 = 퐹(훼1, … , 푎푛 ) A splitting field is a finite extension, and every finite extension is contained in a splitting field.

Splitting Theorem: If K is the splitting field of some 푓 푥 ∈ 퐹[푥], then any irreducible polynomial 푔 푥 ∈ 퐹[푥] with one root in K splits completely in K. (A field satisfying the latter condition is called a normal extension of F.) Conversely, any finite normal extension is a splitting field. Pf. Suppose 푔(푥) has the root 훽 ∈ 퐾. Then 푝1 훼1, … , 훼푛 = 훽 for some 푝1 ∈ 퐹[푥1, … , 푥푛 ]. Let 푘 푝1, … , 푝푘 be the orbit of 푝1 under the symmetric group. Then ∏푖=1 푥 − 푝푖 훼1, … , 훼푛 ∈ 퐹[푥] by symmetry so it is divisible by 푔(푥), the irreducible polynomial of 훽. If K is an extension of F, then an intermediate field satisfies 퐹 ⊆ 퐿 ⊆ 퐾. A proper intermediate field is neither F nor K. Note 퐹 ⊆ 퐿 ⇒ 퐺 퐾 퐿 ⊆ 퐺(퐾 퐹).

The order of 퐺 = 퐺(퐾 퐹) divides [퐾: 퐹], since 퐺 = [퐾: 퐾퐺] and 퐾: 퐹 = 퐾: 퐾퐺 [퐾퐺: 퐹].

Characteristic Properties of Galois Extensions: For a finite extension K, the following are equivalent. 1. 퐾 퐹 is a Galois extension. 2. 퐾퐺 = 퐹 where 퐺 = 퐺(퐾 퐹). 3. 퐾 is a splitting field over F. Pf. (1)⇔(2): By the Fixed Field Theorem, 퐺 = [퐾: 퐾퐺]. (1)⇔(3): Let 훾1 be a primitive element for K, with irreducible polynomial 푓. Let 훾1, … , 훾푟 be the roots of 푓 in K. There is a unique F-automorphism 휎푖 sending 훾1 ⇝ 훾푖 for each i, and these make up the group 퐺(퐾 퐹). Thus the order of 퐺(퐾 퐹) is equal to the number of 퐺 conjugates of 훾1 in K. So 퐾 퐹 Galois⇔ 푟 = 퐺 = 퐾: 퐾 ⇔ 푓 (degree r) splits completely in K⇔ K is a splitting field.

If 퐾 퐹 is a Galois extension, and 푔 ∈ 퐹[푥] splits completely in K with roots 훽1, … , 훽푟 , then  G operates on the set of roots {훽푖}.  G operates faithfully if K is a splitting field of 푔 over 퐹.  G operates transitively if 푔 is irreducible over F.  If K is the splitting field of irreducible 푔, then G embeds as a transitive subgroup of 푆푟.

6-6 Fundamental Theorem

Fundamental Theorem of Galois Theory: Let K be a Galois extension of a field F, and let 퐺 = 퐺(퐾/퐹). Then there is a bijection between subgroups of G and intermediate fields, defined by 퐻 ⇝ 퐾퐻 퐺(퐾 퐿) ⇜ 퐿 Let 퐿 = 퐾퐻 (the fixed field of a subgroup H of G). 퐿 퐹 is a Galois extension iff H is a normal subgroup of G. If so, 퐺 퐿 퐹 ≅ 퐺 퐻. K } 퐻 = 퐺(퐾 퐿) operates on K fixing L. 퐺 = 퐺(퐾 퐹) operates on K fixing F. L { } 퐻 normal: 퐺 퐻 ≅ 퐺 퐿 퐹 operates here. F Pf. Let 훾1 be a primitive element for 퐿 퐹 and let 푔 be the irreducible polynomial for 훾1 over −1 F. Let the roots of 푔 in 퐾 be 훾1, … , 훾푟. For 휎 ∈ 퐺, 휎 훾1 = 훾푖, the stabilizer of 훾푖 is 휎퐻휎 . −1 퐻 Thus 휎퐻휎 = 퐻 ⇔ 훾푖 ∈ 퐿 = 퐾 . H is normal⇔All 훾푖 ∈ 퐿 ⇔ 퐿 퐹 Galois. Restricting 휎 to L gives a homomorphism 휑: 퐺 → 퐺(퐿 퐹) with kernel H.

6-7 Roots of Unity

2휋푖 Let 휁푛 = 푒 푛 . 퐹(휁푛 ) is a cyclotomic field. For p prime, the Galois group of 퐹(휁푝 ) is × isomorphic to 픽푝 , of order 푝 − 1. Ex. For 푝 = 2푟 + 1, G is cyclic of order 2푟 . Let 휉 be a primitive root modulo 푝, and 휎 휁 = 휁휉 . 2 2푟−1 The degree of each extension in the following chain is 2: 퐹 = 퐾⌌휎⌍ ⊂ 퐾⌌휎 ⌍ ⊂ ⋯ ⊂ 퐾⌌휎 ⌍ ⊂ 2푟 2휋 2푟−1 퐾⌌휎 ⌍ = 퐾. cos generates 퐾⌌휎 ⌍ so the regular p-gon can be constructed. 푝

Kronecker-Weber Theorem: Every Galois extension of ℚ whose Galois group is abelian is contained in a cyclotomic field ℚ(휁푛 ). Ex. If p is prime and L is the unique quadratic extension of ℚ in ℚ(휁푝 ), 1. If 푝 ≡ 1 (mod 4) then 퐿 = ℚ( 푝). 2. If 푝 ≡ 3 (mod 4) then 퐿 = ℚ(푖 푝). Show this by letting 휎 be a generator as before, take the orbit sums of the roots for ⌌휎2⌍.

Kummer Extensions Let 퐹 be a subfield of ℂ containing 휁 = 푒2휋푖 푝 , and let 퐾 퐹 be a Galois extension of degree p. Then K is obtained by adjoining a pth root (some 훽 with 훽푝 ∈ 퐹). Pf. 1. For 푏 ∈ 퐹, 푔 푥 = 푥푝 − 푏 is either irreducible in F, or splits completely in F. (Take 퐼 ≠ 휎 ∈ 퐺(퐾 퐹); then 휎푘 훽 = 휁푘푣 훽 for some 푣 ≠ 0, so G operates transitively on the roots of 푔.) 2. K is a vector space over F; each 휎 ∈ 퐺 is a linear operator. Choose a generator 휎. 휎푝 = 퐼 implies that the matrix is diagonalizable and all eigenvalues are powers of 휁. Let 훽 be an eigenvector with eigenvalue 휆 ≠ 1; then 휎 훽푝 = 훽푝 ⇒ 훽푝 ∈ 퐾퐺 = 퐹, but 훽 ∉ 퐹, so 퐹 훽 = 퐾.

6-8 Cubic Equations

Let K be the splitting field of an irreducible cubic polynomial 푓 over F with roots 훼1, 훼2, 훼3, let D be the discriminant of 푓, and let 퐺 = 퐺 퐾 퐹 . If… 퐾: 퐹 퐺 ≅ Chain Proper intermediate fields 퐷 is not a square 3 퐴3 ≅ 퐶3 퐹 ⊂ 퐹 훼1 = 퐾 None in F 퐷 is a square in F 6 퐺 ≅ 푆3 퐹 ⊂ 퐹 훼1 퐹 훼1 , 퐹 훼2 , 퐹 훼3 , 퐹(훿) ⊂ 퐹 훼1, 푎2 = 퐾 Pf. Let 훿 = 훼1 − 훼2 훼1 − 훼3 훼2 − 훼3 = ± 퐷. 훿 ∈ 퐹 ⇔ 훿 fixed by every element of G ⇔ only even permutations in G. In general, for an irreducible polynomial of any degree, 훿 = 퐷 ∈ 퐹 ⇔ G contains only even permutations.

Cubic Formula 3 2 To solve 푥 + 푎2푥 + 푎1푥 + 푎0 = 0 first substitute 푥 = 푦 − 푎2/3 (Tschirnhausen transformation) to put it in the form 푥3 + 푝푥 + 푞 = 0. 2 ′ 2 For roots 훼1, … , 훼푛 , 푧 = 훼1 + 휔훼2 + 휔 훼3, 푧 = 훼1 + 휔 훼2 + 휔푎3 are eigenvectors for 3 2 2 휎 = (123). Let 퐴 = cyc 훼1 , 퐵 = cyc 훼1 훼2, 퐶 = cyc 훼1훼2 . Then 퐵 − 퐶 = 퐷. Express 퐴, 퐵 + 퐶 in terms of elementary symmetric polynomials, apply Vieta’s formula, and solve for 퐴, 퐵, 퐶. Find 푧3, then take the cube root:

3 푞 푝3 푞2 3 푞 푝3 푞2 푥 = − + + − + + 2 27 4 2 27 4

6-9 Quartic Equations

Let 푓 ∈ 퐹[푥] be an irreducible quartic polynomial. Let 훽1 = 훼1훼2 + 훼3훼4, 훽2 = 훼1훼3 + 훼2훼4, 훽3 = 훼1훼4 + 훼2훼3. 푔 푥 = 푥 − 훽1 푥 − 훽2 (푥 − 훽3) is the resolvent cubic of 푓. What is 퐺 = 퐺(퐾/퐹)? D square D not square 푔 reducible 퐷2 (푔 splits completely) 퐷4 or 퐶4 (푔 has 1 root in F) 푔 irreducible 퐴4 푆4 퐷2 = {퐼, 12 34 , 13 24 , 14 23 } For the ambiguous case, let 훾 = 훼1훼2 − 훼3훼4, 휖 = 훼1 + 훼2 − 훼3 − 훼4. 훿훾 or 훿휖 is a square in F iff 퐺 = 퐶4. Pf. The 훽푖 are distinct since 훽1 − 훽2 = 훼1 − 훼4 (훼2 − 훼3). 푆4 operates on 퐵 = {훽푖}, giving 휑: 푆4 → 푆3. If 푔 irreducible, G operates transitively on B, so 3| 퐺 . −푏± 푏2−4푐 Special case: 푓 푥 = 푥4 + 푏푥2 + 푐 = 0. Then the roots are 훼 , 훼 = , 1 2 2 훼3 = −훼1, and 훼4 = −훼2, so 퐺 ⊆ 퐷4. Look at expressions such as 훼1훼2.

Quartics are solvable in the following way: 1. Adjoin 훿 = 퐷. 2. Use Cardano’s formula to solve for a root of the resolvent cubic 푔(푥) and adjoin it. 3. The Galois group over the field extension K is a subgroup of 퐷2. At most 2 more square root extensions suffice.

6-10 Quintic Equations and the Impossibility Theorem

Quintic Equations Impossibility Theorem: The general quintic equation is not solvable by radicals. Pf. 1. The following are equivalent: (We say that 훼 is solvable [by radicals] over F.) a. There is a chain of subfields 퐹 = 퐹0 ⊂ 퐹1 ⊂ ⋯ ⊂ 퐹푟 = 퐾 ⊂ ℂ such that 훼 ∈ 퐹푟, 퐹푗+1 = 퐹푗 (훽푗 +1) where a power of 훽푗 +1 is in 퐹푗 . b. There is a chain of subfields 퐹 = 퐹0 ⊂ 퐹1 ⊂ ⋯ ⊂ 퐹푟 = 퐾 ⊂ ℂ such that 훼 ∈ 퐹푟, and 퐹푗+1 is a Galois extension of 퐹푗 of prime degree. (Equivalent to (a) by Kummer’s Theorem.) c. There is a chain of subfields 퐹 = 퐹0 ⊂ 퐹1 ⊂ ⋯ ⊂ 퐹푟 = 퐾 ⊂ ℂ such that 훼 ∈ 퐹푟, and 퐹푗+1 is an abelian Galois extension of 퐹푗 . 2. Let 푓, 푔 ∈ 퐹[푥], and let 퐹′ be a splitting field of 푓푔. 퐾′ contains a splitting field 퐾 of 푓, and a splitting field 퐹′ of 푔. Let 퐺 = 퐺 퐾 퐹 , 퐻 = 퐺 퐹′ 퐹 , 풢 = 퐺(퐾′ /퐹). Then 퐺, 퐻 are quotients of 풢 (Fundamental Theorem) and 풢 is isomorphic to a subgroup of 퐺 × 퐻. 퐾′ 퐾 풢 퐹′ 퐺 퐻 퐹 a. Let the canonical maps 풢 → 퐺, 풢 → 퐻 send 휎 ⇝ 휎푓, 휎 ⇝ 휎푔 . Then 풢 → 퐺 × 퐻 defined by 휎 ⇝ (휎푓 , 휎푔 ) is injective. 3. Let 푓 ∈ 퐹[푥] be a polynomial whose Galois group G is simple and nonabelian. Let 퐹′ be a Galois extension of F with Galois group of prime order, and let 퐾′ be a splitting field of 푓 over 퐹′. Then 퐺 퐾′ 퐹′ ≅ 퐺. In other words, we cannot make progress solving for the roots by replacing F by a prime extension. a. From (3), |퐺| divides |풢| and |풢| divides 퐺 × 퐻 = 푝|퐺|. 2 cases: i. 풢 = 퐺 : Then 퐾′ = |퐾|, 퐻 is a quotient of |풢|, contradicting simplicity. ii. 풢 = 퐺 × 퐻 : Then 퐺 = 퐺(퐾′ 퐹′ ). 4. If 푓 is an irreducible quintic polynomial with Galois group 퐴5 or 푆5 then the roots of 푓 are not solvable over F.

a. Adjoin 훿 = 퐷 to reduce to 퐴5 case. b. 퐴5 is simple. c. By (3), in the “solvable series,” each field extension has Galois group 퐴5. 푓 remains irreducible after each extension. 5. There exists a polynomial with Galois group 푆5. a. A subgroup of 푆5 containing a 5-cycle and transposition is the whole group. b. If 푓 is a quintic irreducible polynomial with exactly 3 real roots, then the Galois group is 푆5 by (a). c. Example: 푥5 + 16푥 + 2.

In general, a polynomial equation over a field of characteristic 0 is solvable by radicals iff its Galois group is a solvable group. In (4) above, the Galois group after adjoining an element is a subgroup of the previous one, with factor group prime cyclic.

6-11 Transcendence Theory

Lindemann–Weierstrass Theorem: If 훼1, . . . , 훼푛 are algebraic numbers linearly independent over the rational numbers ℚ, then 푒훼1 , . . . , 푒훼푛 are algebraically independent over ℚ; in other words the extension field ℚ 푒훼1 , . . . , 푒훼푛 has transcendence degree n over ℚ.

[Incomplete]

7 Algebras

An algebra 풜 over a ring R is a module 풜 over R with multiplication defined so that for all 푥, 푦, 푧 ∈ 풜, 푐 ∈ 푅, 1. Associative 푥 푦푧 = 푥푦 푧 2. Distributive 푥 푦 + 푧 = 푥푦 + 푥푧, 푥 + 푦 푧 = 푥푧 + 푦푧 3. 푐 푥푦 = 푐푥 푦 = 푥(푐푦) If there is an element 1 ∈ 풜 so that 1푥 = 푥1 = 푥, then 1 is the identity element. 풜 is commutative if 푥푦 = 푦푥. 풜 is a division algebra if each element has an inverse (푥푥−1 = 1).

7-1 Division Algebras

Frobenius Theorem: The only finite-dimensional division algebras D over the real numbers are ℝ (the real numbers), ℂ (the complex numbers) and 퐻 (the quaternions). Pf. Associate with each 푑 ∈ 퐷 the linear transformation 푇푑 푥 = 푑푥. 1. Lemma: The set V of all 푎 ∈ 퐷 such that 푎2 ∈ ℝ, 푎2 ≤ 0 forms a subspace of codimension 1 (i.e. dimℝ(푉/퐷) = 1). Proof: Let 푝 be the characteristic polynomial of 푇푎 . By Cayley-Hamilton, 푝 푇푎 = 0. Since there are no zero factors in D, one irreducible real factor of 푝 must be 0 at 푎. If the factor is linear (푥 − 푟), then 푎 ∈ ℝ, 푎 = 0. If the factor is irreducible quadratic (푥2 − 2ℜ 푟 푥 + 푟 2), then this is the minimal polynomial. Since the minimal polynomial has the same factors as the characteristic polynomial, 푝 푥 = 푥2 − 2 푘 2ℜ 푟 푥 + 푟 . Since our minimal polynomial has no repeated complex root, 푇푎 is diagonalizable over ℂ (see Linear Algebra notes, 8-2). 2 a. If trace 푇푎 = 0, then the eigenvalues are pure imaginary ±푟푖, and 푇푎 is 2 2 2 2 2 diagonalizable with eigenvalues – 푟 . Thus 푇푎 = −푟 퐼푉 ⇒ 푎 = −푟 ≤ 0. 2 2 2 b. Conversely, if 푎 ≤ 0, then 푇푎 = −푟 퐼푉 for some 푟, and the eigenvalues of 푇푎 can only be ±푟푖. Then trace 푇푎 = 0. c. Hence 푉 = {푎| trace 푇푎 ≤ 0}. 푇: 푎 → trace⁡(푇푎 ) is a linear operator with range ℝ (dimension 1). Since V is the kernel, V has codimension 1. 2. From (1), 퐷 = 푉 ⊕ ℝ. Since 푉2 = {푣2|푣 ∈ 푉} gives the reals in (−∞, 0], 푉4 gives the reals in [0, ∞) and V generates D as an algebra. −푎푏 −푏푎 푎2+푏2− 푎+푏 2 3. Let 퐵 푎, 푏 = = . From the definition of V, B is an inner product 2 2 (positive definite, symmetric). Choose a minimal subspace W of V generating D as an algebra, and let 푒푖 1 ≤ 푖 ≤ 푛 be an orthonormal basis with respect to B. Then 2 i – 푒푖 = 1, ii 푒푖푒푗 = −푒푗 푒푖(푖 ≠ 푗). a. If 푛 = 0, 푉 ≅ ℝ. 2 b. If 푛 = 1, 푉 ≅ ℂ. (푒1 = −1) c. If 푛 = 2, 푉 ≅ 퐻. (Check the relations.) 2 d. If 푛 > 2, then using then from (ii) 푒1푒2푒푛 = 1 ⇒ 푒1푒2푒푛 = ±1 ⇒ 푒푛 = ±푒1푒2. So span 푒푖 1 ≤ 푖 ≤ 푛 − 1 ⊂ 푊 with basis 푒푖 1 ≤ 푖 ≤ 푛 − 1 also generates D as an algebra, contradicting minimality.

Simple and Semisimple Algebras

[See group theory notes.] [Add some stuff here.]

References

[A] Algebra by Michael Artin [EoAA] Elements of Abstract Algebra by Richard Dean [PftB] Proofs from the Book by Titu Andreescu and Gabriel Dospinescu [TBGY] Abstract Algebra: The Basic Graduate Year, by Robert Ash Wikipedia