Further Algebra Lectures Notes

8/13/2019 Further Algebra Lectures Notes

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Further Algebra

Catarina [email protected]

University of Hertfordshire

Semester A 2013/ 2014

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Useful Information

Program:1. Polynomials

2. Rings of polynomials

3. Groups

4. building up to

5. Galois theory

Assessment:Exam at the end of term 80% + Coursework 20%.

You need to pass at each component!

Coursework hand-in date: tbd, what coursework would you

like?

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What is this?

2x3

3x2 + x+ 5

What isx?

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1. IfuZ is s.t. u

|1 thenuis called aunit.

2. Ifa Z\{0}is not a unit we say that it is irreducibleiffa= bc implies that eitherborcis a unit.

3. Ifa Z\{0}is not a unit we say that it is primeiffa|bcimplies thata

|bora

|c(or both).

In 2) can bothb, cbe units?

Irreducibles and primes coincide in Z, but not in every number

system!

Greatest common divisor?

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Polynomials

Apolynomialinx, with rational coefficients, is an expression

f = a0 + a1x+ a2x2 + + amxm

for somem Z andai Q. Ifam= 0 thenmis called thedegreeof the polynomialf, deg f.

We might also call itf(x).

The set of all polynomials with coefficients in Q is denoted Q[x](Watch the square brackets!!!!!!)

We can define in a similar way Z[x],R[x], . . .

Example

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Addition

(a0 + a1x+ + amxm

) (b0 + b1x+ + bnxn

) =(a0 + b0) + (a1 + b1)x+ (a2 + b2)x2 +

Example

Multiplication

(a0 + a1x+ + amxm) (b0 + b1x+ + bnxn) =a

0b

0+ (a

0b

1+ a

1b

0)x+ (a

0b

2+ a

2b

0+ a

1b

1)x2 +

Example

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(Q[x],,) forms a ring, thering of rational polynomialsProof.

Exercise. What is the zero polynomial?

What is a ring?

Definition

A ring is an algebraic structure (R,+, ) with two binaryoperations, "addition" + and "multiplication", that satisfy

(R,+) is an abelian group,

(R, ) is a semigroup (is associative), a (b+ c) = a b+ a c

(b+ c) a= b a+ c a for alla,b, cR.

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Does the ring (Q[x],,) have an identity?What is it?

The idea is that (Q[x],,) shares a lot of properties withanother very important ring: (Z,+, ), the ring of integers.

Let 1 = 1 + 0x+ 0x2 + c

+ 0xn be theidentitypolynomial.

Theorem

Given a polynomial f Q[x] there exists a polynomial g Q[x]s.t. f

g= 1iff a0

= 0and ai= 0for all i>0.

Proof.

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Theorem

Q is isomorphic to a subring ofQ[x]

isomorphic? subring?

We will get to it, let us first see how similar Q[x] is to Z.

Definition

Givenf,g Q[x] we say thatfdividesg, and writef|g, if thereexistsh

Q[x] s.t. g= f

h. Iffdoes not dividegwe write

f|g.

We might dropf htofhif and when there is no confusion.

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Example

where we do long division

Definition

Iff

Q[x] is such thatf

|1 thenf is called aunit.

Example

The units in Q[x] are precisely the non-zeroconstant

polynomials.

follows from theorem on page 9

What are the units in Z[x]?

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Definition

Iff Q[x] is neither the zero polynomial, 0, nor a unit we saythat it isirreducibleif wheneverf = gh, withg,h Q[x], itfollows that eithergorhis a unit. A non-zero polynomial that is

not irreducible is calledreducible.

Example

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Definition

Iff Q[x] is neither 0 nor a unit we say that it isprimeifwhenever f|gh, withg,h Q[x], it follows thatf|gorf|h(orboth).

Irreducible polynomials, for our purposes, are exactly the prime

polynomials.

We will call them irreducible.

Remember that they do for Q[x] what the prime numbers do forZ.

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Which polynomials are irreducible?

All these notions can be naturally extended (from Q) to anycommutative ring with identityR, i.e. we can speak of

reducibility of a polynomial inR[x], whereRis a commutativering with 1.

Example

2x2 + 2 is irreducible in Q[x] and R[x], and reducible in Z[x]and C[x]

Why doesRneed to be commutative and have a 1? How does

this affect the polynomials?

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Theorem (Gauss)If F Z[x] is s.t. F = gh with g, h Q[x] then we can writeF = GH with G,H Z[x],deg G= deg g anddeg H= deg h.

Corollary

If f Z[x] is irreducible inZ[x] then it is irreducible inQ[x].

Proof.

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Theorem (Eisensteins Test)

Let f = a0 + a1x+ + anxn

Z

[x]. If there exists a primep Z s.t.1. p|aifor all i= 0, . . . , n 12. p|an3. p2

|a

0then f is irreducible inQ[x].

Proof.

Example

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Definition

A polynomiala0 + a1x+ + anxn is calledmonicifan= 1.

The following holds for any field K in place of Q:

Theorem (Unique Factorization of Polynomials)Let f Q[x] be non-zero. Then either f is a unit or it can beexpressed as a product of a unit and finitely many monic

irreducible polynomials. Moreover, this factorization is unique

up to the product by constants.

we get this by finding the "roots" of the polynomial

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Definition

A field is an algebraic structure (F,+, ) with two binaryoperations, "addition" + and "multiplication", that satisfy (F,+, ) is a ring, (F\{0}, ) is an abelian group.

Equivalently, a field is a commutative ring with identity whichcontains a multiplicative inverse for every nonzero element, e.g.

R,Q,C.

DefinitionLetKbe a field andf K[x]. An elementr K is arootoff inK iff(r) = 0.

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Lemma

IfK is non-zero, r is a root of f K[x] iff it is a root off .

Proof.

Example

Lemma

Let K be a field and f K[x]. An element r K is a root of f iff(x r)|f .

Proof.

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Lemma (The Rational root test)Let f = a0 + a1x+ + anxn Z[x] with f= 0. If r/s is a rootof f, with r, s Z, s= 0andgcd(r, s) = 1, then r|a0 and s|an.

Remark:Iff K[x] is s.t. deg f = 2,3, thenfis reducible inKiff it has a polynomial of degree 1 as a factor iff fhas a root in

K.

Example

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Definition

LetKbe a field andf K[x]. An elementr K is asimplerootoff if (x

r)

|f but (x

r)2

|f. We say thatris a root off of

multiplicitymif (t r)m|f but (t r)m+1 |f.

Lemma

Let K be a field and f K[x] s.t. f= 0. Then f has a finitenumber of roots in K. If these roots are r1, . . . , rtwith multiplicitym1, . . . ,mt, respectively, then

f = (x

r1)

m1 (x

r2)

m2 . . . (x

rt)

mtg,

where gK[x] has no roots in K . Conversely, iff = (x r1)m1 (x r2)m2 . . . (x rt)mtg,and g has no roots in K,then the roots of f in K are r1, . . . , rtwith multiplicitiesm1, . . . ,mt, respectively.

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Theorem

Let K be a field and f a non-zero polynomial in K[x].Thenumber of roots of f in K , with multiplicities, cannot exceed the

degree of f.

Proof.Immediate consequence of previous lemma.

ExampleWhat can we say of polynomials in Q of degree 3...

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Theorem (Fundamental Theorem of Algebra)

Let f C[x] be s.t. deg f 1. Then f has a root inC. It followsthat f factorises completely into

f = (x r1)m1 (x r2)m2 . . . (x rt)mt

with r1, . . . , rt C and m1 + + mt= deg f .

Example

Applying this to R[x] we get:

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Theorem

If f R[x] then f can be expressed as a product of polynomialsof degree at most2 inR[x].

Proof.

Corollary

An irreducible polynomial inR[x

]is either of degree1or2.

How to factorize it falls into the numerical analysis field...

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Definition

LetKbe a field and

f = a0 + a1x+ + an1xn1 + anxn K[x]

witha0, an= 0. Thereciprocalpolynomial offis the polynomial

g= an + an1x+ + a1xn1 + a0xn K[x].

LemmaA polynomial f[x] K[x] withdeg f>0and f(0) = 0, isirreducible in K iff its reciprocal is irreducible in K .

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break

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Groups

What you have already seen:

A group (G, ) is a setGtogether with a binary operation: G GGsatisfying...

subgroupHG order of an element and of a subgroup homomorphism of groups, kernel, image normal subgroup; quotient group first isomorphism theorem

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A groupGiscyclicif there existsgGs.t.

G= {1, g,g2, . . . ,gn1}

in this case we writeG= g, and we know that|g| = |G| = nisthe order ofg, i.e. gn = 1.

ExampleZ5 with addition mod 5, or as you might have called itC5.

In fact, for any g

G, thesubgroup generated byg is

g = {gk : k Z}.And, given any subsetSG, thesubgroup generated byS is

S = {sk11 sk22 . . . sknn : n N, si S, ki Z}.

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Definition

LetGbe a group andx, yG. Thecommutatorofxandy isthe element

[x, y] = x1y1xyG.

Lemma

We have xy= yx if and only if[x, y] = 1.

Proof.

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Definition

Thederived subgroupof the groupGis the subgroup

generated by the commutators,

G = [x, y] : x, yG.

Lemma

The group G is abelian iff G = {1}.

Proof.

Example

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Remember

TheoremIf G is a finite group with|G| = p, where p is a prime, then G isa cyclic group. Furthermore G= Zp.

Where Zpis an additive group, working under addition modulo

p.

The infinite cyclic group is (Z,+).

Cyclic groups are abelian, why is that?

Zn= Z/(nZ), remember cosets and factor groups?

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Let G be a group and H K G Define the product of HK to be


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LetGbe a group andH,K G. Define the product ofHKto beHK = {hk : hH, k K}.

Proposition

Given H,K subgroups of the group G, we have HK G iffHK = KH.

Proof.

Proposition

Let G be a group and H,K G, with H,K finite. Then

|HK| = |H||K||H K| .

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TheoremA finite cyclic group, of order m, has a unique subgroup of order

q for every divisor q of m. Furthermore, that subgroup is cyclic.

Proof.

What can we say aboutC6, the cyclic group of order 6?

What about Z7?

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Let (G ) be a group and g1 the inverse of g G For any


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Let (G, ) be a group, andg the inverse ofgG. For anyg, xG, the elementsgxg1 Gis theconjugateofxbyg.IfHis a subgroup ofG, then

gHg1 = {ghg1 : hH} Gis theconjugate subgroupofHbyg.

We say that a subgroupHofGis anormal subgroupif for all

gGwe havegHg1 H, or equivalentlygHg1 = H; and inthis case we writeHG.

Proposition

Let G be a group and H,K G. If H G or K G thenHK G.

Proof.

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Th (S d I hi Th )


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Theorem (Second Isomorphism Theorem)

Let G be a group. Let HG and N G. Then H N H, and

HNN

= HH N.

Proof.

Theorem (Third Isomorphism Theorem)

Let G be a group, and let H,N G be s.t. N

H. Then

H/N G/N andG/N

H/N= G

H.

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Definition

A groupGis calledsimpleif its only normal subgroups are{1}andG itself.

Example

Every cyclic group of prime order is simple.

permutation recall

Snis the group of all permutations on the elements{1, . . . , n},under composition a.k.a multiplication

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Theorem

Two permutations in Snare conjugate iff have the same cycle

structure.

Proof.

A permutation isevenif it can be written as a product of an

even number of transpositions (i.e. cycles of length two).

An=

{

Sn : is even

}is theAlternating groupof degreen.

Lemma

AnSnwith index|Sn: An| = 2, and order|An| = n!/2.

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Proposition

Given n>1, we have

Sn= (1 2), (1 2 n).

Proof.

Finite simple groups form the building blocks of finite groups,they have only recently been all classified. We look at a family

of them:

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Theorem

A5 is simple.

Proof.

Theorem

Anis simple for all n5.

What is going on withA4 andA3?

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Soluble groups were first studied by Galois while working on

the solution of equations by radicals

DefinitionA group issolvable(orsoluble) if it has a finite chain of

subgroups

{1}

= G0

G1

G2

Gn= G

satisfying the following for alli= 0, . . . , n 1: Gi Gi+1 and Gi+1/Gi is abelian.

Example

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Solubility is inherited by subgroups and preserved under

quotients.

Theorem

Let G be a group, let HG and N G. Then

1. if G is soluble then H is soluble;2. if G is soluble then G/N is soluble;

3. if N and G/N are soluble then G is soluble.

Proof.

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We can say that the groupGis anextensionof the groupAby

a groupB ifG/A= B. (Remark on the quotient by A!)

The class of Abelian groups is closed under taking subgroups

and quotients, but not under taking extensions, but the class of

soluble groups is also closed under taking extensions, one of

the reasons why they are important.

Theorem

A soluble group is simple iff it is cyclic of prime order.

Proof.

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Theorem


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Theorem

The symmetric group Snis not soluble for n5.

Proof.

Reminder:

Definition

We say that two elements of a group a,bGareconjugateifthere existsgGs.t. a= g1bg.Conjugacy is an equivalence relation (check it as an exercise).

The conjugacy classes ofGareC1= {1},C2, . . . ,Cn, andbecause these form a partition we have

|G| = 1 + |C2| + + |Cn|.

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Definition

LetGbe a group andxG. Thecentralizerofx inGis thefollowing subgroup ofG

CG(x) = {gG: xg= gx}.

Can we check that it actually forms a subgroup.

Lemma

Let G be a group and xG. The number of elements in theconjugacy class of x is the index of C

G(x

)in G,

|G

:C

G(x

)|.

Proof.

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Corollary


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Corollary

The number of elements in any conjugacy class of a finite

group G divides the order of G.

Definition

Thecentreof a groupGis the set of elements of Gthat

commute with every element

Z(G) = {xG: xg= gxfor allgG}.

LemmaZ(G) G

Proof.

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Definition

Letpbe a prime. A finite groupGis ap-groupif|

G|

= pk forsomek N.

Example

Theorem

If G is a p-group then it has non-trivial centre.

Proof.

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Lemma


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If G is a finite p-group of order pn then G has a chain of normal

subgroups

G0 = {1} G1G2 Gn= G

s.t.|Gi| = pi for all i= 0, . . . , n.

Proof.

Corollary

Every finite p-group is soluble.

Proof.

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We will need the first part of the following result

Theorem (Sylows Theorem(s))

Let G be a finite group of order pnr where p is a prime that

does not divide r. Then

1. G has at least one subgroup of order pn;

2. all such subgroups are conjugate in G;

3. any p-subgroup of G is contained in one of order pn;

4. the number of subgroups of G of order pn leaves

remainder1on division by p.

Proof.

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Definition


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Definition

LetGbe a group with|G| = pnr, wherepis a prime that doesnot divider. A subgroup ofGof orderpn is called aSylow

p-subgroup.

Theorem (Cauchy)

If a prime p divides the order of a finite group G, then G has asubgroup of order p.

Proof.

Example

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break

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Rings


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g

AringR= (R,+, ) is a non-empty setRwith two binaryoperations + (addition) and(multiplication), with the followingproperties for alla,b, cR:

1. (a+ b) + c= a+ (b+ c), associative law for addition;

2. a+ b= b+ a, commutative law for addition;3. there exists 0Rs.t. a+ 0 = a, thezeroof the ring;4.a as.t. a+ (a) = 0, inverses for addition,negatives;5. (ab)c= a(bc), associative law for multiplication;

6. a(b+ c) = ab+ acand (a+ b)c= ac+ bc, distributivelaws.

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Acommutative ringRsatisfies the following extra property

ab= ba

for alla,bR. (We dropped the signfor multiplication)

A ringR is aring with unityif there exists 1R(the unityelement), 1= 0, s.t. for allaR

a1 = 1a= a.

Example

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A commutative ring with identity, R, is called afieldif all


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non-zero elements have multiplicative inverses, i.e. for all a= 0inRthere existsa1 Rs.t. aa1 = 1.Definition

A subsetTof a field (K,+, ) is asubfieldif the following hold1. Tis closed with respect to + and2. (T,+,

) is a field.

Theorem

Let T be a non-empty subset of a field K. Then T is a subfield

of F iff T has at least two elements and the following hold

1. If a,bT then a+ b,abT .2. If aT thenaT and if a= 0then a1 T .

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D fi iti


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Definition

LetRandSbe rings. A mappingf : RS is a(ring)homomorphismif for allx, y Rwe have

f(x+ y) = f(x) + f(y), f(xy) = f(x)f(y).

Example

The mappingf : Z Zn,f(x) = x( mod n) is ahomomorphism.

Example

The mappingf : Z Z,f(x) = 2xis not a homomorphism.

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Example


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Example

For any two ringsRandS, the mappingf : RS,f(x) = 0(the zero ofS) is a homomorphism.

Theorem

Let R and S be rings with zero elements0Rand0Srespectively, and let f : RS be a homomorphism. Then

1. f(0R) = 0S;

2. f(a) = f(a) for all aR;3. f(R) is a subring of S.

Proof.

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Corollary

If f :

R

S is a homomorphism of rings and a,

b

R then

f(a b) = f(a) f(b).

If the ring homomorphismf : RSis one-one we call it amonomorphism, or anembedding. Iffis one-one and onto we

call it anisomorphismand we say thatRandSareisomorphic

and writeR= S.Example

An isomorphism fromRinto itself,f : RR, is called anautomorphism.

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Theorem


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Znis a field iff n is prime.

Definition

LetFbe a field. Theprime subfieldofFis the intersection of all

subfields ofF.

Proposition

Let F be a field and H its prime subfield. Then H= Q orH= Zpfor some prime p.

LetFbe a field andH its prime subfield. IfH= Q we say thatF hascharacteristic0; ifH= Zpthen we say that F hascharacteristicp.

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Example

Q,R,C all have characteristic 0, and Zphas characteristicp.

We denote bycar(F) the characteristic of the field F.

Proposition

Let K be a subfield of the field F. Then the car(K) = car(F).

Proof.

It follows from the fact thatK andFhave the same prime

subfield.

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Field extensions


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Take the polynomialsf(t) = t2 2,g(t) = t2 + 1 Q[t].

We know that neither of them has roots in Q, butf(t) has roots

2 R, andg(t) has rootsi C.

Hence, given a fieldFand a polynomialh(t) F[t] with noroots inF, it makes sense to look for a "bigger" fieldK whereh(t) has roots. And, it might be useful to find such a fieldK assmall as possible. This yields two questions:

givenh(t) F[t] can we find a fieldKcontainingF inwhichhhas at least oneroot;

and, can we find a fieldL, containingF, in whichhcan bedecomposed into linear factors, i.e. wherehhasallits

roots.

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Example


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Example

IfFis a field s.t. Q F C then we know thath(t) F[t]always has roots in C, a field we already know well.

If this is not the case, we need to construct an abstract field

which containsF in whichhhas roots.

ClassicGalois Theorydeals with the first case, and thats what

we are sticking to!

He constructed a theory based on polynomials over thecomplex field, from a "modern" point of view this was

generalized to arbitrary fields. From a polynomial we obtain a

field extension related to it.

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Definition

IfK, Lare fields and : K

Lis a monomorphism, we say

thatL is anextensionofK, and write "L : K is a field extension".

This is equivalent to saying (up to isomorphism) thatK is a

subfield ofL, sinceK= (K), so we can also writeK L.

We can think ofLas a vector space overK, with respect to the"natural" operations:

addition as the addition in L: x+ yfor allx, yL; scalar multiplication as multiplication inL: kxfor allk K

and allxL.We can now use tools from linear algebra...

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If L is a vector space there is a basis of L over K i e elements


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IfLis a vector space there is abasisofLoverK, i.e. elements

we need to add to Kto obtain the entire fieldL

Example

Definition

LetL : Kbe a field extension. Thedegreeof this extension isthe cardinality of the basis ofLoverK, i.e. the dimension ofL

considered as a vector space over K. We denote it [L : K]. If[L : K] is finite we say thatLis afinite extensionof K.

Example

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Theorem

Let L K be a field e tension Then L K iff [L K ] 1


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Let L : K be a field extension. Then L = K iff[L : K] = 1.

Proof.

Reminding Lagranges Theorem, we have:

TheoremLet L : K and K : M be field extensions. Then

[M : L][L : K] = [M : K].

Proof.

Proof idea: if{a1, . . . , ar}is a linearly independent subset of MoverL (i.e. a basis), and{b1, . . . , bs}is a l.i. subset ofL overK,then aibj forms a l.i. subset of M over K . 63/106

Corollary


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Let K1, . . . ,Knbe fields, and suppose that Ki+1 : Ki is an

extension for i= 1, . . . , n 1. Then[Kn : K1] = [Kn: Kn1][Kn1: Kn2] [K2: K1].

Definition

LetFbe a field, and X L. We denote byXthesubfieldof Fgenerated by X, i.e. the smallest subfield of FcontainingX.

Example

Taking the usual addition and multiplication,1 = Q. So, Q isthe subfield of C generated by 1.

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Definition

LetKbe a subfield of the fieldF, and

F. We denote by

K() the subfieldK {}, i.e. the smallest subfield thatcontainsK and.

In fact, we have

K() ={M : Mis a subfield ofF, s.t.K M, M}.

Also, ifX F thenK(X) is the smallest subfield ofFcontainingbothK andX. And ifX = {x1, . . . , xn}we writeK(x1, . . . , xn).

Example

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Definition


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LetL : Kbe a field extension, andL. We say that isalgebraic overK if there exists a non-zero polynomial

p(t) K[t] s.t. p() = 0. Otherwise we say thatistranscendental over K.

Example

Definition

LetL : Kbe a field extension. We say that Lis analgebraic

extensionofK if is algebraic overK for allL. Otherwisewe say thatLis atranscendental extensionof K.

Example

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Lemma

Let L : K be a finite field extension. Then L is algebraic over K .

Proof.

Definition

LetL : Kbe a field extension. We say that Lis asimple

extensionofK if there existsL. s.t. L = K().

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Example

Theorem

Let L : K and M : L be field extensions, and let

M. If isalgebraic over K , then it is also algebraic over L.

Proof.

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Th


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Theorem

Let L : K be a field extension, and

L algebraic over K .

There exists a unique monic polynomial mK[t] of smallestdegree s.t. m() = 0. Furthermore,

m is irreducible over K , and

if f

K[t] is s.t. f() = 0then m divides f.

Proof.

The monic polynomialm, defined above, is called the

minimum polynomialof the element.

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The following result gives the basis of the theory of finite


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extensions: allows us to calculate [K() : K]; show that thefinite simple extensionK() is unique (up to isomorphism);and, later, to calculate the corresponding Galois group.

Theorem

Let K() be a simple extension of the field K , where isalgebraic with minimum polynomial m over K . Then eachK() has auniqueexpression as = f(), where f K[t]is a polynomial with degree less than the degree of m.

In other words, if the degree of m is n, there existunique

ki K s.t. = kn1

n1 + + k1 + k01.

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CorollaryLet K() be a simple extension of the field K , where isalgebraic with minimum polynomial m over K . Assume that

m has degree n. Then{n1, n2, . . . , ,1}forms a basis forK() over K and so[K() : K] = n.

Proof.

Immediate consequence of the previous theorem.

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Galois group


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Definition

LetKbe a subfield of the fieldL. An automorphism : LLis aK-automorphismofLif(k) = kfor allkK, i.e. actsas the identity inK.

Theorem

If L : K is a field extension then the set of all K -automorphismsof L forms a groups under composition of maps.

Proof.

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Definition


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Definition

TheGalois group (L : K) of the extensionL : Kis the group ofallK-automorphisms ofL, under composition of maps.

Example

Example

Under certain conditions, there is a correspondence between

the subgroups of the Galois group ofL : Kand the subfieldsMs.t. K ML, which we will look into.

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What we are aiming towards

Theorem

A polynomial f with coefficients in a field K of characteristic

zero is soluble by radicals if and only if its Galois group is

soluble.

And then show thatf(t) = t5 8t+ 2 is not soluble by radicals.

For this we still need...

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Splitting Fields


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Examplex2 + 2 extends Q to Q(i

(2))...

DefinitionLetKbe a field andfa polynomial ofK. We say thatf splits

overKif it can be written as a product of linear factors

f(t) = k(t

1)

(t

n)

wherek, 1, . . . , nK.

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Definition


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Definition

LetKbe a field and

K[t]. We say thatT is asplitting field forfover the field K ifK is a subfield ofT and

1. fsplits overT;

2. IfKis a subfield ofT T andfsplits overT thenT = T.

Condition 2 is equivalent to saying that T = K(1, . . . , n),where1, . . . , nare the roots off inT.

We construct a splitting field by adjoining to Kthe roots off, we

already know how to do this for irreducible polynomials, so we

writefas a product of irreducibles and work on each factor

separately.

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We can now calculate the order of the corresponding Galois

group:

Corollary

If L : K is a finite separable normal extension of degree n, then

there are precisely n distinct K -automorphisms of L, so that|(L : K)| = n.

All our extensions will be separable and normal!

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Solutions of equations by radicals


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Definition

An extensionL : K isradicalifL = L(1, . . . , m) where foreachi= 1, . . . ,mthere exists an integer ni s.t.

ni

i K(1, . . . , i1).The elementsiare said to form aradical sequenceforL : K

Example

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Definition


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e t o

Letfbe a polynomial over a field Kof characteristic 0, and let

Tbe a splitting field for f overK. We say thatf issolvable by

radicalsif there exists a fieldMcontainingT s.t. M : K is aradical extension.

Zeros

Theorem

If K is a field of characteristic 0 and K

L

M where M : K is

a radical extension, then the Galois group of L : K is a solublegroup.

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D fi iti


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Definition

Letfbe a polynomial over a field K, with a splitting fieldT overK. TheGalois group off overKis the Galois group (T : K).

Groups of permutations! Restating a seen theorem.

Theorem

Let f be a polynomial over a field K of characteristic 0. If f is

solvable by radicals then the Galois group of f over K is a

soluble group.

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An insoluble quintic


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Lemma

Let p be a prime, and f an irreducible polynomial of degree p

overQ. Suppose that f has precisely 2 non-real zeros inC.

Then the Galois group of f overQ is the symmetric group Sp.

Theorem

The polynomial t5 6t+ 3overQ is not soluble by radicals.

Proof.

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"There are more ways of killing a quintic thanchoking it with radicals"

Ian Stewart inGalois Theory

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Finite FieldsWhere we classify all finite fields


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Theorem

Let K be a field of characteristic p. Then, for all x, yK wehave

(x+ y)p = xp + yp and (x y)p = xp yp.

Proof.

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Definition

Let K be a field The formal derivative of a polynomial


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LetKbe a field. Theformal derivativeof a polynomial

f = a0 + a1t+

+ antn

K[t] is

Df = a1 + 2a2t+ + nantn.

Theorem

Let K be a field and f K[t], and let T be a splitting field for fover K. Then the roots of f in L are all distinct if and only if f

and Df have no non-constant common factor.

Example

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Theorem

1 L t K b fi it fi ld Th |K | n f i d


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1. Let K be a finite field. Then|K| = pn for some prime p andsome n N. Every element of K is a root of the polynomialtp

n t , and K is a splitting field of this polynomial over theprime subfieldZp.

2. Let p be a prime and let n N. There exists, up toisomorphism, exactly one field of order pn.

Proof.

WOW

Only fields of prime power order exist, and there is only one of

each!

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LetGF(pn) be the finite field of orderpn, we call it the Galoisfield of orderpn.


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TheoremThe group of non-zero elements of the Galois field GF(pn) iscyclic.

Proof.

We constructGF(pn) by finding and irreducible polynomial f of

degreenin Zp[t], thenGF(pn

) = Zp[t]/f.

Example

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Notation

N is the set of natural numbers{1,2, 3, . . .}


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N0= N

{0

} Z is the set of integers Q is the set of rational numbers, i.e.{pq : p Z,q Z\{0}} R is the set of real numbers

R+ is the set of positive real numbers

R+0 = R+ {0} R is the set of negative real numbers R0 = R {0}

means "for all", e.g. "x R" means "for allxin the set ofreals"means "there exists some", e.g. "x R s.t." means "thereexists somexin the set of reals such that"

1 means "there exists a unique"

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Greek alphabet


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alpha iota rhobeta kappa sigmagamma lambda taudelta mu upsilonepsilon nu phi

zeta xi chi eta oomicron psi theta pi omega

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Further Algebra Lectures Notes

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Transcript of Further Algebra Lectures Notes