Algebra through Examples

Lesson 1

General DetailsE-mail: Josie.Shamash@weizmann.ac.ilRecommended reading:

- Basic Algebra 1/2 by Jacobs- TODO: Fill from others

Administrative Details:- There will be 5 assignments. Each around 5%- 1 home exam – usually around 80% (best 4 assignments out of the 5 are chosen)

The Axiums of a FieldA field F has two binary operations: +, ∙ such that ∀a ,b , c ,d∈F :F is closed under them

Addition(1a) Commutativity: a+b=b+a(1b) Associativity: (a+b )+c=a+(b+c )(1c) Neutral element: a+0F=a(1d) Inverses ∀a∃a ,a+ (a )=0F

Multiplication(1m) Commutativity: a ∙b=b ∙a(2m) Associativity: (a ∙b ) ∙ c=a ∙(b ∙ c )(3m) Identity: a ∙1F=a

(4m) Inverses: ∀a≠0F∃a1 . a∙ (a1 )=1F

We also demand that 0F ≠1F

DistributivityTo connect the two definitions (as they can be independent according to the current definition) we add distributivity, which states that:a ∙ (b+c )=a ∙b+a ∙ c

NamingAny set satisfying (¿) is called a group (an additive group)If also commutatibity is satisfied, we denote it as a commutative (abelian) group.If the operation is denoted by multiplication, we call it a multiplication group.(2m, 3m, 4m is satisfied).Usually denote operation by +¿ only for abelian groups.

A RingA ring is any structure that satisfies (1-4a), (2m), (3m) & Distribution.If the multiplication is commutative, it is called a commutative ring.

If (4m) holds (not necessarily with(1m)), then it is called a division ring.

A ring without (3m) is sometimes referred to as a rng. (a ring without the i).

Examples

Fields- Q- R- C- Zp={0,1 ,…, p−1 } with respect to addition and multiplication mod p. For instance, in Z5 – 2 ∙3=1 (mod p )

RingsSince fields support additional properties than ring, any field is a ring.For instance - Z

And in addition, here are a few "pure" rings:- R [ x ]=¿ Ring of polynomials with real coefficients- M n (R )=¿ Ring of n× n matrices over R - Not commutative!

- M n (F )=¿ Ring of n× n matrices over some field F - Not commutative!- F [ x ]=¿ Ring of polynomials over some field F- Z [ x ]=¿ Ring of polynomials over Z- Z ×Z= {( a ,b )|a ,b∈Z } with coordinate-wise addition and multiplication:

(a1 , b1 )+( a1+b1 )=(a1+a2 , b1+b2 ) - If R ,S are Rings →R ×S is a Ring.- Z [ x , y ]=¿ polynomials in x∧ y with coefficients in Z.

Commutative Rings- A sub-Ring if R is a Ring.

S is a sub-Ring if 1F ,0F∈S and S is a Ring in respect of operations in Rfor instance, M n (R ) is a sub-Ring of M n (Q )

IdealsIf R is a Ring, I⊆R is an Ideal if and only if:

- I is an additive subgroup of R- ∀a∈R ,b∈ I . a∙ b , b ∙ a∈ I

(R ∙ I ⊆ I∧I ∙R⊆ I )Note that if 1F∈ I →R=I

ExamplesIn any Ring R:

- {0 }, R are Ideals (Trivial)In a commutative Ring, if b∈R → R ∙bis an Ideal. Is also called principal Ideal and is denoted by (b)

- a1b+a2b=(a1+a2 ) b+R ∙b

- a ' (b ∙a )= (a ∙b ) a'=(a ' ∙ a ) b∈R ∙b

In case of a non commutative Ring, a left Ideal is an additive subgroup satisfying multiplication on the left. In the same way, a Right Ideal satisfies multiplications on the right.

Ideals in Z- 2Z- 7 Z- n Z(∀n∈Z )

In fact, every Ideal in Z is a principal Ideal!

ProofLet I be an Ideal in Z (notation: I⊲R)If I={0F } it is a principal!

So assume I ≠ {0F }. Let n be the smallest positive integer in I .(I is closed under addition inverse so must have one!).Let m∈ I .We can find q ,r∈Z s.t. m=q ∙n+r ,0≤ r<n

m⏟∈ I

−q ∙n⏟∈I

=r∈ I

But we know r<n→ Contradiction by minimality in choice of n. So r must be 0!Therefore:

m=q ∙n∈nZSo we proved that ∀m∈ I .m∈nZ→ I⊆n ZBut also n Z⊆ I since n∈ I !Therefore n Z=I .

More Ideal ExamplesM 2 ( R ) is a non-commutative Ring

k={[a bc d ]|a ,b , c∈R} is a subring but not a left or right Ideal.

e.g.

[1 11 1] ∙[a b

0 c ]=[a b+ca b+c]∈ k only if a≠0

[a b0 c ] ∙[1 1

1 1]=[a+b a+bc c ]∈ k only if c≠0

However, I={[a b0 0]|a , v∈R} is a right Ideal!

e.g.

[a b0 0] ∙[ x y

u v ]=[¿ ¿0 0 ]∈ I

It is not, however, a left Ideal:

[ x yu v ] ∙[a b

0 0]=[ax ¿ua ¿ ]if ua≠0→∉ I

Fields have no non-trivial ideals.

Quotients of RingsLet R be a Ring and I an Ideal.∀ a∈R define:I+a= { x+a|x∈ I }−¿ co-set or I determined by a.RI={ I+a|a∈R } (equality sets)

Quotient Ring – we define operations +, ∙ to get a ring(Note: co-sets are disjoint or equal. Proving it would be an assignment).

Define ( I+a )+( I +b )=I+(a+b)Define ( I+a ) ∙ ( I +b )=I+(a ∙b)

Must show the definition does not depend on co-sets representatives:Suppose I+a=I+a ' and I+b=I+b 'Need to show: I+ (a'+b' )=I+(a+b) and I+a' ∙ b '=I +a∙b

∃ x∈ I a'=x+a∃ y∈ I b'=x+b

So - I+ (a'+b' )=I+( x+a+ y+b )=I +( x+ y )⏟∈I

+(a+b )=I +(a+b)

Note: I+ z=I , ∀ z∈ I

Lets look at I+a' ∙ b 'I+a' ∙ b '=I +( x+a ) ( y+b )=I+ xy⏟

∈ I

+ay⏟∈I

+ xb⏟∈I

+ab=I +a ∙b

In the RI quotient ring, the 0F element is I .

Since I+ (I+a )=I +aThe 1F element is I+1 etc…

Examples

1.Z

nZFor instance, when n=6

(6 Z+2 )+ (6Z+3 )=6Z+5(6 Z+3 )+ (6Z+4 )=6Z+7=6Z+1TODO: Had a multiplication I did not have time to copy

We can actually think of Z

nZ as {0 ,1 ,…,n−1 } wrt +, ∙mod n

2.F [x ]

f ( x ) F [ X ]where F is a field

for instance, when f ( x )=x2−3 x+2 ,F=ZSo in fact:

R [ x ]x2−3 x+2

= {I +ax+b|a ,b∈ R }

Since addition and multiplication are in polynomials mod ( x2−3 x+2 )Same as before (with numbers) - ∀ f , g∈R [ x ] . (I+ f (x ) )+( I +g ( x ) )=I +f ( x )+g (x ).Any polynomial f ( x ) can be written in the form:

f ( x )=q ( x ) ( x2−3 x+2 )+r ( x )

where q ( x ) , r (x )∈R [ x ]∧ [ degree ( r (x ) )<2∨r ( x )=0 ]

Also, since x2−3 x+2=( x−1 ) ( x−2 )→( I +( x−1 ) ) ∙ ( I +( x−2 ) )=I

( I +(2x+1 ) )+( I +(3 x−5 ) )=I +(5 x−4 )( I +(2 x+1 ) ) ∙ ( I +(3 x−5 ) )=I+ (2x+1 ) (3 x−5 )=¿

I+6 x2−2x−5=I+6 ( x2−3 x+2 )+ (−16 x−17 )=¿I−16 x−17

(2 x+1 ) (3 x−5 ) ≡−16 x−17 (mod I )a≡ b (mod I )↔I +a=I+b

------End of lesson 1

Homo-morphisms of ringsIf R ,S are Rings, then the function ϕ :R →S is a ring homomorphism if

1) ∀ a ,b∈R ϕ (a+b )=ϕ (a )+ϕ (b )2) ∀ a ,b∈R ϕ (a ∙ b )=ϕ (a ) ∙ ϕ (b)3) ϕ (1R )=1R

If ϕ satisfies (1) and (2) then: if ϕ (1 )=x→ ϕ (1 )=ϕ (1∙1 )=ϕ (1 )2

x=x2 so ( x−1 ) x=0If R is a domain (ab=0→a=0∨b=0¿ then it follows that either x=0 or x−1=0.

If x=0 then:

ϕ (a )=ϕ (a ∙1 )=ϕ ( a ) ∙ ϕ (1 )=ϕ (a ) ∙ x=0Otherwise, get ϕ (1 )=1If R is not a domain, (1)&(2) ϕ ≠0 do not in general imply ϕ (1 )=1.

Claim: If ϕ :R →S homomorphism, then kerϕ {a∈ R|ϕ (a )=0} is an ideal in R.Proof – in assignment 1.

Imϕ {ϕ (a )|a∈R }

Homomorphism theorem for Rings

1) If ϕ :R →S is onto S then R

kerϕ≅ S (≅ is isomorphic!)

& isomorphism (homomorphism which is 1-1 & onto) is given by:kerϕ+a→ϕ (a)

2) If I⊲R ideal then the map a→ I+a is a homomorphism from R to RI & its kernel

is I .

Proofs: VerificationIn (1) you need to check that the map is well-definedi.e. if kerϕ+a=kerϕ+a ' then ϕ (a )=ϕ(a ')If this holds, then a−a'∈kerϕAs a '=a'∈kerϕ+a'=kerϕ+a

Proof:∃ x∈kerϕ :a '=x+a

ϕ (a' )=ϕ (x+a )=ϕ ( x )+ϕ (a )=ϕ (a)Note: kerϕ={0 } ↔ϕis 1−1.

Our note:Lets prove the note!→Suppose we have s1∈S s.t. ∃ x1 , x2∈ R ϕ (x1 )=ϕ (x2 )=s1.However: ϕ ( x1−x2 )=ϕ ( x1 )−ϕ (x2 )=0→ x1−x2∈kerϕ→ x1−x2=0→x1=x2→ Contradiction!←First lets prove that 0 is in the kerϕ:a=a+0→ϕ (a )=ϕ (a+0 ) →ϕ ( a )=ϕ (a )+ phi (0 ) → phi (0 )=0Now, since ϕ is 1-1, there can only be one element of R going to 0. And we just found it.So kerϕ={0 }.

ExampleR [ x ]

( x2+1 )≅C

∑j=0

k

a j xj

Look at homomorphism: f ( x ) →f ( i)from R [ x ]ϕ→

C

What is the kernel?

kerϕ={ f ( x )∈R [ x ]|f (i )=0}={f ( x )∈ R [x ]|f ( x ) is amultipleof x2+1by another polynom }(we shall see that later)

Example2ϕ :Z → {0 ,1 ,…,n−1 } that sends x∈Z to x (mod n )= remainder of x (mod n).

kerϕ=nZ so Z

nZ= Zn

From now on we’re going to look at commutative Rings!

Commutative RingsDefinition: R is a domain if ab=0→a=0∨b=0 for all a ,b∈R.Domain – תחום שלמות

ExamplesR [ X ] , F [x ] (F some field )ZZ [ x ]Z X Z (not a domain!)Z5 X Z5 (not a domain!)

−ring of nxn matricesa

field(not a domain!)

PIDDefinition: R is a principal ideal domain (תחום ראשי)If it is a domain & every ideal in it is a principal(i.e. of the form (a )=Ra , for some a∈ R )

ExamplesF [ X ] ← Assignment 1

Counter example?Z [ x ] is not a PID! But it’s a domain…Look at the ideal generated by x and 2 (the set of polynomials over Z with an even constant term)

x ∙ Z [ x ]+2 ∙ Z [x ]

For the sake of contradiction, suppose it were a principal ideal. Then there would exist some polynomial g ( x ) which generated the ideal. But since 2 is in the ideal, it must be a multiple of g ( x ), so g ( x ) must be a constant, say n. But x is also in the ideal, so it must be the product of n with some f ( x ) in Z [ x ]: x=nf ( x ). Since the coefficient of x on the left hand side is 1, the coefficient of x on the right hand side must also be 1. On the other hand, the coefficient of x on the right hand side is a multiple of n. So n=±1. But this means that our ideal is actually generated by 1 or 1, which means it is all of Z [ x ]. But this is not true, since there are elements of Z [ x ] which are not in our ideal – x+1 for instance. Thus, our ideal must not be a principal ideal!

3 More properties of Z(1) Euclidean property

If a ,b∈Z non-zero, then ∃g , r∈Z s.t. 0≤r<|b| and a=bq+r.(2) Every 2 non-zero elements have a greatest common divisor

if a ,b∈Z .gcd (a ,b )=d, is a number in Z s.t. d∨a ,d∨b and if d ' is also a common divisor then d '∨d . (unique up o a sign).

(3) Unique Factorization into primes

Proof of (2):In Z. If a ,b∈ZLook at the ideal Za+Z b = principal ideal!So ∃ d∈Z .Za+Z b=Z da=1∙ a+0 ∙b∈Z d so a multiple of d, d∨a.Similarily, b∈Z a+Zb so d∨b.Now let d '∈Z .d '∨a∧d '∨b.

d '∨a→a∈Z d ' so Za⊆Zd 'a∨b→Z b∈Zd '

And so also Za+Z b⊆Z d 'So d∈Zd ' →d '∨d.

Note: Suppose d∧d ' are both gcd’s of a∧b in Z.d∨d ' so ∃ x∈Z .dx=d 'd '∨d so ∃ y∈Z .d ' y=dd ' yx=d '

d ' ( yx−1 )=0d ' ≠0 , so yx−1=0

yx=1→ y, x∈ {±1 }So the GCD in Z is unique up o a sign.

In general: in any domain, we get uniqueness of the GCD up o an invertible element.

In Rings – invertible elements are referred to as units.

Bezout’s Theorem(In Z)Let a ,b≠0 in Z & let d=gcd (a ,b).Then, ∃u , v∈Z .au+bv=dThis follows trivially from the fact that Za+Z b=Z d.

Theorem:Let R be a PID, then if a ,b≠0 then a ,b have a gcd (unique up to multiplication by a unit)And Bezout’s theorem holds in R.Bezout’s theorem holds – if d=gcd (a , b ) then ∃u , v∈R .au+bv=d .

Definition: 1) If R is a Ring and p≠0∈ R is a prime element, whenever p∨a ∙b (a ,b∈R) then

p∨a∨p∨b.2) If R is a Ring and x≠0∈R is an irreducible element then if x=a ∙b for some

a ,b∈R then a or b must be a unit.

In Z: prime=irreducible.

Claim: If R is a domain then pprime→ pirreducible.Proof: Suppose p is prime and that p=a ∙b so also p∨a ∙b so p∨a or p∨b. Wlog, We might as well assume that p∨a. So ∃u∈R such that pu=a. So abu=a→a (bu−1 )=0∧a≠0.Sobu−1=0→bu=1 and bis a unit.

However, irreducible not → prime in general.

Example:Z [√−5 ]= {a+b√−5|a ,b∈Z } subring of C

This contains irreducible elements that are not prime.It does contain prime elements!First, recall that if x+iy∈C →‖ x+ iy ‖2=x2+ y2

And if z1 , z2∈C, then ‖z1‖2 ∙‖z2‖

2=‖z1 ∙ z2‖2.

Use this to show √−5 is a prime element in the ring.

Assume √−5∨r ∙ s∈Z [−5 ]We then got ‖√−5‖2∨(‖r‖2 ∙‖s‖2) so 5∨‖r‖2‖s‖2 and ‖r‖2 ,‖s‖2 are integers

And so 5∨‖r‖2 or 5∨‖s‖2

Wlog, 5∨‖r‖2

And write r=a+b√−5 , a , b∈Z5∨a2+5b2→a2(¿hencealso a) are integer multiples of 5.So write a=5a' , a'∈Z .

And r=5a'+b√−5=√−5⏟∈Ring

(−√−5a '+b )⏟∈Z [√−5 ]

So √−5∨r in the ring.

We now show that Z [√−5 ] contains irreducible elements that are not prime.Look at:

2 ∙3=6=(1+√−5 )(1−√−5)First note that 2 is irreducible.Suppose 2=r ∙ s

4=‖2‖2=‖r‖2 ∙‖s‖2

Case 1:

‖r‖2=2=‖s‖2

But on the other hand, if r=a+b√−5 then we get: a2+5b2=2 which has no solutions with a ,b∈Z .Case 2: wlog, ‖r‖=1 and ‖s‖2=4 then get a2+5b2=1→a2=1∧b=0→a=±1 and r=±1 and so is a unit.

Note: Can show in a similar way that units of Z [√−5 ] are ±1.

We now show that 2 is not prime in Z [√−5 ].By (*) we have that 2∨(1+√−5 ) (1−√−5 )Suppose 2∨1+√−5.Then we have a+b√−5 ,a ,b∈Z :2 ( a+b√−5 )=1±√−5→2a=1 - impossible.So 2 divides neither of the factors and so is not prime.

We shall show that In a PID, all irreducibility implies primeness.Conclusion: Z [√−5 ] I not a PID!

------- end of lesson 2

R=Z [−5 ] not a PID.

Take I=2 R+(1+√−5 ) R6=2∙3=(1+√−5 ) (1−√−5 )

2 irreducible but not prime.Also 1+√5

If I was principal, then we would have r such that R ∙ r=2 R+(1+√−5 ) R

Giving – r|2 , r|1+√−5So ∃ s . rs=2Case 1: r is a unit→R ∙r=R → I=R. We will show this is impossible.Suppose ∃ a ,b ,c , d∈Z .1=2 (a+b√−5 )+( c+d√−5 ) (1+√−5 )1=2a+c−5d+√−5 (2b+c+d ) So that: 2a+c−5d=1 ,⇒ c+d=1 (mod 2 ) 2b+c+d=0⇒c+d=0(mod 2) Contradiction!

Case 2: s is a unit.r s−1=2 and r s−1 s∨1+√−5So 2∨1+√−5 - contradiction!

Future Assignments:The grader is Niv Sarig. And he will put the assignments in his web page:http://www.wesdom.weizmann.ac.il/~nivmoss/ate.html

There is a mailbox for the course!

Claim: In a PID all irreducibles are prime.Proof: Suppose a is irreducible and a∨b ∙ c in a ring R (Assuming b ∙ c≠0).Since R is a PID, a & b have a gcd.gcd ( a ,b )=d . Assume a=d ∙a'. As a is irreducible & d∨a then either d is invertible or a ' is invertible.Case 1: d is a unit. Wlog d=1.By bezout: ∃u , v . au+bv=1a∨b ∙ c so ∃r∈R .ax=bc

aux+bxv=x aux=buc

So

bxv+buc=x b ( xv+uc )=x⇒b∨x

So ∃b '∈R .bb;=xax=bc

abb '=bc

b ( ab'−c )=0

R is a domain and b≠0 so ab '−c=0⇒ ab'=c∧a∨c

Case 2: a ' is a unit.

a ( a−1 )−1=d

So, a∨d and d∨b so a∨b.

Unique FactorizationDefinition: A domain R (a commutative ring) is a unique factorization domain (UFD) if any non-unit a ,a≠0 can be written as a product of irreducible elements uniquely (up to order of the factors and units).e .g .6=2∙3=3 ∙2=(−3 ) ∙(−2)

Example: Z , F [ x ] , any field ,Z [ x ]- which is not a PID!

UFD does not imply PID!But PID⇒UFD.

We showed that Z [√−5 ] is NOT a PID.

Euklidian PropertyDefinition: A domain R is Euclidean if we can define a map δ :R ¿{0¿}→ N (called the Euclidean norm) s.t. for a ,b≠0∈R ,∃q ,r∈Rsuch that:a=bq+r and δ (r )<δ (b ) or r=0.And ∀ x , y∈R .δ ( x )≤δ ( xy )(definition – Herstein, Jacobson does not require δ (x ) ≤δ ( xy ))

Examples:1) Z .δ=||2) F [ x ] , F is a field, δ=¿ degree of a polynomial3) F is a field, δ (a )=0 ,∀ a≠0

Theorem: In a Euclidean domain, every 2 non-zero elements have a gcd.Proof: Uses Euclid’s algorithm. Write: a=bq1+r1 , δ ( r1 )<δ (b )

If r1=0 then a=bq and gcd ( a ,b )=bIf not: write b=r1q2+r2, δ ( r2 )<δ ( r1 ) or r2=0

If r2=0 then gcd ( a ,b )=r1Otherwise, I can write r1=r2q3+r3 , δ (r3 )<δ (r2 ) or c3=0If r3=0 then gcd a ,b=r2…

Since δ (b )>δ ( r1 )>δ (r 2)>…Is a proper decreasing sequence of units we getFor k , δ ( rk )=0, the last non-zero zk is the GCD.

Note: Z [√−5 ] is not Euclidean!

And in assignment 2 you show 6+2 (1+√−5 ) have no GCD.

Theorem: If R is Euclidean then R is a PID.Proof: If I is an ideal in R , I ≠0Pick a∈ I and minimal Euclidean norm. And then I=Ra.

Theorem(use for PID→UFD!)In a PID any increasing chain of Ideals stabilizes.I.e. Given I 1⊆ I 2⊆…⊆ I n⊆ I n+1⊆…⊆RI j Ideals ∃ k s . t . I k=I k+1… etc…

Proof:Look at the union of all the Ideals: ¿n=1¿ ∞ I n=J . J is an ideal and so principal.So ∃ a∈ R .J=Ra.a∈ J so ∃ k .a∈ I k

I k⊇Ra=J So ∀ t ≥0. I k +t⊂I k etc. But given I k+t⊇ I k∀ t ≥0So we get equality…

Example: Z [ i ]=¿ring of Gaussian integers ¿ {a+bi|a ,b∈Z }Turns out – this ring is Euclidean.Proof: Define δ (x+iy )=x2+ y2=‖x+iy‖2.δ is multiplicative. Need to show Euclidean property holds.Take a ,b∈Z [i ] a ,b≠0

Z [ i ]⊆Q [i ]= {r+si|r , s∈Q } - which is a field!

(r+si )−1 , ,= r−isr2+s2

r+si≠0

So a ∙b−1∈Q [ i ].

So write: a ∙b−1=α+ βi ,α , β∈Q .∃u , v∈Z :|u−α|≤ 12

,|u−β|≤ 12

Let q=u+iv∈Z [i ]ab−1=u+iv+ (α−u )+i (β−v )∈Q

ab−1=q+(α−u )+(β−v )So α=bq+[ (α−u )+( β−v ) ]br=a−bq∈Z [ i ]

Remains to show that δ (i )<δ (b ).

δ (r )=‖(α−u )+i ( β−v )‖2 ∙‖b‖2

‖( α−u )+ i ( β−v )‖2= (α−u )2+ ( β−v )2≤ 14+ 14=12

So that δ (r )≤ 12

δ (b )<δ (b )

Euclidean ⇒ PID.But PID does not imply Euclidean!

Counter Example:

Z [ 12+ √−192 ] a PID but not Euclidean. Check…

In 2004 it was shown that Z [√14 ] is Euclidean.

It is easy to show that: Z [√−n ] (0>n∈N ) is Euclidean ⇔n=1∨2

In Euclidean domains: we used the Euclidean property to construct the GCDs.In UFD: Use factorization to construct GCD’s.

a=p1 ,… , pk

b=q1 ,…,ql

Where they are irreducible.GCD=product of common factors.

It turns out: Irreducible implies prime in a UFD.

Sum upEuclidean⇒PID⇒UFD

But the arrows don’t go the other way!

Example:

R=Z [ x , x2

, x3

,… , xn

,… ]=x ∙Q [ x ]+Z

56

x5+ 23

x4+3=5x4 ∙ x6+2∙ x

3∙ x3+3

R is a subring of Q [x ].

R ≠Q [ x ] as 12∉R.

There are very interesting properties:1) R is a bezout Ring (and in particular, every 2 elements ≠0 have a GCD)2) Any finitely generated is principal3) But R is not a PID!

4) Ideals generated by {x , x2

,…,…} is not principal!

5) R not a UFD. x is divisable in this ring, by every integer ≠0. So x cannot be factored as products of individuals.

--End of lesson 3

Commutative Rings

Chinese Remainder Theoremx≡2 (mod 3 ) x≡3 (mod 5 ) x≡(mod 7) e .g . x=23

This is 4th century china

Lady with the eggsx≡ (mod 2 ) x≡1 ( mod3 ) 𝑥≡1(mod 4 ) ⋮ 𝑥≡0(𝑚𝑜𝑑7) x=301

CRT in ZLet n1 ,…,nk be pair-wise mutually prime integers. (gcd (ni ,n j )=1 ∀ i , j)

And let a1 ,…,ak be arbitrary integers.Then there exists an integer x s . t .x≡ ai (mod ni )

Note: There will be no solution x s . t . x≡1(mod 2) and x≡0(mod 6)

CRT in a commutative ring RLet I 1 ,…, ik be pair-wise co-prime ideals in R.(The ideal generated by a sum of any two ideals is R: I j+ I k=R ∀ j≠ k)

And a1,…,an∈R arbitrary elements.

Then, there exists x∈ R such that x≡ a j ( mod I j )Or in other words x+ I j=a j+ I j∀ j

Derive CRT for Z from the general theorem:If gcd ( ni ,n j )=1 then ni Z+n j Z=Z so conditions on ideals ni Z hold etc…

Prove for n=2We have I 1+I 2=RSo we have b j∈ I j s . t . b1+b2=1Let x=a2b1+a1b2

x+ I1=a2b1⏟∈ I1

+a1b2+ I 1=a1b2+ I 1=a1 (1−b1 )+ I 1=a1−a1b1+ I1⏟∈I1

=a1+ I 1

x≡ a1 (mod I 1) Similarly x≡ a2 (mod I 2 )

If I , J ideals in RDenote I ∙ J=¿the additive subgroup generated by the products {ab|a∈ I ,b∈ J }{a1b1+…+an bn|ai∈ I , b j∈J n≥0 }Note: {ab|a∈ I ,b∈ J } is closed under multiplication by elements of R.Not necessarily closed under addition.

And then I ∙ J will be an ideal. I ∙ J ⊆I , J and in fact I ∙ J ⊆ I ∩J ideal

Examples:In Z

3Z ∙3 Z=9 ZBut 3Z ∩3Z=3ZNote: If p ,q mutually prime then: pZ ∙qZ=pq Z=pZ ∩q Z

In general:I 1 ∙ I 2 ∙…∙ I k- smallest ideal containing set of products.We start by writing

I 1+ I 2=R⇒∃ c2∈ I 1 ,b2∈ I 2 :c2+b2=1⋮

I 1+ I n=R⇒∃ cn∈ I 1 , bn∈ I 2:cn+bn=1

Look at the product: ∏i=2

n

c i+bi=1

Let J1=I 2 ∙…∙ I n

The product has elements that has a multiplication of some c, except for the b’s.multiplesof some c⏟

∈ I1

+b1 ∙…∙bn⏟∈ J1

=1

So that I 1+J 1=R

By the CRT for case n=2 have y1∈ R s. t .

{ y1≡1 (mod I 1 )y1≡0 ( mod J 1 )

Since J1⊆ I 2∩ I 3∩…∩ I n we also get y1≡0 (mod I j ) j>1

Repeat for each i: J i=∏k ≠i

I k

Form I i+J i=R

And get y i∈R s .t .y i ≡1 ( mod Ii )y i ≡0 (mod J i )

And so also y i ≡0 (mod I k )k ≠iLet x=a1 y1+a2 y2+..+an yn

mod I 1: x ≡a1+0+¿ similarly for all j x≡a j (mod I j )

In ZNote that x≡ ai (mod ni ) ∀ i not unique.

x+∏ ni will solve all the congruences.

Corollaries:Let R be a commutative ring. I 1 ,…, I n mutually coprime ideals in R.Then

R( I1∩…∩ I n )

≅ ( RI 1 )×( R

I 2 )×…×( RI n )

(actually equivalent to CRT)

Proof: Define a homomorphism f :R →( RI1 )×…×( R

I n )By f ( a )=(a+ I 1,…,a In

)=(a (mod I 1 ) ,…,a (mod I n) )Clearly this is a homomorphism. (not so clear. TODO go over it)Clearly f is additive and multiplicative.

f (1 )=(1 (mo d1) ,…,1 (mod I n) )

We calculate ker f :a∈ ker f ⇔a≡ (mod I j ) for all j ⇔a∈ I 1∩…∩ I n

ker f=I 1∩ …∩ I n

We need to show f is onto ( RI1 )×( R

I 2 )× …×( RI n ) to get isomorphism

(by homomorphism theorem)

Let (a1+ I 1 ,…,an+ In )∈( RI1 )×( R

I 2 )×…×( RI n )

We want x s . t . f ( x )=( a1+ I1 ,…,an+ I n )

Or x≡ ai (mod I i ) for all i.Existence of such an x is guaranteed by the CRT.

Special case of corollary1<m∈Z

m=∏i=1

k

piri p i distinct primes. I i=pi

ri Z

( Zm Z )≅( Z

p1r1Z )× …×( Z

pkrk Z )

Isomorphism of ringsFor a commutative ring R, denote by R¿=¿ set of units (invertible elements) of RThen R¿=¿multiplicative abelian group.

e.g. ( Z6Z )

¿

= {1 ,5 }=¿group of two elements

Looking at the group of units on both sides we get:

( Z6Z )

¿

≅isomorphism∨unit groups( Z

p1r 1Z )

¿

×…×( Zpk

rk Z )¿

Denote by φ (m )=¿ {k|0<k<m s. t . gcd (k ,m )=1 }(euler phi function)

E.g. φ (6 )=2

Clearly ( Zm Z )

¿

has φ (m) elements.

From (*) we get the formula: φ (m )=φ ( p1r1 ) ∙…∙φ ( pkrk )

Application to public key encoding RSA (1975)Encoding – publicDecoding – secret

Let p1 , p2 “very large” prime numbers.Let d=p1 ∙ p2Let e=φ (d )=φ ( p1 ) ∙ φ ( p2 )=( p1−1 ) ( p2−1 )Let r be any large number co-prime to e.By Bezout, we have s , t s .t . sr+te=1sr≡1(mode)

We publish only d and r (and not s , e , p1 , p2).

Let a be a positive integer smaller than d .

We encode a as ar (mod d )=b

Claim: bs≡ a (mod d ) !Note: This determines a uniquely as a was chosen to be less than d .Proof: First case: gcd ( a ,d )=1

( Zd Z )

¿

≅ ( Zp1Z )

¿

∙( Zp2Z )

¿

has φ (d )=e elements.

Recall in a group G of order nxn=1 for all x∈G.Follows from Lagraunge’s theorem – shall prove later.

So that ae ≡1(mod d) a=a+d Z elements of ( Zd Z )

¿

rs≡1 (mod e )

bs≡ (ars ) (mod d ) ≡ a¿+1≡ (ae )l ∙a≡a (mod d ) - as required.

Second case: gcd ( a ,d )≠1

Then wlog can assume q1∨a and gcd ( a , p2)=1Z

d Z ≅ψ ( Z

p1Z )×( Zp2Z )

ψ (a+d Z )=(a (mod p1 ) , a (mod p2 ))=(0 (mod p1 ) , a (mod p2 ))

Another corollary from Cauchy’s theoremFerma’s little theorem: For a prime p , x ≠0x p−1≡1(mod p)

So we have a p2−1≡1 (mod p2 )ae=a( p2−1) (p1−1) ≡1 (mod p2 )

ψ is an isomorphism so we have:

ψ (ae+d Z )=(ψ (+d Z ) )e=(0 (mod p1 ) ,1 (mod p2 )) Again, writing: rs=¿+1 we get

ψ (bs+d Z )=ψ (ars+d Z )=ψ ( a¿+1+d Z )=ψ (a¿+d Z ) ∙ψ (a+d Z )=¿

ψ (ae+d Z ) ∙ (0 (mod p1 ) , a (mod p2 ))=¿

(0 ( mod p1 ) ,1 (mod p2 )) ∙ (0 (mod p1) , a (mod p2 ))=(0 (mod p1 ) , a (mod p2 ))=¿

ψ (a+d Z )Since ψ is an isomorphism we get a≡ bs (mod d )

Short introduction to Group TheoryH subgroup of G if ∀a ,b∈H a ,b−1∈H (¿ H ≠0 )Cosets of subgroup in GHa right coset = {ha|h∈H }aH left coset = {ah|h∈H }

Properties: Cosets are disjoint or equal.Suppose Ha∩ Hb ≠∅So have h ,h'∈H s. t . ha=h' b(h ')−1ha=b and b∈HaHb⊆Ha And similarly Ha⊆Hb.

Definition:N is a normal subgroup of G if ∀g∈G :Ng=gN .(does not imply ng=gn∀ N !!!)

If G is Abelian, all subgroups are normal!

Example: G=S3: group of permutations on {1,2,3 }

p=(1 2 32 1 3)

{ Id , r } is a subgroup of G. Which is not normal!

H ∙(1 2 33 2 1)={(1 2 3

3 2 1) ,(1 2 32 1 3) ∙(1 2 3

3 2 1)=(1 2 32 3 1)}

(1 2 33 2 1) ∙H={(1 2 3

3 2 1) ,(1 2 33 2 1)∙(1 2 3

2 1 3)=(1 2 33 2 1)}

So this is not the same group!

A3=¿ set of even permutations = normal subgroup of order 3

r={Id ,(1 2 32 3 1) ,(1 2 3

3 1 2)}

A3σ=σ A3=S3¿3={(1 2 32 1 3)⏟

σ

,(1 2 33 2 1) ,(1 2 3

1 3 2)}

----- End of lesson 4

TODO: Write it----- end of lesson 5

Theorem: Let p ( x )∈F [x ] be irreducible.Proof

Note: p(u) maximal so F [u ]p (u )

has to be a field!

Consider F⊆K by identifying a∈F with a+ ( p (u ) )It remains to show that p ( x ) has a root in K

Suppose p ( x )=∑i=0

ai xi, a i∈F

Look at the coset u+( p (u ) )=α∈ K

p (α )⊂∑❑

❑

a iui=∑ ai (u+( p (u ) ) )=∑ aiu

i+( p (u ))=¿

Want to show K unique up to isomorphism minimal such that p has a root.Suppose L⊇F , β is a root of p in L.Want to show K ≅ subfield of L.Map: g (u )+( p (u ) ) in K to g ( β )∈ L.

H is independent of choice of coset representative, as if g (u ) ≡h (u ) (mod ( p (u ) ) )Then g (u )=h¿u

----- end of lesson 6

Claim: If f ( x )∈F [ x ] and F⊆K field containing a root of f ( x ) :α

Then if φ∈Gal( KF ) then φ (α ) is a root of f ( x )

In other words, elements of the Galois group permute the roots of f ( x )

Proof: Let f ( x )=∑i=0

k

ai xi , ai∈F

φ ( f (α ) )=φ (0K )=0 , ai∈F

0=φ ( f (α ) )=φ(∑i=0k

ai αi)=∑

i=0

k

φ (ai ) φ (α )i =ai∈F ∑

i=0

k

a iφ ( α )i

Special case:

K splitting field for f ( x )∈F [ x ] then K=F (α1 ,…,αk⏟

roots of F )So any φ∈Gal( K

F ) is determined by images of α 1 ,…,α k under φ

We now know that these are permuted by φ

β∈ K so can be written as a polynomial in α 11 ,…,αk over F

β=∑ ai1…ik∙ α1

i1α2i2…α k

ik

Examples:1) Galois group of the smallest field of x4−2 over Q

Roots of x4−2:

± 4√2 , ± i 4√2x4−2=(x−4√2 ) (x+ 4√2 ) (x−i 4√2 ) (x+i 4√2 )and over k: Q ( 4√2 ,i )

φ∈Gal( KQ )=G will permute 4 roots

So can think of G of being a subgroup of S4We know that |Q ( 4√2 ,i ) :Q|=¿[k :F ] = dimension of K over F.

|Gal (KQ )=G|=8

So G is isomorphic to an 8-element subgroup of S4Possibilities (up to isomorphism) are:C8 ,C4×C2 ,C2×C2× C2 , D8 ,Q8

|S4|=24 (a side note)C8 – is impossible since S4 contains no elements of order 8

Let φ be complex conjugation.obviously φ is an element of order 2. φ∈G

φ ( 4√2 )=4√2φ (− 4√2)=−4√2φ (i 4√2 )=−i 4√2φ (−i 4√2)=−i 4√2Let ψ be the automorphism that permutes roots cyclically:

ψ ( 4√2)=i 4√2 and fixes iψ is of order 4

ψ (−4√2 )=−i 4√2 𝜓(i 4√2 )=ψ ( i )ψ ( 4√2)=ii 4√2=−4√2

¿φ ,ψ>¿ is a group permuted by φ and ψ ≅ D8

Cycle notation in Sn (any permutation can be written as a product of disjoint cycles)Example σ∈S4

σ=(1 2 3 4 5 63 1 4 6 5 2)=(13462 ) (5 )

(1 2 3 4 5 63 5 1 2 4 6)=(13 ) (254 ) (6 )

Can have σ∈S5, σ=(123 ) (45 )Elements of S4 can have orders 1,2,3,4 (again, a side note).The order of the elements is always the least common multiple of the cycles.

Another example:Galois group of p ( x )=x3+2x+1 over QNeed to find the splitting field of the polynomial over Q.We first of all show that p ( x ) has no roots in Q and so is irreducible.

Claim: If f ( x ) is a monic polynomial over Z, then any rational root will be an integer

Proof: f ( x )=xn+an−1 xn−1+…+a1 x+a0 , ai∈Z

r , s∈Z

If rs is a root then: 0=f (rs )= rn

sn +∑i=0

n−1

ai

r i

si

Assume (r , s )=1

rn+∑i=0

n−1

airi sn−i=0

rn=−a0 Sn−a1Sn−1+…−an−1 srn−1

If p is a prime divisor of s, then p∨rn so p∨r.But then, p∨s and p∨r which contradicts the fact that s and r are mutually prime.

So s has no prime divisors. So s=±1. Therefore, rs∈Z

We now show that p ( x ) have no integer roots.p (0 )=1

p (−1 )=−2So there exists α∈ R −1<α <0 and p (α )=0 by continuity of p ( x ) as a real function.

But it’s the only real root, since the derivative is always positive, therefore it’s constantly increasing etc etc…So p ( x ) has no rational roots, and remaining 2 roots are non-real.

Over Q (α )

x3+2x+1= (x−α ) (x2+(2+α ) x+(2+α ) α ) =¿ S . F . ( x−α ) ( x−β ) ( x−β )

Where β and β are nonreal roots.

So the splitting field will be Q (α , β )|Q (α , β ) :Q|=|Q (α ,β ) :Q (α )|⏟

¿2¿¿

Extra fact:If α is a root of some polynomial g ( x ) over a field F.And p ( x ) is the minimal polynomial of α over F, then p ( x )∨g (x ) in F [ x ]Proof: Divide g(x ) by p ( x ) with remainder in F [ x ]

g ( x )=p ( x ) q ( x )+r ( x )degr<deg p or r=0

Substitute x=α : 0=g (α )=p (α ) q (α )+r (α )So α root of r ( x ) of smaller degree than p ( x ) - contradiction!

So |Gal (Q (α ,β )/Q )|=6.

Elements of Galois group permute the set {α ,beta β } and so is isomorphic to a subtgroup of S3 of order 6⇒ Galois group ≅ S3

TOPIC:Cyclotomic fields and their Galois groups over QDefinition Cyclotomic field is one of the form Q ( n√1)n√1=e

2π in positive with root of 1

Note that Q ( n√1) is a splitting field of the polynomial xn−1 over QAs:

xn−1= Πk=0

n−1(x−e2π i

n )We also want to factor xn−1 into irreducible factors over Q.

E.g. x3−1= (x−1 ) ( x2+x+1 )⏟

irreducibleQ

=minimal poly

Definition: Denote by λn ( x )=¿ minimal polynomial of n√1 over QSo λ3 ( x )=x2+x+1λn ( x )=¿ n’th cyclotomic polynomial

λ1 ( x )=x−1

λ2 ( x )=x+1λ3 ( x )=x2+1

4√1=ix4−1=( x2−1 ) ( x2+1 )=(x+1⏟

¿ λ2 )( x−1⏟¿ λ1 )( x2+1⏟

¿ λ3)

Fact: If f ( x ) ∙ g ( x )=xn−1 over Q, then f ( x ) , g ( x )∈Z [ x ](Follows from Gauss’ lemma – Basic algebra 1)

Interesting fact:If we factor xn−1 over Q(i.e. over Z!)Turns out up to n=105 all coefficients are ∈ {0 ,±1 }!For n=105 get coefficients = 2105=3∙5 ∙7

|Q ( n√1 ) :Q|=deg λn=?

Examples: 1)

Q (i )=Q ( 4√1 )Can be thought of a 2 dimensional vector space over Q

a+ ib

(a+ ib ) (c+i d )=ac−bd+i (ad+bc )We can think of them as vectors with regular dot multiplication.

2) Q (ω)=Q ( 3√1)|Q (ω ) :Q|=2 irreducible polynomial λ3 of ω is x2+ x+12 dimensional vector space over Q - addition – as usual

(a+ωb ) (c+ωd )=ac+ω2 (bd )+ω (ad+bc )=ac−bd+ω ( ad+bd−bd )Since:

ω2+ω+1=0ω2=−1−ω3) Q ( 5√1 )

λ5 ( x )=x4+x3+x2+ x+1|Q ( 5√1) :Q|=41 , ρ, ρ2 , ρ3 basis for Q ( 5√1 ) over Q

In general4) p is prime Q ( p√1 )

x p−1= ( x−1 ) (x p−1+x p−2+…+x+1 )The second part is irreducible using einsensteins criterion (lang algebra) = λ p ( x )

|Q ( p√1) :Q|=p−1

5) N=6Let’s factor it over Q:

x6−1= ( x3−1 ) (x3+1 )=( x−1 ) ( x2+x+1 ) ( x+1 ) ( x2−x+1 )6√1=ρω=ρ2ω2=ρ4

dfRoots areL

Roots (Accoringly) 1, ω ,ω2 −1 ρ , ρ5=ρ

What is Q ( ρ )??

2 dimensions over Q. What is the multiplication rule?Notice: −ω is a 6th root of (−ω )2=ωSo can take ρ=−ω

Q ( ρ )=Q (ω )!!!!It’s actually the same field! Not isomorphic – same field!

--- end of lesson

Theorem: [Q ( n√1 ) :Q ]=φ (n )=¿Eular φ-function

Recheck:φ (6 )=|{1,5 }|=2φ (5 )=4φ (4 )=|{1,3 }|=2φ (3 )=2φ ( p )=p−1p is prime

Denote ξ= n√1

Proof: [Q (ξ ) :Q ]=¿degree of the minimal polynomial of ξ over Q=deg λn ( x )Note: ξk is a primitive n’th root of 1 ⇔gcd (k ,n )=1

|{ξk|ξk primen' throot of 1 }|=φ (n )

So in fact, λn ( x )= ∏

gcd ( k ,n)=11≤k<n

(x−ξk )

This is a key fact!

By gauss’ Lemma, xn−1 factors over Q into polynomials in Z [ x ]So in fact, as λn ( x )∨xn−1 over Q (since ξ is a root of xn−1 and λn ( x ) is its root polynomnial)We in fact have that λn ( x )∈Z [ x ]

Suppose d∨n:Then, any d’th root of 1 is also an n’th root of 1.So the roots of λd ( x ) satisfy xn−1=0So λd ( x)∨xn−1 over Q

Conclusion: λd ( x )∨xn−1 for all d∨n.

Conversely:Suppose p ( x ) is an irreducible monic factor of xn−1 (in Q [x ])Any root α of p ( x ) is a root of xn−1 and so α n=1If d minimal such that α d=1 then d∨n.So α is a primitive d’th root of 1. Its minimal polynomial is λd ( x )And so λd ( x)∨p ( x ) but p ( x ) is irreducible and monic and so λd ( x )=p (x ).So every irreducible factor of xn−1 over Z is of the form λd ( x ) for some d∨n.

Conclusion: xn−1=∏

d∨nλd ( x ) over Q. And λd ( x )∈Z [x ]

Example: x6−1= ( x−1 )⏟

¿ λ1 ( x )

( x+1 )⏟λ2 ( x )

( x2+x+1 )⏟λ3 ( x )

( x2−x+1 )⏟λ6 (x )

Corollary from conclusion:From degree of polynomials we get:

n=∑d∨n

deg λd ( x )=∑d∨n

φ (d )

Example:

x12−1=(x6+1 ) ( x6−1 )=¿(x2+1 )⏟

λ4

( x4−x2+1 )⏟λ2( x )

ξ ,ξ 11 ,ξ5 ,ξ7

( x−1 )⏟¿ λ1 ( x )1

( x+1 )⏟λ2 ( x )−1

( x2+x+1 )⏟λ3 ( x )ω, ω2

(x2−x+1 )⏟λ6 ( x )

−ω,−ω2

ξ=12√1

Galois grups of Q ( xi ) over Q, ξ= n√1

Let Gal(Q (ξ )Q )=G

Elements of G permute primitive roots of unity and are determine by the image of ξ .

So G subroup of group of permutations {ξk|gcd1≤k <n

(k ,n )=1} i.e. of Sφ (n )

Let gcd ( k ,n )=1:

ξ ψk→

ξk determines an automorphism of Q (ξ )

Conversely, every automorphism must be of this form.|G|=[Q (ξ ) :Q ]=φ (n )

Suppose gcd ( l , k )=1=gcd (n , k )φk ∙ψ l ( ξ )=ψk ( ξk)=ξkl=ψkl (ξ )

ψ lψ k (ξ )=ψ l (ξk )=ξ lk

So the group is abelian!More precisely:ψk=ψ l=ψm where m≡ kl(mod n)In fact: The map k →ψk

Is group homomorphism between ( ZnZ )

¿

and G

So G≅( Zn Z )

¿

E.g. n=12

( Z12Z )

¿

={1,5,7,11} multiplication mod 12.

ξ=12√1

Note: ξ →ξ11 is complex conjugation

Finite FieldsIf F is finite then its characteristics must be some prime p

And its prime field ≅ZpZ .

So every finite field can be considered to be an extension of Z

p Z .

In fact, it is an algebraic extension. (if α transcendental then 1 , α ,α 2 , α 3 ,… infinitely linearly independent set so any field containing α will be infinite).

First difference between characteristic 0 case and the characteristic p caseWe had quadratic extensions of Q e.g.

Q (√2 ),Q (ω ) ,Q (i) which are isomorphic as fields!

By contrast, Z

p Z has a unique quadratic extension up to isomorphism.

Example: Z2Z clearly unique up to isomorphism. Call it F2 or GF (2 )

Now look at x2+ x+1 which is irreducible over Z2Z

Extend F2 to get a field in which x2+ x+1 has a root.

k=F2 [x ]

x2+x+1{ K :F }=dimF K=2⇒K 2 dimensional vector space over F2 and so has 4 elements.Elements of K can be considered to be remainders of polynomials in x over F2

After division by x2+ x+1 i.e. linear polynomials.0,1 , x , x+1

+¿0 1 x x+10 0 1 ¿

x+1¿1¿1¿0¿ x+1¿x ¿ x¿ x ¿x+1¿0¿1¿ x+1¿ x+1¿ x¿1¿0¿

∙ 0 1 x x+10 0 0 0 01 0 1 x x+1x 0 x x+1 1

x+1 0 x+1 1 x

Very easy to show directly that every field of order 4 is isomorphic to K .

Note: x2+ x+1 is actually the only irreducible quadratic polynomial over F

Theorem: Let F be a finite field then |F|=pk elements for some prime p ,1≤ k∈N .Conclusion: there is no field of order 6,10,15 , etc!

Proof: Let Z

p Z=F p to be the prime field of F then F is a vector space over F p.

And as F is finite, it is finite dimensional over F p. Say dim F=k .

So F≅ F p( k ) as a vector space and so |F|=pk

Example:Look at x4+x3+1 over GF (2 )Claim: x4+x3+1 is irreducible over GF (2 )Clearly it has no roots.

If it factored as 2 irreducible quadratics then we would have x4+x3+1=(x2+x+1 )2

But ( x2+x+1 )2=x4+x2+1

So GF (2 ) [ x ]

( x4+x3+1 ) gives an extension of degree 4 and so a field of order 16!

Its elements can be considered as polynomials of degree less or equal to 3.Or, vectors of length 4 over F2.

Addition is very easy with both notations (mod 2)( x3+x )+( x2+x+1 )=x3+x2+1

a x3+b x2+cx+d↔(abcd)

Multiplication on the other hand, is harder( x3+x )∙ ( x2+x+1 )=x5+ x3+x4+x2+x3+1=x5+ x4+x2+x ≡x2¿

(1010)(0111)=(0100)

Another NotationLet α=x+(x4+x3+1 ) in FSo α root of x4+x3+1 in F. α 4+α3+1=0

1 , α ,α 2, α 3 are linearly independent over Z2Z and so distinct.

Note that F ¿is a group of order 15.So α has order dividing 15⇒ α has order 1 ,3 ,5,15α 4=α3+1α 5=α (α 3+1 )=α 4+α=α3+1+α=α3+α+1≠1. Otherwise, α 3+α=0 and α satisfies polynomials of degree 3 – contradiction.Conclude: α has order 15! So F¿ is cyclic and generated by α .

So F={0,1 , α ,…,α 14 }This notation is convenient for multiplication:

α i ∙ α j=αi+ j (mod 15)

(Addition - problematic!)

Note: Over F x4+x3+1 factors into linear factors and so is a splitting field for this polynomial over F2

Notice that: α 4+α3+1=0

(Over Z

p Z : ( x+ y )p=x p+ y p)

So 0=(α 4+α 3+1 )2=α8+α 6+1⇒ α2 is a root of x4+x3+1

(α 8+α 6+1 )2=α16+α 12+1⇒α 4 is a root of x4+x3+1Same for (α 16+α 12+1 )2 which leads to α 8 is a root as well

So x4+x3+1=( x−α ) ( x−α 2 ) ( x−α 4 ) ( x−α 8 )

Theorem: The multiplicative group of a finite field is cyclic.Proof: next lesson!

Note: If |F|=q then all its nonzero elements will satisfy xq−1=1As |F ¿|=q−1Over a field, the polynomial has at most q−1 different roots. So in this case the set of elements in F ¿ is precisely the set of roots of xq−1

If we take xq−x then every element of F (including 0!) is a root and F is the splitting field of

xq−x .

--- end of lesson 8

Fundemental theorem of Abelian groups:Every Abelian group is a direct product of cyclic groups.(If the group is finite – get a direct product of a finite number of finite cyclic groups).Proof: Jacobson Basic Algebra 1.

For the finite case, you can always write:G=H1×…× H r

H i=¿ direct product of cyclic groups of orders that are powers of a fixed prime pi

p1 ,…, pr direct primes.

Theorem: If F is a finite field, then F ¿ is cyclic.Proof: Assume F ¿=H 1×…× H s as above.

Each H i can be written as a direct product:pi=p-H i=C

pk i1×C

pk i2×…×C

pk ir

Can assume k1≥…≥kr

C k=¿ cyclic of order k

So every element a of H i satisfies Apk1

=1So every element of H i is a root of the polynomial x pk 1

−1=0

H i⊂F and in F there are at most pk1 roots of this polynomial. So |H i|=pk1. Meaning, r=1.

So H i=C pk1 and in general we get:

So F¿=C p1

k1× …×C psk s

p1 ,…, ps are distinct primes!

So F ¿ is cyclic generated by the product of the generators of C p1k1 ,…,Cps

ks.

Corollary: If F is a finite field of order q. Then it is the splitting field of xq−x (where q=pk ,

p is prime) over Z

p Z . And so unique up to isomorphism.

Proof: All the elements of F ¿ are roots of xq−1−1 and so together with 0 all the elements of F are roots of xq− x−x.So every element is a root and the set of roots = F.

We shall show that if F and F ' are both fields of order q=pk then they are isomorphic:Let α∈ F¿ generator.

So a is algebraic over Z

p Z so is a root of an irreducible monic polynomial m (x )∈ ZpZ

[ x ]

So m (x )∨xq−x

F ' is also a splitting field of xq−x over Z

p Z.

So m (x ) has a root β in F '.We map α i to β i∀ i and 0 to 0.We need to show that the map is onto F ' (and so 1-1)And that it is additive! (it is multiplicative by definition).

Suppose βr=1 for r<q−1.Then β is a root of xr−1 in F '.

m (x ) is the minimal polynomial of β so that m (x )∨xr−1 over Z

p ZSo that α r=1 in F.But α is of order q−1 so q−1∨r and r ≥q−1 - contradiction!

We now show the map is additive:a) If α i+α j=αk then need to show βr+βs=βt

b) If α i+α j=0 then need to show βr+βs=0

We shall show (a):

α i+α j=αk implies α is a root of xr+xs−x t so m (x )∨xr+ xs−x t

So then β root of xr+xs−x t and so βr+βs=βt .

Note: It also follows that the roots of xq−x over Z

p Z are distinct.

Theorem: For any prime p and 1≤ k≤ N there exists a field of order pk.

Proof: Take Z

p Z and extend to a splitting field for x pk

−x.

This will be a field of order pk (and will be unique!).

Corollary: For any k ≥1 integer and prime p, there exists an irreducible polynomial of degree

k over Z

p Z .

Proof: Take α a generator of F ¿ where F field of order pk=q. (F=GF (q)¿Z

p Z[α ]=F and

Zp Z

[α ] is a vector space of dimension l over Z

p Z where l is the degree of

the minimal polynomial of α .

So Z

p Z[ α ] is of order pl so k=l and minimal polynomial is irreducible of degree k .

Factorization of X n−1 over finite fieldsExample: GF (16 )=GF (2 ) [α ]α root of x4+x3+1 over GF (2 ).Every element in this field is a root of x16−x.So x4+x3+1∨x16−x over GF (2 ).

Roots of x4+x3+1 in GF (16 ) were: α ,α2 , α4 , α 16

0 root of x. (so x∨x16−x)1 root of x+1 (so x+1∨x16−x)

x16−x=x (x+1 ) (x4+x3+1 )∙ h ( x ) ,h ( x )∈GF (2 ) [ x ] of degree 10.We want to factor h ( x )

Definition:Let f ( x )=¿ polynomial of degree n.

The reciprocal of f ( x ) is g ( x )=xm f (x−1 )

Example:

f ( x )=x5−2x4+3 x2−7 x+19x5 f ( x−1 )=x5 ( x−5−2x−4+3 x−2−7 x−1+19 )=1−2x+3 x2−7 x4+19 x5

Use question 4 in assignment 4 to get the reciprocal of x4+x3+1:

x4+x+1

So x4+x+1 is irreducible and α−1⏟

¿α 14 is a root and also α−2=α13 , α−4=α 11 , α−8=α 7.

We conclude that x4+x+1∨x16−xSo h ( x ) has x4+x+1 as an irreducible factor over GF (2 )Note also: x5−1∨x15−1. Since (x5−1 ) ( x10+ x5+1 )=x15−1.

Over FG (2 ) we have x5−1= (x+1 ) ( x4+x3+ x2+x+1 )So x4+x3+x2+x+1∨x16−x and is irreducible (question 1 in assignment 4).

Note also: 1, α 5 , α10 are roots of x3−1 in GF (16 ): α 3 , α6 , α12 , α24=α9

x3−1 factors to: ( x−1 ) ( x2+x+1 )So x2+ x+1 is the minimal polynomial of α 5 , α10.

So over GF (2 ) :x16−x=x (x−1 ) (x2+x+1 ) ( x4+ x3+1 ) ( x4+x+1 ) ( x4+x3+x2+x+1 )

Roots (in the appropriate order of the factors):0 ,1 , α5 , α10 , α ,α 2 , α 4 , α8 , α 14 , α 13 , α 11 , α7 , α 3 , α 6 , α9 , α12

Note: α ,α−1=α14 are primitives elements (i.e. generators of GF (16 )¿ but the roots of

x4+x3+x2+x+1 are not generators for GF (16 )¿

Though we can use this polynomial to construct GF (16 ) over GF (2 ). And every element of GF (16 ) is a polynomial in α 3 (but not a power of α 3!)

Every element of GF ( pk ) satisfies x pk−1=1.

If xn−1 has a root in GF ( pk ).Must have n∨pk−1

Can see which are the subfields of GF (16 ) by looking at the factorization of x16−x.Possible subfields (are of order 2m ,m≤4):GF (2 ) - prime field and so a subfield!

GF (4 ) – {0,1 , α 5 , α 10 } as GF (4 ) splitting field of x2+ x+1GF (8 ) - Don’t have any irreducible polynomials of degree 3 dividing x16−x! GF (8 ) is the splitting field of an irreducible cubic over GF (2 )! So this is not a subfield of GF (16 ).GF (16 ) (clearly).

Also: GF (16 ) could not be a vector space over GF (8 ) otherwise 16 would equal an integral power of 8.

--- end of lesson

xn−x over GF (2)

- What are the subfields of a given finite field GF (q ) , q=px , p prime.

Lemma: xm−1∨xn−1⇔m∨nProof: Divide = xn−1 by xm−1 with remainder (over Z):

xn−1= ( xm−1 ) (xn−m+ xn−2m+xn−3m+…+xn−km )+xn−km−1⏟remainder

k is such that km ≤n but (k+1 )m>n.

So remainder is 0 ⇔n=km⇔m∨n

Theorem: GF ( pm )⊆GF ( pn ) ⇔m∨nProof: If m∨n then by the lemma xm−1∨xn−1So in particular setting x=p we get pm−1∨pn−1Using the lemma again, we get that x pm−1−1∨xpn−1−1So all the roots of x pm−1−1 are contained in GF ( pn )¿ (which is the set of roots of x pn−1−1)

Meaning GF ( pm )¿⊆GF ( pn )¿ so GF ( pm )⊆GF ( pn )

Now assume GF ( pm )⏟L

⊆GF ( pn )⏟K

So K is a vector space over L, finite. So of finite dimension, say k over L.

|L|k=|K|So pmk=pn so m∨n .

Example:

x16−xn=4 subfields are of order 2m for m∨4n=1 ,n=2 , n=4: GF (2 ) ,GF (4 ) ,GF (16 )

Note: If GF ( pm )⊆GF ( pn ), then φ :GF ( pn ) →GF ( pn ) is frobenius automorphism a→ap

Then φm ( a )=apm

So set if fixed points under

φm={a|φm ( a )=a ,a∈GF (pn )}={a∈GF ( pn )|apm

=a }=¿

{a∈GF ( pn )¿|apn−11=0}∪ {0 }=¿ set of roots of x pn

−x in GF ( pn )

Note: If F finite field |F|=pn and we look at roots of xk−1 in F.Then a is a root ⇔ak=1 in F meaning either: k=0 and a=1 or k∨pn−1.The nontrivial factorizations of polynomials of type xk−1 are only for k∨pn−1(as if gcd ( k , pn−1 )=1 only roots will be 1: (xk−1 )=( x−1 ) ( xk−1+…))

In general, we want to factor x pn

−x or x pn−1−1 over GF ( p ).

Theorem: over Z

p Z=GF ( p ) x pn

−x is a product of all monic irreducible polynomials over

GF ( p ) where degree divides n (each one exactly once as roots are distinct!)

Example:

x16−x=x ( x+1 )⏟irreducibleof degree1

(x2+x+1 )⏟irreducible¿degre e 2

( x 4+ x3+1 ) ( x4+x+1 ) ( x4+x3+x2+x+1 )⏟all irreducibles

of deree 4

Proof: Suppose f ( x )∈ Zp Z

[ x ] monic, irreducible of degree m and m∨n.

Extend GF ( p ) to a field containing a root of f denoted α . This field will have pm elements.

We know by the last theorem, since m∨n this field is contained in a field of GF ( pn ).And so satisfies α pn

=α. If α=0 , f ( x )=x and x∨x pn

−x!

Otherwise α ≠0, α pn−1−1=0 so α root of x pn−1−1And so its minimal polynomial f ( x ) divides x pn−1 and so x pn

−x.

Conversely: Suppose now f ( x )∨x pn

−x ,monic irreducible and its degree is m.

If α is a root of f ( x ), then extending GF ( p ) to a field containing α we get an extension of dimension m over GF ( p ) i.e. a field of order pn.

So α is also a root of x pn

−x.

And so GF ( pm )=GF ( p ) ( α )In other words, every element of GF ( pm ) is a polynomial in α .

α is also a root of X pn

−x as f ( x )∨x pn

−xSo α∈GF ( pn ). Giving that GF ( pm )=GF ( p ) ( α )⊆GF ( pn )But then by the lemma – m∨n.

Error-Correcting Codese.g. spellcheck: eleqhantbed bod

With binary information – location of an error means we can correct it! (0↔1)

Naïve way:Transmit the same message 3 times and take a majority check.The probability of having an error in exactly the same position twice is very low.Very waistul! We might have a more sophisticated way of doing it…

Parity-Check DigitTransmit an extra digit at the end of the message.Send 1 if the message has an odd number of ones.Send 0 if the message has an even number of ones.

e.g. message = 10101 0⏟parity

If we get a message with an odd number of ones we know there’s an error, but we don’t know where it is.If we get an even number we could have had a double error. But this happens with a relatively low probability.

Example: ID with a Sifrat Bikoret03569657121212120+6+5+3+9+3+5+5=2610-last digit = 4!

Hamming Code (7,4)Locates (and so corrects) single errors.

Code words will be of length 7. There will be 4 “information digits” + 3 “redundancy digits”. We call them also parity check digits even though they do not check parity.Assumption: very low probability of double errors.p=¿ probability of error in transmitting a digit. Probability of a correctly transmitted message is (1−p )7

Probability of transmitting exactly one error: 7 p (1−p )6

So if you add them together you get: (1−p )7+7 p (1−p )6

If p=0.1 get 0.853 of a message with ≤1 errors.

Sending 4 digits (with no redundancy) correctly has probability (1−p )4

If p=0.1 get 0.6561.So 0.853 is a big improvement of sending only 4 digits and no errors!

This is a linear code, i .e . our code words are elements of a vector space over GF (2 ): elements of GF (2 )7

Subspace of dimension 4 . i.e. there are going to be 16 possible code words.(same number of code words in GF (2 )4)We define our code by giving a basis: 4 vectors of length 7.(in a 4×7 matrix).

v1 1 0 0 0 0 1 1v2 0 1 0 0 1 0 1v3 0 0 1 0 1 1 0v4 0 0 0 1 1 1 1

Suppose we want to transmit 1101?Send instead v1+v2+v4=1101001

Big advantage: Efficient decoding and locates ≤1 errors.

Use an analog to inner product/scalar multiplication. Induced by matrix multiplication over GF (2 ).

[ x1 … xn ] [ y1⋮yn

]=∑i=1

7

x i y i(mod 2)

It is a bilinear form on GF (2 )4.

Decoding:Suppose we receive y¿=[1 1 0 1 1 1 0 ]v1+v2= y= [1 1 0 0 1 1 0 ]

We compute:y¿ ∙ a=1+1+1=1y¿ ∙ b=1+1=0y¿ ∙ c=1+1=0

The result is sequence 100Which happens to be the binary representation of 4. And the error is in the fourth digit!If there’s no error, we get 0

a=[0001111]

b=[0110011 ]c= [1010101 ]

Hamming matrix:

[1 0 0 0 0 1 10 1 0 0 1 0 10 0 1 0 1 1 00 0 0 1 1 1 1]

The trick is in fact - Orthogonal complements:Recall: V is a vector space overF.B:V × V → F Is a bilinear form if it is linear in both variables:

B (a1 v1+a2 v2 ,w )=a1B ( v1 ,w )+a2+B (v2 ,w )B (v ,a1w2+a2w2 )=a1B (v ,w )+a2 (v ,w2 )

And for any subspace W of V we can define

W ⊥⏟OrthogonalComplementof W wrt B

= {w∈V|B (u ,w )=0 for all w∈W }

W⊥ is a subspace of V .

If F has charactaristics 0 and B is non-degenerate bilinear form.e.g. If F=R and B is dot product.If F=C and B is inner product ( v ,w )=vT ∙ w Then we have that:

W ⊕W⊥=VFor V finite dimension.Proof: uses fact that W ∩W⊥={0 } so that the union of base for W and a base for W⊥ is a base for V .

In general, for F or characteristic p and arbitrary bilinear form this is not true!

e.g. Taking product defined in GF (27 ) can see that [1 1 0 0 0 0 0 ] is orthogonal to itself!

E.g.If W =span {[1 1 0 0 0 0 0 ] } then W ⊊W⊥

e.g.[0 0 1 1 0 0 0 ]∈W ⊥¿

And W⊥≠ GF (2 )7

But: dim W +dimW ⊥=dim V ← proof in Basic Algebra 1 (Jacobson)E.g. dim W⊥ above will be 6!Take as a basis for W⊥:

[0 0 1 0 0 0 00 0 0 1 0 0 00 0 0 0 1 0 00 0 0 0 0 1 00 0 0 1 0 0 01 1 0 0 0 0 0

]--- end of lessonThe parity check matrix is defined to be a matrix whose columns are a basis for the orthogonal complement of the code.

Correcting Errors in linear codes over GF(2)Given a vector which contains errors, we want to correct it to the code word that differs from it in the fewest digits. Define -Hamming distance: d (v ,w )=¿ # of digits which v and w differ.e.g.

v=(1 0 1 1 0 0 1 1 ) ,w=(0 1 1 1 1 0 1 0 )d (v ,w )=4

Turns out, that in the hamming code, every 2 words/vectors are at distance ≥3.

TODO: Draw words in the code in a schematic way

Circle of radius 1 around w=¿ all vectors v such that d (w , v )=1.

So any vector with one error can only be corrected in one way o a codeword.General: We can correct r errors if the minimal distance between two code words ≥2r+1

Note: In the hamming code we have 16 elements. In the whole space, we have 27=128 elements. The elements at distance exactly 1 from a codeword ¿7 ∙16.So in fact, every element in the space is either in the code or at distance 1 from a codeword as 7 ∙16+16=128.

BCH CodeBose-Chandhuri-HocquenghemDouble error correcting code that uses GF (16 ) and has a nice decoding algorithm similar to that of the hamming code.

Construct by starting with the parity check matrix H (and then the code will be orthogonal complement of its rows).

The elements will be vectors in GF (2 )15

(need minimal hamming distance to be at least 5!)

GF (16 )¿={1 , α ,…,α14 } where α is the root of x4+x3+1 over GF (2 ).

Use: representation of GF (16 ) as vectors over GF (2 ) of length 4.

Form of H is going to be as follows:8×15 matrix over GF (2 )

H=[b1 b2 … b15c1 c2 … c15 ]

Where b i , ci∈GF (2 )4 row vectors.

We think of also as elements of GF (16 ).

Take b i=¿ vector of length 4 corresponding to α i−1 in the table.So we have 1 , α ,…,α14 in the top half of the matrix.c i’s will be defined later…

We want: If x=( x1 … x15 ) codeword, we want:

(1) H ∙xT=0⇔x in code(2) If x has at most 2 errors, want it to detect by multiplication by H .

Suppose x has exactly 2 errors in positions i and j. Then we can write:x=xc+ei+e j

And then:

H ∙x=Hx+ H ei+H e j=H e i+ H e j=(bi+b j

c i+c j)

So we want to choose the c i’s so we can recover from this vector.

Bad choice: c i=b i. Get Hx=(bb) - in this case we cannot recover i and j.

If b=(1001). We could have had: (

0001)

⏟b1

+(1000)

⏟b 4

But also: (0111)

⏟b8

+(1110)

⏟b 0

And a lot of other possibilities.

Another bad choice: define c i=( bi )2 (thinking of b i as an element of GF (16 ) so that c i

corresponding to α 2 i−2

So we should then get:

Hx=( bi+b j

bi2+b j

2)=( b i+b j

(bi+b j )2)=( b

b2)If you square you get the same thing….

Definition: Take c i=b i3.

( bi+b j

b i3+b j

3)=(bc ) want to show i and j determined uniquely and how to find them.

c=bi3+b j

3=(bi+b j ) (b i2+bi b j+b j

2 )=b(bi2+b ib j+b j

2)=b (b2+bi b j )(regarding the elements of GF (16 ))We first assume we have exactly 2 errors. So i≠ j and b≠0. Get c b−1+b2=bi b j

So b i and b j are roots in GF (16 ) of the quadratic equation:

(x−b i ) (x−b j )=x2−( bi+b j )x+b ib j=x2−bx+cb−1+b2

So given b and c, construct this polynomial.b i and b j are its unique solutions (in the field GF (16 )).

For convenience write: H '=H with α notation.

H '=[1 α α 2 … α 14

1 α 3 α 6 … α 12]Suppose y is a received message with errors in positions i and j.

And suppose H ' y=( αi−1+α j−1

α3 i−3+α 3 j−3)=(α5α7)

Equivalently: H ∙ y=(10110111) polynomial will be: x2+α5 x+α 8

Since: c b−1+b2=α 7 ∙ α−5+α 10=α 2+α10=α3

Need i and j such that: α i−1+α j−1=α 5 and α i−1∙ α j−1=α 8

i+ j−2≡8 (mod 15 )i+ j ≡10 (mod 15 )Checking possibilities: Get only i=3 , j=7 satisfies α i−1+α j−1=α 5 as well.Note: If the quadratic polynomial has no roots, then it cannot result from a double error.Meaning in fact that some triple errors are detectable but not correctable.

Single errors are also correctable using H :

It is the only case where we get a vector of the form: ( bb3) and then determine b i=b by

checking.So the polynomial will be x (x−b ).

We want to determine the dimension of the code and how to calculate a matrix for the code.

Claim: rankH=8Conclusion: dim code=7We shall show, that the first eight columns are linearly independent.

Suppose ∑i=1

8

ai(b i

b i3)=(00) and a i∈GF (2 )

Then we also get ∑i=1

8

ai( α i−1

α 3 i−3)=0⇒∑i=0

7

ai+1( α i

α 3i)=0⇔

∑i=0

7

ai+1αi=0 and ∑

i=0

7

ai+1α3i=0

Look at the polynomial ∑i=0

7

ai+1 x i=0 over GF (2 ) And α and α 3 are both roots.

So their minimal polynomials both divide ∑i=0

7

ai+1 xi

x4+x3+1 , x4+x3+x2+ x+1∨∑i=0

7

ai+1 xi

The product ( x4+x3+1 ) ( x4+x3+ x2+x+1 ) which is a polynomial of degree 8 divides

∑i=0

7

ai+1 x i which is of degree less or equal to 7! So ∑i=0

7

ai+1 xi is the zero polynomial!

Therefore all coefficients are zero and therefore linearly independent.Thus are also a basis for our vector space.

We construct C=¿ matrix for the code.H will be of the form: 7×15Where the first 8 columns are are the redundancy digits and the last 7 columns are the information digits.

Take (11) ,( αα 3) ,…,( α7

α21) first 8 columns of H '.

The 9’th column ( α 8

α24) is a linear combination of the first 8 columns: ∑i=0

7

si( αi

α3 i)

So the row vector ( s0 s1 … s7 1 0 … 0 ) orthogonal to all rows of H ' and H !

Take as the first row of c.

Similarly, column 10: ( α 9

α27)=¿ linear combination of 8 columns of H '.

t 0(11)+…+t 7(α 7

α 21)=( α9

α 27)So t 0(11)+…+t 7(α 7

α 21)+( α 9

α27)=(00)So take the vector (t 0 … t 7 0 1 0 … 0 ) orthogonal to rows of H ' take to be row 2 of C etc.