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Algebra through Examples
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Lesson 1
General DetailsE-mail: [email protected] 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.
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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
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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
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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…
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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:
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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 }.
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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…
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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 )
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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 )
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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 .
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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 )
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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
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TODO: Write it----- end of lesson 5
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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
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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
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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.
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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
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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
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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
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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 ]
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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
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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
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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
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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
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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.
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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 ).
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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
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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
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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+…))
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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.
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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.
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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]
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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⊥:
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[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).
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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:
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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.
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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)
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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.