CIMPASchoolofNumberTheoryinCryptographyandItsApplicationsSchoolofScience,KathmanduUniversity,Dhulikhel,NepalJuly19th-July31th,2010
Finite fields
Michel W alds chmidt
Course3:July23,2010These notes are extracted from the full text, the p df of which is ava ilable from the w eb site
http:
//www.math.jussieu.fr/∼miw/1/88
Gauss fields
A field with finitely many elements is also called a Gauss Field . F or instance, given a prime numb er p , the quotient
Z/p
Zis a Gauss field. Given tw o fields F and F
"with p elements, p prime, there is a unique isomo rph ism F
→F
". Hence, w e denote by
Fpthe unique field with p elements.
The cha racteristic of finite field F is a prime numb er p , hence, its prime field is
Fp. Mo reover, F is a finite vecto r space over
Fp; if the dimension of this space is s , which means that F is a fi nite extension of
Fpof degree [ F :
Fp]= s , then F has p
selements. Therefo re, the numb er of elements of a finite field is alw ays a p ow er of a prime numb er p , and this prime numb er is the cha racteristic of F .
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Gauss fields
The multiplica ti ve group F
×of a field with q elements has order q
−1 , hence, x
q−1=1 fo r all x in F
×, and x
q= x fo r all x in F . Therefo re, F
×is the set of ro ots of the p olynomial X
q−1−1 , while F is the set of ro ots of the p olynomial X
q−X :
(1) X
q−1−1=
!x∈F×
( X
−x ) ,X
q−X =
!x∈F
( X
−x ) .
Exercise 2. Pro ve that if F is a finite field with q elemen ts, then the p olynomial X
q−X + 1 has no ro ot in F . Deduce that F is not algebraically closed.
3
Subgroups of the multiplicative group of a field
Prop osition 3. Any finite subgroup of the multiplicative group of a field K is cyclic. If n is the order of G , then G is the set of ro ots of the p olynomial X
n−1 in K .
Pro of. Let K b e a field an d G a finite subgroup of K
×of order n and exp onent e . By Lagrange’s theo rem, e divides n . Any x in G is a ro ot of the p olynomial X
e−1 . Since G has order n , w e get n ro ots in the field K of this p olynomial X
e−1 of degree e
≤n . Hence e = n . W e conclude by using the fact that there exists in G an element of order e , hence, G is cyclic and is the set of ro ots of the p olynomial X
n−1 in K .
4
Lemma 4
Lemma 4. Let K b e a field of cha racteristic p . F or x and y in K , w e have ( x + y )
p= x
p+ y
p.
Pro of. When p is a prime numb er and n an integer in the range 1
≤n < p , the binomial co e
fficient
"p n
#= p ! n !( p
−n )!
is di visible by p .
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Lemma 5: f ∈ F
q[ X ] ⇐⇒ f ( X
q)= f ( X )
qW e shall use rep eatedly the follo wing fact: Lemma 5. Let
Fqb e a finite field with q elements, F an extension of
Fqand f
∈F [ X ] a p olynomial with co e
fficients in F . Then f b elongs to
Fq[ X ] if and only if f ( X
q)= f ( X )
q.
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Pro of of f ∈ F
q[ X ] ⇐⇒ f ( X
q)= f ( X )
qPro of of Lemma 5. Acco rding to (1), fo r a
∈F , the relation a
q= a holds if and only if a
∈Fq. Since q is a p ow er of the cha ra cteristic p of F , if w e write
f ( X )= a
0+ a
1X +
···+ a
nX
n,
then, by Lemma 4,
f ( X )
p= a
p0+ a
p1X
p+
···+ a
pnX
npand by induction
f ( X )
q= a
q0+ a
q1X
q+
···+ a
qnX
nq. Therefo re, f ( X )
q= f ( X
q) if and only if a
qi= a
ifo r all i =0 , 1 ,. .. ,n .
Prop os ition 6
F rom Lemma 4, w e deduce: Prop osition 6. If F b e a finite field of cha racteristic p , then
F rob
p: F
→F x
'→x
pis an automo rphism of F .
The F rob enius automo rphis m
Pro of of prop osition 6. Indeed, this map is a mo rphism of fields since, by Lemma 4, fo r x and y in F ,
F rob
p( x + y ) = F rob
p( x ) + F rob
p( y ) and F rob
p( xy ) = F rob
p( x )F rob
p( y ) .
It is injective since x
p=0 implies x =0 . It is surjective b ecause it is injective and F is fi nite.
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F rob enius
This automo rphism of F is called the F rob enius of F over
Fp. It extends to an automo rphism of the algeb raic closure of F . If s is a non–negative integer, w e denote by F rob
spor by F rob
psthe iterated automo rphism F rob
0p=1 , F rob
ps= F rob
ps−1◦F rob
p( s
≥1) ,
so that, fo r x
∈F ,
F rob
0p( x )= x, F rob
p( x )= x
p, F rob
p2( x )= x
p2,. .. ,
F rob
ps( x )= x
ps( s
≥0) .
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F rob enius If F
has p
selements, then the automo rphism F rob
sp= F rob
psof F is the identit y.
If F is a finite field with q elements and K a finite extension of F , then F rob
qis a F –automo rphism of K called the F rob enius of K over F .
11
F rob enius
Let F b e a finite field of cha racteristic p with q elements. Acco rding to Prop osition 3, the multiplicative group F
×of F is cyclic of order q
−1 . Let
αb e a generato r of F
×, that means an element of order q
−1 . F or 1
≤"<s , w e have 1
≤p
!−1 <p
s−1= q
−1 , hence,
α p!−1*=1 and F rob
!p(
α)
*=
α. Therefo re, F rob
phas order s in the group of automo rphisms of F . It follo ws that the extens ion F/
Fpis Galois, with Galois group the cyclic group of order s generated by F rob
p.
As a consequence, if F is a field with q elements and K a finite extension of F , then the extension K /F is Galois with Galois group the cyclic group generated by the F rob enius F rob
qof K over F .
12
Galois theo ry fo r finite fields Theo rem 7.
Let F b e a finite field with q elements and K a fin ite extension of F of degree s . Then there is a bijection b et w een the subfields E of K containing F and the diviso rs d of s . K
s/d
$|E
d
$|F
%s
•
If E is a subfield of K containing F , then the numb er of elements in E is of the fo rm q
dwhere d divides s .
•Conversely , if d divides s , then K has a unique subfield E with q
delements, which is the fixed field by F rob
pdand this field E contains F :
E =
{α∈K ; F rob
qd(
α)=
α}.
13/88When do es X
n− 1 divides X
m− 1 ?
Exercise 8. Let F b e a field, m and n tw o p ositiv e in tegers, a and b two in tegers
≥2. Pro ve that the follo wing conditions are equiv alen t. (i) n divides m . (ii) In F [ X ], the p olynomial X
n−1 divides X
m−1. (iii) a
n−1 divides a
m−1. (ii’) In F [ X ], the p olynomial X
an−X divides X
am−X . (iii’) b
an−b divides b
am−b .
HintDenote r the remainder of the Euclidean division of m by n . Prove that a
r−1 is the remainder of the Euclidean division of a
m−1 by a
n−1 . See also [3], Theorems 19.2, 19.3, 19.4.
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Existence of finite fields with p
selements
W e no w prove that fo r any prime numb er p and any integer s
≥1 , there exists a finite field with p
selements. Theo rem 9. Let p b e a prime numb er and s a p ositive integer. Set q = p
s. Then there exists a field with q elements. Tw o finite fields with the same numb er of elements are isomo rphic. If
Ωis an algeb raically closed field of cha racteristic p , then
Ωcontains one and only one subfield with q elements.
Pro of of Theo rem 9
Pro of. Let F b e a splitting fi eld over
Fpof the p olynomial X
q−X . Then F is the set of ro ots of this p olynomial, hence, has q elements. If F
"is a field with q elements, then F
"is the set of ro ots of the p olynomial X
q−X , hence, F
"is the splitting field of this p olynomial over its prime field, and, therefo re, is isomo rphic to F . If
Ωis an algeb raically closed field of cha racteristic p , then the unique subfi eld of
Ωwith q elements is the set of ro ots of the p olynomial X
q−X .
Finite subfields of F
pFix an algeb raic closure
Fpof
Fp. F or each s
≥1 , denote by
Fpsthe unique subfield of
Ωwith p
selements. F or n and m p ositive integers, w e have the follo wing equivalence:
(10)
Fpn⊂Fpm⇐⇒n divides m.
If these conditions are satisfi ed, then
Fpm/
Fpnis cyclic, with Galois group of order m /n generated by F rob
pn.
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Finite subfields of F
p(continued)
Let F
⊂Fpb e a finite field of cha racteristic p with q elements, and let x b e an element in
Fp. The conjugates of x over F are the ro ots in
Fpof the irreducible p olynomial of x over F , and these are exactly the images of x by the itera ted F rob enius F rob
qi, i
≥0 .
Tw o fields with p
selements are isomo rphic (cf. Theo rem 9), but if s
≥2 , there is no unicit y of such an isomo rphic, b ecause the set of automo rphisms of
Fpshas mo re than one element (indeed, it has s elements).
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Rema rks
•
The additiv e group ( F, +) of a finite field F with q elemen ts is cyclic, generated b y 1, hence, is isomorphic to
Z/q
Z.
•
The m ultiplicativ e group ( F
×,
×) of a finite field F with q elemen ts is cyclic, hence, is isomorphic to the additiv e group
Z/ ( q
−1)
Z.
•
A finite field F with q elemen ts is isomorphic to the ring
Z/q
Zif and only if q is a prime n um b er (whic h is equiv alen t to sa ying that
Z/q
Zhas no zero divisor).
19
Simples t example of a finite field * = F
pA field F with 4 elements has tw o elements b esides 0 and 1 . These tw o elements pla y exactly the same role: the map which p ermutes them and sends 0 to 0 and 1 to 1 is an automo rphism of F : this is nothing else than F rob
2. Select one of these tw o elements, call it
α. Then
αis a generato r of the multiplicative group F
×, which means that F
×=
{1 ,
α,
α 2}and F =
{0 , 1 ,
α,
α 2}. Here is the addition table of this field F :
( F, +) 01
αα20 01
αα 21 10
α 2αααα201
α 2α 2α10
20
Theo rem of the primitive element
Recall (Theo rem 7) that any finite extension of a finite field is Galois. Hence, in a fi nite field F , any irreducible p olynomial is sepa rable: finite fields are p erfect . Prop osition 11. Let F b e a finite field and K a fi nite extension of F . Then there exist
α∈K such that K = F (
α) .
Pro of.
Letq=psbethenumberofelementsinK,wherepisthecharacteristicofFandK;themultiplicativegroupK×iscyclic(Proposition3);letαbeagenerator.ThenK= &0,1,α,α 2,...,α q−2 '=Fp(α),
and,therefore,K=F(α).
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Exercises
Exercise 12. Pro ve the no rmal basis Theo rem : giv en a finite extension F
1⊂F
2of finite fields, there exists an elemen t
βin F
×2suc h that the conjugates of
βov er F
1form a basis of the vector space F
2ov er F
1. Pro ve that, with suc h a basis, the F rob enius map F rob
q1(where q
1is the n um b er of elemen ts in F
1) b ecomes a shift op erator on the co ordinates.
Exercise 13. Let F b e a finite field, E an extension of F and
α,
βtwo elemen ts in E whic h are algebraic ov er F of degree resp ectiv ely a and b . Assume a and b are relativ ely prime. Pro ve that F (
α,
β)= F (
α+
β) .
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F undamental res ult One of the main res ults of the theo ry of finite fields is the follo wing : Theo rem 14. Let F b e a finite field with q elements,αan element in an algeb raic closure of F . There exist integers
"≥1 such that
α q!=
α. Denote by n the smallest:
n = min
{"≥1 ; F rob
!q(
α)=
α}. Then the field F (
α) has q
nelements, which means that the degree of
αover F is n , and the minimal p olynomial of
αover F is
(15)
n−1!!=0 (
X
−F rob
!q(
α)
)=
n−1!!=0 $
