A Short Study of Galois Field
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Transcript of A Short Study of Galois Field
Presented By
Exam Roll: 1820
Registration No: 2010-712-250
Department Of Mathematics
University Of Dhaka
A SHORT STUDY OF GALOIS FIELD
Objectives Of The Project
To introduce Galois Field.
To discuss related theorems.
Computational approach of Galois Field.
Applications of Galois Field.
What is Galois field ?
A Galois field(so-named in honor of Évariste Galois) is a
field in which the number of elements is always a
positive integer power of a prime number.
That is, the number of elements of a
Galois field is of the form pn , where p is
a prime and n is a positive integer.
Generally it is denoted by GF(pn ).
Évariste Galois(1811-1832)
Examples
1. GF(31)={0,1,2} forms a Galois field.
Addition table
Where 0 is the additive identity element and
0 is the additive inverse of 0
1 is the additive inverse of 2
2 is the additive inverse of 1
Addition 0 1 2
0 0 1 2
1 1 2 0
2 2 0 1
Multiplication 1 2
1 1 2
2 2 1
Multiplication Table
Where 1 is the multiplicative identity element and
1 is the multiplicative inverse of 1
2 is the multiplicative inverse of 2
2. GF-13:The elements are 0,1,2,3,4,5,6,7,8,9,a,b and c.
Here a=10, b=11, c =12.
Addition Table
Additi
on
0 1 2 3 4 5 6 7 8 9 a b c
0 0 1 2 3 4 5 6 7 8 9 a b c
1 1 2 3 4 5 6 7 8 9 a b c 0
2 2 3 4 5 6 7 8 9 a b c 0 1
3 3 4 5 6 7 8 9 a b c 0 1 2
4 4 5 6 7 8 9 a b c 0 1 2 3
5 5 6 7 8 9 a b c 0 1 2 3 4
6 6 7 8 9 a b c 0 1 2 3 4 5
7 7 8 9 a b c 0 1 2 3 4 5 6
8 8 9 a b c 0 1 2 3 4 5 6 7
9 9 a b c 0 1 2 3 4 5 6 7 8
a a b c 0 1 2 3 4 5 6 7 8 9
b b c 0 1 2 3 4 5 6 7 8 9 a
c c 0 1 2 3 4 5 6 7 8 9 a b
Multipl
ication
1 2 3 4 5 6 7 8 9 a b c
1 1 2 3 4 5 6 7 8 9 a b c
2 2 4 6 8 a c 1 3 5 7 9 b
3 3 6 9 c 2 5 8 b 1 4 7 a
4 4 8 c 3 7 b 2 6 a 1 5 9
5 5 a 2 7 c 4 9 1 6 b 3 8
6 6 c 5 b 4 a 3 9 2 8 1 7
7 7 1 8 2 9 3 a 4 b 5 c 6
8 8 3 b 6 1 9 4 c 7 2 a 5
9 9 5 1 a 6 2 b 7 3 c 8 4
a a 7 4 1 b 8 5 2 c 9 6 3
b b 9 7 5 3 1 c a 8 6 4 2
c c b a 9 8 7 6 5 4 3 2 1
Multiplication Table
Theorem 1: A multiplicative group of GF(Pn) is cyclic.
Theorem 2: Let F be a finite field with Pn elements and let
F. Then there exists elements and in F such that
= 2+ 2.
Theorem 3: GF( pn) has a subfield F′ with pm elements if
and only if m | n. Moreover, F' is unique.
Theorem 4: The number of elements in a finite field of
characteristic p is of the form Pn, n being some positive
integer.
Theorem 5: Each element of a finite field with p elements
satisfies the equation 𝑥𝑝𝑛− 𝑥 = 0.
Theorem 6: lf a finite field F has pn elements , then F is
the decomposition field of the polynomial 𝑥𝑝𝑛− 𝑥.
CHECKING
Now we illustrate the example GF-17 by a programming
language MATHEMATICA 5.2.We know that
𝑍17={0, 1,2,3,4,5,…….14, 15, 16 } is a field. By using this
program we can check any GF-n, where n is any prime
number.
Here we start with EXIT to avoid the disturbance of those
symbols which was used in previous time and to show all
the input and output of every line.
Mathematica program for Addition Table
In[1]:= Exit
In[2]:= n=17
Out[2]= 17
In[3]:= v=Table[i,{i,0,n-1}]
Out[3]= {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}
In[4]:= For[i=1,i<= n,i++,
For[j=1,j<= n,j++,
p=v[[i]]+v[[j]];
If[p>= n,r[j]=p-n,r[j]=p]];
q[i]=Table[r[k],{k,1,n}]];
rslt=Table[q[l],{l,1,n}];
In[5]:= TableForm[rslt,TableHeadings->
{{"0","1","2","3","4","5","6","7","8","9","10","11","12","13
","14","15","16"},{"0","1","2","3","4","5","6","7","8","9","
10","11","12","13","14","15","16"}},TableSpacing->{1,1}]
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0
2 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0 1
3 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0 1 2
4 4 5 6 7 8 9 10 11 12 13 14 15 16 0 1 2 3
5 5 6 7 8 9 10 11 12 13 14 15 16 0 1 2 3 4
6 6 7 8 9 10 11 12 13 14 15 16 0 1 2 3 4 5
7 7 8 9 10 11 12 13 14 15 16 0 1 2 3 4 5 6
8 8 9 10 11 12 13 14 15 16 0 1 2 3 4 5 6 7
9 9 10 11 12 13 14 15 16 0 1 2 3 4 5 6 7 8
10 10 11 12 13 14 15 16 0 1 2 3 4 5 6 7 8 9
11 11 12 13 14 15 16 0 1 2 3 4 5 6 7 8 9 10
12 12 13 14 15 16 0 1 2 3 4 5 6 7 8 9 10 11
13 13 14 15 16 0 1 2 3 4 5 6 7 8 9 10 11 12
14 14 15 16 0 1 2 3 4 5 6 7 8 9 10 11 12 13
15 15 16 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
16 16 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Out[5]=
Mathematica program for Multiplication Table
In[1]:= Exit
In[1]:= n=17
In[2]:= 17
In[3]:= v=Table[i,{i,0,n-1}]
In[3]:= {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}
In[4]:= For[i=2,i<=n,i++,
For[j=2,j<=n,j++,
p=v[[i]]*v[[j]];
If[p>n,r[j]=Mod[p,n],r[j]=p]];
q[i]=Table[r[k],{k,2,n}]];
rslt=Table[q[l],{l,2,n}];
In[5]:= TableForm[rslt,TableHeadings-
>Automatic,TableSpacing->{1,1}]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
2 2 4 6 8 10 12 14 16 1 3 5 7 9 11 13 15
3 3 6 9 12 15 1 4 7 10 13 16 2 5 8 11 14
4 4 8 12 16 3 7 11 15 2 6 10 14 1 5 9 13
5 5 10 15 3 8 13 1 6 11 16 4 9 14 2 7 12
6 6 12 1 7 13 2 8 14 3 9 15 4 10 16 5 11
7 7 14 4 11 1 8 15 5 12 2 9 16 6 13 3 10
8 8 16 7 15 6 14 5 13 4 12 3 11 2 10 1 9
9 9 1 10 2 11 3 12 4 13 5 14 6 15 7 16 8
10 10 3 13 6 16 9 2 12 5 15 8 1 11 4 14 7
11 11 5 16 10 4 15 9 3 14 8 2 13 7 1 12 6
12 12 7 2 14 9 4 16 11 6 1 13 8 3 15 10 5
13 13 9 5 1 14 10 6 2 15 11 7 3 16 12 8 4
14 14 11 8 5 2 16 13 10 7 4 1 15 12 9 6 3
15 15 13 11 9 7 5 3 1 16 14 12 10 8 6 4 2
16 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
Out[5]:=
VERIFICATION:
We verify the theorem 2 for p=23, n=1, i.e. for Galois field
GF(23).
We know that 𝑍23={0, 1,2,3,4,5,…….22} is a field. Thus we
may consider GF(23)=𝑍23.
Now using the following computer program, we verify that
for every αGF(23),there exists µ, GF(23) satisfying
α=µ2+𝜐2.Here we use MATHEMATICA 5.2 programming
tools.
In[1]:= Exit
In[1]:= n=23;
In[2]:= v=Table[i,{i,0,n-1}];
In[3]:= For[i=1,i<= n,i++,
For[j=0,j<= n,j++,
For[k=0,k<= n,k++,
mat=v[[j]]^2+v[[k]]^2;
rem=Mod[mat,n];
If[v[[i]]==Šrem,
p[i]=v[[i]];q[i]=v[[j]];r[i]=v[[k]]]]]
Print[p[i],"=",q[i],"^2 +",r[i],"^2"]]
Out[3]:=
0 = 0 ^2 + 0 ^2
1 = 22 ^2 + 0 ^2
2 = 22 ^2 + 22 ^2
3 = 22 ^2 + 18 ^2
4 = 22 ^2 + 16 ^2
5 = 22 ^2 + 21 ^2
6 = 21 ^2 + 18 ^2
7 = 22 ^2 + 12 ^2
8 = 21 ^2 + 21 ^2
9 = 22 ^2 + 13 ^2
10 = 22 ^2 + 20 ^2
11 = 20 ^2 + 18 ^2
12 = 21 ^2 + 13 ^2
13 = 22 ^2 + 14 ^2
14 = 22 ^2 + 17 ^2
15 = 20 ^2 + 12 ^2
16 = 21 ^2 + 14 ^2
17 = 22 ^2 + 19 ^2
18 = 20 ^2 + 20 ^2
19 = 22 ^2 + 15 ^2
20 = 21 ^2 + 19 ^2
21 = 20 ^2 + 14 ^2
22 = 21 ^2 + 15 ^2
Since it is possible to write α=µ2 +𝜐2 where α, µ, 𝑍23then we conclude that every root can be expressed as the
sum of two squares.
1. Galois field is used in Cryptography
2. Galois field is used in Coding theory and Combinatorial
design and Quantum error correction.
3. The most commonly used Galois Field is GF(256) i.e
GF(28).
4. All CD and DVD players use computation in Galois
Field, as do many disk storage system.
5. Galois Field is used to develop some Mathematical
Theories , which have a lot of real life application.
REFERENCES[1] Hiram Palely and Paul M. Weichsel: “A First Course in Abstract Algebra”
New York, Holt, 1996.
[2] J.S. Milne:,”Fields and Galois Theory”, The photograph is of Sabre Peak,
Moraine Creek, New Zealand
[3] R. S. Aggarwal: A text book on modern algebra.
[4] Mary Gray: “A radical approach to algebra”, Addison-Wesley publishing
Co. London,1970.
[5] Professor Abdur Rahman : “ Abstract Algebra”,Dhaka,1995.
[6] Bhattacharya, P.B. adds Jain, S.K., and Naipaul: “A first course in rings,
fields and vector spaces, Halsted Press, New York, 1977.
[7] John.B Fraleigh , “A First Course in Abstract Algebra”. Copyright by
Pearson Education, Inc
Website : www.mathworld.wolfarm.com
www.google.com
www.encyclopedia.com
http://members.aol.com/gmtsgibbs/galois.html
https://en.wikipedia.org