means is a constant function. Since must be the zero function. But then is it linear?
But of course, it's not a field homomorphism.
let GF(2^n) be the extension of the binary field.Let " F " be a linear function, which means F(X+Y)=F(X)+F(Y) and F(0)=0.
Assume that X,Y are any two elements in the GF(2^n) -the finite field- . The question is can we define a linear function "F" such that F(X)=F(Y), for any X,Y belong to GF(2^n)?? in another word "what kind of function that will look like this?? will do this F(X)=F(Y)???
First of all, Thank you so much for helping.. I really appreciate that.
Well, remember that a finite field or Galois field is a field of order p elements under modulo-p addition and multiplication is denoted by GF(p).
GF(p) = {0,1,2,... ,p-1} modulo p.
For p=2, we obtain the binary field GF(2) which has only two elements. In binary arithmetic we use modulo 2 addition and multiplication. This arithmetic is actually equivalent to ordinary arithmetic , except that we consider 2 to be equal 0 which means (1+1=2=0). Note that since (1+1=0) ,then (1=-1).
GF(2)= {0,1}.
The extension of the binary field is GF(2^n), where 2^{2^n-1}=1. Therefore ,the set GF(2^n)= {0,1, alpha, alpha^2, alpha^3,..., alpha^{2^n-2}} is a finite field of 2^n elements. Note that the addition and multiplication definied on GF(2^n) imply modulo-2 addition and multiplication . Hence, the binary field GF(2) forms a subfield of GF(2^n)$.
For exmaple , let n=4 .The polynomial p(X)=1+X+X^4 is the primitive polynomial over GF(2).Set the p(alpha)=1+ alpha + alpha^4 = 0 Then, alpha^4 = 1 + alpha .We can construct GF(2^4) using this relation. The identity (alpha^4=1+alpha) is used repeatedly to form the polynomial representations for the elements of GF(2^4) , and every element could be constructed from the basis (1, alpha,alpha^2,alpha^3).
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The only thing I know is that "F" is a linear function, saying that F(0)=0 that's one of the properties of being linear. "F" is defined as:
F: GF(2^n)->C2*C2*...C2 'n- times'.
Linear properties are:
1) F(X+Y) = F(X)+F(Y).
2) F(aX) = a F(X),where 'a' is a constant.
this implies that -> F(0)=0
The question is : What kind of linear function "F" that will satisfy F(X)=F(Y) for any X,Y in the GF(2^n). How to define that linear function??
I am not sure, simply because -> The range of "F" is not a Finite field , it is not GF(2^n).
F is defined like this :
F: GF(2^n)-> C2 x C2 x C2... n times
The range looks like this in case of n=3 {000,010,100,001,110,101,011,111}, note that this is a cyclic group and it is not a finite field.
Now the problem if I define sets like this:
Sa={X in GF(2^n) | F(X)= a), where a is a group element belongs to the (C2 x C2 x C2... n times). so for example we may have S101={alpha, alpha^3} in case n=3. SO, there exist a function "F" that can group the two elements from GF(2^n) . MY question is : I want to define a function "F" of course a linear function that will take any X,Y to the same set "S"..the whole idea is I need to know what kind of function that will look like this F(X)=F(Y).
You said a constant, in may range which is C2 xC2 xC2 how that will look like.
Thank you so much for your help, I know you are smart
Let me make the problem clear:
Assume that given any two elements X,Y from GF(2^n), I know in advance that I have these two elements. I want to find a linear function "F" such that F(X)=F(Y). SO, it is not for all X,Y in GF(2^n). the actual term is the two elements from the GF(2^n), how can I define a linear function such that F(X)=F(Y) knowing that:
F: GF(2^n) -> C2 x C2 xC2... n times.
I think the F function should be based and depends on both X, and Y.
So for example:
n=3 , which means I know a finite field GF(2^3)= {0, alpha, alpha^1, ..,alpha^6}. If given (X=alpha^4) and my (Y= alpha^2) the problem is how to define a linear function F based on (alpha^4, alpha^2) such that F(alpha^4)=F(alpha^2) ???
What can and be considered as? As vector spaces.
Let be a linear map and be two different vectors such that .
The line belongs to its kernel.
Let and be two non zero vectors in , can they be linearly independent? and are dependent iff , so and are independent. can be completed in a basis of , to determine , you have to determine the range of a basis.
That let you choices, .
If and , you build a base , and you also have choices (what is demanded is )