How to prove {f(x)}^2＝f(x^2) in finite field GF(2)?

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b97704035

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As is in the title, for polynomial f(x) in finite field GF(2), how to prove {f(x)}^2＝f(x^2) ?
 
b97704035 said:
As is in the title, for polynomial f(x) in finite field GF(2), how to prove {f(x)}^2＝f(x^2) ?
Click to expand...

I have tried several examples like
f(x) = x^0 + x^1
f(x) = x^0 + x^1 + x^2
f(x) = x^0 + x^1 + x^2 + x^3
f(x) = x^0 + x^1 + 0 + x^3
and {f(x)}^2＝f(x^2) works for them. However I am stuck trying to prove that it works for all cases of f(x).
 
b97704035 said:
I have tried several examples like
and {f(x)}^2＝f(x^2) works for them. However I am stuck trying to prove that it works for all cases of f(x).
Click to expand...
Surely you recall that in \(\displaystyle GF(2)\) it is the case that \(\displaystyle (\forall x)[x^2=x]~?\)
 
pka said:
Surely you recall that in \(\displaystyle GF(2)\) it is the case that \(\displaystyle (\forall x)[x^2=x]~?\)
Click to expand...

Hi pka,
Thanks for the response.
By your hint, I researched again and realized that I may have unclear understanding of the scope of GF(2) so far.

Is my solution in following logical to solve this problem as is titled?

Since all polynomials in GF(2) are : 1, x, 1+x, x^2, 1+x^2, x+x^2, 1+x+x^2,
I just substitute all polynomials above one by one as "x" into the equation {f(x)}^2＝f(x^2) ,
and if all of them are proved to be correct, the proof is done.
 
b97704035 said:
.
By your hint, I researched again and realized that I may have unclear understanding of the scope of GF(2) so far.
Since all polynomials in GF(2) are : 1, x, 1+x, x^2, 1+x^2, x+x^2, 1+x+x^2,
Click to expand...
Why is that the case. Is it not the case that \(\displaystyle x^2=x~?\).
 
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