# Multiplying in GF(2)

• Feb 1st 2010, 10:35 AM
Migotek84
Multiplying in GF(2)
We are working in GF(2). Consider following polynomial:
$\displaystyle A: x_1+x_2+1=0$

Now multiply it by all monomials, that mean $\displaystyle x_1, x_2, x_3, x_4$ to get:
$\displaystyle A_1: x_1x_2=0$
$\displaystyle A_2: x_1x_3+x_2x_3+x_3=0$
$\displaystyle A_3: x_2x_4+x_2x_4+x_4=0$

I know that $\displaystyle A_2:=A*x_3, A_3:=A*x_4$ but from where we have $\displaystyle A_1$? We should multiply $\displaystyle A$ by $\displaystyle x_1$ and then by $\displaystyle x_2$. From this two new equations we have $\displaystyle A_1$ but how we could get it?
• Feb 8th 2010, 09:10 AM
Migotek84
I found the solution. Many thanks for tonio(Handshake)
• Feb 8th 2010, 09:20 AM
tonio
Quote:

Originally Posted by Migotek84
We are working in GF(2). Consider following polynomial:
$\displaystyle A: x_1+x_2+1=0$

Now multiply it by all monomials, that mean $\displaystyle x_1, x_2, x_3, x_4$ to get:
$\displaystyle A_1: x_1x_2=0$
$\displaystyle A_2: x_1x_3+x_2x_3+x_3=0$
$\displaystyle A_3: x_2x_4+x_2x_4+x_4=0$

I know that $\displaystyle A_2:=A*x_3, A_3:=A*x_4$ but from where we have $\displaystyle A_1$? We should multiply $\displaystyle A$ by $\displaystyle x_1$ and then by $\displaystyle x_2$. From this two new equations we have $\displaystyle A_1$ but how we could get it?

This is the second time you ask this question and this is the second time I don't understand what you mean: you call $\displaystyle x_1+x_2+1 = 0$ "A polynomial", which it is not since it is equalled to zero. It would then be an equality.

Next, you want this equality to be multiplied "by the monomials $\displaystyle x_1,x_2,x_3,x_4$ "....so you're working in the ring of polynomials in 4 indeterminates ?!

Anyway, you get 4 weird expressions that you call $\displaystyle A_1, A_2, A_3$...and I don't have the faintest idea from where did you get these expressions!. For example , multiplying the forementioned equality by the monomial $\displaystyle x_1$ we get:

$\displaystyle x_1(x_1+x_2+1) = x_1\cdot 0\Longrightarrow x_i^2+x_1x_2+x_1=0$....so ?? Of course , if you already know that $\displaystyle x_1^2+x_1=0$ no matter what VALUE in $\displaystyle \mathbb{F}_2$ we choose for $\displaystyle x_1$ then you can EVALUATE and get $\displaystyle x_1x_2=0$...both AS POLYNOMIALS this is not so!

I think you must first understand well what you want, then decide how to ask it in a clear way and then post again.

Tonio
• Feb 8th 2010, 10:02 AM
Arczi1984
I agree with tonio.
These are equations not polynomials.
I think that this is some part of Yours previous post http://www.mathhelpforum.com/math-he...imination.html - 4 variables over GF(2).
So You consider equation $\displaystyle A$ and then You multiplied it by 4 monomials:
$\displaystyle x_1(x_1+x_2+1)=0 => x_1x_2=0$
$\displaystyle x_2(x_1+x_2+1)=0 => x_1x_2=0$
$\displaystyle x_3(x_1+x_2+1)=0 => x_1x_3+x_2x_3+x_3=0$
$\displaystyle x_4(x_1+x_2+1)=0 => x_1x_4+x_2x_4+x_4=0$
Am I right?

In last part You used word 'equations' - good! :)

PS. I knew that this problem/example I saw before. This is a toy example from http://math.uc.edu/~aac/pqcrypto2008...008mohamed.pdf
• Feb 8th 2010, 10:20 AM
Migotek84
I'm sorry for my mistake:(
Of course it is not a polynomial but polynomial equation.
Yes Arczi1984, it is from this paper.
Many thanks for tonio beacuse You help me very much.
Once more I'm so sorry for this mess.