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$F_2$ is polynomial field of group of integer modulo $2.f(x)$ is $x^2 + x + 1$.

I didn't got how the multiplication is happening in the table.I referred to many sources related to this topic but still i am facing difficulty in understanding it.I will be very thankful if someone explains the concept behind it.

This is the multiplication table for the field ${Bbb F}_4 = {Bbb F}_2[x]/langle x^2+x+1rangle$ consisting of the residue classes of the elements $0,1,x,x+1$ which are the remainders of ${Bbb F}_2[x]$ modulo $x^2+x+1$.

For instance, $[x] cdot [x+1] = [xcdot(x+1)] = [x^2+x]$ and the residue class of $x^2+x$ modulo $x^2+x+1$ is $[1]$, i.e., $x^2+x = 1cdot (x^2+x+1) + 1$ with quotient $q(x)=1$ and remainder $r(x)=1$. This is an elementary way to view this field extension.

Here are all 4x4 operations done to fill in the multiplication table.
I will write $X$ for the transcendent variable of the polynomial ring $Bbb F_2[X]$ over the field $Bbb F_2$ with two elements, and $x$ for the class $[X]$ which is $X$ modulo $X^2+X+1$ (irreducible=prime in the polynomial ring).

(  0  )*(  0  ) = [  0  ] * [  0  ] = [   0   ] = 0

(  0  )*(  1  ) = [  0  ] * [  1  ] = [   0   ] = 0

(  0  )*(  x  ) = [  0  ] * [  X  ] = [   0   ] = 0

(  0  )*(x + 1) = [  0  ] * [X + 1] = [   0   ] = 0



(  1  )*(  0  ) = [  1  ] * [  0  ] = [   0   ] = 0

(  1  )*(  1  ) = [  1  ] * [  1  ] = [   1   ] = 1

(  1  )*(  x  ) = [  1  ] * [  X  ] = [   X   ] = x

(  1  )*(x + 1) = [  1  ] * [X + 1] = [ X + 1 ] = x + 1



(  x  )*(  0  ) = [  X  ] * [  0  ] = [   0   ] = 0

(  x  )*(  1  ) = [  X  ] * [  1  ] = [   X   ] = x

(  x  )*(  x  ) = [  X  ] * [  X  ] = [  X^2  ] = x + 1

(  x  )*(x + 1) = [  X  ] * [X + 1] = [X^2 + X] = 1



(x + 1)*(  0  ) = [X + 1] * [  0  ] = [   0   ] = 0

(x + 1)*(  1  ) = [X + 1] * [  1  ] = [ X + 1 ] = x + 1

(x + 1)*(  x  ) = [X + 1] * [  X  ] = [X^2 + X] = 1

(x + 1)*(x + 1) = [X + 1] * [X + 1] = [X^2 + 1] = x

In the few ($2times 2=4$) cases where $[X^2+dots]$ appears as a result of computing the product of two polynomials of degree one (representing thus $x,x+1$ in $Bbb F_2[X]$) we replace above $X^2$ by $X^2-(X^2+X+1)=-X-1=X+1$ (working modulo $X^2+X+1$.)

P.S. The above was produced by computer, it is good to know that such computation can be done, assisted and learned in this way.
Used sage code:

sage: F = GF(2)

sage: R.<X> = PolynomialRing(F)

sage: K.<x> = R.quotient( X^2 + X + 1 )

sage: elements = [ K(0), K(1), x, x+1 ]



sage: for a in elements:

....:     for b in elements:

....:         A, B = a.lift(), b.lift()

....:         print( "({:^5})*({:^5}) = [{:^5}] * [{:^5}] = [{:^7}] = {}"

....:                .format(a, b, A, B, A*B, a*b) )

....:     print

....: