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

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.

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marked as duplicate by user21820, Joshua Mundinger, Jyrki Lahtonen abstract-algebra Apr 22 at 3:55

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    $\begingroup$ Are you familiar with quotient rings? If so, where are you stuck? $\endgroup$ – Bill Dubuque Dec 8 '18 at 17:46
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This is the multiplication table for the field ${\Bbb F}_4 = {\Bbb F}_2[x]/\langle x^2+x+1\rangle$ 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, $[x] \cdot [x+1] = [x\cdot(x+1)] = [x^2+x]$ and the residue class of $x^2+x$ modulo $x^2+x+1$ is $[1]$, i.e., $x^2+x = 1\cdot (x^2+x+1) + 1$ with quotient $q(x)=1$ and remainder $r(x)=1$. This is an elementary way to view this field extension.

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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 ($2\times 2=4$) 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
....:     
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