# How the following multiplication table is solved ( related to $F_2[X]/f(x)$ ) [duplicate]

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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.

## marked as duplicate by user21820, Joshua Mundinger, Jyrki Lahtonen abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 22 at 3:55

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• Are you familiar with quotient rings? If so, where are you stuck? – Bill Dubuque Dec 8 '18 at 17:46

## 2 Answers

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 $$$$, i.e., $$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.

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
....: