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Multiplication and addition table for $\mathbb{Z}_2[x]/I$, where $I = 〈x^3 + x^2 + 1〉 $

how to construct multiplication and addition table any one can explain

  • The addition is the one of polynomials of degree $2$. The multiplication is obtained by replacing all the occurences of $x^{3}$ by $ -x^2-1$ until you get a polynomial of degree $2$. Try with $\mathbb{C} = \mathbb{R}[x]/(x^2+1)$ to really see how it works. – reuns Nov 06 '17 at 06:13

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This will be precisely like the multiplication table for $\mathbb{Z}[x]$, but with the additional rule that $x^3 = -x^2-1$.

Because of this rule, every element will be of the form $a+bx+cx^2$ (if it involved higher powers of $x$, we could reduce them according to the rule).

So, we just need know what the following product looks like: \begin{align*} (a_0+b_0x+c_0x^2)(a_1+b_1x+c_1x^2)&= a_0a_1+a_0b_1x+a_0c_1x^2+b_0a_1x+b_0b_1x^2 \\ &+b_0c_1x^3+c_0a_1x^2+c_0b_1x^3+c_0c_1x^4 \\ & = a_0a_1+(a_0b_1+b_0a_1)x+(a_0c_1+b_0b_1+a_1c_0)x^2 \\ &+ (b_0c_1+c_0b_1)(-x^2-1)+c_0c_1x(-x^2-1) \\ & = (a_0a_1-b_0c_1-c_0b_1+c_0c_1)+(a_0b_1+b_0a_1-c_0c_1)x+(a_0c_1+b_0b_1+a_1c_0-b_0c_1-c_0b_1+c_0c_1)x^2 \\ \end{align*} Now, just plug in your given values of $a_i,b_i,c_i$.

Note that a multiplication table itself for $\mathbb{Z}[x]/I$ isn't super well defined as it's very infinite, so in some sense the above is the "best you can hope for".

Mark Schultz-Wu
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  • @mark...i edited now is this same as your answer –  Nov 06 '17 at 06:28
  • Now, use the same multiplication rule I wrote down, but compute all of the coefficients mod 2. Besides that everything is the same (note that one of the consequences of computation mod 2 is that all of the $-$'s I wrote are to be treated at $+$'s). – Mark Schultz-Wu Nov 06 '17 at 06:30
  • @Mark..is that necessary to $(a_0+b_0x+c_0x^2)(a_1+b_1x+c_1x^2)$..in z2 only 1, x are there –  Nov 06 '17 at 06:33
  • $x^2$ is there as well. $\mathbb{Z}_2[x]/I$ has $8$ elements. You could compute each of the 64 products (it can be reduced if you're smart, but there's still 15+ unique ones approximately), or you could compute one product in general like the above. Even though each coefficient can only be 0 or 1, it's still likely easier to do this computation than each ~15 I'd guess you'd have to do otherwise. – Mark Schultz-Wu Nov 06 '17 at 06:35