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Construct a field of 9 elements – construct the addition and multiplication tables. Begin with polynomials having coefficients $0$, $1$, and $2$ (integers modulo $3$) and use the modulus $X^2+X+2$

I really am not sure what its asking. I know how to do the multiple and addition of a field of $9$ but not sure about the rest at all.

Any help? Or even a similar example? Help me understand. Thanks!

Am I being asked to mod the addition and multiplication tables by $\ X^2+X+2 \quad (8) \ $? Am I supposed to write out all possible polynomials with modulo $3$ coefficients and $\mathrm{mod} \ X^2+X+1$?

Make no sense to me.

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  • $\begingroup$ I think the question is asking you for the splitting field of the polynomial $x^{2}+x+1$ over $\mathbb{Z}_{3}$. It should have 9 elements. $\endgroup$
    – TheNumber23
    Nov 26, 2013 at 5:35
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    $\begingroup$ Something is wrong. In the field of three elements $+1=-2$, so $$x^2+x+1=x^2-2x+1=(x-1)^2$$ is not irreducible!? Thus the quotient ring $\Bbb{F}_3[x]/\langle x^2+x+1\rangle$ is not a field. $\endgroup$
    – Jyrki Lahtonen
    Nov 26, 2013 at 5:51
  • $\begingroup$ Sorry it suppsed to be X^2+X+2 $\endgroup$
    – user111532
    Nov 26, 2013 at 22:14
  • $\begingroup$ Does that make it make sense? I'm still clueless $\endgroup$
    – user111532
    Nov 26, 2013 at 22:15
  • $\begingroup$ Yes, that will fix the problem. $\endgroup$
    – Jyrki Lahtonen
    Nov 27, 2013 at 12:56

1 Answer 1

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Since $x^2+x+2=0$ it follows that also $x^2=-x-2\equiv 2x+1$ since working mod 3. The elements of the field are only the nine linear terms $ax+b$ where each of $a,b$ can be independently any of $0,1,2$ giving the desired total of $9$ elements. Addition is no problem since the overflow can be done componentwise, e.g. $2x+2+x+0=3x+2=0x+2.$

For multiplication you use the above $x^2=2x+1$ when needed, for example $$(2x+1)\cdot (x+2)=2x^2+5x+2=2x^2+2x+2,\tag{1}$$ since $5=2$ mod 3, and now put in $2x+1$ for the $x^2$ so that $2x^2=4x+2=x+2$ replaces the $2x^2$ term in $(1)$, so it comes out $(x+2)+2x+2=3x+4=3x+1=0x+1.$

Other multiplications are similar, using the $x^2=2x+1$ as needed.

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  • $\begingroup$ Wheres the accept option? $\endgroup$
    – user111532
    Nov 30, 2013 at 7:47

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