0
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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 '13 at 5:35
  • 2
    $\begingroup$ 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 '13 at 5:51
  • $\begingroup$ Sorry it suppsed to be X^2+X+2 $\endgroup$ – user111532 Nov 26 '13 at 22:14
  • $\begingroup$ Does that make it make sense? I'm still clueless $\endgroup$ – user111532 Nov 26 '13 at 22:15
  • $\begingroup$ Yes, that will fix the problem. $\endgroup$ – Jyrki Lahtonen Nov 27 '13 at 12:56
3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Wheres the accept option? $\endgroup$ – user111532 Nov 30 '13 at 7:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.