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Consider the field $E:= \frac{\mathbb{Z}_3[X]}{\langle x^2 + x + 2\rangle}$. If I'm right the elements of the quotient ring can be found as: $$a_0 + a_1x + \langle x^2 + x + 2\rangle.$$ So we got the possibilities in $\mathbb{Z}_3$: $$\{0,1,2,\beta, 1+\beta , 2+\beta, 2\beta, 1+2\beta ,2+2\beta \}.$$ Here $\beta = \overline{x} = x + \langle x^2 + x + 2\rangle$ is a root of $x^2 + x+2$. (Correct me if my notation is wrong.)

So how do we get the elements of unit of $E^{\times},\cdot$. I assume $1$ is in it, but don't know how to calculate the other elements. With the elements, what would be the Cayley table of $E^{\times},\cdot$?

Other little question: we know that $\beta$ is a solution of $x^2 + x+2$, what is the other root?

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    $\begingroup$ As you are moding out an irreducible quadratic, the Vieta relations are the easiest way to find the other root (see Iefuneste's answer). When/if you learn more about finite fields and/or Galois theory, another method (that generalizes to more complicated situations) is to use the Frobenius automorphism: If $\beta$ is one root of an irreducible polynomial in $\Bbb{Z}_p[x]$, then $\beta^p$ is always another. Here $p=3$, and $$\beta^3=\beta^2\cdot\beta=-(\beta+2)\beta=-\beta^2-2\beta=-\beta+2$$ is the other root. This can be used recursively, but observe that $(-\beta+2)^3$ brings back $\beta$. $\endgroup$ – Jyrki Lahtonen Aug 18 '19 at 8:41
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    $\begingroup$ And, yes, your notation is fine. Too many engineers/programmers (with an interest in coding theory and/or crypto) get stuck denoting $\beta$ by $x$, IMO hampering the development of their understanding. $\endgroup$ – Jyrki Lahtonen Aug 18 '19 at 8:43
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After I figured out how to proper multiplicate in a quotient ring via: Constructing a multiplication table for a finite field, I managed to find the unit elements by calculating every possible combination. I found for instance: \begin{split} \beta(1+\beta)& = x^2 + x + \langle x^2 + x + 2\rangle \\ & =x^2 + x + \langle x^2 + x + 2\rangle + (0 + \langle x^2 + x + 2\rangle)\\ &= x^2 + x + \langle x^2 + x + 2\rangle + 2x^2+2x+4+ \langle x^2 + x + 2\rangle\\ &= 3x^2+ 3x +4 +\langle x^2 + x + 2\rangle\\ &=0+0+1+\langle x^2 + x + 2\rangle\\ &=1 \end{split} If I do this for the other elements, I find that $(2+\beta)(1+2\beta)=1$ and $(2\beta)(2+2\beta)=1$.

So the elements of unit become: $E^{\times},\cdot = \{1,\beta,1+\beta,2+\beta,1+2\beta,2\beta,2+2\beta\}$. The Cayley table is found by multiplying all the elements with each other. They are calculated similar as above.

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    $\begingroup$ $-1=2$ is missing from your list of elements of $E^*$. $E$ is a field, so all the $9-1$ non-zero elements are units. $\endgroup$ – Jyrki Lahtonen Aug 18 '19 at 8:34
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1) After line 1 you keep writing $x^2+x+1$ instead of $x^2+x+2$ (and ditto for capital letters).

2) The other root of the equation $X^2+X+2=0$ is $2\beta +2$, in order that the sum $\beta+ (2\beta+2)$ of the roots of that equation be $2=-1$, the opposite of the coefficient of $X$.

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  • $\begingroup$ Ddited the typos, my bad, do you know how to calculate the elements of unit? $\endgroup$ – Belgium_Physics Aug 17 '19 at 21:12
  • $\begingroup$ The invertible elements of $E$ ( which I suppose is what you mean by "elements of unit" ) are $E^{\times}=E\setminus \{0\}$, an equality valid for any field. (As mentioned by @Jyrki you forgot $2=-1$ in your description of that multiplicative group) $\endgroup$ – lefuneste Aug 18 '19 at 8:44
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    $\begingroup$ In your comment the first word of "Edited the typos" has a typo. I see the beginning of an interesting infinite recursion process... $\endgroup$ – lefuneste Aug 18 '19 at 8:51
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    $\begingroup$ Haha, I think I was possessed by the devil of typos back then $\endgroup$ – Belgium_Physics Aug 18 '19 at 8:58

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