# Elements of $E^{\times},\cdot$ of the quotient ring $E:= \frac{\mathbb{Z}_3[X]}{\langle x^2 + x + 2\rangle}$

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?

• 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 $\beta$. – Jyrki Lahtonen Aug 18 '19 at 8:41
• 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. – Jyrki Lahtonen Aug 18 '19 at 8:43

## 2 Answers

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.

• $-1=2$ is missing from your list of elements of $E^*$. $E$ is a field, so all the $9-1$ non-zero elements are units. – Jyrki Lahtonen Aug 18 '19 at 8:34

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 $$X$$.

• Ddited the typos, my bad, do you know how to calculate the elements of unit? – Belgium_Physics Aug 17 '19 at 21:12
• 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) – lefuneste Aug 18 '19 at 8:44
• In your comment the first word of "Edited the typos" has a typo. I see the beginning of an interesting infinite recursion process... – lefuneste Aug 18 '19 at 8:51
• Haha, I think I was possessed by the devil of typos back then – Belgium_Physics Aug 18 '19 at 8:58