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I have a polynomial ring $\mathcal{R} = \mathbb{Z}_3[x]/(x^2+x+2)$ and I am supposed to find a generator of the multiplicative group of $\mathcal{R}$.

Well, I know that the $\vert \mathcal{R} \vert = 9$, therefore the number of elements of the multiplicative group is $9-1 = 8$ and that the elements $f(x)\in$ $\mathcal R$ are $[f(x)]=\{g(x) \in \mathbb Z_3 \mid g(x) \equiv f(x)$ mod $(x^2+x+2) \}$. Though I do not know how to find the generator instead of brute-forcing. Is there a smart way to do it? Thank you in advance.

mathew7k5b
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2 Answers2

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Write down explicitly:

$$\mathcal R^*=\left\{\,1,\,2,\,w,\,w+1,\,w+2,\,2w,\,2w+1,\,2w+2\,\right\}\;,\;\;\text{with}$$

$$w^2=-w-2=2w+1\pmod 3$$

Now, for example (all the operations are carried on modulo $\;3\;$ ):

$$\begin{align*}&w\\ &w^2=2w+1\\ &w^3=w(2w+1)=2w^2+w=2(2w+1)+w=w+2+w=2w+2\\ &w^4=w(w^3)=2w^2+2w=2(2w+1)+2w=w+2+2w=2=-1\end{align*}$$

Thus, we have already an element of order $\;8\;$ (why?) . You can now find all four elements of order 8 (how?)

DonAntonio
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The ring is the splitting field of $x^9-x$; since $$ x^9-x=x(x^8-1)=x(x-1)(x+1)(x^2+1)(x^4+1) $$ and $x^4+1=(x^2+x+2)(x^2+2x+2)$, the complete factorization is $$ x^9-x=x(x^8-1)=x(x-1)(x+1)(x^2+1)(x^2+x+2)(x^2+2x+2) $$

Note that if $\alpha$ is a root of $x^2+1$, then $\alpha^4=1$, so it is not good for the purpose. Thus we remain with the roots of the other degree two factors: indeed the group is cyclic of order $8$, so it must have $\varphi(8)=4$ elements which are generators.

egreg
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