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Let the polynomial $$k(x) = x^3 + 2x + 1 \in \mathbb{Z}_3[x]$$ Let $R = \mathbb{Z}_3[x]/(k(x))$ be the ring of equivalence classes in $\mathbb{Z}_3[x]$ modulo $k(x)$. As usual, we represent each equivalence class by the polynomial of lowest degree and positive coefficients (so, for example, $[x + 2]$ denotes the equivalence class that contains the polynomial $x + 2$). The answer must be represented in this way.

Is $[x]$ a primitive element in $R$?

In the solution they found that the number of elements in the multiplicative group is 26 and thus concluded that the possible orders are $x^{k}$ where $k=1,2,13,26$ because of Langrange theorem.

My question is how does $x^{k}=1$ guarantee that it is the primitive element since the definition of primitive element is an element who can generate the entire multiplicative group except $0$. I don't really see the correlation.

Need_MathHelp
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  • @xxxxxxxxx I meant the order of a subgroup needs to divide the order of a group. But the point is why $x^{k}≠1for k=1,2,3,13$? Is there a theorem that says so cuz I can't find it, if it supposed to be intuition well I don't have it either. Basically I don't understand how come showing $x^{k}$ is not k solves the problem. – Need_MathHelp Oct 14 '24 at 10:02
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    @SassatelliGiulio The theorem you mentioned is not mentioned in our course and according to exam rules, I cannot use it them unless I can prove it. I am studying for the exam so I would rather understand how they are solving it in old exams. – Need_MathHelp Oct 14 '24 at 11:06
  • @xxxxxxxxx It is edited now, I hope it makes more sense. – Need_MathHelp Oct 14 '24 at 11:07
  • @xxxxxxxxx Okay so if x has an order of 3 then it won't be primitive element because it cannot generate the entire subgroup. Because in a cyclic group the order of an element is the number of elements that can be generated by repeatedly applying the group operation to g. One more thing tho is it true that if G is a group and |G|>2 then G is cyclic? – Need_MathHelp Oct 14 '24 at 11:55
  • @Need_MathHelp Ok, I've removed it so you that you won't run the risk of learning from it. – Sassatelli Giulio Oct 14 '24 at 13:08
  • @SassatelliGiulio I didn't mean that way at all but I meant that it is strictly forbiden to use theorem not in mentioned in my course unless we give a full proof and I just wanna solve questions like they do in the old exams. But thanks for trying to help anyways, I truly appreciate it. – Need_MathHelp Oct 14 '24 at 14:36
  • @xxxxxxxxx thanks for the help but I also wanna make sure because there is a theorem that states |G|>2 then G is cyclic and when I google that I find confusing info so I just wanted to see if you could clear it up for me. – Need_MathHelp Oct 14 '24 at 14:39
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    It is not the case that every group of order greater than 2 is cyclic. It is true that the multiplicative group of nonzero elements in a finite field is always cyclic. – Arturo Magidin Oct 14 '24 at 15:38
  • @Need By Lagrange $x^{26}=1$ so $x$ has order $26!\iff! x^2\neq 1,, x^{13}\neq 1,$ by the Order Test. $\ \ $ – Bill Dubuque Oct 15 '24 at 06:02

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