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Notation: ⟨x⟩ = Ideal generated by x.

From my understanding, the result should look like: $$\{(a_0+a_1X+a_2X^2+\cdots+a_nX^n)+\langle X^3+2X+2\rangle\mid a_0, a_1,\dots, a_n \in\mathbb{Z}_3\},$$

but i need a more specific form and i'm not even sure about this one.

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  • $\begingroup$ Can you prove that every polynomial in $\mathbb Z_3[x]$ is equivalent to a polynomial of degree at most $2$? (Hint: polynomial division). When are two polynomials of this kind equivalent to each other? Now use such polynomials as the representatives you are looking for. $\endgroup$ – Mark Bennet Sep 5 '17 at 20:37
  • $\begingroup$ So $\mathbb{Z_3}{[X]}/⟨X^3+2X+2⟩$ = $$\{(a_0+a_1X+a_2X^2\rangle\mid a_0, a_1,a_2 \in\mathbb{Z}_3\}$$. There are 27 possible polynomials of degree 2. $\endgroup$ – Surreal Everything Sep 6 '17 at 7:38
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It is a vector space of dimension $3$ (basis is $\{1,x,x^2\}$) over $\Bbb Z_3$, so there are $3^3 = 27$ equivalence classes.

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  • $\begingroup$ What if you mod out by a different cubic polynomial? .. $\endgroup$ – Chris Custer Sep 6 '17 at 3:43
  • $\begingroup$ For any cubic polynomial, the vector space has the same basis so the answer is 27. $\endgroup$ – Surreal Everything Sep 6 '17 at 7:28

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