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So I get that $\mathbb{Z}_3[x]/\langle x^2+1\rangle$ is a field since $x^2+1$ is irreducible. I don't get how to show it only has $9$ elements. And is that the same thing as saying it has order $9$?

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    $\begingroup$ Yes, it has 9 elements, and that's the same as saying it has order $9$. You can explicitly list them, too: there are three polynomials of degree $0$ and six of degree $1$. $\endgroup$ – T. Bongers Feb 23 at 20:50
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    $\begingroup$ Yes it is the same thing as saying there are 9 elements. Hint: The elements of the field are of the form, $a + bx + (x^2+1) \mathbb{Z}_3[x]$ where $a, b \in \mathbb{Z}_3$. $\endgroup$ – symchdmath Feb 23 at 20:53
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Yes, it has $9$ elements, namely $$ \{ 0,1,2,x,x+1,x+2,2x,2x+1,2x+2\} $$ considered as equivalence classes modulo $(x^2+1)$. Because of $x^2=-1$, there are no higher-degree polynomials in it.

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  • $\begingroup$ So I don't understand why it's only these 9 elements. Since all elements have form $g(x) + \langle x^2 + 1 \rangle$, how come we only consider the different options for $g(x)$? How come we don't take into account the $(x^2 + 1) \mathbb{Z}_3[x]$ and why must the degree of $g(x)$ be less than 2? $\endgroup$ – Chubbles Feb 23 at 21:22
  • $\begingroup$ @Chubbles, because any polynomial $g(x)$ is congruent to the remainder $r(x)$ of degree less than 2 where $g(x)=q(x)(x^2+1)+r(x)$ so only the various possible $r(x)$ of degree $0$ or $1$ give different elements of the quotient ring (field). $\endgroup$ – Ned Feb 23 at 21:39
  • $\begingroup$ OH okay. Got it! Very helpful thank you! $\endgroup$ – Chubbles Feb 23 at 23:52
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Hint $:$ $\Bbb Z_3 [x] / \langle x^2 + 1 \rangle \cong \left (\Bbb Z [x] / \langle x^2 + 1 \rangle \right ) / \langle 3 \rangle \cong \Bbb Z[i] / \langle 3 \rangle.$

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