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In the polynomial ring $\mathbb Z_3[x]$, the ideal generated by $x^6+1$ is a prime ideal.

My attempt:

Theorem: Let $F$ be a field and $f(x)\in F[x]$. The ideal is $\langle f(x) \rangle$ is maximal iff $f(x)$ is irreducible over $F$.

Here $Z_3$ is a field, and the ideal generated by $x^6+1$ is maximal since $x^6+1$ is irreducible over $\mathbb Z_3$. So, $\langle x^6+1 \rangle$ is maximal and hence prime as every maximal ideal is prime ideal. Hence, the ideal generated by $x^6+1$ is a prime ideal.

Method 2:

Theorem:$F[x]$ is a PID iff $F$ is a field.

Theorem: Let $R$ be a PID. For an non-zero ideal $I$ such that $I \neq R$, then $I$ is prime ideal iff $I$ is maximal ideal. Using the above proof again, $I=\langle x^6+1 \rangle$ is maximal and hence prime.

But my answer key says this is false, am I doing something wrong here? Please try to explicitly point out where I am wrong and also what would be the correct approach. Thanks !

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    $\begingroup$ Over the field of three elements: $x^6+1=(x^2+1)^3$. $\endgroup$ – Angina Seng Jul 9 '20 at 10:37
  • $\begingroup$ Okay. This would imply it being not prime ideal, but where have I gone wrong, could you help me with that? $\endgroup$ – NAVI - s1mpleo Jul 9 '20 at 10:38
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    $\begingroup$ You said $x^6+1$ is irreducible.. $\endgroup$ – justadzr Jul 9 '20 at 10:39
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    $\begingroup$ Even over $\Bbb Z$, $x^6+1=(x^2+1)(x^4-x^2+1)$ is reducible. $\endgroup$ – Angina Seng Jul 9 '20 at 10:39
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    $\begingroup$ Thanks @AnginaSeng and Yourong for the help. I understand my mistake now. (sorry I am unable to tag more than 1 person) $\endgroup$ – NAVI - s1mpleo Jul 9 '20 at 10:43
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We know ideal $I$ is prime iff $\frac{R}{I}$ is integral domain

Here $I= <x^6+1> $

We have. $a= (x^2+1)\neq0 $ and$b=(x^4-x^2+1) \neq 0$ in $\frac{Z_3[x]}{<x^6+1>}$

But $ab =(x^2+1)(x^4-x^2+1) =0$

$\implies \frac{Z_3[x]}{<x^6+1>}$ is not an integral domain

$\implies <x^6+1> $is not prime

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    $\begingroup$ Another easy way to factor it is that by the Freshman dream theorem, $x^6+1=(x^2+1)^3$. Oh, ooop i see in the comments above it was covered as well.. $\endgroup$ – rschwieb Jul 9 '20 at 12:36

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