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Is every ideal in $\Bbb Z[x]/(x^3-1)$ principal?

Edit: $\Bbb Z[x]/(x^3-1)$ is not isomorphic to $\Bbb Z[x]/(x-1) \oplus \Bbb Z[x]/(x^2+x+1)$ (by a modulo $3$ argument)

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    $\begingroup$ Neither of the rings is integral, so they cannot be PIDs. $\endgroup$ – asdq Aug 8 '18 at 9:01
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    $\begingroup$ @lisyarus All the ideals of $\Bbb Z\times \Bbb Z$ are principal and, in general, the product of two principal ideal rings is a principal ideal ring: let $I_1\times I_2$ be a generic ideal of $R_1\times R_2$, with $\langle a_j\rangle =I_j$ in $R_j$. Then, $(xa_1,ya_2)\in I_1\times I_2$ can be written as $(x,y)\cdot (a_1,a_2)$, and thus $I_1\times I_2=\langle (a_1,a_2)\rangle$. $\endgroup$ – Saucy O'Path Aug 8 '18 at 9:48
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    $\begingroup$ @SaucyO'Path Somehow I thought that an ideal of the form $I_1 \times I_2$ cannot be principal. Apparently, I need more coffee. Thank you for proving me wrong! $\endgroup$ – lisyarus Aug 8 '18 at 11:57
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    $\begingroup$ lisyarus Your isomorphism $\mathbb Z [x] / (x^2-1) \cong \mathbb Z \times \mathbb Z$ is also wrong! $\endgroup$ – user26857 Aug 10 '18 at 17:03
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    $\begingroup$ Relatively prime doesn't mean comaximal, that is, $(x-1,x+1)$ is not the whole ring. This means that you can not use CRT. $\endgroup$ – user26857 Aug 16 '18 at 5:56
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The ideal $I=(3,x-1)$ of the ring $R=\Bbb{Z}[x]/(x^3-1)$ is not principal.

Assume contariwise that the coset of $p=a+bx+cx^2$ would generate $I$. Let $J$ be the ideal generated by $p$. As an abelian group it is generated by $p=a+bx+cx^2$, $xp=c+ax+bx^2$ and $x^2p=b+cx+ax^2$. Consider the matrix $$ M=\left(\begin{array}{ccc}a&b&c\\c&a&b\\b&c&a\end{array}\right). $$ The index of $J$ in $R$ (if finite) is $|\det M|=|a^3+b^3+c^3-3abc|$ (by the basic theory of finitely generated abelian groups, most notably invariant factors and such).

A brute force check modulo $9$ shows that whenever $3\mid \det M$ we also have $9\mid\det M$. Therefore $|\det M|\neq3$, $|R/J|\neq3$.

But $R/I\simeq \Bbb{Z}_3$, so $J\neq I$.

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  • $\begingroup$ I'm sure we have studied this homogeneous cubic earlier, and those observations show that it cannot attain the value $\pm3$ at an integral point. I need to go now, so cannot search. $\endgroup$ – Jyrki Lahtonen Aug 8 '18 at 12:35
  • $\begingroup$ Suppose the ideal $(3,x-1)$ is principal (in $R$). Then there is $f\in\mathbb Z[X]$ such that $(3,X-1)=(f,X^3-1)$. Since $X-1\in(f,X^3-1)$ we get $1\in(f,X^2+X+1)$ (why?). But $f\in(3,X-1)$ and therefore $1\in(3,X-1,X^2+X+1)$, a contradiction. $\endgroup$ – user26857 Aug 20 '18 at 7:54

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