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Let $\alpha$ be a root of $x^3+x^2-2x+8$. As part of another problem, I am trying to show that, in the ring $\Bbb Z[\alpha,4/\alpha]$, the ideal $(2)$ splits as a product of three distinct primes. This is equivalent to the fact that $\Bbb Z[\alpha,4/\alpha]/(2)$ is a product of three domains.

A useful fact is that the minimal polynomial of $4/\alpha$ is $x^3-x^2+2x+8$. The following all seems correct to me: \begin{align} \Bbb Z[\alpha,4/\alpha]/(2)&\cong \Bbb Z[x,y]/(2,x^3+x^2-2x+8,y^3-y^2+2y+8,xy-4)\\ &\cong \Bbb F_2[x,y]/(x^3+x^2,y^3+y^2,xy)\\ &\cong \Bbb F_2[x,y]/(x^2,y^3+y^2,xy)\cap(x+1,y^3+y^2,xy)\\ &=\Bbb F_2[x,y]/(x^2,y^3+y^2,xy)\oplus \Bbb F_2\\ &=\Bbb F_2\oplus \Bbb F_2[x,y]/(y^2,x^2,xy)\cap (y+1,x^2,xy)\\ &=\Bbb F_2\oplus \Bbb F_2\oplus \Bbb F_2[x,y]/(x^2,xy,y^2). \end{align} If it is unclear, I'm using the Chinese remainder theorem in a few places. Clearly the third component is not a domain, so something has gone wrong. Its possible that the first line is incorrect, and that you need another condition relating $x$ and $y$. For example, in the quotient, there is no difference between $2/\alpha$ and $4/\alpha$, which seems wrong.

The bigger context here is that $\Bbb Z[\alpha,4/\alpha]$ is the ring of integers of $\Bbb Q(\alpha)$. This ring is interesting because Dedekind came up with a proof it is not monogenic. However, I am stuck on this detail, which is really just basic ring theory.

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    $\begingroup$ Do you have an integral basis? Is it just $\{1,\alpha,4/\alpha\}$ ? $\endgroup$ – Lubin Feb 16 at 6:36
  • $\begingroup$ @Lubin I’m pretty sure that’s correct; you can compute the discriminant to show that $\Bbb Z [\alpha, 4/\alpha]$ is the ring of integers and also show that $1,\alpha,4/\alpha$ is a basis. $\endgroup$ – Elliot G Feb 16 at 6:43
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    $\begingroup$ Great question. Since $1, \alpha, 4/\alpha$ is a $\mathbb{Z}$-basis for $\mathbb{Z}[\alpha, 4/\alpha]$, it seems like you must be missing some quadratic relations in the ideal defined in line $1$. In particular, $\alpha^{2}$ (for example) must be expressible as a $\mathbb{Z}$-linear combination of $1, \alpha, 4/\alpha$, though it's not obvious to me what that linear combination should be. (One can deduce the relation $4\alpha^{2}+4\alpha-8 = 8(4/\alpha)$, e.g., but this vanishes mod $2$, and isn't strong enough.) $\endgroup$ – Alex Wertheim Feb 16 at 6:49
  • $\begingroup$ @AlexWertheim that sounds right. There’s gotta be something missing here. I will have to come back to this tomorrow though. $\endgroup$ – Elliot G Feb 16 at 6:50
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    $\begingroup$ It hasta be true, ’cause the polynomial obviously factors into linears over $\Bbb Z_2$, courtesy of the Newton Polygon. $\endgroup$ – Lubin Feb 16 at 14:58
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The missing relations are $\alpha^2 = 2 - 2(4/\alpha) - \alpha$ and $(4/\alpha)^2 = 4/\alpha - 2\alpha - 2$.

These translate into $x^2 = 2 - 2y - x$ and $y^2 = y - 2x - 2$.

Modulo $2$ they become $x^2 = x$ and $y^2 = y$ (and these make the cubic relations superfluous).

So you have $$\begin{align*}\mathbb{F}_2[x,y]/(x^2 - x, xy, y^2 - y) &= \mathbb{F}_2[x,y]/(x,xy,y^2-y) \cap (x-1,xy,y^2-y) \\&\cong \mathbb{F}_2[x,y]/(x,xy,y^2-y)\oplus \mathbb{F}_2[x,y]/(x-1,xy,y^2-y)\\ & \cong \mathbb{F}_2[y]/(y^2-y)\oplus \mathbb{F}_2[y]/(y,y^2 - y)\\ &\cong \mathbb{F}_2 \oplus\mathbb{F}_2 \oplus\mathbb{F}_2.\end{align*}$$

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    $\begingroup$ Indeed, you have found two orthogonal idempotents in $R/(2)$, where $R$ is the ring of integers in the field, namely $e_1=\alpha$ and $e_2=4/\alpha$. The third is, of course, $e_3 = 1-e_1-e_2$. Thanks for this — I was going the long way ’round Robin Hood’s barn to try to find the idempotents, and it never occurred to me that they might be staring me in the face. $\endgroup$ – Lubin Feb 16 at 15:56
  • $\begingroup$ Thanks for writing this down! I wrote down these relations last night in a comment, but then deleted the comment after incorrectly computing $\mathbb{F}_{2}[x, y]/\langle x^{2}-x, xy, y^{2}-y \rangle$. (I incorrectly reduced both summands to $\mathbb{F}_{2}[y]/\langle y, y^{2}-y\rangle$.) It is a relief to see that things work out as expected after all! $\endgroup$ – Alex Wertheim Feb 16 at 23:05

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