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The notes of a module I'm doing has an example I don't quite understand.

If $\theta$ is the root of $t^3-4t+2$, then $2\theta-1$ is a unit of $\mathcal{O}_{\mathbb{Q}(\theta)}$. First note that $2\theta-1$ is a root of the monic integer polynomial $\frac{1}{8} \left[(t+1)^3-4(4(t+1))+16\right]=t^3+3t^2-13t+1$.

How did they find the polynomial above? I understand why they did it (the $+1$ in the end is useful later one to determine $2\theta -1$ is indeed a unit).

Could you please explain? thanks in advance.

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Short answer: that comes out from trying to get a $\pm 1$ as the constant term.

The general $y=a\theta^2+b\theta+c\in\mathbb{Z}[\theta]\subset\mathbb{Q}(\theta)$ has minimal polynomial $$ y^3 - (3 c + 8 a) y^2 + (16 a^2 + 6 a b + 16 a c - 4 b^2 + 3 c^2) y - (4 a^3 + 8 a^2 b + 16 a^2 c + 6 a b c + 8 a c^2 - 2 b^3 - 4 b^2 c + c^3) $$ So you want, for a unit, $$ -4 a^3 - 8 a^2 b - 16 a^2 c - 6 a b c - 8 a c^2 + 2 b^3 + 4 b^2 c - c^3 = \pm 1 $$

For simplicity try $a=0$ and by switching signs we can assume RHS is $1$: $$ 2 b^3 + 4 b^2 c - c^3 = 1 $$ so $(a,b,c)=(0,2,-1)$ is a choice, as are $(0,-1,1)$, $(0,-1,-1)$ or $(0,-3,5)$ (and of course $(a,b,c)=(0,0,-1)$ which is trivial). I think they are all the solutions with $a=0$ but haven't checked it.

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