TITLE: Which one is the ring of integers of $K$; $\mathbb Z[\alpha,\frac{\alpha^2}{2}]$ or $\mathbb Z[\alpha,\frac{\alpha+\alpha^2}{2}]$?
QUESTION [5 upvotes]: If $f := T^3 - T^2 + 2T + 8$ and $\alpha$ is the unique real root of $f$, If $K := \mathbb Q(\alpha)$, which one of the following rings are the ring of integers of $K$:
$$\mathbb Z\left[\alpha, \frac{\alpha^2}{2}\right]\quad \text{or}\quad \mathbb Z\left[\alpha,\frac{\alpha + \alpha^2}{2}\right]?$$

If I could compute the discriminant of them, the one with the smaller value would be a larger order, but I know neither computing the discriminant of a ring extended by $2$ elements, nor I can justify that there is no other order with a smaller discriminant.

REPLY [2 votes]: You can try to compute $(\alpha^2/2)^2$ and see that it isn't in the $\Bbb Z$-module of basis $\langle 1,\alpha,\alpha^2/2 \rangle$, which hints that $\alpha^2/2$ may not be an algebraic integer
$(T^3-T^2+2T+8)(T^3+T^2+2T-8) = T^6+3T^4+20T^2-64$
thus $T^3+3T^2+20T-64$ is the minimal polynomial of $\alpha^2$,
and then $T^3+\frac 32T^2+5T-8$ is that of $\alpha^2/2$.
Hence $\alpha^2/2$ is not an algebraic integer, so $\Bbb Z[\alpha,\alpha^2/2]$ can't be the ring of integer of $\Bbb Q(\alpha)$.
On the other hand, the $\Bbb Z$-module of basis $\langle 1,\alpha,(\alpha+\alpha^2)/2 \rangle$ is stable by multiplication, so it's a subring of the ring of integers.