I was trying to find the ring of integers of $\mathbb{Q}(\theta)$, where $\theta^3 -2\theta + 2 = 0$. I compute the discriminant of the basis $\{1, \theta, \theta^2\}$, but unfortunately it is $-4*19$, which is not square-free. I can check by hand that $(a + b\theta + c\theta)/2$ is not in the ring of integers of $\mathbb{Q}(\theta)$, where $a,b,c \in \mathbb{Z}_2$. Is there a way to avoid all these computations?

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    $\begingroup$ The polynomial is Eisenstein at $p=2$, so locally at $2$, $\theta$ generates the ring of integers. $\endgroup$ – Angina Seng Dec 2 '19 at 4:17
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    $\begingroup$ @LordSharktheUnknown can you elaborate a bit more? $\endgroup$ – Philomeno Dec 2 '19 at 4:18

To avoid talking about $p$-adic fields and ramifications, I give here a more accessible proof. It is, of course, equivalent to the comment by @Lord Shark the Unknown.

Let $\mathcal O$ be the integer ring of $\mathbb Q(\theta)$. We consider the quotient ring $R = \mathcal O / 2\mathcal O$.

Let $t$ be the image of $\theta$ in $R$. From $\theta^3 -2\theta + 2 = 0$, we deduce that $t^3 = 0$ in $R$.

It follows that $t^2 \neq 0$ in $R$. Otherwise, $\theta^2$ lives in $2\mathcal O$ and we have $(\theta^2 / 2 - 1)\theta + 1 = 0$, which means that there exists $s\in R$ such that $st = 1$ in $R$. This is impossible, since it would lead to $0 = (st)^3 = 1$ in $R$.

Now if $a, b, c$ are integers such that $(a + b\theta + c\theta^2)/2$ is an element of $\mathcal O$, then we have $a + bt + ct^2 = 0$ in $R$.

Multiplying by $t^2$, we see that $at^2 = 0$ in $R$. Therefore $a = 0$ in $R$, which means $a$ is an even integer.

The equation becomes $bt + ct^2 = 0$ in $R$. Multiplying by $t$, we get $b = 0$ in $R$ and hence $b$ is an even integer.

Finally from $ct^2 = 0$ in $R$ we see that $c$ is an even integer.

Hence $a, b, c$ are all even, and the element $(a + b\theta + c\theta^2)/2$ lives in $\mathbb Z[\theta]$.

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