# Find the ring of integers of $\mathbb{Q}(\theta)$

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?

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