4
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @LordSharktheUnknown can you elaborate a bit more? $\endgroup$ – Philomeno Dec 2 '19 at 4:18
1
$\begingroup$

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 $\mathbb Z[\theta]$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.