I've been trying to solve the following problem:

Find the ring of integer of $\mathbb{Q}(\theta)$ when $\theta^3 + \theta + 1 = 0$.

I started by computing the discriminant, which is $-31$. Then I said since the discriminant is square free and $\theta$ is clearly integral over $\mathbb{Q}$, then the ring of integers of $\mathbb{Q}(\theta)$ is $\mathbb{Z}[\theta]$. Am I right?


I will write $t$ instead of $\theta$. So we work in the field $K=\Bbb Q[T]/(T^3+T+1)=\Bbb Q[t]$, where $t$ is the image of $T$ modulo the ideal $(T^3+T+1)$. Consider the basis as a $\Bbb Q$ vector space of $K$ given by the algebraic integers $$ 1,\ t,\ t^2\ . $$ We fix an embedding of $K$ into $\Bbb C$, let $t_1$ be the image of $t$. The polynomial $X^3+X+1\in K[X]$ has three roots in $\Bbb C$, one is $t_1$, and let $t_2,t_3$ be the other two. There are thus three embeddings $\sigma_k:K \to \Bbb C$, given by $\sigma_k:t\to t_k$. The discriminant of the basis $1,t,t^2$ is now (in my very ad-hoc computation): $$ \begin{aligned} \Delta[1,t,t^2] &=\det{}^2(\ (\sigma_k(t_j))_{1\le j,k\le 3}\ )\\ &= \begin{vmatrix} 1&1&1\\ t_1&t_2&t_3\\ t_1^2&t_2^2&t_3^2 \end{vmatrix}^2 \\ &=(\ (t_1-t_2)(t_1-t_3)(t_2-t_3)\ )^2\text{ (Vandermonde)} \\ &=-(t_1-t_2)(t_1-t_3) \cdot (t_2-t_1)(t_2-t_3)\cdot(t_3-t_1)(t_3-t_2) \\ &=-\prod(t_1-t_2)(t_1-t_3)=-\prod(t_1^2-(t_2+t_3)t_1+t_2t_3) \\ &=-\prod(2t_1^2-(t_1+t_2+t_3)t_1+t_2t_3)=-\prod(2t_1^2+t_2t_3) \\ &=-\prod\frac 1{t_1}(2t_1^3+t_1t_2t_3) =-\frac 1{t_1t_2t_3}\prod(2t_1^3-1) \\ &=-\frac 1{-1}\prod(-2t_1-2-1)=-(2t_1+3)(2t_2+3)(2t_3+3) \\ &=-(27+18\sum t_1+12\sum t_1t_2+8t_1t_2t_3) \\ &=-(27+0+12-8)=-31\ . \\[3mm] &\text{Alternatively one can comuute the resultant} \\ \Delta[1,t,t^2] &=\pm\operatorname{Resultant}(\ X^3+X+1\ ,\ (X^3+X+1)'\ ) \\ &= \pm \begin{vmatrix} 1 & 0 & 1 & 1 & \\ & 1 & 0 & 1 & 1\\ 3 & 0 & 1 & & \\ & 3 & 0 & 1 & \\ & & 3 & 0 & 1 \end{vmatrix} \end{aligned} $$ (The sign factor above is $(-1)$ to the power $3(3-1)/2$, so it is $-1$.)

Now we come to your question. We have the discriminant computed for the integral basis $1,t,t^2$, it is $-31$ and it is squarefree. Well, in this case the base is integral. In the hypothetical case of an integral system $(v_1,v_2,v_3)$ with a discriminant $\Delta[v_1,v_2,v_3]\in\Bbb Z$ where there is a prime factor $p$ that appears to the power $\ge $ two in the discriminant, then we may search for an integral linear combination of the shape $\frac 1p(n_1 v_1+n_2v_2+n_3v_3)$, which is guaranteed to exist.

Reference: Computing integral bases, John Paul Cook, see for instance Theorems 3.3. and 3.4.


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