# Integral basis of $\mathbb{Q}(\theta)$, where $\theta^3-\theta-4=0$

I am working on the text book "algebraic number theory" by Jurgen Neukirch(P15, exercise 6). To prove the integer basis is$\{1, \theta, \frac{\theta^2+\theta}{2}\}$. After a long and tedious calculation, I still get nothing.(Following the method which can be used in exercise 5: the $\mathbb{Q}(\sqrt[3]{2})$'s case, which is typical and you can find in stackexchange!).

Any help is going to be appreciated.

First of all, you have to show that $\frac{\theta^2+\theta}{2}$ is integral over $\mathbb{Z}$. Then, show that $1,\theta,\frac{\theta^2+\theta}{2}$ are linearly independent over $\mathbb{Z}$. Compute the discriminant $d(1,\theta,\frac{\theta^2+\theta}{2})$. The result will be -107 (at least according to what I have found out so far). Since 107 is a prime number and thus square-free, you can use Theorem 2.12, ch.1 of Neukirch's book to conclude that $1,\theta,\frac{\theta^2+\theta}{2}$ is a basis of the ring of integers of $\mathbb{Q}(\theta)$.
• Sorry these are going to be very basic questions but could you explain how you showed $(\theta^2 + \theta)/2$ is integral over $\mathbb Z$? (I've been trying to find an integral polynomial for it but got bogged down in algebra and am wondering if there's a smarter way than brute force trying to construct a cubic that works). Also, how were you able to compute the traces of e.g. $(\theta^3 + \theta^2)/2$ for the trace form of the discriminant? Many thanks! Jan 1, 2020 at 19:10
• Oh I figured it out. For anyone else: the characteristic polynomial of the matrix representation of $(\theta^2+\theta)/2$ will lie in $\mathbb Z[x]$, and since $(\theta^2+\theta)/2$ is a zero of its own char poly this shows it's integer. To compute the traces just use the ${1, \theta, \theta^2}$ basis of $\mathbb Q(\theta)$. I found Keith Conrad's notes here: kconrad.math.uconn.edu/blurbs/galoistheory/tracenorm.pdf very helpful. Jan 1, 2020 at 22:33
• An alternative way to see that it is integral is this one: start with the equation $X=(\theta^2+\theta)/2$. Take both sides of the equation to the power of $3$ and then, use the relation $\theta^3-\theta-4=0$ to eliminate all powers $\geq 3$ of $\theta$ on the right hand side. The result should then be $X^3=2\theta^2+3\theta+4$. So, $\theta$ is a root of $f(X)=X^3-2\theta^2-3\theta-4$. Hence, it is integral over $\mathbb{Z}[\theta]$ and therefore integral over $\mathbb{Z}$. Jan 19, 2020 at 5:34