Due to the primitive element theorem we can consider a number field as $\mathbf{L} = \mathbf{Q}[X]/(f)$ where $f$ is a monic polynomial in $\mathbf{Q}$.
Often the ring of integers $\mathcal{O}$ of this number field is not monogenic: there is no algebraic integer $\alpha \in \mathcal{O}$ so that its powers are a $\mathbf{Z}$-basis of $\mathcal{O}.$
I have two questions:
- Is there always an algebraic integer $\alpha \in \mathcal{O}$ so that its powers are a $\mathbf{Q}$-basis of $\mathbf{L}$? Ie can we always take the $f$ above to be a monic integer polynomial. (Obviously, for monogenic fields such an element exists.)
- If so, given a number field in the form above, can this element be constructed ?
Best Answer
The answer to your first question is yes: Assume $a \in L$ and let $\sum_{i=0}^n k_iX^i$ be a polynomial with root $a$, $k_n \neq 0$ and such that all the $k_i \in \mathbb{Z}$. Then, the element $k_na$ is integral because it is a root of $$\bigg(\sum_{i=0}^{n-1}k_n^{n-1-i}k_iX^i\biggr)+X^n$$ This shows that for each $a \in L$ there is a $k \in \mathbb{Z} \setminus\{0\}$ such that $ka$ is integral. Now, apply this to the case where $a$ is a primitive element.
Answer to your second question: If you already have a primitive element $a$, the chances are good that you will be able to transform it into an integral primitive element: Just find a nonzero polynomial $f(X) \in \mathbb{Z}[X]$ with $f(a)=0$ and apply the above procedure.
(However, there is no general procedure for finding a primitive element as far as I know.)