Let $K$ be a number field and let $L$ be an extension of $K$ with respective rings of integers $\mathcal{O}_L$ and $\mathcal{O}_K$.
I am aware of something called the relative discriminant of $L$ over $K$. I'm interested in looking at the ramification properties of a particular $L$.
If I am looking at a number field $K$ and wish to calculate its discriminant over $\Bbb Q$ I can take a $\Bbb Z$-basis for $\Bbb Q$ and calculate the discriminant of those elements (how the absolute discriminant is defined).
If I have an extension $L$ of a number field $K$ and I know an $\mathcal{O}_K$-basis for $L$, can I calculate the relative discriminant in this way, or does this give me something completely different?
Edit: I know that the relative discriminant actually gives an ideal of $\mathcal{O}_L$, is there an analogous numerical invariant?
In general you can't find a basis for $O_L$ over $O_K$ because it isn't always free, but what you can do is this:
Find a basis $b_1, ... , b_n$ for $L$ over $K$ and ideals $I_1, ... , I_n$ of $K$, such that $O_L = b_1 I_1 + ... + b_n I_n$ (direct sum).
Then the relative discriminant is the product
$\Delta\{b_1,..., b_n\} \cdot (I_1 ... I_n )^2$.
The easiest example is this: let $P$ be a maximal ideal of $K$ whose order in the class group is $2$, so $P^2$ is the principal ideal generated by an element $\alpha$. Then you can define $L= K (\alpha^{1/2})$.
In most cases $O_L$ can be written in the form
$1\cdot O_K + \alpha^{1/2}\cdot P^{-1}$.
so the relative discriminant is $(4\alpha)\cdot P^{-2}$.