Exponent of a ramified prime in discriminant Let $L/K$ be a number field extension, and let $\mathfrak p$ be a prime of $K$ ramified in $L$, with ramification index $e$. Then $\mathfrak p$ divides the discriminant $\partial_{L/K}$ of $L/K$. How is $e$ related to the exponent of $\mathfrak p$ in $\partial_{L/K}$, that is the highest power of $\mathfrak p$ dividing $\partial_{L/K}$ ?
If necessary I can assume $L/K$ is Galois and tamely ramified.
 A: Assuming that $L/K$ is Galois, the prime $\mathfrak{p}$ splits as
$$\prod_{i=1}^{r} \mathfrak{P}^e_i$$
where $\mathfrak{P}$ has inertial degree $f$ and $erf = [L:K]$. Assuming that the extension is tamely ramified, the different $\mathcal{D}_{L/K}$ coming from the primes above $\mathfrak{p}$ is equal to $\prod_{i=1}^{r} \mathfrak{P}^{e-1}_i$, and the discriminant is the norm of the different from $L$ to $K$, which is
$$\prod_{i=1}^{r} \mathfrak{p}^{f(e-1)} = \mathfrak{p}^{{rf(e-1)}} = \mathfrak{p}^{m},$$
where $m = [L:K]\left(1 - \frac{1}{e} \right)$.
If you don't assume that $L/K$ is Galois, then "ramification degree $e$" doesn't really mean anything --- some primes above $\mathfrak{p}$ may be ramified and some not.
If you don't assume that $L/K$ is tamely ramified then the answer will depend on more than just the invariant $e$. For example, if $K = \mathbf{Q}$, then $L = \mathbf{Q}(\sqrt{-1})$ and $\mathbf{Q}(\sqrt{2})$ are both Galois with $e = 2$ but the power of $2$ dividing the discriminant is $2^2$ or $2^3$.
On the other hand, you can determine the exponent if you know the orders of the higher ramification groups. In particular, one will have
$$m = [L:K] \left(\sum_{n=0}^{\infty} \frac{|I_n| - 1}{|I_0|}\right).$$
Here $|I_0| = e$ is the inertia group, and $|I_1|$ is the wild inertia group whose order is the largest power of $p$ dividing $|I_0|$. Since $|I_n| = 1$ for sufficiently large $n$, this is a finite sum. The difference between  $L = \mathbf{Q}(\sqrt{-1})$ and $\mathbf{Q}(\sqrt{2})$ is that $\mathbf{Z}/2\mathbf{Z} = I_0 = I_1 \ne I_2$ in the first case, and  $\mathbf{Z}/2\mathbf{Z} = I_0 = I_1 = I_2 \ne I_3$ in the second. 
