# 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.

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.

• How do you define the different above $\mathfrak{p}$ ideal Commented Feb 11, 2019 at 22:40
• @reuns, since this is a local question, it is clear from the context that what is meant is the factors of the different consisting of primes dividing $\mathfrak{p}$. Commented Feb 12, 2019 at 6:08
• Thanks for the answer. I read in Lang's Algebraic number theory that, in my case, $\mathfrak P_i^{e-1}$ divides the different, but it wasn't stated that this was the exact power. Is there any upper bound on $\sum_{n=0}^{\infty} \frac{|I_n| - 1}{|I_0|}$ in the general case ? Commented Feb 12, 2019 at 8:54
• There will be an upper bound depending on $G$ AND on $[K:\mathbf{Q}]$, but not on $G$ alone. For example, consider the case when $G = \mathbf{Z}/p \mathbf{Z}$. Then the exponent will be $|G|(1-1/p)m$, where $m$ is the smallest integer such that $I_m$ is trivial. If $K$ varies, there is no upper bound on $m$. For example, if $K = \mathbf{Q}(\zeta_{p^n})$ and $L = \mathbf{Q}(\zeta_{p^{n+1}})$, then $m = p^n$. Commented Feb 12, 2019 at 20:55
• On the other hand, there is an upper bound which depends only on the localization of $K$ at $\mathfrak{p}$. If $A$ denotes the completion of the ring of integers of $K$ at $\mathfrak{p}$, and $\pi$ is a uniformizer of $A$, then one can take (still under the assumption the extension is cyclic of degree $p$) $n$ to be the smallest integer such that $1 + \pi^n A$ contains all $p$th powers, and in fact this bound is acheived. For example, if $A = \mathbf{Z}_p$, then $n = 3$ if $p = 2$ and $n = 2$ otherwise. The same bounds hold for the ring of integers of an unramified extension of $\mathbf{Q}_p$. Commented Feb 12, 2019 at 20:56