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.


1 Answer 1


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.

  • $\begingroup$ How do you define the different above $\mathfrak{p}$ ideal $\endgroup$
    – reuns
    Commented Feb 11, 2019 at 22:40
  • $\begingroup$ @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 $\mathfrak{p}$. $\endgroup$ Commented Feb 12, 2019 at 6:08
  • $\begingroup$ 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 ? $\endgroup$ Commented Feb 12, 2019 at 8:54
  • $\begingroup$ 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 $m = p^n$. $\endgroup$ Commented Feb 12, 2019 at 20:55
  • $\begingroup$ 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 $\mathbf{Q}_p$. $\endgroup$ Commented Feb 12, 2019 at 20:56

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