# Example of number field with certain conditions on ramification index and degree

I am looking for a number field with degree $$n$$ over $$\mathbb{Q}$$ and with a ramified prime $$p$$ with ramification index $$e$$ such that $$\textrm{gcd}(n, p-1) = 1$$ and $$\textrm{gcd}(e, p-1)>1$$. I would also be interested in a slightly stronger condition, namely where $$\textrm{gcd}(n, p-1) = 1$$ and $$e|(p-1)$$.

I would prefer the number field to be as simple as possible. Simple here could mean small degree, or small absolute value of the discriminant of the extension. So far, I have had no luck with trying simple cases for quadratic, cubic and quartic extensions. For those cases I either get complete ramification ($$e = n$$) or for $$\mathbb{Q}[\sqrt{p}, \sqrt{q}]$$ I get $$e=2$$ and $$n=4$$. Cyclotomic extensions also do not work, since there we have complete ramification as well.

• It is indeed, edited my post for clarification. Aug 2, 2020 at 11:25
• If the discriminant of a number field of degree $> 2$ is divisible by a prime $p$ but not by $p^2$, then $e = 2$ for exactly one prime above $p$. If you want, say, a quintic field with $e = 3$, start from $f(x) \equiv x^3(x-1)(x-2) \bmod 5$, pick an irreducible $f$ with this reduction and check that everything works. Aug 2, 2020 at 12:26

Such a number field $$K$$ cannot be Galois over $$\mathbb Q$$. Indeed, if $$K$$ is Galois, then every prime ideal above $$p$$ has the same ramification index and residue degree, so $$e\mid n$$. That rules out all quadratic, biquadratic and cyclotomic extensions.
For an example when $$K$$ is not Galois, one place to start would be to try to find an odd extension in which $$p=3$$ has ramification index $$2$$. An example from LMFDB is $$K = \mathbb Q(\alpha)$$ where $$\alpha$$ is a root of $$x^3 - x^2 + 2x + 1$$. In this extension, there are two primes $$\mathfrak p_1, \mathfrak p_2$$ above $$3$$. One is unramified, and the other has ramification index $$2$$.