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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.

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  • $\begingroup$ It is indeed, edited my post for clarification. $\endgroup$
    – Kenneth Goodenough
    Aug 2, 2020 at 11:25
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    $\begingroup$ 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. $\endgroup$
    – franz lemmermeyer
    Aug 2, 2020 at 12:26

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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 $2$.

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