According to my hand computation, the claim is not true for $K = \mathbb{Q}(\sqrt[3]{3})$. Have you checked this case?

It is true that a prime ramifies in $K$ if and only if it ramifies in the Galois closure of $K$. There might be a formula like the one you are describing that is true for primes greater than $d$.

Warning: this part of the answer was written without access to my usual number theory references, so there may be errors. It isn't meant to be read by itself anyway; it is meant to be a guide to a good book on algebraic number theory, like Neukirch or Janusz.

# How you should approach a question like this

Your fundamental strategy to be to relate the order of ramification of $p$ in $L$ and in the Galois closure of $L$. This does not precisely determine the power of $p$ dividing the discriminant, but we have the following

**Fact:** Let $L/K$ be an extension of number fields and $p$ a prime of $K$ with ramification indices $e_1$, $e_2$, ... $e_r$ and residue field extensions of degree $f_1$, $f_2$, ..., $f_r$. Then the power of $p$ dividing $D_{L/K}$ is at least $\sum (e_i-1) f_i$, with equality if and only if all the $e_i$ are ~~less than~~ relatively prime to the characteristic of $p$.

From now on, I will address the question of how to relate the orders of ramification in $L$ and in the Galois closure. If you need to deal with the case that some of the $e_i$ are greater than or equal to the characteristic of $p$, you should read about higher ramification groups.

Let $M$ be the Galois closure of $L$, let $G$ be $\mathrm{Gal}(M/K)$, and let $H$ be the fixed field of $L$. Fix a prime $p$ of $K$, and^{1} $\mathfrak{p}$ a prime over $p$. Let $D \subseteq G$ be the decomposition group of $\mathfrak{p}$ and $I \subset D$ the inertia group. $I$ is normal in $D$; the quotient $D/I$ is cyclic and has a canonical generator $F$ called the Frobenius.

I believe that FC's computation was for the case $G=S_n$, $D=I=\{ e, (12) \}$.

In $M$, the primes lying over $p$ are in bijection with $G/D$. Each of them has $f=|D/I|$ and $e=|I|$.

Let $X=G/H$ (a set with $G$-action). In $L$, the primes above $p$ are in bijection with the $D$-orbits in $X$. Let $O$ be such a $D$-orbit, corresponding to a prime $q_O$. Because $I$ is normal, $O$ breaks up as a union of $I$-orbits all of the same cardinality. Then $q_O$ has $e$ equal to the cardinality of these $I$-orbits, and $f$ equal to the number of them.

Using these ideas, you should be able to relate ramification in $L$ and $M$ for any $G$ and $H$ which interest you.

^{1} I fix $\mathfrak{p}$ only for expositional purposes. Changing $\mathfrak{p}$ will conjugate $(D,I,F)$. Probably the right way to think of all of this stuff is working up to conjugacy.