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Changed answer: Even in the strongest interpretation of the slightly vague question, the The answer is no: Let $L$ be a root field of $X^3-2$ over $\mathbb Q$, and $M$ the Galois closure of $L/\mathbb Q$. Then the primes $2$ and $3$ are both totally ramified in $L$, yet in both cases there is only one prime $P$ above $p=2$ or $p=3$, with $e(P)=3$ if $p=2$, and $e(P)=6$ if $p=3$.

3 deleted 116 characters in body

I'm pretty sure that

Changed answer: Even in the strongest interpretation of the slightly vague question, the answer is no. : Let $I\le D$ L$be the inertia group. Then a root field of$I$and X^3-2$ over $D/I$ are cyclic\mathbb Q$, and the question amounts to extracting$M$the orders Galois closure of$I$(ramification index) and$D/I$(inertia degree) just from L/\mathbb Q$. Then the orbit lengths of primes $D$.

There 2$and$3$are certainly number fields both totally ramified in$L/K$which have a L$, yet in both cases there is only one prime $P$ of above $M$ such that p=2$or$I>1$and p=3$, with $D$ is cyclic. But then, by Chebotarev's density theorem, there are infinitely many unramified primes e(P)=3$if$P'$with decomposition group p=2$, and $D$, so e(P)=6$if$e(P')=1\lt e(P)$.p=3$.

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I'm pretty sure that the answer is no. Let $I\le D$ be the inertia group. Then $I$ and $D/I$ are cyclic, and the question amounts to extracting the orders of $I$ (ramification index) and $D/I$ (inertia degree) just from the orbit lengths of $D$.

There are certainly number fields $L/K$ which have a prime $P$ of $M$ such that $I>1$, I>1$and$D$is cyclicand$D\le G$. (The latter does not follow automatically from Dedekind, but still should hold normally.) But then, by Chebotarev's density theorem, there are infinitely many unramified primes$P'$with decomposition group$D$, so$e(P')=1\lt e(P)\$.

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