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 $p=3$.
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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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