Let $K$ be an algebraic number field. Let $A$ be the ring of integers in $K$. Let $L$ be a cyclic extension of $K$ of degree $l$, where $l$ is a prime number. Let $B$ be the ring of integers in $L$. Let $G$ be the Galois group of $L/K$. Let $\sigma$ be a generator of $G$. Let $\mathfrak{D}_{L/K}$ be the relative different of L/K.

My question: Is the following proposition true? If yes, how would you prove this?

Proposition Let $\mathfrak{P}$ be a prime ideal of $B$. Let $\mathfrak{p} = \mathfrak{P} \cap A$. Then $\mathfrak{P}$ divides $\mathfrak{D}_{L/K}$ if and only if $\sigma(\mathfrak{P}) = \mathfrak{P}$ and $\mathfrak{p}B \neq \mathfrak{P}$.

Related question: Selfconjugate prime ideal of a cyclic extension of an algebraic number field of prime degree.


General theory about splitting of primes shows that either $\mathfrak p B$ is prime (inert case), that $\mathfrak p B$ splits as a product of $l$ distinct primes which are acted on simply transitively by $G$ (split case), or that $\mathfrak p$ is the $l$th power of a single prime ideal in $B$ (ramified case).

By assumption $\mathfrak P$ divides $\mathfrak p$, and it is fixed by $\sigma$, so we are not in the split case. The assumption that $\mathfrak p B \neq \mathfrak P$ shows that we are not in the inert case either. Thus we are in the ramified case.

General ramification theory implies that the ramified primes (i.e. the $\mathfrak P$ which are $l$th roots of the $\mathfrak p$ that are ramified) are precisely the primes dividing $\mathfrak D_{L/K}$. This gives the proposition. [It also gives an answer to the linked question.]

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.