# Selfconjugate prime ideal of a cyclic extension of an algebraic number field of prime degree.

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 $\mathfrak{P}$ be a non-zero prime ideal of $B$ such that $\sigma(\mathfrak{P}) = \mathfrak{P}$. Let $\mathfrak{p} = \mathfrak{P} \cap A$. Suppose $\mathfrak{p}B \neq \mathfrak{P}$.

My question: Is $\mathfrak{p}B = \mathfrak{P}^l$?

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What have you tried so far? –  Old John Jul 25 '12 at 7:24

Yes. $\mathrm{Gal}(L/K)$ acts transitively on the primes above a given prime $\mathfrak{p}$. If $\sigma(\mathfrak{P}) = \mathfrak{P}$ then $\mathfrak{P}$ is fixed by the entire Galois group, and so we see that $\mathfrak{P}$ must be the only prime lying above $\mathfrak{p}$.

In general if we write $\mathfrak{p}B = \prod \mathfrak{P}_i^{e_i}$, then we have the identity $\sum e_i f_i = [L:K]$, where $f_i = [B : \mathfrak{P}_i]$. But in the case of a Galois extension, because the Galois group acts transitively on these primes, we must have $e_i = e_j$ and $f_i = f_j$. Write $e = e_i$, $f = f_i$ for some $i$ in this case.

Returning to our cyclic extension $L/K$, we already know from the above that $\mathfrak{p}B = \mathfrak{P}^e$ for some positive integer $e \geq 1$, and that $ef = l$. Since $l$ is prime, we have either $e = l, f = 1$ or $e = 1, f = l$. Because you required that $\mathfrak{p}$ is not inert, it must the former.

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Yes! There is only one prime lying above $\mathfrak{p}$ since the Galois group acts transitively on primes above $\mathfrak{p}$. This means that $l$ is equal to the product of the ramification degree and the inertia degree. Since $\mathfrak{p}B\not=\mathfrak{P}$ the ramification degree must be greater than one. Then because $l$ is prime we know that the ramification degree must be $l$.

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