I am working through a number theory proof and there is one part whose justification I do not fully understand. I believe the reasoning is merely something related to localization.

Suppose $R$ is a Dedekind domain with fractions $K$. Let $L/K$ be normal with $G$ Galois group. Denote $R'$ the integral closure of $R$ in $L$. Let $\beta$ be a nonzero prime of $R'$, $p = R \cap \beta$ its contraction. Let $N$ denote the norm of an ideal, defined to be $$N(\mathfrak{a}) = \sum_{a \in \mathfrak{a}} RN(a)$$ I wish to show $N(\beta)$ is a power of the prime $p$. Suppose not, that $N(\beta) = qm$ with $q \ne p$ prime and $m \subseteq R$. Let $C$ be the complement of $q$ in $R$ so that $R_q = R_C$. Since $q \ne p$, the prime ideal $\beta$ is not a factor of $qR'$. Thus $\beta_C = R'_C$ as the only prime ideals of $R_C '$ are the extensions of the primes appearing in the factorization of $qR'$.

I understand that if $S$ is a multiplicative set and we localize $S^{-1} A$ a ring with respect to $S$, then the primes in the localization correspond bijectively to the primes of $A$ such that $p \cap S = \emptyset$. In the case $S = R - q$, then this corresponds to the primes contained within $q$. The images then correspond to the primes appearing in the factorization of the image of $q$, namely $qR'$. I understand that since $q \ne p$, then $\beta$ is not a factor of $qR'$. I don't see how one puts this together to get the desired equality.

I am thinking that one shows the set of prime ideals on each side is the same and since $\beta \subset R'$, then this would show that their localizations are the same.

  • $\begingroup$ Your definition of norm is not clear: what is $N(a)$ for $a\in \mathfrak a$? $\endgroup$ – Ferra Sep 16 '16 at 11:01

You just need to note that no prime $q\ne p$ divides $N(\beta)$ because $\Bbb Z$ has unique factorization. What the proof is doing is just saying another prime dividing $N(\beta)$ would imply that $qR'\supseteq N(\beta)$, but then take norms again and you get $N(N(\beta)) = q^{[R':R]}$ is a power of a different prime, $q$. But this is a contradiction to the fact that $(p)\supseteq\beta$.


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