Let $R$ be a Dedekind domain with quotient field $K$, let $L$ be a finite separable extension of $K$, and let $S$ be the integral closure of $R$ in $L$. If $\mathfrak{p}$ is a nonzero prime ideal of $R$ that is contained in the union of the prime ideals of $R$ that split completely in $S$, does it follow that $\mathfrak{p}$ splits completely in $S$? It follows easily if $\mathfrak{p}$ is principal or (by prime avoidance) if only finitely many prime ideals of $R$ split completely in $S$. If necessary, assume that $R$ is the ring of integers in a number field and/or $L/K$ is Galois.

  • $\begingroup$ Isn't it immediate that the only prime ideals contained in $\bigcup_i \mathfrak{p}_i$, where $\mathfrak{p}_i$ are prime, are these $\mathfrak{p}_i$ themselves? $\endgroup$ – Alex B. Aug 31 '11 at 3:36
  • 2
    $\begingroup$ @Alex, not quite - If $\mathfrak{p}$ has infinite order in the class group, then every element of $\mathfrak{p}$ is contained in a prime ideal $\mathfrak{q} \ne \mathfrak{p}$, and so $\mathfrak{p}$ is contained in the union of all the other prime ideals. $\endgroup$ – Michael Aug 31 '11 at 6:24

If the class group is finite, then writing $\mathfrak{p}^h = (\alpha)$, it follows that $\alpha$ is contained in a prime ideal $\mathfrak{q}$ which splits completely in $S$, and thus $\mathfrak{p}^h \subseteq \mathfrak{q} \Rightarrow \mathfrak{p} = \mathfrak{q}$ (because $R$ has dimension one). It sounds like that suffices for your purposes.

| cite | improve this answer | |
  • $\begingroup$ More generally that works if the class group is torsion. That suffices for me. Thanks! $\endgroup$ – Jesse Elliott Aug 31 '11 at 7:33

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.