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.
    Commented Aug 31, 2011 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
    Commented Aug 31, 2011 at 6:24

1 Answer 1


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.

  • $\begingroup$ More generally that works if the class group is torsion. That suffices for me. Thanks! $\endgroup$ Commented Aug 31, 2011 at 7:33

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