Suppose I have a finite field extension of number fields (finite field extensions over $\mathbb{Q}$), say $K\subset L$. Say $P$ is a prime in the number ring contained in $K$ such that $P$ splits completely in every intermediate field strictly between $K$ and $L$, but does not split completely in $L$. I have already shown that the Galois group of such a field has order that is a prime power of a prime $p$ and that it contains a unique smallest normal subgroup of order $p$.

Now, I just want to find an example of such a field extension $L$ over $K$ where $K = \mathbb{Q}$ and the Galois group of $L$ over $K$ is of order $p^{k}$ for $k>1$. Any suggestions?

  • $\begingroup$ Any quadratic extension and any inert prime, having Galois group $\Bbb Z /2 \Bbb Z$? $\endgroup$
    – benh
    Dec 29, 2013 at 14:36
  • $\begingroup$ That does work, but I will edit my question to ask for an extension with Galois group that is not of prime order. $\endgroup$
    – neelp
    Dec 29, 2013 at 14:39

1 Answer 1


Let $\zeta_5 = e^{2\pi i/5}$ be a $5$-th root of unity. Then $L = \Bbb Q(\zeta_5)$ is an abelian extension with Galois group $$(\Bbb Z/5 \Bbb Z)^\times = \Bbb Z/4 \Bbb Z.$$ So we have one intermediate extension corresponding to the one non-trivial subgroup of $\Bbb Z / 4 \Bbb Z$. It is $\Bbb Q(\sqrt{5})=M$, because $\zeta_5+\zeta_5^{-1} = (\sqrt{5}-1)/2 \in \Bbb Q(\zeta_5).$

An odd prime $p$ is completely split in $\Bbb Q(\sqrt{5})=M$ iff $\left(\frac{p}{5} \right)=1$, because $d_M = 5$.

An odd prime $p$ is completely split in $\Bbb Q(\zeta_5)$ iff $p \equiv 1 \bmod 5$.

So choose any prime $p \equiv 4 \bmod 5$ to get an example, e.g. $19 = (8+3\sqrt{5})(8-3\sqrt{5})$.

  • $\begingroup$ Didn't you want to mean "An odd prime $p$ is completely split in $\Bbb Q(\sqrt{5})=M$ iff $\left(\frac{5}{p} \right)=1$" ? $\endgroup$
    – Watson
    Jun 4, 2017 at 15:43

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