Let $L/K$ be a (abelian, Galois) quadratic extension of number fields with $\text{Gal}(L/K)$ generated by $\sigma$ and $\mathfrak{p} = \alpha\mathcal{O}_K$ a principal prime ideal of $\mathcal{O}_K$. Assume $\mathfrak{p}$ splits as $\mathfrak{P} \sigma(\mathfrak{P})$ in $\mathcal{O}_L$ and that $\alpha = \beta \sigma(\beta)$ for $\beta \in L^\times$ (so $\beta$ may not be integral though $\alpha$ is). I would like to show that $\mathfrak{P}$ is principal (and possibly that $\mathfrak{P} = \beta \mathcal{O}_L$).

$\textbf{My attempt}$: It seems clear that if $\beta$ is not integral it must generate a fractional ideal of the form $$\beta\mathcal{O}_L = \frac{\mathfrak{P}I}{\sigma(I)}$$ for some $I \subset \mathcal{O}_L$.

We can assume $\mathfrak{P}I \cap \sigma(I) = \mathcal{O}_L$, i.e. that the numerator and denominator are simplified: we can cancel common factors of $I$ and $\sigma(I)$ so that $I \cap \sigma(I) = \mathcal{O}_L$, and if $\mathfrak{P} \cap \sigma(I) = \mathfrak{P}$ then $\sigma(\mathfrak{P})$ divides $I$ and we get that $$\beta\mathcal{O}_L = \frac{\sigma(\mathfrak{P}) I'}{\sigma(I')}$$ where $I' = I/\sigma(\mathfrak{P})$. So, WLOG we can take the first expression for $\beta\mathcal{O}_L$.

If $\beta$ is not integral we can find an integer $d \ne 1$ such that $d\beta$ is integral ($\beta$ is just a linear combination over the basis for $L$ and $d$ is the lcm of the denominators of the coefficients). Then $$ d\beta\mathcal{O}_L = (d)\frac{\mathfrak{P}I}{\sigma(I)} \subset \mathcal{O}_L.$$

Since this is integral, $\sigma(I)$ divides $(d)\mathfrak{P}I$. As $\mathfrak{P}I \cap \sigma(I) = \mathcal{O}_L$, $\sigma(I)$ divides $(d)$. But so does $I$ (as $\sigma$ fixes $d$). If $I$ is nontrivial then this contradicts $I \cap \sigma(I) = \mathcal{O}_L$, so either $I$ is trivial or $d = 1$. In either case we have $\beta\mathcal{O}_L = \mathfrak{P}$ as desired.

This proof feels awkward and I suspect either it is wrong or just overly complicated. I'd appreciate any feedback!

  • $\begingroup$ It is a bit confusing to use $\mathfrak{p}$ for the prime ideal in the big field. $\endgroup$
    – hans
    Jul 9, 2021 at 17:53
  • $\begingroup$ If $\beta \sigma(\beta) \mathcal{O}_L = \mathfrak{P}\sigma({\mathfrak{P}})$, doesn't it follow from the unique factorization that either $\beta = \mathfrak{P}$ or $\sigma(\beta) = \mathfrak{P}$? $\endgroup$
    – hans
    Jul 9, 2021 at 17:56
  • $\begingroup$ @hans point taken, I edited the question. And no, as $\beta$ is not necessarily integral. It could be that $(\beta) = \mathfrak{P}I/\sigma(I)$, then we still have $(\beta \sigma(\beta)) = \mathfrak{P} \sigma(\mathfrak{P}).$ $\endgroup$
    – wyoumans
    Jul 9, 2021 at 17:59

1 Answer 1


I claim that under the given circumstances $\mathfrak{P}$ is not necessarily principal, i.e., the statement claimed in the question is wrong.

Here is a counterexample. Consider $K = \mathbb{Q}$ and $L = \mathbb{Q}[\sqrt{-47}]$. Let $\alpha = -3$ and $\beta = \frac{1}{4}(1 \pm \sqrt{-47})$. Then the minimal polynomial of $\beta$ is $X^2 - \frac{1}{2} X + 3$. In particular, the norm of $\beta$ is $\beta\sigma(\beta) = -3$. (Note that $\beta$ is not integral, as its minimal polynomial is not integral.)

Then with the help of SAGE one computes the factorization in $L$: $3\mathcal{O}_L =(3, \frac{\sqrt{-47}}{2} - \frac{1}{2})(3, \frac{\sqrt{-47}}{2} + \frac{1}{2})$. Say $\mathfrak{P}=(3, \frac{\sqrt{-47}}{2} - \frac{1}{2})$. Also, SAGE tells that the class group of $L$ is of order $5$ with generator $\mathfrak{a} = (2, \frac{\sqrt{-47}}{2} + \frac{1}{2})$, and that $\mathfrak{P} = \mathfrak{a}^2$ in the class group. Thus $\mathfrak{P}$ is not principal.

I believe that a mistake in the posted proof is contained in the sentense "If $I$ is non-trivial, then ...". Indeed, that $I$ and $\sigma(I)$ divide (d) only means that all $p$-adic valuations of $I$ and $\sigma(I)$ are smaller than those of $(d)$. Why this should imply that $I \cap \sigma(I) = \mathcal{O}_L$?

  • $\begingroup$ Thanks, nice counterexample. Now that you point it out that part is obviously wrong, I had a feeling there was a mistake like that in there. $\endgroup$
    – wyoumans
    Jul 10, 2021 at 16:28

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