# Is it true that this ideal must be principal? (proof verification)

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!

• It is a bit confusing to use $\mathfrak{p}$ for the prime ideal in the big field.
– hans
Jul 9, 2021 at 17:53
• 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 $\sigma(\beta) = \mathfrak{P}$?
– hans
Jul 9, 2021 at 17:56
• @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 $(\beta \sigma(\beta)) = \mathfrak{P} \sigma(\mathfrak{P}).$ Jul 9, 2021 at 17:59

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 $$I \cap \sigma(I) = \mathcal{O}_L$$?