Ideal $\mathfrak p^i$ is not principal

Question

Let c be a postive squarefree integer. Let $K = \mathbb Q(\sqrt{-c})$. Let $p$ be a prime that splits in $K$ and let $\mathfrak p$ be a prime ideal above $p$. I need to prove the following:

Prove that for all integers $i\geq 1$ such that $p^i <\frac{|\text{disc}( K)|}{4}$, the ideal $\mathfrak p^i$ is not principal. I have no idea where to start. Any help would be much appreciated.

Thanks very much for the edit. – Richard Mar 06 '16 at 14:44 — Richard, Mar 06 '16 at 14:44
Also - for future reference, this is a Mathjax tutorial – Mathmo123 Mar 06 '16 at 15:35 — Mathmo123, Mar 06 '16 at 15:35
Thanks, I really need to learn it! – Richard Mar 06 '16 at 15:54 — Richard, Mar 06 '16 at 15:54

score 2 · Answer 1 · edited Apr 13 '17 at 12:19

Here is a hint to start you off.

Suppose that $\mathfrak p^i = (\alpha)$ for some $\alpha\in \mathcal O_K$. Then, taking norms, $$|N_{K/\mathbb Q}(\alpha)| = N\mathfrak p^i=p^i<\frac{|\mathrm{disc}(K)|}4.$$

But we know exactly what the norm of an element of $K$ looks like, and we know exactly what the discriminant of $K$ is (depending on $c$ modulo $4$).

Irrelevant remark:

Here is a cute corollary: take $c=163$. Then $K$ has class number one - i.e. every ideal of $\mathcal O_K$ is principal. It follows that no prime less than $41$ ($|\mathrm{disc}(K)| = \frac{163}4$) can split in $\mathcal O_K$.

The minimal polynomial of $\alpha = \frac{1+\sqrt{-163}}2$ is $f(X) = X^2-X+41$. The fact that every prime less than $41$ is inert in $K$ (only $163$ ramifies) shows that $f$ is irreducible modulo $p$ for all $p<41$. In particular, $$p\nmid f(n)\quad \forall p<41, n<41,$$ from which it follows that $f(n)$ is prime for all $0\le n<41$ (since $f(n)<41^2$ is composite iff it has a prime factor $<41$). This gives a algebraic number theoretic justification for why the polynomial $f$ takes so many consecutive prime values.

Where you wrote Np^i, should it be N(P^i)? i.e. the norm of P^i — Richard, Mar 06 '16 at 15:32
Since the norm is (totally) multiplicative, both of those expressions are the same. — Mathmo123, Mar 06 '16 at 15:33

Arkady · Answer 2 · 2016-03-09T18:20:46.270

2

First let $c\equiv 1 \pmod 4$. Here $\frac{\text{disc}(K)}{4}=c$. Suppose $\mathfrak p^i$ is principal say $(a+b\sqrt{-c})$. Then so is the other guy $(\mathfrak p^i)'=(a-b\sqrt{-c})$ lying above $p^i$. Then the norm of $\mathfrak p^i$ is $$p^i=(a+b\sqrt{-c})(a-b\sqrt{-c})=a^2+cb^2,$$ for $b\neq 0$. So $|b|\geq 1$. So $p^i=a^2+cb^2\geq cb^2\geq c=\frac{|\text{disc}(K)|}{4}$ as needed.

If $c\equiv 3 \pmod 4$ and $\mathfrak p^i$ is principal, then we compute the norm of $\mathfrak p^i$ as above: $$p^i =N (\mathfrak p^i)=(a+b\left(\frac{1+\sqrt{-c}}{2}\right))(a+b\left(\frac{1-\sqrt{-c}}{2})\right))=a^2+ab+(\frac{1+c}{4})b^2$$

As above, we can assume that $|b|\geq 1$ and by switching the sign of $b$ if necessary, we can assume that $a^2+ab$ is non-negative, so that $a^2+ab+(\frac{1+c}{4})b^2\geq \frac{c+1}{4}\geq \frac{\text{disc}(K)}{4}$ as we wanted.

edited Mar 09 '16 at 18:20

answered Mar 06 '16 at 15:12

Arkady

9,727

@Mathmo123 Good catch, Mathmo123. :) – Arkady Mar 06 '16 at 15:20
How did you get that p=a^2 + cb^2? – Richard Mar 06 '16 at 15:41
@Mathmo123 Hey, thank you for your helpful comments! I think that fixes it and makes it even simpler than before? Could you take a quick look? – Arkady Mar 06 '16 at 21:47
Does the same method work for c congruent to 3 mod4? – Richard Mar 08 '16 at 15:21
@Richard See my edit. – Arkady Mar 09 '16 at 18:21

Ideal $\mathfrak p^i$ is not principal

2 Answers2