# possible norms of prime ideals in the class group of $K=\mathbb{Q}(\sqrt{-21})$

I have an example in my notes where we try to compute the class group of the quadratic field $K=\mathbb{Q}(\sqrt{-21})$. My notes then proceed to evaluate the Minkowsk's bound< $\lambda(\sqrt{-21})$ which turns out to be 6. Therefore we know every ideal class has as a representative a prime ideal with a norm at most 6. My notes then claim that the only possible norms for those prime ideals must be 2 3 and 5. It's not clear to me why this should be true. I am aware that the norm of a prime ideal must be a prime power so I guess $4=2^2$ is a possible representative as well. Am I missing out something here does it have something to do with the fact that 2 is ramified in $\mathcal{O}_K$?

• You're right. Two is ramified, so there cannot be prime ideals of norm $4$. I'm not too well versed here, but I usually think of Minkowski's bound as a tool for giving a list of rational primes to be checked. Any prime powers will then be included along the way. Admittedly it is a useful sounding idea to not need to worry about some of the primes with non-trivial inertia. – Jyrki Lahtonen Apr 21 '16 at 13:27
• Thanks for your comment but I am not sure how exactly 2 being ramified implies that there cannot exist prime ideals of norm 4. Can you elaborate a bit? – TheGeometer Apr 21 '16 at 13:34
• @TheGeometer $2$ being ramified means that $2$ splits as $\mathfrak p^2$ in $\mathcal O_K$. $\mathfrak p$ is the only prime ideal with norm a power of $2$, and its norm must be $2$. – Mathmo123 Apr 21 '16 at 13:36
• The key thing you may have been missing is that this a quadratic extension. The $n=efg$ business leaves very few choices, when $n=2$. Here $e=2$ means that $f=1$, so $2^2$ cannot be a norm of a prime ideal. With a higher degree extension (possible non-Galois), there are more choices. – Jyrki Lahtonen Apr 21 '16 at 13:39

You're right that prime ideals can have prime powers for norms, but if a prime in $$\mathbb Z$$ does not ramify or split in an extension, its corresponding ideal is irrelevant to the class number computation.

I ask you to consider for a minute the case of 2 in $$K = \mathbb Q(\sqrt{-35})$$. The ideal $$\langle 2 \rangle$$ might be divisible by $$\langle 2, 1 + \sqrt{-35} \rangle$$. However, assigning $$\theta = \frac{1}{2} + \frac{\sqrt{-35}}{2},$$ which is an algebraic integer on account of the polynomial $$x^2 - x + 9$$, we readily see that $$1 + \sqrt{-35} = 2 \theta$$. Although $$2 \nmid \theta$$, obviously 2 does divide 2.

This suggests that 2 is prime, and not merely irreducible, in this domain. Of course $$\mathcal O_{\mathbb Q(\sqrt{-35})}$$ has class number greater than 1, since, for example, $$\langle 3 \rangle = \langle 3, 1 - \sqrt{-35} \rangle \langle 3, 1 + \sqrt{-35} \rangle$$. The norm of 2 is unnecessary for making this determination.

Now getting back to $$\mathcal O_{\mathbb Q(\sqrt{-21})}$$ and reassigning $$\theta = \frac{1}{2} + \frac{\sqrt{-21}}{2},$$ we see that $$\theta$$ is an algebraic number but not an algebraic integer since the relevant polynomial is $$2x^2 - 2x + 11$$. Thus $$2 \mid (1 - \sqrt{-21})(1 + \sqrt{-21})$$ but 2 divides neither of $$1 \pm \sqrt{-21}$$.

And since $$(1 - \sqrt{-21}) \in \langle 2, 1 + \sqrt{-21} \rangle$$ (and likewise $$(1 + \sqrt{-21}) \in \langle 2, 1 - \sqrt{-21} \rangle$$), we conclude $$\langle 2 \rangle = \langle 2, 1 + \sqrt{-21} \rangle^2$$, and hence $$\mathcal O_{\mathbb Q(\sqrt{-21})}$$ is not a prinicpal ideal domain and must have class number greater than 1.

It is easy to see that 3 ramifies with a corresponding non-principal ideal, and 25 has two distinct factorizations, but the norm of $$\langle 5, 2 \pm \sqrt{-21} \rangle$$ is still 5, right? I'm asking, so correct me if I'm wrong.

So, once again, the norm of 2 has turned out to be irrelevant to the class number computation. I figure that in a biquadratic domain like for $$K = \mathbb Q(\root 4 \of 5)$$ this might be much more fun, since the primes under the bound might ramify or split at the intermediate rings but not all the way down. Or would that be up? Depends on how you want to visualize it, I suppose.