Dear Sir,

In Problem Sheet 6, for k=Q(sq.root 10)

How did the last bit come about? Why is the class group finally {Ok,p2} Why is p3,p*3 and p5 dropped from the class group? The previous lecturer seemed to have
used mod 5 in his notation of the general element of Ok. Can you please give an answer using the methods from “A few class groups”. In particular can you please
give an answer with the same method as you did for Q(\sqrt{-33})? Is this possible?

Reply:

I hope you can see that the methods pretty much the same for all the examples. In that problem, there is a list

O_K,P_2, P_3, P_3', (2), P_5, P_2P_3, P_2P_3'

of ideal of norm less than 7, and every element in the class group is equivalent to one of these. We know that P_2^2=(2), and hence,

P_2^{-1} \sim P_2

The relation

(\sqrt{10})=P_2P_5

shows that

P_5 \sim P_2^{-1} \sim P_2

(2-\sqrt{10}) = P_2P_3

then gives

P_2P_3\sim O_K

and

P_3 \sim P_2^{-1} \sim P_2

Also,

P_3'\sim P_3^{-1} \sim P_2^{-1} \sim P_2

and finally,

P_3'P_2 \sim P_2^2\sim O_K

So it turns out that

O_K, P_2

represent all classes in the class group.

One sees thereby that the class group is either trivial (if P_2 is principal), or Z/2 (if P_2 is not principal). At this point, one checks that there is no element of O_K with norm 2, using the \mod 5 argument, and hence, that P_2 cannot be principal.