3704 Sheet 6

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?


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


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


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


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.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: