## 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

$(\sqrt{10})=P_2P_5$

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.