Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to calculate ideal class group of $\mathbb{Q}(\sqrt{-103})$. By Minkowsky bound every class has an ideal $I$ such that $N(I) \leq 6$. It is enough to consider prime ideals with the same property. Let me denote by $R$ the ring of integers. So we have $2R= (2, \frac{1+\sqrt{-103}}{2})(2, \frac{1-\sqrt{-103}}{2})$, $3R$ and $5R$ are prime. Ideal class group is generated by $P=(2, \frac{1+\sqrt{-103}}{2})$. $N(P)=2$ and thus $P$ is not principal. Absolute norm $N(P^2)=4$ and if $P^2=(a)$ then $(a)=(2)$, but $P^2 \neq 2R$, because $2R=P \overline{P}$, $P \neq \overline{P}$. $N(P^3)$=8 and it is obvious that $P^3$ is not principal. How can I continue this argument for other powers of $P$ and when this stops?

share|improve this question
3  
$$N\left(\frac{5+\sqrt{-103}}2\right)=\frac{5^2+103}4=32=2^5$$ –  Jyrki Lahtonen Nov 21 '11 at 5:19
    
@Jyrki You might as well post this as an answer. –  Alex B. Nov 21 '11 at 5:50
    
@AlexB. I'm thinking about it, but I need to convince myself first that this implies that the predicted power of $P$ is, indeed, principal. For now it just marks the spot at which the OP's argument based on studying norms of individual elements stops working. If you see, how the rest follows from general principles, go ahead and save me the trouble! I'm done with my morning coffee, and need to commute next... –  Jyrki Lahtonen Nov 21 '11 at 6:06
    
@Jyrki Done. $ $ –  Alex B. Nov 21 '11 at 6:59

2 Answers 2

up vote 5 down vote accepted

As Jyrki Lahtonen notes in his comment, the norm of the principal ideal $I=\left(\frac{5+\sqrt{-103}}{2}\right)$ is $2^5$. It will therefore be enough to show that $I=P^5$ or $I=\bar{P}^5$, since 5 is prime and therefore will have to be the exact order of $P$ in the class group.

Now, $P$ and $\bar{P}$ are the only two ideals of $R=\mathbb{Z}\left[\frac{1+\sqrt{-103}}{2}\right]$ above 2. So we immediately know that $I=P^n\bar{P}^{5-n}$ for some $n$ between 0 and 5. But if $n$ is neither 0 nor 5, then $P\bar{P}=2R$ divides (i.e. contains) $I$, which immediately leads to a contradiction, since the generator is not divisible by 2 in $R$. So $I\not\subseteq 2R$, thus $n=0$ or 5 and we are done.

share|improve this answer
1  
+1: Figured this out in the traffic lights :-), thanks! –  Jyrki Lahtonen Nov 21 '11 at 7:06

People do not seem to like the quadratic form description of things, but it is the case that, if the discriminant $\Delta$ is negative, the group of binary quadratic forms of discriminant $\Delta$ is isomorphic to the class group of $\mathbb Q(\sqrt \Delta).$ The conditions for this are either $\Delta \equiv 1 \pmod 4$ is squarefree, or $\Delta$ is divisible by 4, $\Delta /4$ is squarefree, and $\Delta /4 \equiv 2,3 \pmod 4.$

For you, $\Delta = -103 \equiv 1 \pmod 4.$ The group of forms, under Gaussian composition, is cyclic of order 5:

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./classGroup
Absolute value of discriminant?  103   class  number  5

( 1, 1, 26) 
( 2, -1, 13) 
( 2, 1, 13) 
( 4, -3, 7) 
( 4, 3, 7)

Each (A,B,C) refers to $A x^2 + B x y + C y^2.$

If you had a positive $\Delta,$ the congruence restrictions would be the same, we would need to also require that $\Delta$ not be a square, and we would be calculating the narrow class group. In this case, if there is an integral solution to $u^2 - \Delta v^2 = -4,$ narrow class group and class group agree, so we are done. If there is no solution to $u^2 - \Delta v^2 = -4,$ then the class group is the subgroup of squares of the narrow class group. Long story. In this latter case, you are, in effect, keeping the form that represents $1,$ but throwing out the (distinct in this case) form that represents $-1.$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.