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I am a bit confused with the procedure of calculating the ideal class group of a quadratic field.

From what I understood the computation starts by finding the Minkowski's bound say $n$. Then we list all the primes up to $n$ and we claim that every ideal class in the class group will be represented by a prime ideal whose norm is a prime of our list. Then we find all of those and multiply them to figure out the multiplication table and we are done. What I don't like about that is the following:

Minkowski's bound tells us that in each class there is an integral ideal with norm less than $n$. However is not clear that in a given class there will be a prime ideal that will have norm less than $n$.

I am aware that every class contains prime ideals but I don't know what we can say about their norm and the method I described above makes sense to me only if we can always find a prime ideal with norm less than $n$.

So either I misunderstood something or given an integral ideal of norm less than $n$ I can always find a prime ideal in the same ideal class with norm less than $n$ so that it indeed suffices to just consider the prime ideals. Can anyone please clarify?

Edit: I have just read a post of calculating the ideal class group of $\mathbb{Q}(\sqrt{-103})$ (Ideal class group of $\mathbb{Q}(\sqrt{-103})$). Here OP starts as I say by remarking that it suffices to do the calculation just by considering prime ideals of norm $< n = 6$. Then he figures out that the only prime ideals with that property are $P = \left(2, \frac{2 + \sqrt{-103}}{2}\right)$ and its conjugate.He then proceeds to calculate $P^k$ and this turns out to be non trivial for $k<5$. But for me this contradicts the initial assumption that it suffices to do the calculation just for prime ideals. That is because for example we have $[P^2]\neq [P]$(or [$\bar P$]) but norm$(P^2) = 4$. Hence $P^2$ is an integral ideal that dosen't have in it's class a prime ideal with norm less than $n=6$. So as long as I am concerned there might also be an integral (not prime) ideal $I$ of norm say $2$ such that $I$ is not principal and $[I]\neq[P]$. The situation I just described wouldn't happen if we could assume that for each class of ideals there is a prime ideal of norm less than $n=6$. But the example I discussed above with $P^2$ shows that this result does not hold. So what is going on I am getting really confused about this stuff?

Thanks in advance.

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TheGeometer
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  • I hesitate to ask this because I fear it will distract someone who could answer in general from answering only specifically, but: for which quadratic rings have you successfully calculated the ideal class group? Have they all been imaginary? Have they all been unique factorization domains? – Robert Soupe Apr 22 '16 at 18:26
  • You can of course claim that every ideal class in the class group will be represented by a prime ideal whose norm is a prime of your list but I very much doubt that this is true. –  Apr 22 '16 at 19:59
  • Yeah, I just saw calulations of ideal class group and I tried to guess how they do the calculations so I might be wrong. I couldn't find anywhere a rigorous calculation of the ideal class group so I am confused about it.... – TheGeometer Apr 22 '16 at 20:31
  • I think that if I mention http://math.stackexchange.com/questions/84131/ideal-class-group-of-mathbbq-sqrt-103?rq=1 in a comment it'll show up on the right sidebar as a "linked" question. – Mr. Brooks Apr 23 '16 at 21:35
  • As it turns out, all you have to do is paste in the URL of the question and it will get linked nice and neat. Presumably this works on questions from any SE site. – Mr. Brooks Apr 23 '16 at 21:39

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After some thinking I thing I realized what is going on.The result I am talking about that in every ideal class there is a representative that is prime ideal of norm less than $n$ does not hold. The reason we only consider prime ideals of norm less than $n$(therefore of norm $1<p<n$ where $p$ prime) for the computation of the class group must be the following. We know that in every class there is an integral ideal $I$ of norm less than $n$ and we know that this ideal can be written uniquely as a product of prime ideals and since the norm is multiplicative the prime ideals in the factoraisation of $I$ will have norm less than $n$. Therefore $[I]$ is just the product of prime ideal of norm less than $n$. Since $I$ was arbitary, it means every ideal class in the class group can be written as a product of ideals of norm less than $n$. In the example of $\mathbb{Q}(\sqrt{-103})$ we have established that the only prime ideals of norm less than $n=6$ are were just $P$ and $\bar{P}$ who both had norm $2$. Now since every ideal class in the class group contains an integral ideal of norm less than $n=6$ and this can be factored as product of prime ideals it must be that the class group contains at most the following 5 elements $[P],[P^2],[\bar{P}],[\bar{P^2}],e$ then one has to prove those are distinct and non trivial but that is not hard so we are done.

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