Class number of $\mathbb{Q}(\zeta_{11})$

I want to compute the class number of $K=\mathbb{Q}(\zeta_{11})$. The Minkowski bound here is < 59, and looking at the factorisation of primes, we can show that the ideal class group is actually generated by the prime ideals above 23. Also, note that 23 splits completely in $K$. Now, is there an "easy way" to show that all these prime ideals are principal, or do we have to look at them separately?

Solution 1:

If you can find an element $\alpha \in K$ of norm 23, then this gives a factorization of the ideal (23), and since we already have the maximum possible number of conjugates, all of the factors are prime. Of course, finding the $\alpha$ is the hard part :) but since the numbers involved aren't too big, you can probably do this by guessing and checking; I would start with numbers of the form $a + b \zeta_{11}$ ($a,b \in \mathbb{Z}$).