# 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?

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 $a + b \zeta_{11}$ ($a,b \in \mathbb{Z}$).
• Also, $K$ contains the quadratic extension $L = \mathbb{Q}(\sqrt{-11})$. You can easily factor 23 over $L$ first. It turns out to be a product of principal ideals (which is reassuring), so then you only need to find an $\alpha$ whose norm down to $L$ gives one of the prime factors over $L$ (not sure how much that helps, but at least that means you only need half as many multiplications when computing the norm). – Ted Dec 18 '11 at 19:46
• Here is one element of norm $23$: $1+\zeta^7+\zeta^8$. – Álvaro Lozano-Robledo Dec 18 '11 at 19:50
• I would suggest that, to find elements of small norm in $\mathbb{Q}[\zeta_p]$, instead of trying $a+b\zeta$, start instead with numbers of the form $a_0+a_1\zeta+\cdots+a_{p-1}\zeta^{p-1}$, where $a_i=-1$, $0$, or $1$. – Álvaro Lozano-Robledo Dec 18 '11 at 20:10