Let $K=\mathbb Q(\xi_{39})$ be the 39-th cyclotomic field. Pari-GP told me that the prime ideals above $3$ and $13$ are not principal. Is there a way to prove that by hand (no computation made by computer)
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$\begingroup$ There are no prime ideals above 39. You meant prime ideals over 3 and 13. Can you compute the class number by hand? $\endgroup$– KConradCommented Feb 7, 2015 at 19:12
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4$\begingroup$ If you are willing to accept a table entry from a book without checking it yourself, why not accept what a computer tells you too? :) Since you believe $h = 2$, if two ideals are not principal then their product must be principal, so check first that there is an element whose norm is the product of norms of primes over 3 and 39 (norms from a cyclotomic field down to $\mathbf Q$, other than $\mathbf Q$ itself, are never negative). That would show you do not have to check that primes over both 3 and 39 are not principal, but just over one of them. $\endgroup$– KConradCommented Feb 7, 2015 at 19:59
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1$\begingroup$ Here is a way: Assume the prime ideal to be principal say generated by a cyclotomic integer $\alpha$. Calculate the norm of the prime above. Then the $|N(\alpha)|$, is also norma of a suitable integer in a quadratic subfield. Choose an imaginary quadratic subfield in $Q[\zeta_{39}]$, where the norm is a positive definite quadratic form; for a binary form it should be easy to check if it assumes a specific value or not. $\endgroup$– P VanchinathanCommented Feb 8, 2015 at 0:53
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1$\begingroup$ @Vanchinathan:Why is |N(α)| norm of a suitable integer in a quadratic subfield? Moreover, there is only one quadratic subfield in a cyclotomic field. So how do you choose it? $\endgroup$– joaopaCommented Feb 8, 2015 at 2:18
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1$\begingroup$ @joaopa, the norm is transitive, so if you have a tower of finite extensions $L/K/F$ then for $\alpha \in L$ we have ${\rm N}_{L/F}(\alpha) = {\rm N}_{K/F}({\rm N}_{L/K}(\alpha))$. Thus the norm from $L$ to $F$ of any element of $L$ is also the norm from $K$ to $F$ of a related element of $K$ for any field $K$ between $L$ and $F$. (Your previous comment says there is only one quadratic subfield in a cyclotomic field, but surely you meant "possibly more than one" in general, since the Galois group $({\mathbf Z}/n)^\times$ often has more than one subgroup of index two.) $\endgroup$– KConradCommented Feb 8, 2015 at 18:07
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1 Answer
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The quadratic number field $k = {\mathbb Q}(\sqrt{-39})$ has a cyclic class group of order $4$. Its genus field is $K = {\mathbb Q}(\sqrt{-3},\sqrt{13})$, which has class number $2$. The Hilbert class field of $K$ is dihedral over ${\mathbb Q}$, hence not contained in the cyclotomic field $L = {\mathbb Q}(\zeta_{39})$. This already shows that the class number of $L$ is even.
Now verify that one of the prime ideals above $13$ in $K$ is not principal (obvious since its norm down to $k$ is not principal) and that it ramifies completely in $L/K$.
answered Feb 8, 2015 at 17:15