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I saw the following problem:

Let $p_1, . . . , p_n$ be distinct prime numbers ($n\ge 2$) and let $m = -\prod_{i=1}^np_i$. Which primes ramify in $L = \mathbb{Q}(\sqrt{m})$? Use this to show that the class number of $L$ is divisible by $2^{n-1}$.

I have been working on it and cant quite figure it out. I know that $D_L = m$ or $D_L = 4m$ depending on the value of $m\: (\text{mod}\:4)$. So the ramified primes are $p_1, ..., p_n$, plus possibly 2. Now I cant figure out how to make an argument involving the class number.

user26857
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james.pual.11
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    For each $i$, consider $(p_i) = \mathfrak p_i^2 = (p_i, \sqrt{m})^2$. Show that the $\mathfrak p_i$ are non-principal and that no subset of $\mathfrak{p_1, \cdots, p_{n-1}}$ has principal product (of course $\mathfrak{p_1\cdots p_n} = (\sqrt m)$ is principal). Then you're done! $\langle \mathfrak{[p_1],\cdots,[p_{n-1}]}\rangle$ is a subgroup of order $2^{n-1}$. – Mathmo123 Apr 20 '20 at 08:43

1 Answers1

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I give an example of Mathmo123's comment for the case $\mathbb{Z}[\sqrt{-30}]$.

The number $30$ factors as $2×3×5$, three prime factors, so there is a set of eight divisors $d\in\{1,2,3,5,...,30\}$ for which we may define ideals $(d,\sqrt{-30})$. Clearly these have norm $d$ for the divisors $d$ rendered above, but only $1$ and $30$ are actually norms of elements in $\mathbb{Z}[\sqrt{-30}]$ and therefore only $(1,\sqrt{-30})=(1)$ and $(30,\sqrt{-30})=(\sqrt{-30})$ are principal. The remaining six ideals are nonprincipal.

The six nonprincipal ideals pair up to form products equal to the principal ideal $(\sqrt{-30})$, to wit

$(2,\sqrt{-30})×(15,\sqrt{-30})=(30,2\sqrt{-30},15\sqrt{-30},-30)=(\sqrt{-30})$

$(3,\sqrt{-30})×(10,\sqrt{-30})=(\sqrt{-30})$

$(5,\sqrt{-30})×(6,\sqrt{-30})=(\sqrt{-30})$

So, only half of the eight ideals obtained from the divisors of $30=2×3×5$ might generate different classes, and we may define this half by suppressing one of the prime factors -- let us say, $5$. Thus we have $2^{3-1}=4$ ideals that may generate different classes:

$(1,\sqrt{-30})=(1)$

$(2,\sqrt{-30})$

$(3,\sqrt{-30})$

$(6,\sqrt{-30})$

We easily find that the three remaining nonprincipal ideals are squared to give principal ideals

$[(2,\sqrt{-30})]^2=(4,2\sqrt{-30},-30)=(2)$

$[(3,\sqrt{-30})]^2=(3)$

$[(6,\sqrt{-30})]^2=(6)$

This is also sufficient to prove that the three nonprincipal ideals $(2,\sqrt{-30}), (3,\sqrt{-30}), (6,\sqrt{-30})$ do in fact generate distinct ideal classes, because the quotient obtained by dividing any two of $2,3,6$ are not squares of elements in $\mathbb{Q}[\sqrt{-30}]$.

Thereby $\mathbb{Z}[\sqrt{-30}]$ is proved to have $2^{3-1}=2^2=4$ separate classes generated by different ideals, assuring the class number will be a multiple of $4$. Since the identified nonprincipal classes all have order $2$, we find in addition that the class group will contain the Klein four-group $(\mathbb {Z}/2\mathbb{Z})^2$.

In the case of $\mathbb{Z}[\sqrt{-30}]$, the prime factors defined above include all primes within the Monkowski bound, therefore the identified Klein-four class group and its corresponding class number are actually the complete classification of ideals. In most other cases additional primes that lie within the Minkowski bound but do not divide the radicand give rise to additional nonprincipal contributions and thus enlarge the class group. For example, $14=2×7$ and the above argument would predict a single $2$-cycle in the class group of $\mathbb{Z}[\sqrt{-14}]$; but because of the additional prime $3$ within the Minkowski bound this $2$-cycle is embedded in a $4$-cycle of the full class group $\mathbb {Z}/4\mathbb{Z}$. The peculiar case of $30$ having prime factors that saturate the Minkowski bound, pegging the class group to Klein-four, leads to the modified form of unique factorization described here.

Oscar Lanzi
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