# Can class number decrease as we grow over some fixed number field?

Is it possible that a number field $$K$$ has non trivial class group, while one of its finite extension has trivial class group ?

Or may be, is it known Hilbert class field of which number fields have trivial class group ?

• But I think it may be possible. Imagine if h_K (order of class group) is prime p, and h_{K^{Hil}} is also a prime q with p does not divide q-1, then Hilbert class field of K^{Hil} is K^{Hil} itself and hence having trivial class group. In general I think this is close to answering the question when a meta-abelian group is abelian and one of its subgroup appears in such nice fashion (i.e as an extension K^{Hil}/K kind of thing). – dragoboy Nov 19 '19 at 5:43
• Alright, let's say we're talking about $K = \mathbb Z[\sqrt{-5}]$, which has class number 2. What manner of extending it do you think might give the result we're looking for? I tried extending it with $\mathbb Z[\sqrt 2]$. According to LMFDB, $\mathbb Q(\sqrt 2, \sqrt{-5})$ also has class number 2. That's of course just one way out of many to take this. – Robert Soupe Nov 19 '19 at 6:26
• Of course it's possible, $\mathbb{Q}(\sqrt{2},\sqrt{-3})$ has class number $1$ but a subfield $\mathbb{Q}(\sqrt{-6})$ does not. – pisco Nov 19 '19 at 9:25
• @pisco I verify that $\mathbb Q(\sqrt{-6})$ has class number 2. I verify that $\mathbb Q(\sqrt 2 + \sqrt{-3})$ has class number 1 lmfdb.org/NumberField/4.0.576.2 The part that I'm fuzzy on is how to extend the former to get to the latter. Is it adjoining $\mathbb Q(\sqrt{-3})$ or $\mathbb Q(\sqrt 2)$ or will either one of them do the trick? – Robert Soupe Nov 19 '19 at 16:17
• @RobertSoupe Just adjoin either one of them. – pisco Nov 19 '19 at 16:38

According to Paul Pollack, in A Conversational Introduction to Algebraic Number Theory, pp. 176-177, if $$K = {\bf Q}(\sqrt{-5})$$, and $$L = K(i)$$, then $$K$$ has class number $$2$$, while $$L$$ has class number $$1$$. Pollack doesn't give a reference.
Pollack also notes that not every number field has a finite extension with class number $$1$$. He cites a theorem of Golod and Shafarevich, "On the class field tower," (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 28 (1964) 261-272, to the effect that for every $$n$$ there are infinitely many number fields $$K$$ of degree $$n$$ which do not have a finite extension of class number $$1$$. He gives (without proof) the example $${\bf Q}(\sqrt{-3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19})$$ as a quadratic field with no finite extension of class number $$1$$.
The OP first question is whether an algebraic number field $$k$$ with class number $$h_k >1$$ can be embedded in an algebraic number field $$K$$ with class number $$h_K = 1$$. Numerical examples exist (see (1) or the previous answers). The second question is related to the so called "class field tower" problem. The "principal divisor theorem" of CFT states that every ideal of $$k$$ becomes principal in the Hilbert class field $$k_1$$, but the question is, if we repeat the process, i.e. we consider the tower $$k \subset k_1 \subset k_2 ...$$, where $$k_{i+1}$$ is the Hillbert class field of $$k_i$$, is the extension $$k_{\infty}=\cup k_i$$ finite over $$k$$ ? If the above embedding problem with $$h_K =1$$ has a solution, then $$k_{\infty} \subset K$$ and the CF tower is finite. Conversely, if the CF tower is finite, then $$k_{\infty}$$ is the smallest $$K$$ with $$h_K=1$$ in the problem above.
Examples of finite CF towers exist, and it was even believed that all CF towers are finite (see (1)). But in the $$60$$'s, Golod and Shafarevitch constructed examples of infinite CF towers using cohomological methods. The G-S. theorem rests on the following criterion: There exists a function $$\gamma(n)$$, e.g. $$2++2\sqrt {n+1}$$, such that the dimension over $$\mathbf F_p$$ of $$Cl_k /p$$ is < $$\gamma(n)$$ for any algebraic number field $$k$$ of degree $$n$$ whose $$p$$-CF tower (i.e. only $$p$$-extensions are considered in the CF tower) is finite (see (2), chapter 9). Since then, many refinements have been proved, but the approach remains fundamentally the same.