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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 ?

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    $\begingroup$ 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). $\endgroup$
    – dragoboy
    Commented Nov 19, 2019 at 5:43
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    $\begingroup$ 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. $\endgroup$ Commented Nov 19, 2019 at 6:26
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    $\begingroup$ Of course it's possible, $\mathbb{Q}(\sqrt{2},\sqrt{-3})$ has class number $1$ but a subfield $\mathbb{Q}(\sqrt{-6})$ does not. $\endgroup$
    – pisco
    Commented Nov 19, 2019 at 9:25
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    $\begingroup$ @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? $\endgroup$ Commented Nov 19, 2019 at 16:17
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    $\begingroup$ @RobertSoupe Just adjoin either one of them. $\endgroup$
    – pisco
    Commented Nov 19, 2019 at 16:38

3 Answers 3

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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.

(1) F. Lemmermeyer, Class Field Towers, 2010

(2) Cassels-Fhröhlich, Algebraic Number Theory, Acad. Press, 1967

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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 $1$.

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In this answer we look behind the curtain to see how the field expansion enables unique factorization in an otherwise non-UF domain.

Consider the familiar nonunique factorization in $\mathbb{Z}[\sqrt{-5}]$:

$6=2×3=(1+\sqrt{-5})(1-\sqrt{-5})$

When we adjoin $i$ to the corresponding field $\mathbb{Q}[\sqrt{-5}]$, we thereby introduce the Gaussian integers, including $\pm1\pm i$ for all sign choices. Of course we can add, subtract, or multiply these with any $\mathbb{Z}[\sqrt{-5}]$ integers to obtain new integers, such as $\sqrt{5}$. We can also divide some pairs of $\mathbb{Z}[\sqrt{-5}]$ and Gaussian integers. The particular Gaussian integers $\pm1\pm i$ end up playing a major role as divisors.

To wit, we render the following natural primes which are irreducible in $\mathbb{Z}[\sqrt{-5}]$, but now:

$2=(1+i)(1-i)$

$3=[(1+\sqrt{-5})/(1+i)][(1-\sqrt{-5})/(1-i)]\equiv\alpha\overline{\alpha}; \alpha^4-2\alpha^3+2\alpha^2-6\alpha+9=0$

With these reductions the formerly nonunique factorization of $6$ now differ only by order and count as a unique factorization:

$6=2×3=[(1+i)(1-i)][\alpha\overline{\alpha}]$

$6=(1+\sqrt{-5})(1-\sqrt{-5})=[(1+i)\alpha][(1-i)\overline{\alpha}]$

What happens overall: when we adjoin $i$ to $\mathbb{Q}[\sqrt{-5}]$, the formerly irreducible natural primes $\in\{2,3,7\}\bmod20$ are split in the manner above, with the odd primes in that list using $1\pm i$ as divisors. Underying the splitting of the odd primes is the fact that they have the quadratic form $2x^2+2xy+3y^2$ but their doubles have the quadratic form $x^2+5y^2$ (e.g. $3=2(0^2)+2(0)(1)+3(1^2)$ but $6=1^2+5(1^2)$). These are exactly the natural prime factors that would cause nonunique factorization in $\mathbb{Z}[\sqrt{-5}]$ alone, and thereby the nonuniqueness in the factorizations of are resolved (and it turns out no other modes of nonunique factorization are introduced).

Adjoining $i$ to $\mathbb{Q}[\sqrt{-p}]$ also reduces class 2 to class 1 for $p=13$ and $p=37$, through the same reduction process for the appropriate natural primes.

A similar consolidation of ideals and thus of nonunique factorization may be applied to fields with class number greater than 2. Consider the class-4 quadratic domain $\mathbb{Q}[\sqrt{-30}]$. With a class group $\mathbb{Z}/2\mathbb{Z}×\mathbb{Z}/2\mathbb{Z}$, we adjoin the two primitive roots $i=\zeta_4,\omega=\zeta_3$ to the field and proceed to reduce previously irreducible natural primes with appropriate residues $\bmod 120$, for example

$2=(1+i)(1-i)$

$3=(1-\omega)(1-\omega^2)$

$11=\left(\dfrac{6+\sqrt{-30}}{(1+i)(1-\omega)}\right)\left(\dfrac{6-\sqrt{-30}}{(1-i)(1-\omega^2)}\right)$ (these factors solve $z^8-12z^7+72z^6-288z^5+983z^4-3168z^3+8712z^2-15972z+14641=0$)

With these decompositions the nonunique factorizations

$66=2×3×11=(6+\sqrt{-30})(6-\sqrt{-30})$

become consolidated (down to units and order) as

$66=(1+i)(1-i)(1-\omega)(1-\omega^2)\left(\dfrac{6+\sqrt{-30}}{(1+i)(1-\omega)}\right)\left(\dfrac{6-\sqrt{-30}}{(1-i)(1-\omega^2)}\right)$

This, of course, does not prove that the extension $\mathbb{Q}[\sqrt{-30},i,\omega]=\mathbb{Q}[\sqrt{-30},\zeta_{12}]$ is a UFD. But by consolidating the nonunique factorizations in the original quadratic field $\mathbb{Q}[\sqrt{-30}]$ it does appear to be a likely candidate.

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