## Imaginary quadratic fields $$k$$ with $$\text{Cl}_ 2(k)\simeq(2,2^ m)$$ and rank $$\text{Cl}_ 2(k^ 1)=2$$.(English)Zbl 1063.11038

The imaginary quadratic fields $$k$$ such that both $$k$$ and its Hilbert 2-class field have 2-class groups of rank 2 are characterized in this article. For a number field $$F$$, let $$h_2(F)$$, $$\text{Cl}_2(F)$$ and $$F^1$$ denote its 2-class number, 2-class group and Hilbert 2-class field respectively. Also, the pair $$(2,2m)$$ denotes the group $$\mathbb{Z}_2\times\mathbb{Z}_{2^m}$$.
The main theorem is: Let $$k$$ be an imaginary quadratic field with $$\text{Cl}_2(k)\cong (2,2^m)$$. Then rank $$\text{Cl}_2(k^1)= 2$$ if and only if $$k= \mathbb{Q}(\sqrt{-p_1p_2p_3})$$, where $$p_1$$, $$p_2$$ and $$p_3$$ are primes satisfying $$p_1\not\equiv 3\not\equiv p_2\pmod 4$$ and $$p_3\equiv 3\pmod 4$$, $$({p_1\over p_2})= -1$$, $$({p_1\over p_3})= ({p_2\over p_3})= 1$$ and $$h_2(K)= 2$$, where $$K$$ is a non-normal quartic subfield of one of the two unramified cyclic quartic extensions of $$k$$ such that $$\mathbb{Q}(\sqrt{p_1p_2})\subset K$$. For example, the fields $$\mathbb{Q}(\sqrt{-d})$$ for $$d= 310, 406, 598, 1443$$ and $$1615$$ all satisfy these conditions.

### MSC:

 11R29 Class numbers, class groups, discriminants 11R11 Quadratic extensions

### Keywords:

Hilbert 2-class field
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