The following is quoted from the Mathematical Reviews.

MR0544896 (80j:12002) Reviewed

Bhaskaran, M.
Construction of genus field and some applications.
J. Number Theory 11 (1979), no. 4, 488–497.

12A35 (12A65)

Let $k$ be a finite algebraic number field and $K$ its Hilbert class field, i.e., $K/k$ is maximally abelian such that every finite prime divisor of $k$ is unramified in $K$. Let ${\bf Q}$ be the rational number field and $A$ its maximal abelian extension. The author calls the intersection $\tilde{K} = K \cap Ak$ the narrow genus field of $k$, and proves that $\tilde{K}$ is obtained as a compositum field of $k$ and $\Omega^{(p)}$'s, i.e., $\tilde{K}=k\prod_{p} \Omega^{(p)}$, where $p$ runs over every positive rational integer and each $\Omega^{(p)}$ is a cyclic extension of ${\bf Q}$ that has a power of the ideal $(p)$ as its conductor.

{Reviewer's remarks: This seems rather odd, for the following reason. Let $g(x)$ be a polynomial of degree 4 with rational integer coefficients such that $g(x) \equiv x^4 +1 \mod 2^m$, where $m$ is a sufficiently large integer; let $\alpha$ be a root of the equation $g(x)=0$ and set $k={\bf Q}(\alpha)$. If we can take $\alpha$ such that $g(x)$ is irreducible over ${\bf Q}$ and the Galois closure $L$ of $k/{\bf Q}$ has Galois group isomorphic with the symmetric group $S_4$, then clearly $k$ cannot have any quadratic subfield and so $k \cap A = {\bf Q}$, and $[k(\sqrt{i})\colon k] = 4$. Clearly, as $m$ is large, the prime ideal $(2)$ of ${\bf Q}$ is completely ramified in $k$ and $k(\sqrt{i})\subset\tilde{K}$. Since $\mathrm{Gal}(k(\sqrt{i})/k) \cong \mathrm{Gal}({\bf Q}(\sqrt{i})/{\bf Q})$ is a Klein four-group, we can easily obtain a contradiction from the author's theorem stated above. Thus, if it is valid, we must have the conclusion that $\mathrm{Gal}(L/{\bf Q})$ is not isomorphic with $S_4$ for every irreducible $g(x)$ as stated above. This is odd.}

Reviewed by K. Masuda

**Question:** Could anyone explain why $k(\sqrt{i})\subset\tilde{K}$, i.e. why the dyadic prime ideal of $k$ is unramified at the extension $k(\sqrt{i})/k$? Here $i=\sqrt{-1}$.