A question on the genus field of an algebraic number field 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 $i=\sqrt{-1}$.
 A: The setup ensures that the compositum $k(\sqrt{i})$ will be (isomorphic to) $k\otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{i})$.
When $g$ is $2$-adically close to $x^4+1$ -- it turns out that mod $8$ (i.e., $m=3$) is good enough -- then the field $F=\mathbb{Q}_2(\sqrt{i}) = \mathbb{Q}_2(i,\sqrt{2},\sqrt{-2})$ contains four approximate simple roots of $g$ (namely the four eighth roots of unity) and, by Hensel, even four exact roots of $g$. Therefore the $2$-adic completion of $k$ will be (isomorphic to) the local field $F$, with its single fully ramified prime ideal above $2$, and completing $k(\sqrt{i})$ above $2$, which amounts to taking $k(\sqrt{i})\otimes_{\mathbb{Q}} \mathbb{Q}_2$, will produce $F\otimes_{\mathbb{Q}_2}F$, a sum of four copies of $F$. In other words, the dyadic prime decomposes fully from $k$ to $k(\sqrt{i})$. Since $\mathbb{Q}(\sqrt{i})|\mathbb{Q}$ is unramified at all odd finite primes, so is $k(\sqrt{i})|k$; therefore it is unramified at all finite primes, and thus (up to isomorphism) contained in $\tilde{K}$.
Behind all this is the fact that there are only a small handful of ways in which $2$ can ramify in a quadratic extension of $\mathbb{Q}_2$. The argument not only establishes that $2$ is fully ramified in $k|\mathbb{Q}$, but that it is fully ramified in precisely the right way, the same way as in $\mathbb{Q}(\sqrt{i})|\mathbb{Q}$ and $F|\mathbb{Q}_2$.
Examples are easy to come by. E.g., $g(x)=x^4-8x-15$ or $g(x)=x^4-8x+25$ both have Galois group $S_4$, and in both cases we get such an unramified $V_4$ extension $k(\sqrt{i})|k$.
I see no contradiction arising. Of course, since the Klein group $V_4$ isn't a cyclic group, we need at least two nontrivial factors $\Omega^{(p)}$ here, say $\mathbb{Q}(i)$ as $\Omega^{(4)}$ and $\mathbb{Q}(\sqrt{2})$ as $\Omega^{(2)}$, aside from whatever might come from other unramified extensions of $k$ contained in $Ak$.
