Showing that the ray class field of $\Bbb Q(i)$ modulo $(2)$ is equal to $\Bbb Q(i)$ I'm reading Silverman's Advanced Topics in AEC book, and in his review of class field theory he makes the following definition:

Let $K$ be a totally imaginary number field, and $\mathfrak c$ an integral ideal of $K$. A ray class field of $K$ modulo $\frak c$ is a finite abelian extension $K_{\mathfrak c}/K$ with the property that for any finite abelian extension $L/K$, $$\mathfrak c_{L/K}|\mathfrak c\implies L\subseteq K_{\mathfrak c},$$ 

where $\mathfrak c_{L/K}$ denotes the conductor of $L/K$. That is, it is the largest integral ideal $\mathfrak c_0$ of $K$ which is $(1)$ divisible by precisely the primes of $K$ that ramify in $L$, and $(2)$ satisfies
$$((\alpha),L/K)=1\ \text{ for all $\alpha\in K^*$ satisfying $\alpha\equiv1\
 \mathrm{mod}\ \mathfrak c_0$},$$
where $(-,L/K)$ is the Artin map $I(\mathfrak c_0)\to\mathrm{Gal}(L/K)$ defined as follows: 

$I(\mathfrak c_0)$ denotes the fractional ideals of $K$ which are coprime to $\mathfrak c_0$; given $\mathfrak a\in I(\mathfrak c_0)$ write $\mathfrak a=\prod_{\mathfrak p}\mathfrak p^{n_{\mathfrak p}}$. For each prime $\frak p$ dividing $\frak a$, we know that $\frak p$ is unramified in $K$ since it is does not divide $\frak c_0$. For some prime $\mathfrak P$ of $L$ lying over $\frak p$, we get a natural homomorphism (via restriction)
$$\{\sigma\in\mathrm{Gal}(L/K):\mathfrak P^{\sigma}=\mathfrak P\}\to\mathrm{Gal}(R_L/\mathfrak P\mid R_K/\mathfrak p)$$
The latter group is generated by the Frobenius automorphism $x\mapsto x^{N_{K/\Bbb Q}(\mathfrak p)}$, and $(*)$ since $\mathfrak p$ is unramified, there is a unique element $\sigma_{\mathfrak p}\in\mathrm{Gal}(L/K)$, depending only on $\mathfrak p$, mapping to this Frobenius automorphism. We then define
$$(\mathfrak a,L/K)=\prod_{\mathfrak p}\sigma_{\mathfrak p}^{n_{\mathfrak p}}.$$

After the definition of ray class field, Silverman makes the following remark:

It is important to note that the conductor of $K_{\mathfrak c}$ need not actually equal $\frak c$. For example, the ray class field of $\mathbb Q(i)$ modulo the ideal $(2)$ is just $\Bbb Q(i)$ itself, so $\Bbb Q(i)_{(2)}$ has conductor $(1)$.

I'm left with two main questions:

$(1)$ why is the sentence I've marked $(*)$ and place in bold font true?
$(2)$ why is the ray class field of $\Bbb Q(i)$ modulo $(2)$ equal to $\Bbb Q(i)$?

For $(2)$, I've tried naively following definitions, but haven't gotten far: if we let $K=\Bbb Q(i)$, then we want to show that if $L/K$ is any finite abelian extension such that $\mathfrak c:=\mathfrak c_{L/K}$ divides $(2)$, then $L\subseteq K$. But saying $\mathfrak c$ divides $(2)$ is equivalent to having $2\in\mathfrak c$. Therefore if $p\in\Bbb Z$ is any odd prime, then for some $k\in\Bbb Z$ we have $p-1=2k\in\mathfrak c$, i.e. $p\equiv1\ \mathrm{mod}\ \mathfrak c$, and hence $((p),L/K)=1$ if $(p)$ is coprime to $\mathfrak c_0$.
Now, in particular, if $p\equiv_43$, then $(p)$ is a prime ideal of $\mathcal O_K$, and hence $\sigma_{(p)}=((p),L/K)=1$. It follows that $(p)$ splits completely in $L$. Furthermore, since only a finite number of primes of $K$ ramify in $L$, we see we have an infinite number of primes $\mathfrak p=(p)R_K$ with $p\equiv_43$ which split completely in $L$.
Is anything I'm pointing out even going to help me at all? Or is there a simpler way to see this? Also, sorry about the wall of text; I didn't intend for the post to be so long.
 A: In general, if $K$ is a number field and $\mathfrak{c}$ is a cycle in $\mathcal{O}=\mathcal{O}_K$, then $K_{\mathfrak{c}}/K$ has Galois group the ray class group $Cl_{\mathfrak{c}}$, which fits in a short exact sequence of the form
$$
1\to (\mathcal{O}/\mathfrak{c})^{*}/\mathrm{im}[\mathcal{O}^{*}]\to Cl_{\mathfrak{c}}\to Cl_K\to 1.
$$
Now if the map $\mathcal{O}^{*}\to(\mathcal{O}/\mathfrak{c})^{*}$ is surjective, then $|Cl_{\mathfrak{c}}|=h_K$. As $H\subset K_{\mathfrak{c}}$, for $H$ the Hilbert class field of $K$ and $[H:K]=h_K$ as well, we conclude that $H=K_{\mathfrak{c}}$. If also $Cl_{K}=1$ we see that $K_{\mathfrak{c}}=K$.
Now if $K=\mathbb{Q}(i)$ then $\mathcal{O}=\mathbb{Z}[i]$ and $Cl_K=1$, so proving that $\mathbb{Z}[i]^{*}\to(\mathbb{Z}[i]/2)^{*}$ is surjective proves $(2)$. Now $(2)=\mathfrak{p}_2^2$ for $\mathfrak{p}_2=(1+i)$ so $\mathbb{Z}[i]/2$ is a local ring with four elements and maximal ideal of two elements, so $(\mathbb{Z}[i]/2)^{*}$ has order $2$. A generator is $\overline{i}$, which clearly gets hit by $\mathbb{Z}[i]^{*}\to(\mathbb{Z}[i]/2)^{*}$.
