Let $K$ be a number field and let $Ω_K$ be all places of $K$. Let $Ω_f$ be all finite places of $K$ and $Ω_∞ $ be infinite places. Let $S=Ω_∞\cup \{b\in Ω_f \mid v(2)\neq 0 , v(17)\neq 0 \}$.
Define $K(S,2)=\{b \in K^{\times}/{K^{\times}}^2\mid v(b)≡0 \bmod2, \forall v \notin S\}$(This is called a Selmer group of a field).
Then, in the case of $K=\Bbb{Q}$, $K(S,2)$ can be identified with $\{±1,±2,±17,±34\}$.
For general number field $K$, I guess
$K(S,2)$ is generated by primes of $K$ above $2$ and $17$ and units of $K$
Is this correct ? And how can I prove this ?
My try: Any $d\in K(S,2)$ can be written $d=a_1^{n_1}a_2^{n_2}・・・a_m^{n_m}$($a_i$: primes of $K$, $n_1,・・・n_m\in \{0,1\}$) If $\exists i\in \{1,2,・・・,m\}$, $a_i$ is not primes above $2$ and $17$. Then, $ord_{a_i}(d)=1≡0mod2$. This is contradiction. Thus $\forall i$, $a_i$ is primes above $2$ and $17$. Thus $K(S,2)$ is generated by primes above $2$ and $17$ and units of $K$ □
But I worry about in the case class number of $K$ is not trivial, we should take irreducible elements and whether above discussion still works.
Thank you for your help.