I'm trying to prove that class number of the quadratic field with discriminant $17$ is $1$. Let $K =Q[\sqrt{17}]$ be the quadratic field and $\mathfrak{o}_{K}=\mathbb{Z}[\frac{1+\sqrt{17}}{2}]$ be the ring of integers. The Minkowski's bound tells us that the one representative of each ideal class should have norm 1 or 2. Norm 1 $\implies$ $\alpha=\mathfrak{o}_{K}$, so we have the trivial class group. So let's focus on Norm 2. Let $\alpha$ be the element of the ideal class. We have $N(\alpha)=2$. So $\alpha$ lies above 2. Now 2 either ramifies, is inert or splits.
Let $f$ be the degree of the residue field of $\mathfrak{o}_{K}$ over residue field of $\mathbb{Z}$ and $e$ be the ramification index. So if 2 ramifies, we have $f=2$ $\implies$ ($\mathfrak{o}_{K}/\alpha:Z/2Z)=2$. I think this indicates something as we have $(\mathfrak{o}_{K} :\alpha)=2$ but I'm not able to formulate a proper argument. If $f=1$, $(\mathfrak{o}_{K}/\alpha:Z/2Z)=1$ $\implies \mathfrak{o}_{K}/\alpha \cong \mathbb{F}_2$. I think we have the same case for splitting, as $f=1$, in that case too. I will admit my commutative algebra is very much rusty so it could be that I'm missing some silly PID arguments. Any help will be appreciated, thanks!
Also, can this be done without the ramification information? In Neukirch, a similar question is asked while ramification hasn't even been discussed yet.