The Minkowski bound for $\mathbb{Q}(\sqrt{-199})$ (with ring of integers $\mathbb{Z}(\frac{1+\sqrt{-199}}{2})$) is $\frac{2}{\pi}\sqrt{199} < 11$, so we need to consider $2$, $3$, $5$, $7$. Since $|\mathbb{Z}(\frac{1+\sqrt{-199}}{2}):\mathbb{Z}(\sqrt{-199})|=2$, we have to apply Dedekind on $(2)$ with the minimal polynomial $x^2 + x + 50$ (instead of $x^2 + 199$). The latter factorizes mod 2 as $x(x+1)$, so $(2) = (2, \frac{1+\sqrt{-199}}{2})(2, 1 + \frac{1+\sqrt{-199}}{2})$, neither of the factors being principal (and hence both of order $2$), as both have norm $2$ and $2 = (x+\frac{y}{2})^2 + \frac{199y^2}{4}$ is insoluble in $\mathbb{Z}$.
However, http://www.lmfdb.org/NumberField/2.0.199.1 claims that the class group is $C_9$, so there can't be elements of order $2$, which seemingly contradicts the above. Where is the mistake in my approach?
Any help appreciated!
Best Answer
Why do you think that these prime ideals being nonprincipal implies that they have order 2?
Labeling the prime ideals, say $(2) = \mathfrak{p}\mathfrak{p}'$. If the order of $\mathfrak{p}$ is 2, then necessarily $\mathfrak{p} = \mathfrak{p}'$ (hint: you can use that they have the same norm and are equal in the class group). Can you show that isn't the case?
In fact, $\mathfrak{p}^k$ isn't principal for $k = 1,2,\dots,8$, but $\mathfrak{p}^9 = \left(\frac{43\pm\sqrt{-199}}{2}\right)$ (sign depending on choice of $\mathfrak{p}$). Hence the order of $\mathfrak{p}$ in the class group is 9.