Consider $K=\mathbb{Q}(\sqrt{-11})$. It is quite straight forward to show that the Minkowski Bound $M_{K}<3$. It follows that every ideal class contains ideal of norm $\leq 2$.
Now we show that $(2)$ is a prime ideal. From this it follows that there exists no ideal of Norm 2 (and then clearly that the class number is $1$).
I dont understand why this is the case. ($(2)$ is prime ideal $\Rightarrow$ there exists no ideal of norm $2$.)
Best Answer
In a Dedekind domain, every nonzero proper ideal is a product of prime ideals. By the Minkowski bound, every ideal class contains an ideal of norm $\leq 2$, so the class group is generated by (classes of) prime ideals of norm $\leq 2$. If an ideal $I$ has norm $2$ then $I \mid (2)$ (because the norm is contained in the ideal and "contains" = "divides" in a Dedekind domain). So we only have to factor $(2)$ into prime ideals. The ring of integers of $\mathbb{Q}[\sqrt{-11}]$ is $\mathbb{Z}[\frac{1+\sqrt{-11}}{2}]$, so we can factor into primes by using the Kummer-Dedekind theorem: the minimal polynomial of $\frac{1+\sqrt{-11}}{2}$ is $x^2 - x + 3$; modulo $2$ this is $x^2+x+1$ which is irreducible, so $(2)$ is prime, of norm $4$.
An ideal of norm $2$ would be a divisor of $(2)$ different from $(2)$, which does not exist here.