I want to prove that the class number $h_{11}$ of $K=\mathbb{Q}(\sqrt{11})$ is $1$. Here is what I got:
We know that the discriminant $\Delta_K$ of $K$ is $44$ because $11 \equiv 3 \mod 4$. We also know that for every non-zero prime ideal $\mathfrak{p}$ of $\mathcal{O}_K$ there exist an unique rational prime $p$ such that $\mathfrak{p} \mid p\mathcal{O}_K$. For such $p$ we know $\operatorname{Norm}(\mathfrak{p}) = p^m$ for $m \geq 1$. Further, the class group $\operatorname{CL}(K)$ is generated by
$$
\{\overline{\mathfrak{p}} \mid \mathfrak{p} \text{ is a prime ideal and }\operatorname{Norm}(\mathfrak{p}) \leq B_K\},
$$
where
$$
B_K = \left(\frac{2!}{2^2}\right)\sqrt{\Delta_K} = \sqrt{11}
$$
is the Minkowski bound. Therefore, for every prime ideal $\mathfrak{p}$ we get $\operatorname{Norm}(\mathfrak{p}) = 2, 3$.
I don't know how to proceed.
Best Answer
We first compute $\mathfrak O_K$. If $a+b\sqrt{11}$ is integral for $a,b\in \mathbb Q$, then $a-b\sqrt{11}$ is its conjugate and hence their minimal polynomial is $$f\left(x\right)=x^2-2a x+a^2-11b^2.$$ Then $2a$ and $a^2-11b^2$ are both rational integers. Put $a=\frac n 2$ and $b=\frac m 2$, then $$a^2-11b^2=\frac {n^2-11m^2}{4}$$ is a rational integer and hence $$n^2-11m^2\equiv n^2+m^2\equiv 0\mod 4.$$ Therefore $n$ and $m$ are even and hence $\mathfrak O_K=\mathbb Z\left[\sqrt{11}\right]$.
Since $11$ is not a quadratic residue of $3$, $3\mathfrak O_K$ is a prime ideal. Hence if the norm of $\mathfrak p$ is $3$, then $\mathfrak p$ divides $3$ and hence $\mathfrak p$ is exactly $3\mathfrak O_K$. This is impossible since the norm of $3\mathfrak O_K$ is $3^2=9$.
Otherwise the norm of $\mathfrak p$ is $2$. In this case, since $2$ divides the discriminant, $2$ is ramified in $K$, namely $2\mathfrak O_K= P^2$ for a prime ideal $P$. We shall prove that $$P=\left(3+\sqrt{11}\right)\mathfrak O_K.$$ Indeed, since $$\left(\sqrt{11}+3\right)\left(\sqrt{11}-3\right)=2,$$ and $3+\sqrt{11}$ does not belong to $2\mathfrak O_K$, we only need to check that $\left(3+\sqrt{11}\right)\mathfrak O_K\neq \mathfrak O_K$. If for $x+y\sqrt{11}\in\mathfrak O_K$, where $x,y$ are rational integers, $$\left(3+\sqrt{11}\right)\left(x+y\sqrt{11}\right)=1,$$ then $$3x+11y=1$$ and $$x+3y=0.$$ No such integer $x,y$ exists. Hence $P=\left(3+\sqrt{11}\right)\mathfrak O_K$.
Now if $\mathfrak p$ has norm $2$, then it divides $2$ and hence $\mathfrak p$ is exactly $P$, which is integral. This completes the proof.