I have the following question regarding class groups.
Show that $K=\mathbb{Q}(\sqrt{-19})$ is of class number 1.
From what I understand, the Minkowski bound says, for a number field $K$, that any ideal class contains an integral ideal with norm bounded above by \begin{equation} \frac{n!}{n^n} \left(\frac{4}{\pi}\right)^{r_2}\sqrt{|\text{disc}(K)|} \end{equation} In this particular example: \begin{eqnarray} \frac{2!}{2^2} \left(\frac{4}{\pi}\right)^{1}\sqrt{19}&=& \frac{2}{\pi}\sqrt{19}<3 \end{eqnarray} Thus it is suffice to check how (2) and (3) split K.
This is where I am stuck. I was thinking of going about it using quadratic reciprocity to see how 2 and 3 split into $K$. I get split and inert respectively. Because $2$ split in $K$, I was to check and see if there were any integer solution for
\begin{eqnarray} N\left(a+b\left(\frac{1+\sqrt{-19}}{2}\right)\right) & = & a^2+ab+5b^2=\pm2 \end{eqnarray}
Am I on the right track? If so, how do I proceed? If not, where did I go wrong and how do I properly approach this?
Thank you for your time and thank you for your feedback.
Yes, you are on the right track. You can evaluate the Jacobi symbol (or Kronecker symbol) to determine the factorisation of the ideals $(2)$ and $(3)$, see here: How to factor ideals in a quadratic number field?. Then you can proceed in the usual way, like here: Showing that $\mathbb Q(\sqrt{17})$ has class number 1.
There is another way to show that $\mathbb{Q}(\sqrt{-19})$ has class number one, namely the analytic class number formula. In your case the class number is equal to $$ h=\frac{\sqrt{\mid d_k\mid}}{2\pi}L(1,\chi), $$ with the $L$-series attached to a corresponding Dirichlet character, and $d_K=-19$. It is easy to see that this gives $h=1$, because $L(1,\chi)\sim 1.44146156829133589$.