Factorization of (54) in $\mathcal O_{-53}$ In $\mathcal O_{-53}$, we have $54 = 3^3 \cdot 2 = (1+\sqrt{-53})(1-\sqrt{-53})$. What's the factorization of the ideal $(54)$ in $\mathcal O_{-53}$ ? 
 A: As I said in my comment, if you know how $(2)$ and $(3)$ factor, you have the factorization of $(54)$.
Since $-53\equiv3\pmod4$, $2$ is ramified, and so the only ramified primes are $2$ and $53$. Since $-53\equiv1\pmod3$, the prime $3$ splits in $\Bbb Z[\sqrt{-53}\,]$. So we should have $(2)=\mathfrak p_2^2$ and $(3)=\mathfrak p_3\mathfrak p_3'$.
Certainly $\mathfrak p_2=(2,\sqrt{-53}-1\,)=(2,\sqrt{-53}+1\,)$, so that we get
\begin{align}
\mathfrak p_2^2&=(2,\sqrt{-53}-1\,)(2,\sqrt{-53}+1\,)\\
&=(4,2\sqrt{-53}+2,2\sqrt{-53}-2,54)\\
&=(2)\,.
\end{align}
Similarly, you expect that the $\mathfrak p_3$’s  will be $(3,\sqrt{-53}\pm1)$, giving us
\begin{align}
\mathfrak p_3\mathfrak p_3'&=(3,\sqrt{-53}-1)(3,\sqrt{-53}+1)\\
&=(9,3\sqrt{-53}+3,3\sqrt{-53}-3,54)\\
&=(9,3\sqrt{-53}+3,6)\\
&=(3)\,.
\end{align}
So you have your factorization of $(54)$.
Generally speaking, this kind of computation is easy once you identify what the prime ideals upstairs are or look as if they ought to be. Note that we didn’t need to calculate the class number or anything else like that. Of course the story is entirely different in the real quadratic case, ’cause the units make things so much more difficult.
