I want to prove that the Principal Ideal $(6+\sqrt{-13})$ of $ R = \mathbb{Z} + \mathbb{Z} \sqrt{-13}$ is the square of a prime ideal $P$ of $R$
So far I was able to show that 7 is split in R and I believe that the $(7) = (7, \sqrt{-13} -1)(7, \sqrt{-13} +1)$. (Note: $N(6+\sqrt{-13}) = 14$ so by unique prime factorization must equal the product of two prime ideals of norm $7$ which must be unique because 7 is split).
Where I am stuck is showing that $(7, \sqrt{-13} -1)^{2} = (6+\sqrt{-13})$ or $(7, \sqrt{-13} +1)^{2} = (6+\sqrt{-13})$
Also, any suggestions resources, or tricks on how to simplify ideals would be appreciated.
Also, R is the ring of integers.
Best Answer
This ought to do it: \begin{align} (7,1-\sqrt{-13})^2&=\bigl(49,7-7\sqrt{-13},\,(1-\sqrt{-13}\,)^2\bigr)\\ &=\bigl(49,\,(6+\sqrt{-13}\,)(1+\sqrt{-13}\,),\,2(6+\sqrt{-13}\,)\bigr)\\ &=(6+\sqrt{-13}\,)\bigl(6-\sqrt{-13},\,1+\sqrt{-13},\,2\bigr)\,, \end{align} at which point all that remains is to show that $(6-\sqrt{-13},\,1+\sqrt{-13},\,2)$ is the unit ideal, but that’s clear.