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 $R$

Question

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.

Answer 1

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.

Answer 2

Another way is to use the norm of $I = (6 + \sqrt{-13})$, as you intended. Note that $N(I) = 6^2 + 13 = 49$. On the other side $7$ splits in $R$, i.e $(7) = \mathfrak p_1 \mathfrak p_2$, where $\mathfrak p_1 = (7,\sqrt{-13}+1)$ and $\mathfrak p_2 = (7,\sqrt{-13}-1)$ and these are the only prime ideals of $R$ having norm $7$. Thus we must have that

$$I = \mathfrak p_1^2 \quad \quad \text{or} \quad \quad I = \mathfrak p_2^2 \quad \quad \text{or} \quad \quad I = \mathfrak p_1 \mathfrak p_2$$

Note that the last option isn't possible, as $I \not = (7)$. Hence $I$ is a square of prime.

Furthermore, you can explicitly compute that $I \not = \mathfrak p_1^2$. Indeed if that is the case then we have that $\mathfrak p_1 \mid I \implies I \subset \mathfrak p_1$. But then

$$1 = 3\cdot(6 + \sqrt{-13}) - 3\cdot (1+\sqrt{-13}) - 2 \cdot 7\in \mathfrak p_1$$

A contradiction.