While I was reading this paper Lenstra page 15, for d=4729494, he says $2162+\sqrt d$ is divisible by the cube of the prime ideal $(5,2+\sqrt d)$. Can someone please help me how this works?
I see that $(5,2+\sqrt d)^3=(125,150+75 \sqrt d,60+60\sqrt d,8+12 \sqrt d+6d+d\sqrt d)$. How do I proceed from here?
Thanks
The norm of $2162+\sqrt{d}$ is $N(2162+\sqrt{d})=2162^2-d^2=-55250$, which is divisible by 125 (but not 625). This means that the factorisation of $(2162+\sqrt{d})$ into prime ideals would contain 3 factors of ideals with norm 5 (in general, prime ideals have norm a power of $p$, but since we are working in a degree 2 extension over $\mathbb Q$ we can't have a prime ideal with norm $5^3$, and a prime ideal of norm $5^2$ would imply that $5$ is inert and there wouldn't be any prime ideals of norm 5 which makes this factorisation impossible).
What are the ideals of norm 5 in $\mathbb Z[\sqrt{d}]$? Well by Dedekind's criterion they are $\mathfrak p_5=(5,2+\sqrt{d})$ and $\mathfrak q_5=(5,2-\sqrt{d})$, and $\mathfrak p_5\mathfrak q_5=(5)$. Since $5$ does not divide $2162+\sqrt{d}$, this means that $\mathfrak p_5$ and $\mathfrak q_5$ cannot both be factors of $(2162+\sqrt{d})$. i.e. $(2162+\sqrt{d})$ must have a factor of $\mathfrak p_5^3$ or $\mathfrak q_5^3$.
Now it suffices to determine which of these factors is in $(2162+\sqrt{d})$. Well, notice that $2162+\sqrt{d}=432\times5+1\times(2+\sqrt{d})\in \mathfrak p_5$. So $(2162+\sqrt{d}) \subset \mathfrak p_5$ implies that $\mathfrak p_5 | (2162+\sqrt{d})$. Thus we can conclude that $\mathfrak p_5^3$ must divide $(2162+\sqrt{d})$.