(TL;DR) How exactly do we reduce a quaternion ideal into another powersmooth one?

Given a supersingular elliptic curve, it is known that its endomorphism ring is non-commutative. Specifically, there is Deuring correspondence associating supersingular $j$-invariant up to a twist and a maximal order of quaternion algebra $B$ ramified at $p,\infty$.

Suppose $\mathcal O\leftrightarrow j(E)$ is associated by the correspondence, i.e. $\mathsf {End}(E)=\mathcal O$ and there is a left $\mathcal O$-ideal $I\subseteq \mathcal{O}$ with norm $\mathsf {Nrd}(I)=N$ and generators $\alpha,N$. It is said that $I\sim J$ are equivalent if $\exists r\in B^\times$ that $Ir=J$. I was reading this article saying there's an efficient algorithm to find a powersmooth representative (up to $\sim$) of ideal $I$ with large norm. I couldn't understand how exactly it is performed. The main goal in the article is to solve for a $\beta,\lambda$ such that $\beta=\lambda\alpha\mod{N\mathcal O}$ in page 8. But how exactly solving $\beta$ would help one reduce the ideal norm?

Is it possible to write down the explicit procedure? If not, is there any other way to reduce the ideal norm?



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