# Computing the inverse of a full lattice in a quaternion algebra

Let $D$ be quaternion algebra over a number field $F$. Let $\Delta\subseteq D$ be a maximal $\mathcal{O}_{F}$-order. Let $\mathfrak{b}$ be a fractional left $\Delta$-ideal. In his book "Maximal Orders", I. Reiner defines the 'inverse' of a full left $\mathcal{O}_{F}$-lattice $L$ (which every fractional $\Delta$-ideal also is) as

$L^{-1}:=\{ x \in D : L x L \subseteq L \}.$

How can I compute the inverse of $\mathfrak{b}$ according to Reiner's definition using the MAGMA?

• Is there a problem simply using the definition? {x : x in D | forall{l1 : l1 in L : | forall{l2 : l2 in L | l1*x*l2 in L}}} (if $D$ is infinite, you may need to define as a formal set using {! !}.) – Morgan Rodgers May 26 '18 at 16:28
• @MorganRodgers that seems to be working using the formal set version, thank you. I am new to both Magma and Quaternion Algebras, I wasn't aware of this functionality. – Jan Gerrit May 26 '18 at 17:02
• @MorganRodgers as far as I know, this inverse is a fractional Delta ideal itself. Do you know how I can turn this formal set into an ideal (into the type AlgAssVOrdIdl)? – Jan Gerrit May 26 '18 at 17:07
• No I'm not sure there. The formal set construction is really just setting up a method to test if an element is in $L^{-1}$ (so if you ask if $a \in L^{-1}$ it tests $a$ according to the rule we gave). If you want to view it as an ideal, you need to know what generates it. It's not my area of study so I don't know how exactly to find a generating set. – Morgan Rodgers May 26 '18 at 17:39