Invertible lattices in are well behaved: IJ is invertible iff I, J are Context: I'm reading John Voight's book on quaternion algebras (https://math.dartmouth.edu/~jvoight/quat-book.pdf) with the hopes that Brandt matrices will shed some light on a white whale problem of mine; and have a question about what feels like it ought to be a very simple exercise, namely exercise 16.10. (Note that this exercise, or indeed even this chapter, makes no appearance in Voight's hints and solutions document: https://math.dartmouth.edu/~jvoight/hints-solns.pdf.)
Let $B$ be a $\mathbb Q$-algebra, such as a division algebra over $\mathbb Q$.
For a ($\mathbb Z$-)lattice $I \subset B$, we can consider the left and right orders of $I$: namely $\mathcal O_L(I):= \{ \alpha \in B: \alpha I \subset I \}$ and $\mathcal O_R(I) := \{ \alpha \in B: I \alpha \subset I \}$. Note that $I$ is then a left $\mathcal O_L(I)$ and a right $\mathcal O_R(I)$ module. Two lattices $I$ and $J$, and their product lattice $IJ$, are said to be compatible if $\mathcal O_R(I) = \mathcal O_L(J)$: that is, if multiplication in $B$ gives an isomorphism $I \otimes_{\mathcal O} J \cong IJ$, where $\mathcal O = \mathcal O_R(I) = \mathcal O_L(J)$.
A lattice $I$ is said to be invertible if there is another lattice $I'$ so that $II' = \mathcal O_L(I) = \mathcal O_R(I')$ and $I'I = \mathcal O_L(I') = \mathcal O_R(I)$ (so in particular both products $II'$ and $I' I$ are compatible). If $I$ is invertible, then its inverse $I'$ is unique so we call it $I^{-1}$; one can show that $I^{-1} = \{ \alpha \in B: I \alpha I \subset I \}$.
I am trying to show that two compatible lattices $I$ and $J$ are invertible if and only if their product $IJ$ is invertible. One direction is straightforward: if $I$ and $J$ are invertible, then $IJ$ is compatible with $K = J^{-1} I^{-1}$ (which itself is a compatible product) on both sides, and all is as expected. Note that the key appears to be the following statement: if $I$ and $J$ are compatible and each is invertible, then $\mathcal O_L(IJ) = \mathcal O_L(I)$ and $\mathcal O_R(IJ) = \mathcal O_R(J)$. In fact, only left invertibility of $I$ is required for the first statement, right invertibility of $J$ for the second.
However, despite spending several hours, with a collaborator no less!, on the converse, we have made no progress in the other direction. If $IJ$ is invertible, I can certainly propose $K = J (IJ)^{-1}$ as an (a priori, right) inverse for $I$, but although I can compute that $IK = \mathcal O_L(IJ)$, I do not know how to deduce that this last is the same as $\mathcal O_L(I)$ without a priori knowing that $J$ is right invertible. I run into the same problem no matter how I approach. For example, if I knew that the product $K = J (IJ)^{-1}$ is compatible, I would be able to conclude $\mathcal O_L(K) = \mathcal O_L(J (IJ)^{-1}) = \mathcal O_L(J) = \mathcal O_R(I)$, so that the product $IK$ would be compatible, which would be a start. But the compatibility of $J$ and $(IJ)^{-1}$ is equality of $\mathcal O_R(J)$ and $\mathcal O_L( (IJ)^{-1}) = \mathcal O_R(IJ)$ -- same problem yet again!
To be fair, in my quaternion algebra setting, everything would appear to follow from Main Theorem 16.7.7 (in this setting there's a standard involution, so invertible is the same thing as locally principal, etc.). But Voight's exercise 16.10 on p. 265 is stated quite generally, and clearly I am missing something key in the theory. Any help would be much appreciated.
 A: The statement is false!  A counterexample is indicated by Proposition 16.6.15(a), taking $J=\overline{I}$: the product $I\overline{I}$ is compatible by Lemma 16.6.7 but can have larger left (or right) order.  More explicitly, see Example 16.5.12.
I'm very sorry for this mistake!  I traced it back to a very old version of the book, where the extra hypothesis ($\mathcal{O}_{\mathsf{L}}(IJ)=\mathcal{O}_{\mathsf{L}}(I)$) was dropped.  I will make sure it is fixed before publication.  I recall thinking that this extra hypothesis made for a lame exercise, but then I failed to delete the other implication!
Two other comments.

*

*All lattices with left (or right) order maximal are invertible (Proposition 16.6.15).   I think this is what @Kimball is using, and it would finish the job above: your containments of orders become equalities, by maximality.


*About terminology: Reiner in his Maximal Orders indeed uses the term "proper", but I try to resist this overloaded term whenever possible.  Anyway, Kaplansky [Kap69] calls it "concordant", and this predates Reiner (1975).  The farthest I can trace the notion back is to Brandt's composition of quaternary quadratic forms when it is 'allowed', and this ultimately gives rise to groupoids (Chapter 19).  What is the best word to use for the situation where you have two morphisms in a small category and the source of one agrees with the target of the second?  (Is "compatible" not a good word to use?)
