Maximal order in a central simple algebra Suppose $A$ is a central simple algebra over a field $F$, $\mathcal{O}_F$ is an integrally closed subring of $F$ and suppose $\mathcal{O}_F$ is noetherian. Then an order of $A$ is a $\mathcal{O}_F$ lattice of $A$, stable under the algebra operations. In this case, the maximal order always exists( assume dim$_F A$ is coprime to the char of F). My questions:
Are there some nontrivial criterion for an order being maximal? And how to construct a maximal order starting from a given order? (I only know the existence from some ACC argument)
In fact, what I really concern is "why we could always find a maximal order stable under a given involution on $A$" (It is wrong from Eisele's example).
So is there some similar statement (under some condition) which is right?
(I kind of heard of it before...)
Thanks.
 A: Regarding the part of your question about involutions. 
I don't think it's always possible to find a maximal order stable under a given involution. 
For instance, take $F=\mathbb Q$, $A = \mathbb Q^{2\times 2}$ and the involution
$$
  \sigma: A \longrightarrow A: X \mapsto M\cdot X^\top \cdot M^{-1} 
$$
with
$$
  M=\left(\begin{array}{cc} 2&0\newline 0&1\end{array}\right)
$$
Now any maximal order $\Theta$ in $A$ is conjugate to $\mathbb Z^{2\times 2}$, and therefore all $\Theta$-lattices in $\mathbb Q^{1\times 2}$ are of the form $n\cdot L$ for dome $n\in \mathbb Q$ and some fixed full $\mathbb Z$-lattice L in $\mathbb Q^{1\times 2}$. Note that the determinant of a basis matrix for $n\cdot L$ is well-defined up to sign (since two basis matrices differ only by an element of ${\rm GL}(2,\mathbb Z)$), and $\det(n\cdot L) = \pm n^2\cdot \det(L)$.
In what follows assume $\Theta$ is a maximal order such that $\sigma(\Theta) \subseteq \Theta$. This will lead to a contradiction.
For any $S\in{\rm GL}(2,\mathbb Q)$ we can replace $M$ by $M' = S M S^{\top}$, $\Theta$ by $\Theta'=S\Theta S^{-1}$ and $\sigma$ by
$$
   \sigma': A \longrightarrow A: X \mapsto M'\cdot X^\top \cdot M'^{-1} 
$$
 So assume that $\Theta'=\mathbb Z^{2\times 2}$, $L'=\mathbb Z^{1\times 2}$
 and $\sigma'(\Theta') \subseteq \Theta'$.
 Note that the determinant of $M'$ will be $2\cdot \det S^2$, which is not a square in $\mathbb Q$. 
Now here's the problem: $L'\cdot M'$ is not of the form $n\cdot L'$, since
$\det M'$ is not a square in $\mathbb Q$. Therefore $L'\cdot M'$ cannot be stable under
the action of $\Theta'^\top = \Theta'$. Hence
$$
   L' \cdot M' \cdot \mathbb Z^{2\times 2} \neq L'\cdot M'
$$ 
which implies
$$
   L' \cdot M' \cdot \mathbb Z^{2\times 2} \cdot M'^{-1} \neq L'
$$
which tells you that
$$
  \sigma'(\mathbb Z^{2\times 2}) = M'\cdot \mathbb Z^{2\times 2} \cdot M'^{-1}  \nsubseteq {\rm End}_{\mathbb Z}(L')=\mathbb Z^{2\times 2} 
$$
which is a contradiction (as $\Theta'=\mathbb Z^{2\times 2}$ was assumed to be stable under $\sigma'$).
