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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.

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1 Answer 1

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 $\sigma'$).

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Thanks very much! A small mistake: the $\Theta$-lattice in $\mathbb{Q}^{1\times 2}$ might be $rL$, $r \in \mathbb{Q}$:) –  user20421 Feb 9 '12 at 22:40
    
Yes, that was indeed a mistake, thanks. Fixed it. –  Florian Eisele Feb 9 '12 at 22:54
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