I am an arithmetic geometry graduate student, and I find myself needing to learn about factorisation in orders in division algebras.  I know something aout algebraic number theory and commutative algebra, but very little about noncommutative rings.

Let $R$ be a Dedekind domain and $K$ its field of fractions.
The following is Theorem 22.15 from Reiner's _Maximal Orders_:

> **Theorem.** For each maximal left ideal $M$ of a maximal $R$-order $\Lambda$ in a separable $K$-algebra $A$, there is a unique prime (2-sided) ideal $\mathfrak{P}$ of $\Lambda$ such that
>
>> $\mathfrak{P} \subset M \subset \Lambda$ and $\mathfrak{P} = \operatorname{ann}_\Lambda \Lambda/M = \{ x \in \Lambda | x\Lambda \subset M \}.$
>
> We say that $M$ _belongs_ to $\mathfrak{P}$.
Then $\Lambda/M$ is a simple left module over the simple ring $\Lambda/\mathfrak{P}$.
Conversely, each $\mathfrak{P}$ determines a maximal left ideal $M$ of $\Lambda$ which belongs to $\mathfrak{P}$.

In the last sentence, the word "determines" suggests to me that the ideal $M$ belonging to $\mathfrak{P}$ should be unique, but I would also expect the word "unique" to appear in the statement of the theorem.  The proof gives only the existence of such an $M$.

Indeed I think that the ideal need not be unique: let $M$ be a maximal left $\Lambda$-ideal belonging to $\mathfrak{P}$ with right order $\Lambda' \neq \Lambda$.
If $u$ is a unit of $\Lambda$ not in $\Lambda'$, then $uMu^{-1}$ is a maximal left $\Lambda$-ideal belonging to $\mathfrak{P}$ and distinct from $M$.

Is this the only way in which uniqueness can fail?  That is:

> Let $\mathfrak{P}$ be a prime ideal of $\Lambda$ and $M$, $M'$ maximal left $\Lambda$-ideals belonging to $\mathfrak{P}$.
Is there a unit $u$ of $\Lambda$ such that $M' = uMu^{-1}$ ?