# Prime ideals in maximal orders (1- and 2-sided)

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 $$M' = uMu^{-1}$$ ?

## 1 Answer

No. The problem is that $\Lambda$ might have too few units. Here is an example that illustrates this point.

Let $\Lambda$ be the ring of Hurwitz quaternions. This is the subring of the usual quaternions $\mathbb{R} \oplus \mathbb{R} i \oplus \mathbb{R}j \oplus \mathbb{R}k$, freely generated as an abelian group by the four elements

$\frac{1}{2}(1 + i + j + k), i, j, k$.

It is known that $\Lambda$ is a maximal order in its division ring of fractions.

Now let $p$ be any odd prime number and consider the ring $\bar{\Lambda} := \Lambda/p\Lambda$. It turns out that this $4$-dimensional $\mathbb{F}_p$-algebra is always isomorphic to ring of $2 \times 2$ matrices $M_2(\mathbb{F}_p)$ with entries in $\mathbb{F}_p$.

To see this, let $J$ be the Jacobson radical of $\bar{\Lambda}$; then by Wedderburn's Theorem, $\bar{\Lambda} / J$ is a direct sum of matrix rings over division rings. Suppose for a contradiction that all the matrix rings are $1 \times 1$. Since all finite division rings are commutative fields by another theorem of Wedderburn, $\bar{\Lambda} / J$ is commutative. But then $J$ contains the commutator $ij - ji = 2k$ which is a unit in $\bar{\Lambda}$ since $p$ is odd. This is a contradiction because $J$ is necessarily a proper ideal of $\bar{\Lambda}$. So there is at least one $n \times n$ matrix ring $M_n(D)$ with $n > 1$ occurring as a direct summand of $\bar{\Lambda} / J$ for some division ring $D$. A dimension count now shows that $n = 2$, $D = \mathbb{F}_p$ and $J = 0$.

Now consider the action of $\bar{\Lambda}^\times \cong GL_2(\mathbb{F}_p)$ on the set of maximal left ideals in $\bar{\Lambda} \cong M_2(\mathbb{F}_p)$ by conjugation. The stabilizer of the maximal left ideal $\{\begin{pmatrix} \ast & 0 \cr \ast & 0\end{pmatrix} \}$ is the subgroup of lower triangular matrices in $GL_2(\mathbb{F}_p)$ which has index $p+1$ in $GL_2(\mathbb{F}_p)$. So we conclude that there are at least $p+1$ distinct maximal left ideals in $\bar{\Lambda}$ (in fact it can be shown easily that $\bar{\Lambda}^\times$ acts transitively so that there are exactly $p+1$ distinct maximal left ideals in $M_2(\mathbb{F}_p)$.)

However, it is known that $\Lambda^\times$ is a finite group (of order $48$, in fact). So as soon as $p + 1 > 48$ (we can take $p = 53$ for concreteness if you like), $\Lambda^\times$ cannot possibly act transitively by conjugation on the set of maximal left ideals belonging to $p \Lambda$.

One final remark: if you change your hypotheses on $\Lambda$ and assume instead that $\Lambda$ is local (in the sense of having a unique maximal two-sided ideal), then the answer to your question becomes "yes". This is because any two maximal left ideals in a simple Artinian ring are conjugate by a unit, and units can be lifted modulo the Jacobson radical in any ring.