Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.