Let $R$ be a noetherian domain and let $\mathcal{O}$ be an $R$-algebra that is finitely generated and projective as an $R$-module. The set of invertible fractional ideals of $\mathcal{O}$ is a group under multiplication. Is this group always abelian?
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4$\begingroup$ It looks like some of your post got cut off. I also think you meant $R$-module, rather than $\mathcal{O}$-module. Also, what is the definition you use for an invertible ideal in a non-commutative ring? $\endgroup$– Pace NielsenCommented Apr 8, 2015 at 15:25
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4$\begingroup$ What's a fractional ideal at all in a noncommutative ring? $\endgroup$– Fernando MuroCommented Apr 8, 2015 at 21:34
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2$\begingroup$ @FernandoMuro $K$ being the field of fractions of $R$ and $A = K\otimes_R\mathcal{O}$, I would guess a fractional ideal is a lattice $I\subset A$ (finitely generated $R$-submodule with $KI=A$) with left and right order equal to $\mathcal{O}$. Does this work? $\endgroup$– AurelCommented Apr 8, 2015 at 23:09
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3$\begingroup$ @Aurel: conventions differ, but you might ask only that $\mathcal{O}$ is contained in the left and right orders; if $I$ is invertible, then equality holds, so this doesn't matter for the question. (@FernandoMuro: It is also equivalent to take a fractional ideal $I$ to be a $\mathcal{O}$-sub-bimodule of $K \otimes_R \mathcal{O}$ of the form $I=cJ$ where $c \in K^\times$ and $J \subseteq \mathcal{O}$ is a two-sided ideal. Hence the ``fractional''.) $\endgroup$– John VoightCommented Apr 9, 2015 at 2:00
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3$\begingroup$ @FernandoMuro The base ring $R$ is assumed to be commutative: it is a domain. So the field of fractions means what it usually does. (There are notions of rings of quotients for noncommutative rings, but they don't play a role here.) $\endgroup$– John VoightCommented Apr 9, 2015 at 14:24
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1 Answer
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No; it is not abelian in general.
Let $R$ be a discrete valuation ring with maximal ideal generated by an element $\pi$, let $k := R / \pi R$ be the residue field of $R$ and let $\mathcal{O}$ be the inverse image in $M_2(R)$ of the scalar matrices in $M_2(k)$. Note that $\mathcal{O}$ is free of rank $4$ as an $R$-module, with basis $\{e_{11} + e_{22}, \pi e_{11}, \pi e_{12}, \pi e_{21}\}$ say. So $\mathcal{O}$ is, in particular, finitely generated and projective as an $R$-module.
For every $g \in GL_2(R)$, you have the invertible fractional ideal $\mathcal{O}g$. Note that $\mathcal{O}g \cdot \mathcal{O} h = \mathcal{O} gh$ since $GL_2(R)$ normalises $\mathcal{O}$ inside $M_2(R)$. Note also that $\mathcal{O} g = \mathcal{O} h$ if and only if $gh^{-1} \in \mathcal{O}^\times$, which is the preimage in $GL_2(R)$ of the group of scalar matrices in $GL_2(k)$. So we obtain an injective group homomorphism from $PGL_2(k) \cong GL_2(R) / \mathcal{O}^\times$ into the group $I(\mathcal{O})$ of invertible fractional ideals of $\mathcal{O}$.
In the positive direction, if $K$ is the field of fractions of a Dedekind domain $R$ and if $\mathcal{O}$ is a maximal order in some finite dimensional separable $K$-algebra, then $I(\mathcal{O})$ is abelian.
See Theorem 12 and Theorem 13(c) of this paper by Frőhlich, which gives a thorough discussion of the Picard group of a noncommutative ring.
